AMC 10 · 2010 · #13

Grade 11 algebrageometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

trigonometric-ratiosangle-sum-trianglethirty-sixty-ninety-triangle extreme-principlebound-inequality-then-enumerate ↑ Prerequisites: trigonometric-ratiosangle-sum-triangle

📏 Long solution 💡 3 insights

Problem

Two trigonometric terms of a triangle add to their joint maximum, and one side is known. Find another side.

Pick an answer.

(A)

$\sqrt{2}$

(B)

$\sqrt{3}$

(C)

(D)

$2\sqrt{2}$

(E)

$2\sqrt{3}$

How to solve

Strategy Extreme Principle

Tool #14 (Extreme Principle): the equation asks a sum of two quantities that each cap at 1 to equal 2, which is the extreme case — there is no slack anywhere, so one equation splits into two exact ones. Tool #3 (Eliminate Possibilities): sinθ=1 and cosθ=1 each have infinitely many solutions, so the real work is using the triangle's angle bounds to knock out every solution but one; this is the step that makes the answer unique and it is easy to wave through. Tool #4 (Introduce a Variable): once the two exact conditions are in hand they are just a linear system in the angles A and B. Tool #11 (Work Backwards): the bounding argument only proves the angles are forced if such a triangle exists, so I rebuild the triangle from the angles and feed it back into the original equation to confirm the conditions are satisfiable. Tool #1 (Draw a Diagram): with the angles known, the last hazard is naming sides, and a labelled picture is what keeps BC from being confused with AC.

1STEP 1

Force both terms to their maximum

Each term must sit at its own maximum.

cos(2A-B) ≤ 1, sin(A+B) ≤ 1, sum=2 ⟹ cos(2A-B)=1 and sin(A+B)=1

2STEP 2

The angle sum pins A+B

The angle sum makes one corner a right angle.

sin(A+B)=1 and 0° ＜ A+B ＜ 180° ⟹ A+B=90° ⟹ C=90°

3STEP 3

The same bound pins 2A-B

The same bound pins the other expression.

2A-B=3A-90° ∈ (-90°, 180°), cos(3A-90°)=1 ⟹ 3A-90°=0°

4STEP 4

Solve the two-angle system

Solving gives the familiar 30-60-90 shape.

2A-B=0°, A+B=90° ⟹ 3A=90°, A=30°, B=60°, C=90°

5STEP 5

Check the triangle actually exists

Substituting back confirms the triangle exists.

cos(2 · 30°-60°)+sin(30°+60°)=cos 0°+sin 90°=1+1=2 ✓

6STEP 6

Name the sides, then measure BC

Measuring the side gives 2, choice (C).

BC=ABsin A=4sin 30°=4·1/2=2 ⟹ (C) 2

Answer

Size check first, without using the angles. BC is a leg of a right triangle with hypotenuse 4, so the other leg is √(16-BC²). Since BC faces the smallest angle it must be the shortest side, which needs BC² ＜ 16-BC², that is BC ＜ 2√(2)≈ 2.83. That alone kills (E) 2√(3)≈ 3.46 and shows (D) 2√(2) is the isosceles boundary case, which would need A=B=45° and contradicts B=2A. Ratio check next: BC/AB=sin A=sin 30°=1/2, and half of 4 is 2, matching (C) exactly rather than approximately. The choices √(2) and √(3) would need sin A=√(2)/4 or √(3)/4, giving angles near 20.7° and 25.7° that no exact condition here can produce. Finally the found triangle is consistent end to end: sides 2, 2√(3), 4 satisfy 4+12=16, and plugging A=30°, B=60° back in gives cos 0°+sin 90°=2, the original equation. So (C) stands.

💡Key takeaway

When a sum of quantities that each cap at 1 equals 2, every one of them must sit exactly at 1 — and then the triangle's own angle limits pick the single solution out of the infinitely many the trigonometric equation allows.

Force both terms to their maximum
The angle sum pins A+B
The same bound pins 2A-B
Solve the two-angle system
Check the triangle actually exists
Name the sides, then measure BC