AMC 10 · 2024 · #11

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
similar-trianglesarea-trianglespythagorean-theoremarea-rectangles identify-subproblemsconvert-to-algebraarea-difference ↑ Prerequisites: similar-trianglesarea-trianglespythagorean-theoremarea-rectangles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
In rectangle WXYZ with WX = 4 and WZ = 8, point M is on side XY and point A is on side YZ. The angle at M in triangle WMA is a right angle, and the two corner triangles △ WXM and △ WAZ have equal areas. Find the area of △ WMA.

Pick an answer.

(A)
13
(B)
14
(C)
15
(D)
16
(E)
17

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The figure is the whole problem: a rectangle cut by the cevians WM and MA into four triangles whose areas must sum to 32. Tool #1 (Draw a Diagram) makes that decomposition visible at a glance. Tool #7 (Identify Subproblems) turns it into two clean conditions — the equal-area condition links XM and ZA, and the right angle at M links them through the Pythagorean theorem. Tool #13 (Convert to Algebra) names XM = b and ZA = a so those two conditions become a 2 × 2 system that pins down a and b. Once a, b are known, the area falls straight out of 32 - (three corner triangles).

1STEP 1

Name the corner lengths XM = b and ZA = a, so MY = 8 - b and AY = 4 - a; each corner triangle's area then reads straight off its legs.

[△ WXM] = 12\frac{1}{2} · 4 · b = 2b, [△ WAZ] = 12\frac{1}{2} · 8 · a = 4a, [△ MAY] = 12\frac{1}{2}(8-b)(4-a)
2STEP 2

The equal-area condition [△ WXM] = [△ WAZ] becomes 2b = 4a, which tidies to b = 2a.

2b = 4a ⟹ b = 2a
3STEP 3

The right angle at M triggers the Pythagorean theorem WM² + MA² = WA², the second equation tying a and b.

(16 + b²) + ((8-b)² + (4-a)²) = 64 + a²
4STEP 4

Expanding and substituting b = 2a collapses everything to a² - 5a + 4 = 0, so a = 1 or a = 4.

(a-1)(a-4) = 0 ⟹ a = 1 or a = 4
5STEP 5

a = 4 squashes the triangle to a segment, so keep a = 1, b = 2, giving MY = 6 and AY = 3.

a = 1, b = 2, MY = 6, AY = 3
6STEP 6

Subtract the three corner areas 4, 4, 9 from the rectangle's 32 to leave the inner triangle's area, 15.

[△ WMA] = 32 - 4 - 4 - 9 = 15 → (C)
Answer
15
Sanity-check with the right-triangle leg lengths: WM = √(16 + 4) = √(20) and MA = √(36 + 9) = √(45), so 12\frac{1}{2} WM · MA = 12\frac{1}{2}√(20 · 45) = 12\frac{1}{2}√(900) = 15. Same answer. Also the answer 15 is just under half the rectangle area (32), which matches a glance at the picture where △ WMA is the biggest of the four pieces but does not dominate.
💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean theorem plus Grade 6 "rectangle minus the corners" area thinking that you already know!

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