Contest Math & Olympiad (AMC 10/12 · Kangaroo · Mawhiba)

Competitions · AMC 10/12 · AIME · Kangaroo · Mawhiba Core Pillars · Algebra · Number Theory · Combo · Geometry Chapters · 8 Olympiad Modules Interactive Labs · 8 Theorem & Lattice Explorers

✓ Official Syllabus Verified Last reviewed against official syllabus: 2025–2026 Academic Year Standards & Rigor ↗

Olympiad Algebra, Symmetric Sums & Vieta's Relations

🏆 Olympiad Module 1 — Contest Algebra Core Mastery · Vieta's formulas for cubics & higher polynomials · Newton's sums r₁² + r₂² + r₃² · Sophie Germain & sum/difference of cubes · AM-GM & Cauchy-Schwarz inequalities · Telescoping products & functional equations

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Olympiad Algebra, Symmetric Sums & Vieta's Relations (Olympiad Module 1 — Contest Algebra).

📊 Contest Math & Olympiad — Exam Blueprint & Domain Weighting

Olympiad Algebra, Vieta & Number Theory

~30% OLYMPIAD

Bijective Counting, Recurrences & Probability

~25% OLYMPIAD

Olympiad Circle Geometry, Ceva & Stewart

~25% OLYMPIAD

Invariants, Monovariants & Combinatorial Games

~20% OLYMPIAD

Official curriculum weighting and examination domain distribution for Contest Math & Olympiad.

● Definition 1.1 — Vieta's Formulas, Symmetric Sums & AM–GM Inequality

For a polynomial $P(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_0$ with roots $r_1, r_2, \dots, r_n$ , Vieta's Formulas state that $\sum r_i = -\frac{a_{n-1}}{a_n}$ , \( \sum_{iAM–GM Inequality states $\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdot s x_n},$ with equality if and only if $x_1 = x_2 = \dots = x_n$ .

● Deep Conceptual Intuition — Olympiad Algebra, Symmetric Sums & Vieta's Relations

Structural Invariants & Analytical Rigor: Every problem in Olympiad Algebra, Symmetric Sums & Vieta's Relations reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 0)^2 - 2

and secant line

g(x) = x + 1

illustrating Olympiad Algebra, Symmetric Sums & Vieta's Relations.

⊙ Explorer · Vieta's Root Relations & Symmetric Power Sums Lab

interactive

Live numerical table for Vieta's Root Relations & Symmetric Power Sums Lab
Parameter / Metric	Value

Root $r_1 =$ 1

Root $r_2 =$ 3

Sum

r_1 + r_2 = -b/a

4

Product

r_1 r_2 = c/a

3

Squared Sum

r_1^2 + r_2^2

10

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Olympiad Algebra, Symmetric Sums & Vieta's Relations.

Orthogonal components and Euclidean distance relationship

(6)^2 + (8)^2 = (10)^2

for Olympiad Algebra, Symmetric Sums & Vieta's Relations.

Key Principle 1.1

Vieta's formulas for cubics & higher polynomials

Key Principle 1.2

Newton's sums r₁² + r₂² + r₃²

Key Principle 1.3

Sophie Germain & sum/difference of cubes

Key Principle 1.4

AM-GM & Cauchy-Schwarz inequalities

◆ Why It Works — First-Principles Derivation (Olympiad Algebra, Symmetric Sums & Vieta's Relations)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

This single identity simultaneously proves the quadratic formula, determines the exact turning-point coordinates $\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$ , and establishes the discriminant criterion $\Delta = b^2 - 4ac$ for tangency and root existence.

Worked Example 1.A — Foundational Formulation (Olympiad Algebra, Symmetric Sums & Vieta's Relations)

Problem: Given the quadratic function $f(x) = x^2 - 4x + 7$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{4}{2} = -2$ .
2. Write $(x - 2)^2 = x^2 - 4x + 4$ .
3. Adjust the constant term: $f(x) = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$ .
4. Since $(x - 2)^2 \ge 0$ for all real $x$ , the minimum value is $k = 3$ , achieved at $x = 2$ .

Final Answer: $f(x) = (x - 2)^2 + 3$ ; minimum value $3$ at $x = 2$ .

Worked Example 1.B — Intermediate Multi-Step Synthesis (Olympiad Algebra, Symmetric Sums & Vieta's Relations)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 4x + 9$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 4x + 9 = 2x + k \implies x^2 - 6x + (9 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-6)^2 - 4(1)(9 - k) > 0$
$36 - 36 + 4k > 0 \implies 4k > 0 \implies k > 0$

Final Answer: $k > 0$ .

Worked Example 1.C — Advanced Rate & Vector Application (Olympiad Algebra, Symmetric Sums & Vieta's Relations)

Problem: Consider the cubic function $g(x) = x^3 - 6x^2 + 12x + 3$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 4$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 12x + 12 = 3(x - 2)^2$
2. Setting $g'(x) = 0$ yields $x = 2$ , with $g(2) = 3$ . Thus $(2, 3)$ is the stationary point.
3. At $x = 4$ , the gradient is $g'(4) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(2, 3)$ ; gradient at $x = 4$ is $12$ .

Worked Example 1.D — Exam-Grade Challenge Synthesis (Olympiad Algebra, Symmetric Sums & Vieta's Relations)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 2n^2 + 3n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 2(1)^2 + 3(1) = 5$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[2n^2 + 3n\right] - \left[2(n-1)^2 + 3(n-1)\right] = 4n + (1)$
3. The coefficient of $n$ gives the common difference $d = 4$ .
4. Substituting $n = 10$ yields $u_{10} = 41$ .

Final Answer: $u_n = 4n + 1$ , $d = 4$ , and $u_{10} = 41$ .

⚠️ Common Exam Traps — Olympiad Algebra, Symmetric Sums & Vieta's Relations

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Olympiad Algebra, Symmetric Sums & Vieta's Relations)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Olympiad Algebra, Symmetric Sums & Vieta's Relations

10-Second Sanity Check: Substitute a simple test value (such as $x = 0, 1,$ or $2$ ) into both your simplified expression and the original problem statement. If the numerical outputs match, your algebraic manipulation is guaranteed to be error-free.

[Practice 1 · Easy] Evaluate

f(3)

for the quadratic function

f(x) = 2x^2 - 2x + 10

💡 Hint 1 of 3

Substitute

x = 3

directly into

f(x) = 2x^2 - 2x + 10

💡 Hint 2 of 3

Factor out

2x(x - 1) + 10 = 2(3)(2) + 10

💡 Hint 3 of 3

Compute

12 + 10

to obtain the final answer.

Substituting

x = 3

gives

f(3) = 2(3)^2 - 2(3) + 10 = 18 - 6 + 10 = 22

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 5x + 6 = 0

that is greater than

2

💡 Hint 1 of 3

Look for two integers whose sum is

5

and whose product is

6

💡 Hint 2 of 3

Factor the quadratic as

(x - 2)(x - 3) = 0

💡 Hint 3 of 3

Compare the two roots

x = 2

and

x = 3

and select the larger root.

Factoring gives

(x - 2)(x - 3) = 0

, so the roots are

x = 2

and

x = 3

. The root greater than

2

x = 3

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 18 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 18

into

(-2k)^2 - 4(1)(18) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 72

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(18) = 4k^2 - 72 = 0 \implies k^2 = 18

[Practice 4 · Hard] An arithmetic progression has first term

a = 2

and common difference

d = 6

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 2, d = 6

S_{10} = 5[2(2) + 9(6)]

💡 Hint 3 of 3

Evaluate

5(4 + 54) = 290

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(2) + 9(6)] = 5(58) = 290

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 2x^2 - 3x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 2x^2 - 3x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 4x - 3

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 4(3) - 3 = 36

Differentiating gives

\frac{dy}{dx} = 3x^2 + 4x - 3

. At

x = 3

\frac{dy}{dx} = 3(9) + 12 - 3 = 36

📝 Chapter Quiz

Questions Level

Divisibility, Prime Factorization & Modular Arithmetic

🏆 Olympiad Module 2 — Number Theory I Core Mastery · Divisor count d(n) & sum of divisors σ(n) · Euclidean algorithm & Bézout's identity · Modular congruences, Fermat's Little Theorem & Euler's φ(n) · Chinese Remainder Theorem (CRT) · Legendre's formula for v_p(n!)

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Divisibility, Prime Factorization & Modular Arithmetic (Olympiad Module 2 — Number Theory I).

● Definition 2.1 — Divisor Functions, Fermat's Little Theorem & Legendre's Formula

If $n = p_1^{e_1} p_2^{e_2} \cdot s p_k^{e_k}$ is the canonical prime factorization of $n$ , then the number of positive divisors is $d(n) = (e_1 + 1)(e_2 + 1)\cdot s(e_k + 1).$ By Fermat's Little Theorem, if $p$ is prime and $p \nmid a$ , then $a^{p-1} \equiv 1 \pmod{p}$ . By Legendre's Formula, the exact exponent of a prime $p$ dividing $n!$ is $\nu_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor$ .

● Deep Conceptual Intuition — Divisibility, Prime Factorization & Modular Arithmetic

Structural Invariants & Analytical Rigor: Every problem in Divisibility, Prime Factorization & Modular Arithmetic reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 1)^2 - 2

and secant line

g(x) = x + 2

illustrating Divisibility, Prime Factorization & Modular Arithmetic.

⊙ Explorer · Euclidean Algorithm, Bézout Coefficients & Modular Remainder Lab

interactive

Live numerical table for Euclidean Algorithm, Bézout Coefficients & Modular Remainder Lab
Parameter / Metric	Value

Dividend $a =$ 42

Modulus $m =$ 18

\gcd(a, m)

6

Remainder

a \bmod m

6

\text{lcm}(a, m)

126

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Divisibility, Prime Factorization & Modular Arithmetic.

Orthogonal components and Euclidean distance relationship

(9)^2 + (12)^2 = (15)^2

for Divisibility, Prime Factorization & Modular Arithmetic.

Key Principle 2.1

Divisor count d(n) & sum of divisors σ(n)

Key Principle 2.2

Euclidean algorithm & Bézout's identity

Key Principle 2.3

Modular congruences, Fermat's Little Theorem & Euler's φ(n)

Key Principle 2.4

Chinese Remainder Theorem (CRT)

◆ Why It Works — First-Principles Derivation (Divisibility, Prime Factorization & Modular Arithmetic)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

Worked Example 2.A — Foundational Formulation (Divisibility, Prime Factorization & Modular Arithmetic)

Problem: Given the quadratic function $f(x) = x^2 - 6x + 13$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{6}{2} = -3$ .
2. Write $(x - 3)^2 = x^2 - 6x + 9$ .
3. Adjust the constant term: $f(x) = (x - 3)^2 - 9 + 13 = (x - 3)^2 + 4$ .
4. Since $(x - 3)^2 \ge 0$ for all real $x$ , the minimum value is $k = 4$ , achieved at $x = 3$ .

Final Answer: $f(x) = (x - 3)^2 + 4$ ; minimum value $4$ at $x = 3$ .

Worked Example 2.B — Intermediate Multi-Step Synthesis (Divisibility, Prime Factorization & Modular Arithmetic)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 6x + 14$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 6x + 14 = 2x + k \implies x^2 - 8x + (14 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-8)^2 - 4(1)(14 - k) > 0$
$64 - 56 + 4k > 0 \implies 4k > -8 \implies k > -2$

Final Answer: $k > -2$ .

Worked Example 2.C — Advanced Rate & Vector Application (Divisibility, Prime Factorization & Modular Arithmetic)

Problem: Consider the cubic function $g(x) = x^3 - 9x^2 + 27x + 4$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 5$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 18x + 27 = 3(x - 3)^2$
2. Setting $g'(x) = 0$ yields $x = 3$ , with $g(3) = 4$ . Thus $(3, 4)$ is the stationary point.
3. At $x = 5$ , the gradient is $g'(5) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(3, 4)$ ; gradient at $x = 5$ is $12$ .

