AMC 10 · 2024 · #22

Grade 8 geometry-2d

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

area-trianglespythagorean-theoremline-symmetryangle-sum-triangle identify-subproblemsarea-difference ↑ Prerequisites: area-trianglespythagorean-theoremsimilar-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A kite mathcal K is built from two right triangles with legs 1 and √(3) glued along their common hypotenuse. Eight copies of mathcal K tile a polygon, and a large triangle △ ABC is drawn on the tiling. Find the area of △ ABC.

Pick an answer.

(A)

$2+3\sqrt3$

(B)

$\dfrac{9}{2}\sqrt3$

(C)

$\dfrac{10+8\sqrt3}{3}$

(D)

(E)

$5\sqrt3$

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

How to solve

Strategy Draw a Diagram

The whole problem lives inside one picture, so Tool #1 (Draw a Diagram) drives everything: re-mark the figure with the kite's actual side lengths (1, 1, √(3), √(3)) and angles, and the tiling tells you exactly how AB and the altitude to AB are built from kite edges. Tool #7 (Identify Subproblems) then splits "area of △ ABC" into two independent measurements — base AB and height h from C — each of which is a short Pythagorean-theorem calculation on one small piece of the tiling. With base and height in hand, the final area is one Grade 6 formula.

1STEP 1

Each kite half is a right triangle with legs 1 and √(3); by the Pythagorean theorem the shared hypotenuse is 2, fixing all sides and angles.

hypotenuse = √(1² + (√(3))²) = √(4) = 2

2STEP 2

Split area = ½·AB·h into two measurements: by symmetry the altitude hits AB's midpoint, so just find the base AB and the height h.

[△ ABC] =

\frac{1}{2}

· AB · h

3STEP 3

Each of AB's four equal segments is the long leg of a small similar triangle: √(3)·( $\frac{√(3)}{2}$ ) = $\frac{3}{2}$ ; four of them give AB = 6.

one segment of AB =

\frac{√(3)}{2}

· √(3) =

\frac{3}{2}

→ AB = 4 ·

\frac{3}{2}

= 6

4STEP 4

The altitude stacks the small triangle's vertical leg $\frac{√(3)}{2}$ under a vertical kite side √(3), so h = $\frac{3√(3)}{2}$ (namely √(3) + $\frac{√(3)}{2}$ ).

h = √(3) +

\frac{√(3)}{2}

\frac{2√(3) + √(3)}{2}

\frac{3√(3)}{2}

5STEP 5

With the base AB and height h in hand, apply ½·base·height to close out the area.

[△ ABC] =

\frac{1}{2}

· 6 ·

\frac{3√(3)}{2}

= 3 ·

\frac{3√(3)}{2}

\frac{9√(3)}{2}

→ (B)

Answer

\frac{9}{2}

√3

Sanity check the total tiled area. One kite is two right triangles of legs 1 and √(3), so its area is 2 ·

\frac{1}{2}

· 1 · √(3) = √(3), and the eight-kite polygon has area 8√(3) ≈ 13.86. Our △ ABC has area

\frac{9√(3)}{2}

≈ 7.79, which is well under the whole polygon and consistent with the figure (the triangle clearly covers more than half but not all of it). The numerical value 7.79 matches choice (B) exactly.

💡Key takeaway

When a figure is tiled by repeated pieces, label one piece carefully and use similar-triangle scaling to read the base and height of the big triangle straight off the picture — then $\frac{1}{2}$ · base · height closes it out.

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