AMC 10 · 2023 · #3

Grade 7 geometry-2d

area-circlesinteger-pythagorean-triplesratio-proportionsimilar-figures identify-subproblems ↑ Prerequisites: area-circlesinteger-pythagorean-triples

📏 Short solution 💡 2 insights

Problem

A 3-4-5 right triangle is inscribed in circle A, and a 5-12-13 right triangle is inscribed in circle B. Find (area of A)/(area of B).

Pick an answer.

(A)

$frac{9}{25}$

(B)

$frac{1}{9}$

(C)

$frac{1}{5}$

(D)

$frac{25}{169}$

(E)

$frac{4}{25}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

A quick sketch (Tool #1) of each circle with its right triangle makes the key fact pop out: the 90° vertex sits on the circle and the hypotenuse goes straight across, so the hypotenuse is a diameter. Once that is seen, the problem breaks into three easy subproblems (Tool #7): (a) read off each diameter, (b) write each area in terms of its diameter, (c) form the ratio and cancel. No algebra heavier than squaring a fraction is needed.

1STEP 1

Sketch the 3-4-5 triangle inside circle A; its hypotenuse (longest side) spans the circle as a diameter, so d_A = 5.

d_A = 5

2STEP 2

Same reasoning for circle B: the 5-12-13 triangle's hypotenuse is its longest side, so d_B = 13.

d_B = 13

3STEP 3

Write each area as π(d/2)² = πd²/4; forming the ratio cancels π and 4, leaving (d_A/d_B)².

Area_A/Area_B = (π d_A²/4)/(π d_B²/4) = (d_A/d_B)²

4STEP 4

Substitute the diameters and square: ( $\frac{5}{13}$ )² gives the ratio $\frac{25}{169}$ → choice (D).

(

\frac{5}{13}

)² =

\frac{25}{169}

→ (D)

Answer

\frac{25}{169}

Sanity-check magnitudes. Diameter 5 vs diameter 13 means circle A is much smaller than B, so the ratio should be well below 1 — and

\frac{25}{169}

≈ 0.148 fits. Among the choices, the only ones below

\frac{1}{4}

are (B)

\frac{1}{9}

≈ 0.111, (C)

\frac{1}{5}

= 0.2, (D)

\frac{25}{169}

≈ 0.148, (E)

\frac{4}{25}

= 0.16. Only (D) matches the exact squared diameter ratio.

💡Key takeaway

This AMC 10 problem only needs the Grade 7 circle-area formula — a right triangle's hypotenuse is the circle's diameter, so the area ratio is just ( $\frac{5}{13}$ )².