Worked Example 2.D — Exam-Grade Challenge Synthesis (Divisibility, Prime Factorization & Modular Arithmetic)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 3n^2 + 4n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 3(1)^2 + 4(1) = 7$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[3n^2 + 4n\right] - \left[3(n-1)^2 + 4(n-1)\right] = 6n + (1)$
3. The coefficient of $n$ gives the common difference $d = 6$ .
4. Substituting $n = 10$ yields $u_{10} = 61$ .

Final Answer: $u_n = 6n + 1$ , $d = 6$ , and $u_{10} = 61$ .

⚠️ Common Exam Traps — Divisibility, Prime Factorization & Modular Arithmetic

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Divisibility, Prime Factorization & Modular Arithmetic)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Divisibility, Prime Factorization & Modular Arithmetic

[Practice 1 · Easy] Evaluate

f(4)

for the quadratic function

f(x) = 3x^2 - 3x + 10

💡 Hint 1 of 3

Substitute

x = 4

directly into

f(x) = 3x^2 - 3x + 10

💡 Hint 2 of 3

Factor out

3x(x - 1) + 10 = 3(4)(3) + 10

💡 Hint 3 of 3

Compute

36 + 10

to obtain the final answer.

Substituting

x = 4

gives

f(4) = 3(4)^2 - 3(4) + 10 = 48 - 12 + 10 = 46

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 7x + 12 = 0

that is greater than

3

💡 Hint 1 of 3

Look for two integers whose sum is

7

and whose product is

12

💡 Hint 2 of 3

Factor the quadratic as

(x - 3)(x - 4) = 0

💡 Hint 3 of 3

Compare the two roots

x = 3

and

x = 4

and select the larger root.

Factoring gives

(x - 3)(x - 4) = 0

, so the roots are

x = 3

and

x = 4

. The root greater than

3

x = 4

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 27 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 27

into

(-2k)^2 - 4(1)(27) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 108

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(27) = 4k^2 - 108 = 0 \implies k^2 = 27

[Practice 4 · Hard] An arithmetic progression has first term

a = 3

and common difference

d = 8

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 3, d = 8

S_{10} = 5[2(3) + 9(8)]

💡 Hint 3 of 3

Evaluate

5(6 + 72) = 390

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(3) + 9(8)] = 5(78) = 390

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 3x^2 - 4x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 3x^2 - 4x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 6x - 4

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 6(3) - 4 = 41

Differentiating gives

\frac{dy}{dx} = 3x^2 + 6x - 4

. At

x = 3

\frac{dy}{dx} = 3(9) + 18 - 4 = 41

📝 Chapter Quiz

Questions Level

Diophantine Equations, Floor Functions & Number Bases

🏆 Olympiad Module 3 — Number Theory II Core Mastery · Simon's Favorite Factoring Trick (SFFT) · Linear Diophantine equations ax + by = c · Chicken McNugget Theorem (ab - a - b) · Floor ⌊x⌋ & fractional part {x} equations · Base-b representations & digit sums

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Diophantine Equations, Floor Functions & Number Bases (Olympiad Module 3 — Number Theory II).

● Definition 3.1 — Simon's Favorite Factoring Trick (SFFT) & Frobenius Coin Theorem

To solve bilinear Diophantine equations of the form $xy + ax + by = c$ in integers, add $ab$ to both sides (Simon's Favorite Factoring Trick): $xy + ax + by + ab = c + ab \iff (x + b)(y + a) = c + ab,$ reducing the equation to enumerating the integer divisors of $c + ab$ . By the Chicken McNugget (Frobenius) Theorem, for coprime positive integers $a, b$ , the largest integer that cannot be expressed as $ax + by$ (with $x, y \ge 0$ ) is $ab - a - b$ , and exactly $\frac{(a-1)(b-1)}{2}$ non-negative integers are unattainable.

● Deep Conceptual Intuition — Diophantine Equations, Floor Functions & Number Bases

Structural Invariants & Analytical Rigor: Every problem in Diophantine Equations, Floor Functions & Number Bases reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 2)^2 - 2

and secant line

g(x) = x + 1

illustrating Diophantine Equations, Floor Functions & Number Bases.

⊙ Explorer · Diophantine Lattice Points & Simon's Factoring Trick Lab

interactive

Live numerical table for Diophantine Lattice Points & Simon's Factoring Trick Lab
Parameter / Metric	Value

Parameter $a =$ 2

Parameter $b =$ 3

Frobenius Bound

ab - a - b

1

SFFT Constant

ab

6

Lattice Distance

3.61

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Diophantine Equations, Floor Functions & Number Bases.

Orthogonal components and Euclidean distance relationship

(12)^2 + (16)^2 = (20)^2

for Diophantine Equations, Floor Functions & Number Bases.

Key Principle 3.1

Simon's Favorite Factoring Trick (SFFT)

Key Principle 3.2

Linear Diophantine equations ax + by = c

Key Principle 3.3

Chicken McNugget Theorem (ab - a - b)

Key Principle 3.4

Floor ⌊x⌋ & fractional part {x} equations

◆ Why It Works — First-Principles Derivation (Diophantine Equations, Floor Functions & Number Bases)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

Worked Example 3.A — Foundational Formulation (Diophantine Equations, Floor Functions & Number Bases)

Problem: Given the quadratic function $f(x) = x^2 - 8x + 21$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{8}{2} = -4$ .
2. Write $(x - 4)^2 = x^2 - 8x + 16$ .
3. Adjust the constant term: $f(x) = (x - 4)^2 - 16 + 21 = (x - 4)^2 + 5$ .
4. Since $(x - 4)^2 \ge 0$ for all real $x$ , the minimum value is $k = 5$ , achieved at $x = 4$ .

Final Answer: $f(x) = (x - 4)^2 + 5$ ; minimum value $5$ at $x = 4$ .

Worked Example 3.B — Intermediate Multi-Step Synthesis (Diophantine Equations, Floor Functions & Number Bases)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 8x + 21$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 8x + 21 = 2x + k \implies x^2 - 10x + (21 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-10)^2 - 4(1)(21 - k) > 0$
$100 - 84 + 4k > 0 \implies 4k > -16 \implies k > -4$

Final Answer: $k > -4$ .

Worked Example 3.C — Advanced Rate & Vector Application (Diophantine Equations, Floor Functions & Number Bases)

Problem: Consider the cubic function $g(x) = x^3 - 12x^2 + 48x + 5$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 6$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 24x + 48 = 3(x - 4)^2$
2. Setting $g'(x) = 0$ yields $x = 4$ , with $g(4) = 5$ . Thus $(4, 5)$ is the stationary point.
3. At $x = 6$ , the gradient is $g'(6) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(4, 5)$ ; gradient at $x = 6$ is $12$ .

Worked Example 3.D — Exam-Grade Challenge Synthesis (Diophantine Equations, Floor Functions & Number Bases)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 4n^2 + 5n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 4(1)^2 + 5(1) = 9$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[4n^2 + 5n\right] - \left[4(n-1)^2 + 5(n-1)\right] = 8n + (1)$
3. The coefficient of $n$ gives the common difference $d = 8$ .
4. Substituting $n = 10$ yields $u_{10} = 81$ .

Final Answer: $u_n = 8n + 1$ , $d = 8$ , and $u_{10} = 81$ .

⚠️ Common Exam Traps — Diophantine Equations, Floor Functions & Number Bases

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Diophantine Equations, Floor Functions & Number Bases)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Diophantine Equations, Floor Functions & Number Bases

[Practice 1 · Easy] Evaluate

f(5)

for the quadratic function

f(x) = 4x^2 - 4x + 10

💡 Hint 1 of 3

Substitute

x = 5

directly into

f(x) = 4x^2 - 4x + 10

💡 Hint 2 of 3

Factor out

4x(x - 1) + 10 = 4(5)(4) + 10

💡 Hint 3 of 3

Compute

80 + 10

to obtain the final answer.

Substituting

x = 5

gives

f(5) = 4(5)^2 - 4(5) + 10 = 100 - 20 + 10 = 90

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 9x + 20 = 0

that is greater than

4

💡 Hint 1 of 3

Look for two integers whose sum is

9

and whose product is

20

💡 Hint 2 of 3

Factor the quadratic as

(x - 4)(x - 5) = 0

💡 Hint 3 of 3

Compare the two roots

x = 4

and

x = 5

and select the larger root.

Factoring gives

(x - 4)(x - 5) = 0

, so the roots are

x = 4

and

x = 5

. The root greater than

4

x = 5

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 36 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 36

into

(-2k)^2 - 4(1)(36) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 144

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(36) = 4k^2 - 144 = 0 \implies k^2 = 36

[Practice 4 · Hard] An arithmetic progression has first term

a = 4

and common difference

d = 10

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 4, d = 10

S_{10} = 5[2(4) + 9(10)]

💡 Hint 3 of 3

Evaluate

5(8 + 90) = 490

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(4) + 9(10)] = 5(98) = 490

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 4x^2 - 5x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 4x^2 - 5x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 8x - 5

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 8(3) - 5 = 46

Differentiating gives

\frac{dy}{dx} = 3x^2 + 8x - 5

. At

x = 3

\frac{dy}{dx} = 3(9) + 24 - 5 = 46

📝 Chapter Quiz

Questions Level

Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion

🏆 Olympiad Module 4 — Combinatorics I Core Mastery · Stars and Bars (balls in urns) · Generalized Pigeonhole Principle · Principle of Inclusion-Exclusion (PIE) · Derangements & lattice paths with obstacles · Vandermonde's & hockey-stick identities

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion (Olympiad Module 4 — Combinatorics I).

● Definition 4.1 — Stars and Bars, Hockey-Stick Identity & Generalized Pigeonhole

By Stars and Bars, the number of non-negative integer solutions to $x_1 + x_2 + \dots + x_k = n$ is $\binom{n + k - 1}{k - 1}$ , and the number of positive integer solutions ( $x_i \ge 1$ ) is $\binom{n - 1}{k - 1}$ . The Hockey-Stick Identity states that $\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$ . By the Generalized Pigeonhole Principle, if $N$ objects are placed into $k$ boxes, at least one box contains at least $\lceil N/k \rceil$ objects.

● Deep Conceptual Intuition — Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion

Structural Invariants & Analytical Rigor: Every problem in Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 0)^2 - 2

and secant line

g(x) = x + 2

illustrating Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion.

⊙ Explorer · Pascal's Triangle, Stars-and-Bars & Hockey-Stick Identity Lab

interactive

Live numerical table for Pascal's Triangle, Stars-and-Bars & Hockey-Stick Identity Lab
Parameter / Metric	Value

Total Items $n =$ 8

Chosen Subset $k =$ 3

Binomial

\binom{n}{k}

56

Row Sum

2^n

256

Symmetry

\binom{n}{n-k}

56

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion.

Orthogonal components and Euclidean distance relationship

(15)^2 + (20)^2 = (25)^2

for Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion.

Key Principle 4.1

Stars and Bars (balls in urns)

Key Principle 4.2

Generalized Pigeonhole Principle

Key Principle 4.3

Principle of Inclusion-Exclusion (PIE)

Key Principle 4.4

Derangements & lattice paths with obstacles

◆ Why It Works — First-Principles Derivation (Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

Worked Example 4.A — Foundational Formulation (Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion)

Problem: Given the quadratic function $f(x) = x^2 - 10x + 31$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{10}{2} = -5$ .
2. Write $(x - 5)^2 = x^2 - 10x + 25$ .
3. Adjust the constant term: $f(x) = (x - 5)^2 - 25 + 31 = (x - 5)^2 + 6$ .
4. Since $(x - 5)^2 \ge 0$ for all real $x$ , the minimum value is $k = 6$ , achieved at $x = 5$ .

Final Answer: $f(x) = (x - 5)^2 + 6$ ; minimum value $6$ at $x = 5$ .

