Math Competition Prep · Algebra & Geometry

Part I

Algebra I

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ISOLATE → move everything except the variable to the other side. DISTRIBUTE FIRST → expand parentheses before combining like terms. FOIL → First · Outer · Inner · Last. DISCRIMINANT $b^2-4ac$: positive = 2 real roots · zero = 1 repeated root · negative = no real root. CONSECUTIVE ODD → $n,\ n+2,\ n+4$ (not $n+1$).

A·01

Maria has $240 to spend on notebooks and pens. Notebooks cost $8 each; pens cost $3 each. She buys twice as many pens as notebooks. How many notebooks does she buy?

Easy

📌 Key Setup

Let notebooks $= n$. Pens $= 2n$. Total cost: $8n + 3(2n) = 240 \Rightarrow 14n = 240$.

📖 Full Solution

Let $n$ = notebooks, $2n$ = pens.
$8n + 3(2n) = 240 \Rightarrow 14n = 240 \Rightarrow n \approx 17.14$
Since $n$ must be a whole number: $14 \times 17 = 238 \leq 240$ ✓ and $14 \times 18 = 252 > 240$ ✗
Answer: 17 notebooks, spending $238 (leaving $2 unspent).

⚠️ Common trap: Forgetting to multiply pens by 2, writing $8n + 3n = 240$ instead.

A·02

Solve for $x$: $\dfrac{2x-5}{3} = \dfrac{x+4}{2}$

Medium

📌 Cross-Multiply

Multiply both sides by the LCD (6), or cross-multiply: $2(2x-5) = 3(x+4)$. Then distribute carefully — watch the negatives!

📖 Full Solution

Cross-multiply: $2(2x-5) = 3(x+4)$
$4x - 10 = 3x + 12$
$x = 22$ ✓

⚠️ Sign trap: $2 \times (-5) = -10$, not $+10$. One of the most common errors on this problem type.

A·03

A train leaves City A at 60 mph. Two hours later, a second train leaves City A in the same direction at 90 mph. How many hours after the second train departs will it catch the first?

Medium

📌 D = R × T · Catch-Up

At the moment of catch-up, both distances are equal. First train has a 2-hour head start.
Let $t$ = time (hrs) after 2nd train departs: $\ 60(t+2) = 90t$

📖 Full Solution

$60(t+2) = 90t$
$60t + 120 = 90t$
$120 = 30t \Rightarrow t = 4$ hours ✓

Verify: Train 1 travels $60 \times 6 = 360$ mi. Train 2 travels $90 \times 4 = 360$ mi. ✓

⚠️ Forgetting the "+2" in the first train's time is the #1 error here.

A·04

Find the roots of $2x^2 - 7x + 3 = 0$.

Medium

📌 AC Method for Factoring

$a \cdot c = 2 \times 3 = 6$. Find two numbers that multiply to 6 and add to $-7$: those are $-6$ and $-1$.
Rewrite: $2x^2 - 6x - x + 3 = 0$ → group.

📖 Full Solution

Factor: $2x^2 - 6x - x + 3 = 2x(x-3) - 1(x-3) = (2x-1)(x-3) = 0$
$x = \tfrac{1}{2}$ or $x = 3$ ✓

Double-check via formula: $\Delta = 49 - 24 = 25$, so $x = \dfrac{7 \pm 5}{4}$ → $3$ or $\tfrac{1}{2}$ ✓

A·05

The sum of three consecutive odd integers is 57. What is the largest of the three?

Easy

📖 Full Solution

Let the integers be $n,\ n+2,\ n+4$.
$(n) + (n+2) + (n+4) = 57 \Rightarrow 3n + 6 = 57 \Rightarrow n = 17$
The three integers: 17, 19, 21. Largest = 21 ✓

⚠️ Classic trap: Using $n, n+1, n+2$ (consecutive integers) instead of $n, n+2, n+4$ (consecutive odd integers).

A·06

Mixture problem: Solution A is 20% acid; Solution B is 50% acid. How many liters of Solution A must be mixed with 12 liters of Solution B to get a 30% acid mixture?

Hard

📌 Mixture Setup

$\underbrace{0.20x}_{\text{acid from A}} + \underbrace{0.50(12)}_{\text{acid from B}} = \underbrace{0.30(x+12)}_{\text{acid in result}}$

📖 Full Solution

$0.20x + 6 = 0.30x + 3.6$
$6 - 3.6 = 0.10x \Rightarrow 2.4 = 0.10x \Rightarrow x = 24$ liters ✓

Check: $0.20(24) + 0.50(12) = 4.8 + 6 = 10.8$ acid in 36 liters → $10.8/36 = 0.30 = 30\%$ ✓

A·07

Solve the system and find $x+y$:
$3x + 2y = 16$
$x - y = 2$

Medium

📌 Substitution Shortcut

From equation 2: $x = y + 2$. Plug into equation 1. Solve for $y$, then find $x$.

