AMC 10 · 2019 · #12

Grade 11 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

pythagorean-theoremisosceles-right-triangletrigonometric-ratiosdouble-angle-formulaperimeter identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremtrigonometric-ratios

📏 Long solution 💡 3 insights 📊 Diagram

Problem

An isosceles right triangle has both legs equal to 1. On its hypotenuse, a second right triangle is built on the outside, with its right angle at one end of that hypotenuse. The two triangles have the same perimeter. Find the sine of twice the angle at the shared vertex.

Pick an answer.

(A)

$\dfrac{1}{3}$

(B)

$\dfrac{\sqrt{2}}{2}$

(C)

$\dfrac{3}{4}$

(D)

$\dfrac{7}{9}$

(E)

$\dfrac{\sqrt{3}}{2}$

How to solve

Strategy Identify Subproblems

The target angle ∠ BAD is not a single triangle's angle: it is the 45° angle of triangle ABC plus the unknown angle of triangle ACD at A. That splits the work into clean subproblems — first pin down the side lengths of triangle ACD from the perimeter condition, then read its angle at A from those sides, and finally reassemble. Reassembling pays off because doubling a 45° shift turns the sine into a cosine, so only one double-angle step remains.

1STEP 1

Measure the shared hypotenuse

Find the shared hypotenuse first.

AC=√(1²+1²)=√(2), perimeter of △ ABC = 2+√(2)

2STEP 2

Cancel the side they share

The shared side cancels in the perimeter.

CD+DA = BC+BA = 2 ⟹ DA=x, CD=2-x

3STEP 3

Solve for the sides of ACD

Pythagoras fixes the two sides.

2+(2-x)²=x² ⟹ 6-4x=0 ⟹ DA=3/2, CD=1/2

4STEP 4

Read the angle at A in ACD

Read the sine of the angle in the new triangle.

sinθ=CD/DA=1/2/3/2=1/3, cosθ=AC/DA=√(2)/3/2=2√(2)/3

5STEP 5

Split the angle at A

Split the whole angle into forty-five plus that one.

∠ BAD = 45°+θ ⟹ 2∠ BAD = 90°+2θ

6STEP 6

Trade the sine for a cosine

Adding ninety degrees turns sine into cosine.

sin(2∠ BAD)=sin(90°+2θ)=sin 90°cos 2θ+cos 90°sin 2θ=cos 2θ

7STEP 7

Finish with the double angle

The double angle gives seven ninths.

cos 2θ = 1-2sin²θ = 1-2(1/3)² = 1-2/9 = 7/9

Answer

7/9

Check with coordinates: put B = (0, 0), A = (1, 0), C = (0, 1). Then AC = √(2) and D sits on the far side of line AC at D = (1/2√(2), 1+1/2√(2)) ≈ (0.354, 1.354). The distance AD is √(0.646²+1.354²) ≈ 1.5 = 3/2, matching the side found above, and the angle between ray AB and ray AD comes out to about 64.5°. Doubling gives about 128.9°, whose sine is about 0.778 — and 7/9 ≈ 0.778. The sign is right too: 2∠ BAD is obtuse, so its sine must be positive and less than 1, which rules out nothing among the choices but confirms 7/9 over the smaller trap value 1/3.

💡Key takeaway

When two shapes share a side, equal perimeters is really a statement about the other sides only — and when the angle you want is 45° plus something, doubling it turns a sine into a cosine and the rest is one formula.

Measure the shared hypotenuse
Cancel the side they share
Solve for the sides of ACD
Read the angle at A in ACD
Split the angle at A
Trade the sine for a cosine
Finish with the double angle