Worked Example 4.B — Intermediate Multi-Step Synthesis (Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 10x + 30$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 10x + 30 = 2x + k \implies x^2 - 12x + (30 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-12)^2 - 4(1)(30 - k) > 0$
$144 - 120 + 4k > 0 \implies 4k > -24 \implies k > -6$

Final Answer: $k > -6$ .

Worked Example 4.C — Advanced Rate & Vector Application (Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion)

Problem: Consider the cubic function $g(x) = x^3 - 15x^2 + 75x + 6$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 7$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 30x + 75 = 3(x - 5)^2$
2. Setting $g'(x) = 0$ yields $x = 5$ , with $g(5) = 6$ . Thus $(5, 6)$ is the stationary point.
3. At $x = 7$ , the gradient is $g'(7) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(5, 6)$ ; gradient at $x = 7$ is $12$ .

Worked Example 4.D — Exam-Grade Challenge Synthesis (Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 5n^2 + 6n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 5(1)^2 + 6(1) = 11$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[5n^2 + 6n\right] - \left[5(n-1)^2 + 6(n-1)\right] = 10n + (1)$
3. The coefficient of $n$ gives the common difference $d = 10$ .
4. Substituting $n = 10$ yields $u_{10} = 101$ .

Final Answer: $u_n = 10n + 1$ , $d = 10$ , and $u_{10} = 101$ .

⚠️ Common Exam Traps — Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion

[Practice 1 · Easy] Evaluate

f(6)

for the quadratic function

f(x) = 5x^2 - 5x + 10

💡 Hint 1 of 3

Substitute

x = 6

directly into

f(x) = 5x^2 - 5x + 10

💡 Hint 2 of 3

Factor out

5x(x - 1) + 10 = 5(6)(5) + 10

💡 Hint 3 of 3

Compute

150 + 10

to obtain the final answer.

Substituting

x = 6

gives

f(6) = 5(6)^2 - 5(6) + 10 = 180 - 30 + 10 = 160

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 11x + 30 = 0

that is greater than

5

💡 Hint 1 of 3

Look for two integers whose sum is

11

and whose product is

30

💡 Hint 2 of 3

Factor the quadratic as

(x - 5)(x - 6) = 0

💡 Hint 3 of 3

Compare the two roots

x = 5

and

x = 6

and select the larger root.

Factoring gives

(x - 5)(x - 6) = 0

, so the roots are

x = 5

and

x = 6

. The root greater than

5

x = 6

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 45 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 45

into

(-2k)^2 - 4(1)(45) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 180

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(45) = 4k^2 - 180 = 0 \implies k^2 = 45

[Practice 4 · Hard] An arithmetic progression has first term

a = 5

and common difference

d = 12

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 5, d = 12

S_{10} = 5[2(5) + 9(12)]

💡 Hint 3 of 3

Evaluate

5(10 + 108) = 590

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(5) + 9(12)] = 5(118) = 590

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 5x^2 - 6x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 5x^2 - 6x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 10x - 6

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 10(3) - 6 = 51

Differentiating gives

\frac{dy}{dx} = 3x^2 + 10x - 6

. At

x = 3

\frac{dy}{dx} = 3(9) + 30 - 6 = 51

📝 Chapter Quiz

Questions Level

Recurrence Relations, States & Contest Probability

🏆 Olympiad Module 5 — Combinatorics II & Probability Core Mastery · Fibonacci & tiling recurrences a_n = a_{n-1} + a_{n-2} · State-transition Markov equations · Linearity of expectation E[X + Y] = E[X] + E[Y] · Geometric probability (area/length regions) · Conditional symmetry & Bayes in contests

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Recurrence Relations, States & Contest Probability (Olympiad Module 5 — Combinatorics II & Probability).

● Definition 5.1 — Combinatorial Recurrences, State Equations & Linearity of Expectation

Many AMC/AIME counting and probability problems reduce to setting up a recurrence relation $a_n = c_1 a_{n-1} + c_2 a_{n-2}$ by conditioning on the first step (e.g., tiling a $2 \times n$ board with dominoes yields the Fibonacci numbers $F_{n+1}$ ). For random variables $X_1, \dots, X_n$ , Linearity of Expectation $E[X_1 + X_2 + \dots + X_n] = E[X_1] + E[X_2] + \dots + E[X_n]$ holds even when the variables are dependent, making indicator variables the most powerful tool for expected-value problems.

● Deep Conceptual Intuition — Recurrence Relations, States & Contest Probability

Structural Invariants & Analytical Rigor: Every problem in Recurrence Relations, States & Contest Probability reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 1)^2 - 2

and secant line

g(x) = x + 1

illustrating Recurrence Relations, States & Contest Probability.

⊙ Explorer · Recurrence Dynamics, Fixed Points & State Convergence Lab

interactive

Live numerical table for Recurrence Dynamics, Fixed Points & State Convergence Lab
Parameter / Metric	Value

Growth Parameter $r =$ 2.4

Initial State $x_0 =$ 0.8

Equilibrium State

x^*

2.33

Next State

x_1

1.54

StabilityConvergent Attractor

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Recurrence Relations, States & Contest Probability.

Orthogonal components and Euclidean distance relationship

(18)^2 + (24)^2 = (30)^2

for Recurrence Relations, States & Contest Probability.

Key Principle 5.1

Fibonacci & tiling recurrences a_n = a_{n-1} + a_{n-2}

Key Principle 5.2

State-transition Markov equations

Key Principle 5.3

Linearity of expectation E[X + Y] = E[X] + E[Y]

Key Principle 5.4

Geometric probability (area/length regions)

◆ Why It Works — First-Principles Derivation (Recurrence Relations, States & Contest Probability)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

Worked Example 5.A — Foundational Formulation (Recurrence Relations, States & Contest Probability)

Problem: Given the quadratic function $f(x) = x^2 - 12x + 43$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{12}{2} = -6$ .
2. Write $(x - 6)^2 = x^2 - 12x + 36$ .
3. Adjust the constant term: $f(x) = (x - 6)^2 - 36 + 43 = (x - 6)^2 + 7$ .
4. Since $(x - 6)^2 \ge 0$ for all real $x$ , the minimum value is $k = 7$ , achieved at $x = 6$ .

Final Answer: $f(x) = (x - 6)^2 + 7$ ; minimum value $7$ at $x = 6$ .

Worked Example 5.B — Intermediate Multi-Step Synthesis (Recurrence Relations, States & Contest Probability)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 12x + 41$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 12x + 41 = 2x + k \implies x^2 - 14x + (41 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-14)^2 - 4(1)(41 - k) > 0$
$196 - 164 + 4k > 0 \implies 4k > -32 \implies k > -8$

Final Answer: $k > -8$ .

Worked Example 5.C — Advanced Rate & Vector Application (Recurrence Relations, States & Contest Probability)

Problem: Consider the cubic function $g(x) = x^3 - 18x^2 + 108x + 7$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 8$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 36x + 108 = 3(x - 6)^2$
2. Setting $g'(x) = 0$ yields $x = 6$ , with $g(6) = 7$ . Thus $(6, 7)$ is the stationary point.
3. At $x = 8$ , the gradient is $g'(8) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(6, 7)$ ; gradient at $x = 8$ is $12$ .

Worked Example 5.D — Exam-Grade Challenge Synthesis (Recurrence Relations, States & Contest Probability)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 6n^2 + 7n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 6(1)^2 + 7(1) = 13$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[6n^2 + 7n\right] - \left[6(n-1)^2 + 7(n-1)\right] = 12n + (1)$
3. The coefficient of $n$ gives the common difference $d = 12$ .
4. Substituting $n = 10$ yields $u_{10} = 121$ .

Final Answer: $u_n = 12n + 1$ , $d = 12$ , and $u_{10} = 121$ .

⚠️ Common Exam Traps — Recurrence Relations, States & Contest Probability

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Recurrence Relations, States & Contest Probability)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Recurrence Relations, States & Contest Probability

[Practice 1 · Easy] Evaluate

f(7)

for the quadratic function

f(x) = 6x^2 - 6x + 10

💡 Hint 1 of 3

Substitute

x = 7

directly into

f(x) = 6x^2 - 6x + 10

💡 Hint 2 of 3

Factor out

6x(x - 1) + 10 = 6(7)(6) + 10

💡 Hint 3 of 3

Compute

252 + 10

to obtain the final answer.

Substituting

x = 7

gives

f(7) = 6(7)^2 - 6(7) + 10 = 294 - 42 + 10 = 262

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 13x + 42 = 0

that is greater than

6

💡 Hint 1 of 3

Look for two integers whose sum is

13

and whose product is

42

💡 Hint 2 of 3

Factor the quadratic as

(x - 6)(x - 7) = 0

💡 Hint 3 of 3

Compare the two roots

x = 6

and

x = 7

and select the larger root.

Factoring gives

(x - 6)(x - 7) = 0

, so the roots are

x = 6

and

x = 7

. The root greater than

6

x = 7

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 54 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 54

into

(-2k)^2 - 4(1)(54) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 216

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(54) = 4k^2 - 216 = 0 \implies k^2 = 54

[Practice 4 · Hard] An arithmetic progression has first term

a = 6

and common difference

d = 14

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 6, d = 14

S_{10} = 5[2(6) + 9(14)]

💡 Hint 3 of 3

Evaluate

5(12 + 126) = 690

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(6) + 9(14)] = 5(138) = 690

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 6x^2 - 7x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 6x^2 - 7x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 12x - 7

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 12(3) - 7 = 56

Differentiating gives

\frac{dy}{dx} = 3x^2 + 12x - 7

. At

x = 3

\frac{dy}{dx} = 3(9) + 36 - 7 = 56

📝 Chapter Quiz

Questions Level

Power of a Point, Cyclic Quadrilaterals & Angle Chasing

🏆 Olympiad Module 6 — Olympiad Geometry I Core Mastery · Power of a Point (secant-secant, secant-tangent, chord-chord) · Cyclic quadrilateral criteria & Ptolemy's Theorem · Inscribed angle, chord-tangent & radical axis · Homothety & spiral similarity · Shoelace formula & Pick's lattice theorem

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Power of a Point, Cyclic Quadrilaterals & Angle Chasing (Olympiad Module 6 — Olympiad Geometry I).

● Definition 6.1 — Power of a Point, Ptolemy's Theorem & Pick's Lattice Theorem

For a circle of center $O$ and radius $R$ , the Power of a Point $P$ is $|PO^2 - R^2|$ : for any line through $P$ meeting the circle at $A, B$ , $PA \cdot PB = |PO^2 - R^2|$ (and $PT^2 = PA \cdot PB$ for a tangent $PT$ ). For a cyclic quadrilateral $ABCD$ , Ptolemy's Theorem states $AC \cdot BD = AB \cdot CD + BC \cdot DA.$ For a lattice polygon with $I$ interior lattice points and $B$ boundary lattice points, Pick's Theorem gives area $A = I + \frac{B}{2} - 1$ .

● Deep Conceptual Intuition — Power of a Point, Cyclic Quadrilaterals & Angle Chasing

Structural Invariants & Analytical Rigor: Every problem in Power of a Point, Cyclic Quadrilaterals & Angle Chasing reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 2)^2 - 2

and secant line

g(x) = x + 2

illustrating Power of a Point, Cyclic Quadrilaterals & Angle Chasing.

⊙ Explorer · Circle Power of a Point, Chords & Cyclic Geometry Lab

interactive

Live numerical table for Circle Power of a Point, Chords & Cyclic Geometry Lab
Parameter / Metric	Value

Circle Radius $R =$ 3.5

Subtended Angle $\theta =$ 1.8

Chord Length

2R\sin(\theta/2)

5.48

Inscribed Angle

\theta/2

0.90\text{ rad}

Enclosed Segment Area

4.92

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Power of a Point, Cyclic Quadrilaterals & Angle Chasing.

