AMC 10 · 2020 · #12

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglescentroid-2-to-1isosceles-trianglespatial-visualization identify-subproblemsarea-difference ↑ Prerequisites: area-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In isosceles triangle △ AMC with AM = AC, the medians MV (from M to midpoint V of AC) and CU (from C to midpoint U of AM) are perpendicular and each has length 12. Find the area of △ AMC.

Pick an answer.

(A)

(B)

(C)

(D)

144

(E)

192

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): sketch △ AMC, the two medians MV and CU, label their intersection P (the centroid), and mark the right angle at P. Tool #7 (Subproblems): instead of attacking the full triangle directly, (a) use the 2:1 centroid property to get the two leg lengths MP and CP, (b) compute the area of the small right triangle △ MPC, (c) scale up to △ AMC via the fact that the three centroid sub-triangles have equal area. Tool #3 (Eliminate): confirm the final number against the five choices.

1STEP 1

Draw △ AMC with medians MV and CU; they cross at the centroid P, where ∠ MPC = 90° and MV = CU = 12.

P = MV ∩ CU, ∠ MPC = 90°, MV = CU = 12

2STEP 2

Centroid 2:1 rule on each median: the long piece is $\frac{2}{3}$ ·12, so MP = CP = 8 (short pieces PV = PU = 4).

MP =

\frac{2}{3}

(12) = 8, CP =

\frac{2}{3}

(12) = 8

3STEP 3

Triangle △ MPC has legs MP = 8 and CP = 8 meeting at the right angle P, so its area is $\frac{1}{2}$ ·8·8 = 32.

[MPC] =

\frac{1}{2}

· MP · CP =

\frac{1}{2}

· 8 · 8 = 32

4STEP 4

The centroid's segments PA, PM, PC split △ AMC into three equal-area triangles, so [AMC] = 3·32 = 96.

[AMC] = 3 · [MPC] = 3 · 32 = 96

5STEP 5

Only choice (C) equals 96; the others come from skipping the $\frac{2}{3}$ factor (48, 72, 144) or doubling not tripling (192).

[AMC] = 96 → (C)

Answer

Sanity check via a back-of-envelope: a triangle whose medians are each 12 and meet at right angles has the same area no matter how it's oriented. Using the well-known shortcut "area of a triangle equals

\frac{4}{3}

times the area of the triangle formed by its medians", the median triangle here has legs 12 and 12 meeting at 90°, area

\frac{1}{2}

· 12 · 12 = 72, so [AMC] =

\frac{4}{3}

· 72 = 96. ✓ Matches.

💡Key takeaway

This AMC 10 problem only needs Grade 7 area you already know! The two medians meet at the centroid P and the centroid grabs $\frac{2}{3}$ of each median — so MP = CP = $\frac{2}{3}$ · 12 = 8. Because ∠ MPC = 90°, the small triangle MPC has area $\frac{1}{2}$ · 8 · 8 = 32. The centroid splits the whole triangle into three equal-area pieces, so [AMC] = 3 · 32 = 96, answer (C).

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