📖 Full Solution

From eq. 2: $x = y+2$. Substitute into eq. 1:
$3(y+2) + 2y = 16 \Rightarrow 5y + 6 = 16 \Rightarrow y = 2$
Then $x = 4$. So $x + y = 6$ ✓

Verify: $3(4)+2(2)=16$ ✓ and $4-2=2$ ✓

A·08

Expand and simplify: $(3x - 2)^2 - (x+4)(x-4)$

Medium

📌 Two Special Product Identities

$(a-b)^2 = a^2 - 2ab + b^2$ $(a+b)(a-b) = a^2 - b^2$
Then carefully distribute the minus sign between the two expressions.

📖 Full Solution

$(3x-2)^2 = 9x^2 - 12x + 4$
$(x+4)(x-4) = x^2 - 16$
Subtraction: $(9x^2-12x+4) - (x^2-16) = 8x^2 - 12x + 20$ ✓

⚠️ The minus sign distributes to both terms: $-(x^2-16) = -x^2 + 16$. Most errors come from writing $-x^2 - 16$.

A·09

$f(x) = x^2 - 4x + k$. If the graph of $f$ touches the x-axis at exactly one point, find $k$.

Hard

📌 One Touch = Discriminant Zero

"Touches x-axis at one point" means one repeated root → $\Delta = b^2 - 4ac = 0$.

📖 Full Solution

$a=1,\ b=-4,\ c=k$
$(-4)^2 - 4(1)(k) = 0 \Rightarrow 16 - 4k = 0 \Rightarrow k = 4$ ✓

Proof: $x^2-4x+4=(x-2)^2=0 \Rightarrow x=2$ (one solution). The parabola is tangent to the x-axis at $(2,0)$.

A·10

A store has two deals on the same item (original price $P$):
Deal X: 40% off. Deal Y: 25% off, then an additional 20% off the reduced price.
Which deal gives the greater discount?

Hard

📌 Consecutive Discounts ≠ Added Discounts

Two successive discounts of 25% and 20% do not equal 45% off. You must multiply: $0.75 \times 0.80 = \ ?$

📖 Full Solution

Deal X: Final price $= 0.60P$ → 40% off.
Deal Y: $P \times 0.75 \times 0.80 = 0.60P$ → also 40% off!
Both deals give exactly 40% discount. ✓

⚠️ The classic trap: adding $25\% + 20\% = 45\%$. Successive percentage discounts always multiply, never add.

Part II

Geometry

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TRIANGLE SUM → three angles always add to 180°. EXTERIOR ANGLE = sum of the two non-adjacent interior angles. PARALLEL LINES → alternate interior angles are equal; co-interior (same-side) angles sum to 180°. INSCRIBED ANGLE = ½ × central angle (same arc). SIMILAR △ AREA → area ratio = (side ratio)². SOH-CAH-TOA → sin=opp/hyp · cos=adj/hyp · tan=opp/adj.

G·01

In a triangle, one angle is three times the smallest angle, and the third angle is 20° more than the smallest. Find the largest angle.

Easy

📌 Set Up with One Variable

Let smallest angle $= x$. The three angles are $x,\ 3x,\ x+20$. Their sum is 180°.

📖 Full Solution

$x + 3x + (x+20) = 180 \Rightarrow 5x + 20 = 180 \Rightarrow x = 32°$
The three angles: $32°,\ 96°,\ 52°$. Largest = 96° ✓
Check: $32+96+52=180$ ✓

G·02

Two parallel lines are cut by a transversal. One co-interior (same-side interior) angle measures $(3x+15)°$; the other measures $(2x+5)°$. Find $x$.

Easy

📌 Co-Interior Angles Are Supplementary

Also called "consecutive interior" or "C-angles." When lines are parallel: co-interior angles sum to 180°.
Do NOT confuse with alternate interior angles, which are equal.

📖 Full Solution

$(3x+15)+(2x+5)=180 \Rightarrow 5x+20=180 \Rightarrow x=32$ ✓
Verify: $3(32)+15=111°$ and $2(32)+5=69°$. Sum $=180°$ ✓

G·03

A right triangle has legs of length 7 and 24. What is the area of the circle whose diameter equals the hypotenuse?

Medium

📌 Two Steps

Step 1 — Pythagorean Theorem: $c = \sqrt{7^2 + 24^2}$.
Step 2 — Circle: radius = hypotenuse ÷ 2. Then area $= \pi r^2$.

📖 Full Solution

$c = \sqrt{49+576} = \sqrt{625} = 25$
Radius $= 25 \div 2 = 12.5$
Area $= \pi(12.5)^2 = 156.25\pi$ ✓

⚠️ The #1 mistake: using the hypotenuse (25) as the radius, getting $625\pi$. Always halve the diameter!

