Competition · AMC preparation · step 4 of 4

AMC 8 · 2023 · #24

Grade 8 geometry-2d

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

area-trianglessimilar-figuresratio-proportion area-differenceidentify-subproblemsconvert-to-algebra ↑ Prerequisites: area-trianglessimilar-figures

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

An isosceles triangle ABC (with AB = BC) has total height h from base AC up to the apex B. Two copies of the same triangle are drawn. In the first copy, one segment parallel to AC is drawn so that a small triangle of height 11 is left UNshaded at the top, and the trapezoid below it is shaded. In the second copy, a different parallel segment is drawn so that the small triangle near the apex is shaded and the trapezoid at the bottom (height 5) is UNshaded. The two shaded regions have the same area. Find h.

Pick an answer.

(A)

14.6

(B)

14.8

(C)

(D)

15.2

(E)

15.4

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The two figures are independent area calculations that must be made equal — that's a textbook Tool #7 (Subproblems) setup: solve each shaded area separately, then equate. Tool #1 (Diagram) keeps the geometry honest: each small top triangle (heights 11 and h-5) is similar to △ ABC, so its area is A_total · (height ratio)². Once each shaded area is written symbolically, Tool #13 (Algebra) finishes it: the A_total and h² cancel, leaving a simple linear equation in h. Younger tools (Guess & Check on the five answer choices) also work and we mention it in Review.

1STEP 1

Label both figures

Parallel cuts make each top triangle similar to △ABC: Figure 1's unshaded top has height 11, Figure 2's shaded top has height h - 5.

Fig 1: top triangle height = 11. Fig 2: top triangle height = h - 5.

2STEP 2

Find Figure 1 shaded area

Figure 1's shaded trapezoid is the whole triangle minus the top triangle, whose area share is ( $\frac{11}{h}$ )² = $\frac{121}{h^2}$ .

A_shaded,1 = A_total - A_total · 121/h² = A_total (1 - 121/h²)

3STEP 3

Find Figure 2 shaded area

Figure 2's shaded region IS the top triangle with height ratio $\frac{h-5}{h}$ , so its area share is ( $\frac{h-5}{h}$ )².

A_shaded,2 = A_total · ((h-5)/h)² = A_total · ((h-5)²)/h²

4STEP 4

Set the two areas equal

Set the shaded areas equal, divide out A_total, and multiply by h² to clear denominators: h² - 121 = (h-5)².

1 - 121/h² = ((h-5)²)/h² ⟹ h² - 121 = (h-5)²

5STEP 5

Cancel the squared terms

Expanding the right side cancels h² on both sides, so the quadratic is really linear: 10h = 146.

h² - 121 = h² - 10h + 25 ⟹ -121 = -10h + 25 ⟹ 10h = 146

6STEP 6

Divide by 10 and pick the choice

Divide by 10 to isolate h: h = 14.6, matching choice (A).

h = 146/10 = 14.6 ⟹ (A)

Answer

14.6

Sanity check the answer. With h = 14.6, the small UNshaded triangle in Figure 1 has area-fraction

\frac{121}{14.6²}

\frac{121}{213.16}

≈ 0.568, so the shaded trapezoid is about 1 - 0.568 = 0.432 of the whole triangle. The shaded triangle in Figure 2 has height 14.6 - 5 = 9.6 and area-fraction (

\frac{9.6}{14.6}

)² =

\frac{92.16}{213.16}

≈ 0.432 — the two match. Also h = 14.6 is just barely bigger than 11 (required so the Figure 1 cut sits below the apex), so the geometry is consistent.

💡Key takeaway

This AMC 8 problem only needs Grade 8 similarity (area scales with the square of the height ratio) plus a one-line linear equation you already know!

Label both figures
Find Figure 1 shaded area
Find Figure 2 shaded area
Set the two areas equal
Cancel the squared terms
Divide by 10 and pick the choice

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