Orthogonal components and Euclidean distance relationship

(21)^2 + (28)^2 = (35)^2

for Power of a Point, Cyclic Quadrilaterals & Angle Chasing.

Key Principle 6.1

Power of a Point (secant-secant, secant-tangent, chord-chord)

Key Principle 6.2

Cyclic quadrilateral criteria & Ptolemy's Theorem

Key Principle 6.3

Inscribed angle, chord-tangent & radical axis

Key Principle 6.4

Homothety & spiral similarity

◆ Why It Works — First-Principles Derivation (Power of a Point, Cyclic Quadrilaterals & Angle Chasing)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

Worked Example 6.A — Foundational Formulation (Power of a Point, Cyclic Quadrilaterals & Angle Chasing)

Problem: Given the quadratic function $f(x) = x^2 - 14x + 57$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{14}{2} = -7$ .
2. Write $(x - 7)^2 = x^2 - 14x + 49$ .
3. Adjust the constant term: $f(x) = (x - 7)^2 - 49 + 57 = (x - 7)^2 + 8$ .
4. Since $(x - 7)^2 \ge 0$ for all real $x$ , the minimum value is $k = 8$ , achieved at $x = 7$ .

Final Answer: $f(x) = (x - 7)^2 + 8$ ; minimum value $8$ at $x = 7$ .

Worked Example 6.B — Intermediate Multi-Step Synthesis (Power of a Point, Cyclic Quadrilaterals & Angle Chasing)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 14x + 54$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 14x + 54 = 2x + k \implies x^2 - 16x + (54 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-16)^2 - 4(1)(54 - k) > 0$
$256 - 216 + 4k > 0 \implies 4k > -40 \implies k > -10$

Final Answer: $k > -10$ .

Worked Example 6.C — Advanced Rate & Vector Application (Power of a Point, Cyclic Quadrilaterals & Angle Chasing)

Problem: Consider the cubic function $g(x) = x^3 - 21x^2 + 147x + 8$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 9$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 42x + 147 = 3(x - 7)^2$
2. Setting $g'(x) = 0$ yields $x = 7$ , with $g(7) = 8$ . Thus $(7, 8)$ is the stationary point.
3. At $x = 9$ , the gradient is $g'(9) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(7, 8)$ ; gradient at $x = 9$ is $12$ .

Worked Example 6.D — Exam-Grade Challenge Synthesis (Power of a Point, Cyclic Quadrilaterals & Angle Chasing)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 7n^2 + 8n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 7(1)^2 + 8(1) = 15$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[7n^2 + 8n\right] - \left[7(n-1)^2 + 8(n-1)\right] = 14n + (1)$
3. The coefficient of $n$ gives the common difference $d = 14$ .
4. Substituting $n = 10$ yields $u_{10} = 141$ .

Final Answer: $u_n = 14n + 1$ , $d = 14$ , and $u_{10} = 141$ .

⚠️ Common Exam Traps — Power of a Point, Cyclic Quadrilaterals & Angle Chasing

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Power of a Point, Cyclic Quadrilaterals & Angle Chasing)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Power of a Point, Cyclic Quadrilaterals & Angle Chasing

[Practice 1 · Easy] Evaluate

f(8)

for the quadratic function

f(x) = 7x^2 - 7x + 10

💡 Hint 1 of 3

Substitute

x = 8

directly into

f(x) = 7x^2 - 7x + 10

💡 Hint 2 of 3

Factor out

7x(x - 1) + 10 = 7(8)(7) + 10

💡 Hint 3 of 3

Compute

392 + 10

to obtain the final answer.

Substituting

x = 8

gives

f(8) = 7(8)^2 - 7(8) + 10 = 448 - 56 + 10 = 402

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 15x + 56 = 0

that is greater than

7

💡 Hint 1 of 3

Look for two integers whose sum is

15

and whose product is

56

💡 Hint 2 of 3

Factor the quadratic as

(x - 7)(x - 8) = 0

💡 Hint 3 of 3

Compare the two roots

x = 7

and

x = 8

and select the larger root.

Factoring gives

(x - 7)(x - 8) = 0

, so the roots are

x = 7

and

x = 8

. The root greater than

7

x = 8

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 63 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 63

into

(-2k)^2 - 4(1)(63) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 252

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(63) = 4k^2 - 252 = 0 \implies k^2 = 63

[Practice 4 · Hard] An arithmetic progression has first term

a = 7

and common difference

d = 16

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 7, d = 16

S_{10} = 5[2(7) + 9(16)]

💡 Hint 3 of 3

Evaluate

5(14 + 144) = 790

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(7) + 9(16)] = 5(158) = 790

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 7x^2 - 8x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 7x^2 - 8x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 14x - 8

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 14(3) - 8 = 61

Differentiating gives

\frac{dy}{dx} = 3x^2 + 14x - 8

. At

x = 3

\frac{dy}{dx} = 3(9) + 42 - 8 = 61

📝 Chapter Quiz

Questions Level

Ceva's, Menelaus', Stewart's Theorems & Triangle Centers

🏆 Olympiad Module 7 — Olympiad Geometry II Core Mastery · Stewart's Theorem man + dad = bmb + cnc · Ceva's concurrency & Menelaus' collinearity theorems · Angle Bisector Theorem & length formula · Inradius r = A/s, Circumradius R = abc/(4A) & Heron's formula · Mass points & area-ratio chasing

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Ceva's, Menelaus', Stewart's Theorems & Triangle Centers (Olympiad Module 7 — Olympiad Geometry II).

● Definition 7.1 — Stewart's Theorem, Ceva's Theorem & Triangle Radii Formulas

In $\triangle ABC$ with sides $a, b, c$ and semiperimeter $s = \frac{a+b+c}{2}$ : (1) Heron's Formula & Radii: $[ABC] = \sqrt{s(s-a)(s-b)(s-c)} = rs = \frac{abc}{4R}$ ; (2) Stewart's Theorem: for a cevian of length $d$ dividing side $a$ into segments $m$ and $n$ , $b^2 m + c^2 n = a(d^2 + mn);$ (3) Ceva's Theorem: cevians $AD, BE, CF$ are concurrent if and only if $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$ .

● Deep Conceptual Intuition — Ceva's, Menelaus', Stewart's Theorems & Triangle Centers

Structural Invariants & Analytical Rigor: Every problem in Ceva's, Menelaus', Stewart's Theorems & Triangle Centers reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 0)^2 - 2

and secant line

g(x) = x + 1

illustrating Ceva's, Menelaus', Stewart's Theorems & Triangle Centers.

⊙ Explorer · Triangle Cevians, Angle Bisector Ratio & Area Partition Lab

interactive

Live numerical table for Triangle Cevians, Angle Bisector Ratio & Area Partition Lab
Parameter / Metric	Value

Base Ratio $k =$ 1.5

Altitude Scale $h =$ 2.5

Linear Scale

k

1.50

Area Ratio

k^2

2.25

Perimeter Ratio

1.50

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Ceva's, Menelaus', Stewart's Theorems & Triangle Centers.

Orthogonal components and Euclidean distance relationship

(24)^2 + (32)^2 = (40)^2

for Ceva's, Menelaus', Stewart's Theorems & Triangle Centers.

Key Principle 7.1

Stewart's Theorem man + dad = bmb + cnc

Key Principle 7.2

Ceva's concurrency & Menelaus' collinearity theorems

Key Principle 7.3

Angle Bisector Theorem & length formula

Key Principle 7.4

Inradius r = A/s, Circumradius R = abc/(4A) & Heron's formula

◆ Why It Works — First-Principles Derivation (Ceva's, Menelaus', Stewart's Theorems & Triangle Centers)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

Worked Example 7.A — Foundational Formulation (Ceva's, Menelaus', Stewart's Theorems & Triangle Centers)

Problem: Given the quadratic function $f(x) = x^2 - 16x + 73$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{16}{2} = -8$ .
2. Write $(x - 8)^2 = x^2 - 16x + 64$ .
3. Adjust the constant term: $f(x) = (x - 8)^2 - 64 + 73 = (x - 8)^2 + 9$ .
4. Since $(x - 8)^2 \ge 0$ for all real $x$ , the minimum value is $k = 9$ , achieved at $x = 8$ .

Final Answer: $f(x) = (x - 8)^2 + 9$ ; minimum value $9$ at $x = 8$ .

Worked Example 7.B — Intermediate Multi-Step Synthesis (Ceva's, Menelaus', Stewart's Theorems & Triangle Centers)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 16x + 69$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 16x + 69 = 2x + k \implies x^2 - 18x + (69 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-18)^2 - 4(1)(69 - k) > 0$
$324 - 276 + 4k > 0 \implies 4k > -48 \implies k > -12$

Final Answer: $k > -12$ .

Worked Example 7.C — Advanced Rate & Vector Application (Ceva's, Menelaus', Stewart's Theorems & Triangle Centers)

Problem: Consider the cubic function $g(x) = x^3 - 24x^2 + 192x + 9$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 10$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 48x + 192 = 3(x - 8)^2$
2. Setting $g'(x) = 0$ yields $x = 8$ , with $g(8) = 9$ . Thus $(8, 9)$ is the stationary point.
3. At $x = 10$ , the gradient is $g'(10) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(8, 9)$ ; gradient at $x = 10$ is $12$ .

Worked Example 7.D — Exam-Grade Challenge Synthesis (Ceva's, Menelaus', Stewart's Theorems & Triangle Centers)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 8n^2 + 9n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 8(1)^2 + 9(1) = 17$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[8n^2 + 9n\right] - \left[8(n-1)^2 + 9(n-1)\right] = 16n + (1)$
3. The coefficient of $n$ gives the common difference $d = 16$ .
4. Substituting $n = 10$ yields $u_{10} = 161$ .

Final Answer: $u_n = 16n + 1$ , $d = 16$ , and $u_{10} = 161$ .

⚠️ Common Exam Traps — Ceva's, Menelaus', Stewart's Theorems & Triangle Centers

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Ceva's, Menelaus', Stewart's Theorems & Triangle Centers)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Ceva's, Menelaus', Stewart's Theorems & Triangle Centers

[Practice 1 · Easy] Evaluate

f(9)

for the quadratic function

f(x) = 8x^2 - 8x + 10

💡 Hint 1 of 3

Substitute

x = 9

directly into

f(x) = 8x^2 - 8x + 10

💡 Hint 2 of 3

Factor out

8x(x - 1) + 10 = 8(9)(8) + 10

💡 Hint 3 of 3

Compute

576 + 10

to obtain the final answer.

Substituting

x = 9

gives

f(9) = 8(9)^2 - 8(9) + 10 = 648 - 72 + 10 = 586

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 17x + 72 = 0

that is greater than

8

💡 Hint 1 of 3

Look for two integers whose sum is

17

and whose product is

72

💡 Hint 2 of 3

Factor the quadratic as

(x - 8)(x - 9) = 0

💡 Hint 3 of 3

Compare the two roots

x = 8

and

x = 9

and select the larger root.

Factoring gives

(x - 8)(x - 9) = 0

, so the roots are

x = 8

and

x = 9

. The root greater than

8

x = 9

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 72 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 72

into

(-2k)^2 - 4(1)(72) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 288

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(72) = 4k^2 - 288 = 0 \implies k^2 = 72

[Practice 4 · Hard] An arithmetic progression has first term

a = 8

and common difference

d = 18

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 8, d = 18

S_{10} = 5[2(8) + 9(18)]

💡 Hint 3 of 3

Evaluate

5(16 + 162) = 890

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(8) + 9(18)] = 5(178) = 890

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 8x^2 - 9x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 8x^2 - 9x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 16x - 9

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 16(3) - 9 = 66

Differentiating gives

\frac{dy}{dx} = 3x^2 + 16x - 9

. At

x = 3

\frac{dy}{dx} = 3(9) + 48 - 9 = 66

📝 Chapter Quiz

Questions Level

Invariants, Monovariants, Game Strategy & Extremal Principle

🏆 Olympiad Module 8 — Olympiad Strategy Core Mastery · Parity & modular invariants in board operations · Chessboard coloring proofs (domino/tetromino tilings) · Monovariants & termination proofs · Winning vs losing positions in impartial games (Nim/symmetry) · The Extremal Principle

Mastering the essential concepts, analytical methods, and exam-grade problem-solving techniques for Invariants, Monovariants, Game Strategy & Extremal Principle (Olympiad Module 8 — Olympiad Strategy).

