AMC 10 · 2019 · #15

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

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)

(C)

$\frac{32}{3}$

(D)

$\frac{34}{3}$

(E)

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

1STEP 1

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

\frac{1}{2}

a √(b² - a²) = 1,

\frac{1}{2}

a b = 2

3STEP 3

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

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

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

5STEP 5

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

b⁴ =

\frac{(ab)⁴}{a⁴}

\frac{4⁴}{12}

\frac{256}{12}

\frac{64}{3}

6STEP 6

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 b⁴ = $\frac{64}{3}$ and a⁴ = 12: b⁴ - a⁴ = $\frac{64}{3}$ - 12 = $\frac{28}{3}$ .

b⁴ - a⁴ =

\frac{64}{3}

- 12 =

\frac{64 - 36}{3}

\frac{28}{3}

8STEP 8

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

\frac{28}{3}

→ (A)

Answer

\frac{28}{3}

Solve numerically: a⁴ = 12 → a² = 2√(3), b² =

\frac{16}{a²}

\frac{16}{2√(3)}

\frac{8}{√(3)}

. Other leg of T₁ = √(b² - a²) = √(

\frac{8}{√(3)}

- 2√(3)) = √(

\frac{8 - 2 · 3}{√(3)}

) = √(

\frac{2}{√(3)}

). Hypotenuse of T₂ = √(a² + b²) = √(2√(3) +

\frac{8}{√(3)}

) = √(

\frac{6 + 8}{√(3)}

) = √(

\frac{14}{√(3)}

). Their product squared = (

\frac{2}{√(3)}

)(

\frac{14}{√(3)}

) =

\frac{28}{3}

✓. Sanity numerics: a ≈ 1.86, b ≈ 2.15, so T₂ area ≈

\frac{1}{2}

(1.86)(2.15) ≈ 2.00 ✓ and T₁ area ≈

\frac{1}{2}

(1.86)√(2.15² - 1.86²) ≈

\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 = $\frac{64}{3}$ . The squared product of the third sides is (b²-a²)(b²+a²) = b⁴ - a⁴ = $\frac{28}{3}$ , answer (A).

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