G·04

In circle O, the inscribed angle $\angle ABC = 38°$ intercepts arc $\widehat{AC}$. What is the central angle $\angle AOC$?

Medium

📌 Inscribed Angle Theorem

Central angle = 2 × inscribed angle (same intercepted arc).
Memory trick: the center "sees" the arc from the inside — a wider view — so it's always twice as large.

📖 Full Solution

$\angle AOC = 2 \times \angle ABC = 2 \times 38° = 76°$ ✓

⚠️ The trap is doing the reverse (dividing by 2). The central angle is always the larger of the two.

G·05

Two similar triangles have corresponding sides in ratio 3 : 5. The area of the smaller triangle is 36 cm². What is the area of the larger triangle?

Medium

📌 Area Scale Factor

For similar figures: Area ratio = (Side ratio)²
If sides are $3:5$, then areas are $3^2:5^2 = 9:25$.

📖 Full Solution

Area ratio $= \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25}$
$\dfrac{36}{A} = \dfrac{9}{25} \Rightarrow A = \dfrac{36 \times 25}{9} = 100$ cm² ✓

⚠️ The most common error: using the side ratio directly $\left(\frac{5}{3} \times 36 = 60\right)$ instead of squaring it.

G·06

A regular hexagon has a perimeter of 48 cm. What is its area?

Hard

📌 Regular Hexagon = 6 Equilateral Triangles

Each equilateral triangle has side $s$. Area of one $= \dfrac{\sqrt{3}}{4}s^2$. Total: multiply by 6.

📖 Full Solution

Side $= 48 \div 6 = 8$ cm
Area $= 6 \times \dfrac{\sqrt{3}}{4}(8^2) = 6 \times \dfrac{64\sqrt{3}}{4} = 6 \times 16\sqrt{3} = 96\sqrt{3}$ cm² ✓

G·07

An exterior angle of a triangle measures 125°. One of the two non-adjacent interior angles is 72°. Find the other non-adjacent interior angle.

Easy

📌 Exterior Angle Theorem

Exterior angle = sum of the two non-adjacent (remote) interior angles.

📖 Full Solution

Exterior angle = sum of two non-adjacent interior angles:
$125° = 72° + \theta \Rightarrow \theta = 53°$ ✓

Third angle (adjacent to exterior): $180° - 125° = 55°$
Verify: $72° + 53° + 55° = 180°$ ✓

G·08

A cone and a cylinder share the same base and height. The cone's volume is what fraction of the cylinder's volume?

Medium

📌 Volume Formulas

Cylinder: $V = \pi r^2 h$ · Cone: $V = \dfrac{1}{3}\pi r^2 h$
Notice: the only difference is the factor of $\dfrac{1}{3}$ in the cone formula.

📖 Full Solution

$\dfrac{V_\text{cone}}{V_\text{cyl}} = \dfrac{\frac{1}{3}\pi r^2 h}{\pi r^2 h} = \dfrac{1}{3}$ ✓

This is always $\frac{1}{3}$, regardless of the actual radius or height. Three cones always fill exactly one cylinder of the same base and height — a beautiful geometric fact!

G·09

In right triangle $ABC$ with right angle at $C$, $\tan(A) = \dfrac{3}{4}$. Find $\sin(A) + \cos(A)$.

Hard

📌 SOH-CAH-TOA → 3-4-5

$\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{4}$. Use the Pythagorean Theorem to find the hypotenuse, then read off sin and cos.

📖 Full Solution

opp $= 3$, adj $= 4$, hyp $= \sqrt{9+16} = 5$
$\sin A = \dfrac{3}{5},\quad \cos A = \dfrac{4}{5}$
$\sin A + \cos A = \dfrac{3}{5} + \dfrac{4}{5} = \dfrac{7}{5}$ ✓

G·10

A square is inscribed in a circle of radius 6 cm. Find the area of the region inside the circle but outside the square. (Express in terms of $\pi$.)

Hard

📌 Inscribed Square: Diagonal = Diameter

The diagonal of the square = diameter of the circle $= 2r = 12$.
Side of square: if diagonal $= d$, then side $= \dfrac{d}{\sqrt{2}}$.

📖 Full Solution

Circle area $= \pi(6^2) = 36\pi$
Square diagonal $= 12$. Side $= \dfrac{12}{\sqrt{2}} = 6\sqrt{2}$.
Square area $= (6\sqrt{2})^2 = 72$
Shaded region $= 36\pi - 72$ cm² ✓

⚠️ Common mistake: using 6 directly as the side length (giving area = 36). Always derive the side from the diagonal.

Algebra I & GeometryMastery Problems

Algebra I

Geometry

Algebra I & Geometry
Mastery Problems