● Definition 8.1 — Invariants, Coloring Proofs & Game-Theoretic Winning Strategies

An invariant is a quantity (such as parity $\bmod 2$ , remainder $\bmod k$ , algebraic product, or alternating color sum) that remains unchanged under every legal operation. If the initial state and target state have different values of the invariant, the target state is provably unreachable. In two-player impartial games, a position is winning (N-position) if at least one legal move leads to a losing (P-position), and losing if every legal move leads to a winning position; symmetry and invariant-pairing strategies force the opponent into P-positions.

● Deep Conceptual Intuition — Invariants, Monovariants, Game Strategy & Extremal Principle

Structural Invariants & Analytical Rigor: Every problem in Invariants, Monovariants, Game Strategy & Extremal Principle reduces to identifying the underlying algebraic or geometric structure before computing. Whether isolating parameters, verifying domain constraints, or exploiting symmetry, writing down the governing identity first eliminates arithmetic traps and guarantees full method marks.

Coordinate representation of

f(x) = (x - 1)^2 - 2

and secant line

g(x) = x + 2

illustrating Invariants, Monovariants, Game Strategy & Extremal Principle.

⊙ Explorer · Invariant Sums, Monovariant Decay & Game State Lab

interactive

Live numerical table for Invariant Sums, Monovariant Decay & Game State Lab
Parameter / Metric	Value

Initial Invariant $I_0 =$ 4

Step Contraction $r =$ 0.5

Asymptotic Limit

I_0/(1-r)

8.00

Partial State

S_5

7.75

Monovariant Remainder

0.25

Drag the sliders or directly manipulate the canvas to explore how changes in the parameters govern Invariants, Monovariants, Game Strategy & Extremal Principle.

Orthogonal components and Euclidean distance relationship

(27)^2 + (36)^2 = (45)^2

for Invariants, Monovariants, Game Strategy & Extremal Principle.

Key Principle 8.1

Parity & modular invariants in board operations

Key Principle 8.2

Chessboard coloring proofs (domino/tetromino tilings)

Key Principle 8.3

Monovariants & termination proofs

Key Principle 8.4

Winning vs losing positions in impartial games (Nim/symmetry)

◆ Why It Works — First-Principles Derivation (Invariants, Monovariants, Game Strategy & Extremal Principle)

First-Principles Justification: Any quadratic or bilinear relation $ax^2 + bx + c = 0$ can be transformed by completing the square:

a\!\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

Worked Example 8.A — Foundational Formulation (Invariants, Monovariants, Game Strategy & Extremal Principle)

Problem: Given the quadratic function $f(x) = x^2 - 18x + 91$ , express $f(x)$ in the completed-square form $(x - h)^2 + k$ and state its minimum value and the value of $x$ at which it occurs.

Step-by-Step Solution:
1. Halve the coefficient of $x$ : $-\frac{18}{2} = -9$ .
2. Write $(x - 9)^2 = x^2 - 18x + 81$ .
3. Adjust the constant term: $f(x) = (x - 9)^2 - 81 + 91 = (x - 9)^2 + 10$ .
4. Since $(x - 9)^2 \ge 0$ for all real $x$ , the minimum value is $k = 10$ , achieved at $x = 9$ .

Final Answer: $f(x) = (x - 9)^2 + 10$ ; minimum value $10$ at $x = 9$ .

Worked Example 8.B — Intermediate Multi-Step Synthesis (Invariants, Monovariants, Game Strategy & Extremal Principle)

Problem: The line $y = 2x + k$ intersects the curve $y = x^2 - 18x + 86$ at two distinct points. Find the range of possible values for the constant $k$ .

Step-by-Step Solution:
1. Equate the line and curve equations:
$x^2 - 18x + 86 = 2x + k \implies x^2 - 20x + (86 - k) = 0$
2. For two distinct real intersections, the discriminant must satisfy $\Delta = b^2 - 4ac > 0$ :
$(-20)^2 - 4(1)(86 - k) > 0$
$400 - 344 + 4k > 0 \implies 4k > -56 \implies k > -14$

Final Answer: $k > -14$ .

Worked Example 8.C — Advanced Rate & Vector Application (Invariants, Monovariants, Game Strategy & Extremal Principle)

Problem: Consider the cubic function $g(x) = x^3 - 27x^2 + 243x + 10$ . Find the coordinates of the stationary point of inflection and the gradient of the curve at $x = 11$ .

Step-by-Step Solution:
1. Differentiate term by term:
$g'(x) = 3x^2 - 54x + 243 = 3(x - 9)^2$
2. Setting $g'(x) = 0$ yields $x = 9$ , with $g(9) = 10$ . Thus $(9, 10)$ is the stationary point.
3. At $x = 11$ , the gradient is $g'(11) = 3(2)^2 = 12$ .

Final Answer: Stationary point $(9, 10)$ ; gradient at $x = 11$ is $12$ .

Worked Example 8.D — Exam-Grade Challenge Synthesis (Invariants, Monovariants, Game Strategy & Extremal Principle)

Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression where $S_n = 9n^2 + 10n$ . Find an expression for the $n$ -th term $u_n$ , determine the common difference $d$ , and evaluate $u_{10}$ .

Step-by-Step Solution:
1. For $n = 1$ , $u_1 = S_1 = 9(1)^2 + 10(1) = 19$ .
2. For $n \ge 2$ , use the fundamental identity $u_n = S_n - S_{n-1}$ :
$u_n = \left[9n^2 + 10n\right] - \left[9(n-1)^2 + 10(n-1)\right] = 18n + (1)$
3. The coefficient of $n$ gives the common difference $d = 18$ .
4. Substituting $n = 10$ yields $u_{10} = 181$ .

Final Answer: $u_n = 18n + 1$ , $d = 18$ , and $u_{10} = 181$ .

⚠️ Common Exam Traps — Invariants, Monovariants, Game Strategy & Extremal Principle

1. Sign & Domain Omissions: Always verify whether negative roots or boundary values violate denominator, square-root, or logarithmic domain restrictions before writing your final answer.

2. Premature Rounding: Keep exact fractions, surds, or $\pi$ expressions throughout intermediate steps and round only at the final line if requested.

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Invariants, Monovariants, Game Strategy & Extremal Principle)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Canceling terms across a sum in $\frac{x^2 + a}{x + a}$	Division only cancels common multiplicative factors of the entire numerator and denominator.	Factorize numerator and denominator completely before canceling common binomial factors.
Writing $(a + b)^2 = a^2 + b^2$ when squaring binomials	Omits the cross-product term $2ab$ required by the distributive law.	Always expand as $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Solving quadratic inequalities by treating $ < $ like $ = $ without a sign chart	A parabola changes sign at its distinct real roots, so the solution is an interval or union of rays.	Find the critical roots first, then sketch the parabola or test a point in each interval.

▲ Exam Pacing & Verification Strategy — Invariants, Monovariants, Game Strategy & Extremal Principle

[Practice 1 · Easy] Evaluate

f(10)

for the quadratic function

f(x) = 9x^2 - 9x + 10

💡 Hint 1 of 3

Substitute

x = 10

directly into

f(x) = 9x^2 - 9x + 10

💡 Hint 2 of 3

Factor out

9x(x - 1) + 10 = 9(10)(9) + 10

💡 Hint 3 of 3

Compute

810 + 10

to obtain the final answer.

Substituting

x = 10

gives

f(10) = 9(10)^2 - 9(10) + 10 = 900 - 90 + 10 = 820

[Practice 2 · Easy] Find the positive root of the equation

x^2 - 19x + 90 = 0

that is greater than

9

💡 Hint 1 of 3

Look for two integers whose sum is

19

and whose product is

90

💡 Hint 2 of 3

Factor the quadratic as

(x - 9)(x - 10) = 0

💡 Hint 3 of 3

Compare the two roots

x = 9

and

x = 10

and select the larger root.

Factoring gives

(x - 9)(x - 10) = 0

, so the roots are

x = 9

and

x = 10

. The root greater than

9

x = 10

[Practice 3 · Medium] If the quadratic equation

x^2 - 2kx + 81 = 0

has equal (repeated) real roots and

k > 0

, find the exact value of

k^2

💡 Hint 1 of 3

For equal real roots, the discriminant

\Delta = b^2 - 4ac

must equal zero.

💡 Hint 2 of 3

Substitute

a = 1, b = -2k, c = 81

into

(-2k)^2 - 4(1)(81) = 0

💡 Hint 3 of 3

Simplify

4k^2 = 324

and divide both sides by 4.

Setting the discriminant to zero:

\Delta = (-2k)^2 - 4(1)(81) = 4k^2 - 324 = 0 \implies k^2 = 81

[Practice 4 · Hard] An arithmetic progression has first term

a = 9

and common difference

d = 20

. Find the sum of the first

10

terms,

S_{10}

💡 Hint 1 of 3

Recall the arithmetic series sum formula

S_n = \frac{n}{2}[2a + (n - 1)d]

💡 Hint 2 of 3

Substitute

n = 10, a = 9, d = 20

S_{10} = 5[2(9) + 9(20)]

💡 Hint 3 of 3

Evaluate

5(18 + 180) = 990

Using

S_n = \frac{n}{2}[2a + (n - 1)d]

, we have

S_{10} = \frac{10}{2}[2(9) + 9(20)] = 5(198) = 990

[Practice 5 · Hard] Find the instantaneous rate of change

\frac{dy}{dx}

of the curve

y = x^3 + 9x^2 - 10x + 7

x = 3

💡 Hint 1 of 3

Differentiate

y = x^3 + 9x^2 - 10x + 7

term by term with respect to

x

💡 Hint 2 of 3

The derivative is

\frac{dy}{dx} = 3x^2 + 18x - 10

💡 Hint 3 of 3

Substitute

x = 3

3(3)^2 + 18(3) - 10 = 71

Differentiating gives

\frac{dy}{dx} = 3x^2 + 18x - 10

. At

x = 3

\frac{dy}{dx} = 3(9) + 54 - 10 = 71

📝 Chapter Quiz

Questions Level

Practice set

Fifty graded Contest Math & Olympiad problems spanning all eight core chapters. Click any problem for the complete step-by-step worked solution.

[Olympiad Algebra, Symmetric Sums & Vieta's Relations] Find the equation of the tangent line to the curve $y = x^3 - 3x^2 + 6x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Basic

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 3(2)^2 + 6(2) + 2 = 8 - 12 + 12 + 2 = 10.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 6x + 6.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 6(2) + 6 = 6$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (10) = 6(x - 2) \implies y = 6x + (-2).$
Final Answer: $y = 6x + (-2)$ , crossing the $y$ -axis at $(0, -2)$ .

[Divisibility, Prime Factorization & Modular Arithmetic] An arithmetic sequence has first term $u_1 = 4$ and common difference $d = 9$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Basic

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 4$ and common difference $d = 9$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 4 + 11(9) = 4 + 99 = 103.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(4 + 103) = 6(107) = 642.$
Final Answer: $u_{12} = 103$ and $S_{12} = 642$ .

[Diophantine Equations, Floor Functions & Number Bases] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 15\text{ cm}$ and $AC = 20\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Basic

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(15)^2 + (20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(15)(20) = 150\text{ cm}^2$ , and the semiperimeter is $s = \frac{15 + 20 + 25}{2} = 30\text{ cm}$ .

Step 3 (Inradius): Using the identity $[ABC] = r \cdot s$ :
$r = \frac{[ABC]}{s} = \frac{150}{30} = 5\text{ cm}.$
Final Answer: $AB = 25\text{ cm},\; \text{Area} = 150\text{ cm}^2,\; r = 5\text{ cm}$ .

[Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion] The function $f(x) = \frac{6x + 7}{x - 6}$ is defined for $x \ne 6$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 7$ .

Basic

Step 1 (Algebraic Inversion): Set $y = \frac{6x + 7}{x - 6}$ and multiply both sides by $(x - 6)$ :
$y(x - 6) = 6x + 7 \implies xy - 6y = 6x + 7 \implies x(y - 6) = 6y + 7.$

Step 2 (Domain of Inverse): Dividing by $(y - 6)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{6x + 7}{x - 6}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 6.$

Step 3 (Solving $f(x) = 7$ ): Since $f$ is self-inverse, $x = f^{-1}(7) = \frac{6(7) + 7}{(7) - 6} = 49$ .
Final Answer: $f^{-1}(x) = \frac{6x + 7}{x - 6} \; (x \ne 6)$ , and $x = 49$ .

[Recurrence Relations, States & Contest Probability] Solve the logarithmic equation $\log_2(x - 7) + \log_2(x + 7) = \log_2(147)$ for $x > 7$ , showing why the negative root must be rejected.

Basic

Step 1 (Combine Logarithms): Using the logarithmic product identity $\log_2(u) + \log_2(v) = \log_2(uv)$ on the domain $x > 7$ :
$\log_2\big((x - 7)(x + 7)\big) = \log_2(147) \implies x^2 - 49 = 147.$

Step 2 (Solve Quadratic & Check Domain): Rearranging gives $x^2 = 196 \implies x = \pm 14$ . Since $\log_2(x - 7)$ requires $x - 7 > 0 \iff x > 7$ , the negative root $x = -14$ is extraneous and must be rejected.
Final Answer: $x = 14$ .

[Power of a Point, Cyclic Quadrilaterals & Angle Chasing] Find the equation of the tangent line to the curve $y = x^3 - 8x^2 + 5x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Basic

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 8(2)^2 + 5(2) + 2 = 8 - 32 + 10 + 2 = -12.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 16x + 5.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 16(2) + 5 = -15$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (-12) = -15(x - 2) \implies y = -15x + (18).$
Final Answer: $y = -15x + (18)$ , crossing the $y$ -axis at $(0, 18)$ .

[Ceva's, Menelaus', Stewart's Theorems & Triangle Centers] An arithmetic sequence has first term $u_1 = 2$ and common difference $d = 8$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Basic

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 2$ and common difference $d = 8$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 2 + 11(8) = 2 + 88 = 90.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(2 + 90) = 6(92) = 552.$
Final Answer: $u_{12} = 90$ and $S_{12} = 552$ .

[Invariants, Monovariants, Game Strategy & Extremal Principle] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 9\text{ cm}$ and $AC = 12\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Basic

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(9)^2 + (12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(9)(12) = 54\text{ cm}^2$ , and the semiperimeter is $s = \frac{9 + 12 + 15}{2} = 18\text{ cm}$ .

Step 3 (Inradius): Using the identity $[ABC] = r \cdot s$ :
$r = \frac{[ABC]}{s} = \frac{54}{18} = 3\text{ cm}.$
Final Answer: $AB = 15\text{ cm},\; \text{Area} = 54\text{ cm}^2,\; r = 3\text{ cm}$ .

[Olympiad Algebra, Symmetric Sums & Vieta's Relations] The function $f(x) = \frac{4x + 6}{x - 4}$ is defined for $x \ne 4$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 5$ .

Basic

Step 1 (Algebraic Inversion): Set $y = \frac{4x + 6}{x - 4}$ and multiply both sides by $(x - 4)$ :
$y(x - 4) = 4x + 6 \implies xy - 4y = 4x + 6 \implies x(y - 4) = 4y + 6.$

Step 2 (Domain of Inverse): Dividing by $(y - 4)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{4x + 6}{x - 4}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 4.$

Step 3 (Solving $f(x) = 5$ ): Since $f$ is self-inverse, $x = f^{-1}(5) = \frac{4(5) + 6}{(5) - 4} = 26$ .
Final Answer: $f^{-1}(x) = \frac{4x + 6}{x - 4} \; (x \ne 4)$ , and $x = 26$ .

[Divisibility, Prime Factorization & Modular Arithmetic] Solve the logarithmic equation $\log_2(x - 5) + \log_2(x + 5) = \log_2(75)$ for $x > 5$ , showing why the negative root must be rejected.

Basic

Step 1 (Combine Logarithms): Using the logarithmic product identity $\log_2(u) + \log_2(v) = \log_2(uv)$ on the domain $x > 5$ :
$\log_2\big((x - 5)(x + 5)\big) = \log_2(75) \implies x^2 - 25 = 75.$

Step 2 (Solve Quadratic & Check Domain): Rearranging gives $x^2 = 100 \implies x = \pm 10$ . Since $\log_2(x - 5)$ requires $x - 5 > 0 \iff x > 5$ , the negative root $x = -10$ is extraneous and must be rejected.
Final Answer: $x = 10$ .

[Diophantine Equations, Floor Functions & Number Bases] Find the equation of the tangent line to the curve $y = x^3 - 6x^2 + 4x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Basic

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 6(2)^2 + 4(2) + 2 = 8 - 24 + 8 + 2 = -6.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 12x + 4.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 12(2) + 4 = -8$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (-6) = -8(x - 2) \implies y = -8x + (10).$
Final Answer: $y = -8x + (10)$ , crossing the $y$ -axis at $(0, 10)$ .

[Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion] An arithmetic sequence has first term $u_1 = 7$ and common difference $d = 7$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Basic

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 7$ and common difference $d = 7$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 7 + 11(7) = 7 + 77 = 84.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(7 + 84) = 6(91) = 546.$
Final Answer: $u_{12} = 84$ and $S_{12} = 546$ .

[Recurrence Relations, States & Contest Probability] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 24\text{ cm}$ and $AC = 32\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Basic

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(24)^2 + (32)^2} = \sqrt{576 + 1024} = \sqrt{1600} = 40\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(24)(32) = 384\text{ cm}^2$ , and the semiperimeter is $s = \frac{24 + 32 + 40}{2} = 48\text{ cm}$ .

Step 3 (Inradius): Using the identity $[ABC] = r \cdot s$ :
$r = \frac{[ABC]}{s} = \frac{384}{48} = 8\text{ cm}.$
Final Answer: $AB = 40\text{ cm},\; \text{Area} = 384\text{ cm}^2,\; r = 8\text{ cm}$ .

[Power of a Point, Cyclic Quadrilaterals & Angle Chasing] The function $f(x) = \frac{2x + 5}{x - 2}$ is defined for $x \ne 2$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 3$ .

Basic

Step 1 (Algebraic Inversion): Set $y = \frac{2x + 5}{x - 2}$ and multiply both sides by $(x - 2)$ :
$y(x - 2) = 2x + 5 \implies xy - 2y = 2x + 5 \implies x(y - 2) = 2y + 5.$

Step 2 (Domain of Inverse): Dividing by $(y - 2)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{2x + 5}{x - 2}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 2.$

Step 3 (Solving $f(x) = 3$ ): Since $f$ is self-inverse, $x = f^{-1}(3) = \frac{2(3) + 5}{(3) - 2} = 11$ .
Final Answer: $f^{-1}(x) = \frac{2x + 5}{x - 2} \; (x \ne 2)$ , and $x = 11$ .

[Ceva's, Menelaus', Stewart's Theorems & Triangle Centers] Solve the logarithmic equation $\log_2(x - 3) + \log_2(x + 3) = \log_2(27)$ for $x > 3$ , showing why the negative root must be rejected.

Basic

Step 1 (Combine Logarithms): Using the logarithmic product identity $\log_2(u) + \log_2(v) = \log_2(uv)$ on the domain $x > 3$ :
$\log_2\big((x - 3)(x + 3)\big) = \log_2(27) \implies x^2 - 9 = 27.$

Step 2 (Solve Quadratic & Check Domain): Rearranging gives $x^2 = 36 \implies x = \pm 6$ . Since $\log_2(x - 3)$ requires $x - 3 > 0 \iff x > 3$ , the negative root $x = -6$ is extraneous and must be rejected.
Final Answer: $x = 6$ .

[Invariants, Monovariants, Game Strategy & Extremal Principle] Find the equation of the tangent line to the curve $y = x^3 - 4x^2 + 3x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Basic

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 4(2)^2 + 3(2) + 2 = 8 - 16 + 6 + 2 = 0.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 8x + 3.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 8(2) + 3 = -1$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (0) = -1(x - 2) \implies y = -1x + (2).$
Final Answer: $y = -1x + (2)$ , crossing the $y$ -axis at $(0, 2)$ .

[Olympiad Algebra, Symmetric Sums & Vieta's Relations] An arithmetic sequence has first term $u_1 = 5$ and common difference $d = 6$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Intermediate

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 5$ and common difference $d = 6$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 5 + 11(6) = 5 + 66 = 71.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(5 + 71) = 6(76) = 456.$
Final Answer: $u_{12} = 71$ and $S_{12} = 456$ .

[Divisibility, Prime Factorization & Modular Arithmetic] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 18\text{ cm}$ and $AC = 24\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Intermediate

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(18)^2 + (24)^2} = \sqrt{324 + 576} = \sqrt{900} = 30\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(18)(24) = 216\text{ cm}^2$ , and the semiperimeter is $s = \frac{18 + 24 + 30}{2} = 36\text{ cm}$ .

Step 3 (Inradius): Using the identity $[ABC] = r \cdot s$ :
$r = \frac{[ABC]}{s} = \frac{216}{36} = 6\text{ cm}.$
Final Answer: $AB = 30\text{ cm},\; \text{Area} = 216\text{ cm}^2,\; r = 6\text{ cm}$ .

[Diophantine Equations, Floor Functions & Number Bases] The function $f(x) = \frac{7x + 4}{x - 7}$ is defined for $x \ne 7$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 8$ .

Intermediate

Step 1 (Algebraic Inversion): Set $y = \frac{7x + 4}{x - 7}$ and multiply both sides by $(x - 7)$ :
$y(x - 7) = 7x + 4 \implies xy - 7y = 7x + 4 \implies x(y - 7) = 7y + 4.$

Step 2 (Domain of Inverse): Dividing by $(y - 7)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{7x + 4}{x - 7}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 7.$

Step 3 (Solving $f(x) = 8$ ): Since $f$ is self-inverse, $x = f^{-1}(8) = \frac{7(8) + 4}{(8) - 7} = 60$ .
Final Answer: $f^{-1}(x) = \frac{7x + 4}{x - 7} \; (x \ne 7)$ , and $x = 60$ .

[Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion] Solve the logarithmic equation $\log_2(x - 8) + \log_2(x + 8) = \log_2(192)$ for $x > 8$ , showing why the negative root must be rejected.

Intermediate

Step 1 (Combine Logarithms): Using the logarithmic product identity $\log_2(u) + \log_2(v) = \log_2(uv)$ on the domain $x > 8$ :
$\log_2\big((x - 8)(x + 8)\big) = \log_2(192) \implies x^2 - 64 = 192.$

Step 2 (Solve Quadratic & Check Domain): Rearranging gives $x^2 = 256 \implies x = \pm 16$ . Since $\log_2(x - 8)$ requires $x - 8 > 0 \iff x > 8$ , the negative root $x = -16$ is extraneous and must be rejected.
Final Answer: $x = 16$ .

[Recurrence Relations, States & Contest Probability] Find the equation of the tangent line to the curve $y = x^3 - 2x^2 + 10x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Intermediate

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 2(2)^2 + 10(2) + 2 = 8 - 8 + 20 + 2 = 22.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 4x + 10.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 4(2) + 10 = 14$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (22) = 14(x - 2) \implies y = 14x + (-6).$
Final Answer: $y = 14x + (-6)$ , crossing the $y$ -axis at $(0, -6)$ .

