Competition · AMC preparation · step 4 of 4

AMC 10 · 2019B · #15

Grade 8 geometry-2d
pythagorean-theoremarea-trianglesdifference-of-squaressystems-of-equations caseworkidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-trianglessystems-of-equations
📏 Long solution 💡 4 insights
Problem
Two right triangles T₁, T₂ have areas 1 and 2. One side of T₁ equals one side of T₂, and a different side of T₁ equals a different side of T₂. Find the square of the product of the two not-shared (third) sides.

Pick an answer.

(A)
$\frac{28}{3}$
(B)
10
(C)
$\frac{32}{3}$
(D)
$\frac{34}{3}$
(E)
12

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw): sketch two right triangles with labeled legs/hypotenuse and discover the only role-swap consistent with both areas. Tool #9 (Easier Problem): once we recognize the shared values are a leg of T₁ paired with the same as a leg of T₂, and the hypotenuse of T₁ paired with a leg of T₂ — the rest is two area equations in two unknowns. Tool #13 (Algebra): set up ab = 4 and a²(b²-a²) = 4, then evaluate b⁴ - a⁴ — no heavy machinery needed. Tool #3 matches 28/3 to choice (A).

1STEP 1

Draw both triangles

Draw both triangles, shared sides a < b. Only one role-swap works: a and b are legs of T₂; in T₁, a is a leg and b the hypotenuse.

T₁: legs a, √(b² - a²); hypotenuse b. T₂: legs a, b; hypotenuse √(a² + b²).
2STEP 2

Write the two area equations

Write the area equations: T₁ has legs a and √(b² - a²) so ½a√(b² - a²) = 1; T₂ has legs a, b so ½ab = 2.

1/2 a √(b² - a²) = 1, 1/2 a b = 2
3STEP 3

Simplify the second area

Simplify: T₂'s area gives ab = 4; squaring T₁'s area gives a²(b² - a²) = 4.

ab = 4 and a²(b² - a²) = 4
4STEP 4

Substitute to find a fourth power

Expand a²b² - a⁴ = 4 and substitute a²b² = (ab)² = 16, so a⁴ = 12.

a² b² - a⁴ = 4 → 16 - a⁴ = 4 → a⁴ = 12
5STEP 5

Find the other fourth power

From b = 4a\frac{4}{a}, b⁴ = 256a4\frac{256}{a⁴} = 25612\frac{256}{12} = 643\frac{64}{3}.

b⁴ = (ab)⁴/a⁴ = 4⁴/12 = 256/12 = 64/3
6STEP 6

Multiply the two unshared sides

The third sides are √(b² - a²) and √(a² + b²); their product squared is (b² - a²)(b² + a²) = b⁴ - a⁴.

(√(b² - a²) · √(a² + b²))² = (b² - a²)(b² + a²) = b⁴ - a⁴
7STEP 7

Substitute the fourth powers

Substitute b⁴ = 643\frac{64}{3} and a⁴ = 12: b⁴ - a⁴ = 643\frac{64}{3} - 12 = 283\frac{28}{3}.

b⁴ - a⁴ = 64/3 - 12 = (64 - 36)/3 = 28/3
8STEP 8

Match against the choices

Match 283\frac{28}{3} to choice (A).

28/3 → (A)
Answer
28/3
Solve numerically: a⁴ = 12 → a² = 2√(3), b² = 16a2\frac{16}{a²} = 1623\frac{16}{2\sqrt{3}} = 83\frac{8}{\sqrt{3}}. Other leg of T₁ = √(b² - a²) = √(83\frac{8}{\sqrt{3}} - 2√(3)) = √(8−2⋅33\frac{8 - 2 · 3}{\sqrt{3}}) = √(23\frac{2}{\sqrt{3}}). Hypotenuse of T₂ = √(a² + b²) = √(2√(3) + 83\frac{8}{\sqrt{3}}) = √(6+83\frac{6 + 8}{\sqrt{3}}) = √(143\frac{14}{\sqrt{3}}). Their product squared = (23\frac{2}{\sqrt{3}})(143\frac{14}{\sqrt{3}}) = 283\frac{28}{3} ✓. Sanity numerics: a ≈ 1.86, b ≈ 2.15, so T₂ area ≈ 12\frac{1}{2}(1.86)(2.15) ≈ 2.00 ✓ and T₁ area ≈ 12\frac{1}{2}(1.86)√(2.15² - 1.86²) ≈ 12\frac{1}{2}(1.86)(1.07) ≈ 1.00 ✓.
💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagoras you already know! The shared sides a, b have to play different roles in the two triangles: legs of T₂, but leg-and-hypotenuse of T₁. The two area equations give ab = 4 and a²(b² - a²) = 4, which yield a⁴ = 12 and b⁴ = 643\frac{64}{3}. The squared product of the third sides is (b²-a²)(b²+a²) = b⁴ - a⁴ = 283\frac{28}{3}, answer (A).

  • Draw both triangles
  • Write the two area equations
  • Simplify the second area
  • Substitute to find a fourth power
  • Find the other fourth power
  • Multiply the two unshared sides
  • Substitute the fourth powers
  • Match against the choices

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