[Power of a Point, Cyclic Quadrilaterals & Angle Chasing] An arithmetic sequence has first term $u_1 = 3$ and common difference $d = 5$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Intermediate

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 3$ and common difference $d = 5$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 3 + 11(5) = 3 + 55 = 58.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(3 + 58) = 6(61) = 366.$
Final Answer: $u_{12} = 58$ and $S_{12} = 366$ .

[Ceva's, Menelaus', Stewart's Theorems & Triangle Centers] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 12\text{ cm}$ and $AC = 16\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Intermediate

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(12)(16) = 96\text{ cm}^2$ , and the semiperimeter is $s = \frac{12 + 16 + 20}{2} = 24\text{ cm}$ .

Step 3 (Inradius): Using the identity $[ABC] = r \cdot s$ :
$r = \frac{[ABC]}{s} = \frac{96}{24} = 4\text{ cm}.$
Final Answer: $AB = 20\text{ cm},\; \text{Area} = 96\text{ cm}^2,\; r = 4\text{ cm}$ .

[Invariants, Monovariants, Game Strategy & Extremal Principle] The function $f(x) = \frac{5x + 3}{x - 5}$ is defined for $x \ne 5$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 6$ .

Intermediate

Step 1 (Algebraic Inversion): Set $y = \frac{5x + 3}{x - 5}$ and multiply both sides by $(x - 5)$ :
$y(x - 5) = 5x + 3 \implies xy - 5y = 5x + 3 \implies x(y - 5) = 5y + 3.$

Step 2 (Domain of Inverse): Dividing by $(y - 5)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{5x + 3}{x - 5}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 5.$

Step 3 (Solving $f(x) = 6$ ): Since $f$ is self-inverse, $x = f^{-1}(6) = \frac{5(6) + 3}{(6) - 5} = 33$ .
Final Answer: $f^{-1}(x) = \frac{5x + 3}{x - 5} \; (x \ne 5)$ , and $x = 33$ .

[Olympiad Algebra, Symmetric Sums & Vieta's Relations] Solve the logarithmic equation $\log_2(x - 6) + \log_2(x + 6) = \log_2(108)$ for $x > 6$ , showing why the negative root must be rejected.

Intermediate

Step 1 (Combine Logarithms): Using the logarithmic product identity $\log_2(u) + \log_2(v) = \log_2(uv)$ on the domain $x > 6$ :
$\log_2\big((x - 6)(x + 6)\big) = \log_2(108) \implies x^2 - 36 = 108.$

Step 2 (Solve Quadratic & Check Domain): Rearranging gives $x^2 = 144 \implies x = \pm 12$ . Since $\log_2(x - 6)$ requires $x - 6 > 0 \iff x > 6$ , the negative root $x = -12$ is extraneous and must be rejected.
Final Answer: $x = 12$ .

[Divisibility, Prime Factorization & Modular Arithmetic] Find the equation of the tangent line to the curve $y = x^3 - 7x^2 + 9x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Intermediate

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 7(2)^2 + 9(2) + 2 = 8 - 28 + 18 + 2 = 0.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 14x + 9.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 14(2) + 9 = -7$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (0) = -7(x - 2) \implies y = -7x + (14).$
Final Answer: $y = -7x + (14)$ , crossing the $y$ -axis at $(0, 14)$ .

[Diophantine Equations, Floor Functions & Number Bases] An arithmetic sequence has first term $u_1 = 8$ and common difference $d = 4$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Intermediate

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 8$ and common difference $d = 4$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 8 + 11(4) = 8 + 44 = 52.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(8 + 52) = 6(60) = 360.$
Final Answer: $u_{12} = 52$ and $S_{12} = 360$ .

[Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 6\text{ cm}$ and $AC = 8\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Intermediate

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(6)(8) = 24\text{ cm}^2$ , and the semiperimeter is $s = \frac{6 + 8 + 10}{2} = 12\text{ cm}$ .

Step 3 (Inradius): Using the identity $[ABC] = r \cdot s$ :
$r = \frac{[ABC]}{s} = \frac{24}{12} = 2\text{ cm}.$
Final Answer: $AB = 10\text{ cm},\; \text{Area} = 24\text{ cm}^2,\; r = 2\text{ cm}$ .

[Recurrence Relations, States & Contest Probability] The function $f(x) = \frac{3x + 10}{x - 3}$ is defined for $x \ne 3$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 4$ .

Intermediate

Step 1 (Algebraic Inversion): Set $y = \frac{3x + 10}{x - 3}$ and multiply both sides by $(x - 3)$ :
$y(x - 3) = 3x + 10 \implies xy - 3y = 3x + 10 \implies x(y - 3) = 3y + 10.$

Step 2 (Domain of Inverse): Dividing by $(y - 3)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{3x + 10}{x - 3}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 3.$

Step 3 (Solving $f(x) = 4$ ): Since $f$ is self-inverse, $x = f^{-1}(4) = \frac{3(4) + 10}{(4) - 3} = 22$ .
Final Answer: $f^{-1}(x) = \frac{3x + 10}{x - 3} \; (x \ne 3)$ , and $x = 22$ .

[Power of a Point, Cyclic Quadrilaterals & Angle Chasing] Solve the logarithmic equation $\log_2(x - 4) + \log_2(x + 4) = \log_2(48)$ for $x > 4$ , showing why the negative root must be rejected.

Intermediate

Step 1 (Combine Logarithms): Using the logarithmic product identity $\log_2(u) + \log_2(v) = \log_2(uv)$ on the domain $x > 4$ :
$\log_2\big((x - 4)(x + 4)\big) = \log_2(48) \implies x^2 - 16 = 48.$

Step 2 (Solve Quadratic & Check Domain): Rearranging gives $x^2 = 64 \implies x = \pm 8$ . Since $\log_2(x - 4)$ requires $x - 4 > 0 \iff x > 4$ , the negative root $x = -8$ is extraneous and must be rejected.
Final Answer: $x = 8$ .

[Ceva's, Menelaus', Stewart's Theorems & Triangle Centers] Find the equation of the tangent line to the curve $y = x^3 - 5x^2 + 8x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Intermediate

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 5(2)^2 + 8(2) + 2 = 8 - 20 + 16 + 2 = 6.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 10x + 8.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 10(2) + 8 = 0$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (6) = 0(x - 2) \implies y = 0x + (6).$
Final Answer: $y = 0x + (6)$ , crossing the $y$ -axis at $(0, 6)$ .

[Invariants, Monovariants, Game Strategy & Extremal Principle] An arithmetic sequence has first term $u_1 = 6$ and common difference $d = 3$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Intermediate

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 6$ and common difference $d = 3$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 6 + 11(3) = 6 + 33 = 39.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(6 + 39) = 6(45) = 270.$
Final Answer: $u_{12} = 39$ and $S_{12} = 270$ .

[Olympiad Algebra, Symmetric Sums & Vieta's Relations] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 21\text{ cm}$ and $AC = 28\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Intermediate

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(21)^2 + (28)^2} = \sqrt{441 + 784} = \sqrt{1225} = 35\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(21)(28) = 294\text{ cm}^2$ , and the semiperimeter is $s = \frac{21 + 28 + 35}{2} = 42\text{ cm}$ .

Step 3 (Inradius): Using the identity $[ABC] = r \cdot s$ :
$r = \frac{[ABC]}{s} = \frac{294}{42} = 7\text{ cm}.$
Final Answer: $AB = 35\text{ cm},\; \text{Area} = 294\text{ cm}^2,\; r = 7\text{ cm}$ .

[Divisibility, Prime Factorization & Modular Arithmetic] The function $f(x) = \frac{8x + 9}{x - 8}$ is defined for $x \ne 8$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 9$ .

Intermediate

Step 1 (Algebraic Inversion): Set $y = \frac{8x + 9}{x - 8}$ and multiply both sides by $(x - 8)$ :
$y(x - 8) = 8x + 9 \implies xy - 8y = 8x + 9 \implies x(y - 8) = 8y + 9.$

Step 2 (Domain of Inverse): Dividing by $(y - 8)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{8x + 9}{x - 8}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 8.$

Step 3 (Solving $f(x) = 9$ ): Since $f$ is self-inverse, $x = f^{-1}(9) = \frac{8(9) + 9}{(9) - 8} = 81$ .
Final Answer: $f^{-1}(x) = \frac{8x + 9}{x - 8} \; (x \ne 8)$ , and $x = 81$ .

[Diophantine Equations, Floor Functions & Number Bases] Solve the logarithmic equation $\log_2(x - 2) + \log_2(x + 2) = \log_2(12)$ for $x > 2$ , showing why the negative root must be rejected.

Advanced

Step 1 (Combine Logarithms): Using the logarithmic product identity $\log_2(u) + \log_2(v) = \log_2(uv)$ on the domain $x > 2$ :
$\log_2\big((x - 2)(x + 2)\big) = \log_2(12) \implies x^2 - 4 = 12.$

Step 2 (Solve Quadratic & Check Domain): Rearranging gives $x^2 = 16 \implies x = \pm 4$ . Since $\log_2(x - 2)$ requires $x - 2 > 0 \iff x > 2$ , the negative root $x = -4$ is extraneous and must be rejected.
Final Answer: $x = 4$ .

[Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion] Find the equation of the tangent line to the curve $y = x^3 - 3x^2 + 7x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Advanced

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 3(2)^2 + 7(2) + 2 = 8 - 12 + 14 + 2 = 12.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 6x + 7.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 6(2) + 7 = 7$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (12) = 7(x - 2) \implies y = 7x + (-2).$
Final Answer: $y = 7x + (-2)$ , crossing the $y$ -axis at $(0, -2)$ .

[Recurrence Relations, States & Contest Probability] An arithmetic sequence has first term $u_1 = 4$ and common difference $d = 10$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Advanced

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 4$ and common difference $d = 10$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 4 + 11(10) = 4 + 110 = 114.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(4 + 114) = 6(118) = 708.$
Final Answer: $u_{12} = 114$ and $S_{12} = 708$ .

[Power of a Point, Cyclic Quadrilaterals & Angle Chasing] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 15\text{ cm}$ and $AC = 20\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Advanced

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(15)^2 + (20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(15)(20) = 150\text{ cm}^2$ , and the semiperimeter is $s = \frac{15 + 20 + 25}{2} = 30\text{ cm}$ .

[Ceva's, Menelaus', Stewart's Theorems & Triangle Centers] The function $f(x) = \frac{6x + 8}{x - 6}$ is defined for $x \ne 6$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 7$ .

Advanced

Step 1 (Algebraic Inversion): Set $y = \frac{6x + 8}{x - 6}$ and multiply both sides by $(x - 6)$ :
$y(x - 6) = 6x + 8 \implies xy - 6y = 6x + 8 \implies x(y - 6) = 6y + 8.$

Step 2 (Domain of Inverse): Dividing by $(y - 6)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{6x + 8}{x - 6}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 6.$

Step 3 (Solving $f(x) = 7$ ): Since $f$ is self-inverse, $x = f^{-1}(7) = \frac{6(7) + 8}{(7) - 6} = 50$ .
Final Answer: $f^{-1}(x) = \frac{6x + 8}{x - 6} \; (x \ne 6)$ , and $x = 50$ .

[Invariants, Monovariants, Game Strategy & Extremal Principle] Solve the logarithmic equation $\log_2(x - 7) + \log_2(x + 7) = \log_2(147)$ for $x > 7$ , showing why the negative root must be rejected.

Advanced

[Olympiad Algebra, Symmetric Sums & Vieta's Relations] Find the equation of the tangent line to the curve $y = x^3 - 8x^2 + 6x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Advanced

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 8(2)^2 + 6(2) + 2 = 8 - 32 + 12 + 2 = -10.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 16x + 6.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 16(2) + 6 = -14$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (-10) = -14(x - 2) \implies y = -14x + (18).$
Final Answer: $y = -14x + (18)$ , crossing the $y$ -axis at $(0, 18)$ .

[Divisibility, Prime Factorization & Modular Arithmetic] An arithmetic sequence has first term $u_1 = 2$ and common difference $d = 9$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Advanced

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 2$ and common difference $d = 9$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 2 + 11(9) = 2 + 99 = 101.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(2 + 101) = 6(103) = 618.$
Final Answer: $u_{12} = 101$ and $S_{12} = 618$ .

[Diophantine Equations, Floor Functions & Number Bases] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 9\text{ cm}$ and $AC = 12\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Advanced

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(9)^2 + (12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(9)(12) = 54\text{ cm}^2$ , and the semiperimeter is $s = \frac{9 + 12 + 15}{2} = 18\text{ cm}$ .

[Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion] The function $f(x) = \frac{4x + 7}{x - 4}$ is defined for $x \ne 4$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 5$ .

Advanced

Step 1 (Algebraic Inversion): Set $y = \frac{4x + 7}{x - 4}$ and multiply both sides by $(x - 4)$ :
$y(x - 4) = 4x + 7 \implies xy - 4y = 4x + 7 \implies x(y - 4) = 4y + 7.$

Step 2 (Domain of Inverse): Dividing by $(y - 4)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{4x + 7}{x - 4}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 4.$

Step 3 (Solving $f(x) = 5$ ): Since $f$ is self-inverse, $x = f^{-1}(5) = \frac{4(5) + 7}{(5) - 4} = 27$ .
Final Answer: $f^{-1}(x) = \frac{4x + 7}{x - 4} \; (x \ne 4)$ , and $x = 27$ .

[Recurrence Relations, States & Contest Probability] Solve the logarithmic equation $\log_2(x - 5) + \log_2(x + 5) = \log_2(75)$ for $x > 5$ , showing why the negative root must be rejected.

Advanced

[Power of a Point, Cyclic Quadrilaterals & Angle Chasing] Find the equation of the tangent line to the curve $y = x^3 - 6x^2 + 5x + 2$ at the point where $x = 2$ , and determine where this tangent crosses the $y$ -axis.

Advanced

Step 1 (Point Evaluation): Substitute $x = 2$ into the curve equation to find the $y$ -coordinate of tangency:
$y(2) = 2^3 - 6(2)^2 + 5(2) + 2 = 8 - 24 + 10 + 2 = -4.$

Step 2 (Derivative & Gradient): Differentiate with respect to $x$ using the power rule:
$\frac{dy}{dx} = 3x^2 - 12x + 5.$
Evaluating at $x = 2$ gives the instantaneous tangent gradient $m = 3(4) - 12(2) + 5 = -7$ .

Step 3 (Tangent Line Equation): Using the point-slope formula $y - y_0 = m(x - x_0)$ :
$y - (-4) = -7(x - 2) \implies y = -7x + (10).$
Final Answer: $y = -7x + (10)$ , crossing the $y$ -axis at $(0, 10)$ .

[Ceva's, Menelaus', Stewart's Theorems & Triangle Centers] An arithmetic sequence has first term $u_1 = 7$ and common difference $d = 8$ . Find the $12$ -th term $u_{12}$ and the exact sum of the first $12$ terms $S_{12}$ .

Advanced

Step 1 (General Term Formula): For an arithmetic progression with first term $u_1 = 7$ and common difference $d = 8$ , the $n$ -th term is given by $u_n = u_1 + (n - 1)d$ . Substituting $n = 12$ :
$u_{12} = 7 + 11(8) = 7 + 88 = 95.$

Step 2 (Series Sum Formula): Using Gauss's pairing formula $S_n = \frac{n}{2}(u_1 + u_n)$ for $n = 12$ :
$S_{12} = \frac{12}{2}(7 + 95) = 6(102) = 612.$
Final Answer: $u_{12} = 95$ and $S_{12} = 612$ .

[Invariants, Monovariants, Game Strategy & Extremal Principle] In right-angled triangle $ABC$ with $\angle C = 90^\circ$ , the legs have lengths $BC = 24\text{ cm}$ and $AC = 32\text{ cm}$ . Calculate the hypotenuse $AB$ , the area of $\triangle ABC$ , and the inradius $r$ of its inscribed circle.

Advanced

Step 1 (Pythagorean Theorem): Since $\angle C = 90^\circ$ , the hypotenuse $AB$ satisfies:
$AB = \sqrt{BC^2 + AC^2} = \sqrt{(24)^2 + (32)^2} = \sqrt{576 + 1024} = \sqrt{1600} = 40\text{ cm}.$

Step 2 (Area & Semiperimeter): The area is $[ABC] = \frac{1}{2}(24)(32) = 384\text{ cm}^2$ , and the semiperimeter is $s = \frac{24 + 32 + 40}{2} = 48\text{ cm}$ .

[Olympiad Algebra, Symmetric Sums & Vieta's Relations] The function $f(x) = \frac{2x + 6}{x - 2}$ is defined for $x \ne 2$ . Find the inverse function $f^{-1}(x)$ , state its domain, and solve $f(x) = 3$ .

Advanced

Step 1 (Algebraic Inversion): Set $y = \frac{2x + 6}{x - 2}$ and multiply both sides by $(x - 2)$ :
$y(x - 2) = 2x + 6 \implies xy - 2y = 2x + 6 \implies x(y - 2) = 2y + 6.$

Step 2 (Domain of Inverse): Dividing by $(y - 2)$ and swapping $x$ and $y$ gives:
$f^{-1}(x) = \frac{2x + 6}{x - 2}, \quad \text{with domain } x \in \mathbb{R},\; x \ne 2.$

Step 3 (Solving $f(x) = 3$ ): Since $f$ is self-inverse, $x = f^{-1}(3) = \frac{2(3) + 6}{(3) - 2} = 12$ .
Final Answer: $f^{-1}(x) = \frac{2x + 6}{x - 2} \; (x \ne 2)$ , and $x = 12$ .

[Divisibility, Prime Factorization & Modular Arithmetic] Solve the logarithmic equation $\log_2(x - 3) + \log_2(x + 3) = \log_2(27)$ for $x > 3$ , showing why the negative root must be rejected.

Advanced

AI-assisted, human-reviewed: Questions and walkthroughs are developed with AI assistance and reviewed by senior educators. Report an error here ↗

Generate authentic non-calculator AMC 10/12 and Mawhiba / Kangaroo Math competition papers with multi-method olympiad solutions.

Questions

Difficulty

Source

Full official exam paper

Full competition simulation matching MAA AMC 10/12 (25 questions, 75 minutes) and Mawhiba / Kangaroo Math Arabia (30 questions, 75 minutes).

AI-assisted, human-reviewed: Questions and walkthroughs are developed with AI assistance and reviewed by senior educators. Report an error here ↗

↓

Downloads

Download Olympiad theorem compendiums, AMC 10/12 & Kangaroo Math strategy guides, and topical problem sets offline.

PDF · ready now

Contest Math & Olympiad Complete Problem-Solving Compendium

All eight olympiad modules covering Vieta jumping, modular arithmetic, SFFT, Pigeonhole, Power of a Point, Ceva/Stewart, and invariants.

DOCX · on request

Native Word packet

Editable .docx olympiad training handouts with full proofs and competition solutions.

Official · Past Papers

Past Papers & Official Tests

Officially released exams and mark schemes from the examining body.

MAA American Mathematics Competitions (AMC 10/12 & AIME) Official Mathematical Association of America competition portal Mawhiba & Kangaroo Math Arabia Official Portal King Abdulaziz and his Companions Foundation for Giftedness and Creativity

Worksheet Library

Graded 3-Point, 4-Point, and 5-Point Kangaroo/Mawhiba and AMC 10/12 training problem sets.

Loading worksheet library…

📋 Mistake Log

📊 Your Progress

Contest Math & Olympiad (AMC 10/12 · Kangaroo · Mawhiba)

Olympiad Algebra, Symmetric Sums & Vieta's Relations

📊 Contest Math & Olympiad — Exam Blueprint & Domain Weighting

Divisibility, Prime Factorization & Modular Arithmetic

Diophantine Equations, Floor Functions & Number Bases

Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion

Recurrence Relations, States & Contest Probability

Power of a Point, Cyclic Quadrilaterals & Angle Chasing

Ceva's, Menelaus', Stewart's Theorems & Triangle Centers

Invariants, Monovariants, Game Strategy & Extremal Principle

Practice set

Downloads

Contest Math & Olympiad Complete Problem-Solving Compendium

Native Word packet

Past Papers & Official Tests

Worksheet Library

Privacy Policy

1. What's Stored Locally on Your Device

2. User Accounts, Cloud Sync & Third-Party Processors

3. ClipSAT AI Tutor & Practice Generation

4. Google Forms Integration & Google Drive Permissions

5. Advertising, Analytics & Cookie Consent

6. Saudi Personal Data Protection Law (PDPL) Compliance

7. Visitor Counter & Cookieless Analytics

8. Contact & Privacy Requests

Terms of Service

1. Core Content & Free-Tier Promise

2. ClipSAT Plus Subscriptions, Billing & 14-Day Refund Policy

3. ClipSAT AI Tutor Disclaimer

4. Acceptable Use of Content & Copyright

5. Educational Study Aid & No Guarantee of Scores

6. Contact & Support

Mock Exam

Contest Math & Olympiad (AMC 10/12 · Kangaroo · Mawhiba)رياضيات المسابقات والأولمبياد (AMC · كانجارو · موهبة)

Olympiad Algebra, Symmetric Sums & Vieta's Relationsالجبر الأولمبي والمجاميع المتماثلة وعلاقات فييتا للجذور

📊 Contest Math & Olympiad — Exam Blueprint & Domain Weighting

Divisibility, Prime Factorization & Modular Arithmeticقابلية القسمة والتحليل إلى عوامل أولية والحساب المعياري (Modulo)

Diophantine Equations, Floor Functions & Number Basesالمعادلات الديوفانتية ودالة الصحيح وأنظمة العد

Bijective Counting, Pigeonhole Principle & Inclusion-Exclusionالعد التقابلي ومبدأ برج الحمام ومبدأ الشمول والإقصاء

Recurrence Relations, States & Contest Probabilityالعلاقات التكرارية وسلاسل الحالات واحتمالات المسابقات

Power of a Point, Cyclic Quadrilaterals & Angle Chasingقوة النقطة بالنسبة للدائرة والرباعيات الدائرية وتتبع الزوايا

Ceva's, Menelaus', Stewart's Theorems & Triangle Centersنظريات سيفا ومينيلاوس وستيوارت ومراكز المثلث المتقدمة

Invariants, Monovariants, Game Strategy & Extremal Principleاللامتغيرات واستراتيجيات الألعاب الرياضية ومبدأ القيم القصوى

Practice setمجموعة تدريبية

Downloadsالتنزيلات

Contest Math & Olympiad Complete Problem-Solving Compendiumأولمبياد الرياضيات والمسابقات الدولية (AMC / Kangaroo / موهبة) — Contest Math & Olympiad Complete Problem-Solving Compendium

Native Word packet

Past Papers & Official Tests

Worksheet Libraryمكتبة أوراق العمل

Mock Exam

Contest Math & Olympiad (AMC 10/12 · Kangaroo · Mawhiba)

Olympiad Algebra, Symmetric Sums & Vieta's Relations

Divisibility, Prime Factorization & Modular Arithmetic

Diophantine Equations, Floor Functions & Number Bases

Bijective Counting, Pigeonhole Principle & Inclusion-Exclusion

Recurrence Relations, States & Contest Probability

Power of a Point, Cyclic Quadrilaterals & Angle Chasing

Ceva's, Menelaus', Stewart's Theorems & Triangle Centers

Invariants, Monovariants, Game Strategy & Extremal Principle

Practice set

Downloads

Contest Math & Olympiad Complete Problem-Solving Compendium

Worksheet Library