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 $B$ . What is the ratio of the area of circle $A$ to the area of circle $B$ ?

$\textbf{(A) }\frac{9}{25}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{1}{5}\qquad\textbf{(D) }\frac{25}{169}\qquad\textbf{(E) }\frac{4}{25}$

Pick an answer.

(A)

$frac{9}{25}$

(B)

$frac{1}{9}$

(C)

$frac{1}{5}$

(D)

$frac{25}{169}$

(E)

$frac{4}{25}$

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Toolkit + CCSS Solution

Understand

Restated: A $3$-$4$-$5$ right triangle is inscribed in circle $A$, and a $5$-$12$-$13$ right triangle is inscribed in circle $B$. Find $\frac{\text{area of } A}{\text{area of } B}$.

Givens: Triangle in circle $A$: right triangle with sides $3, 4, 5$; Triangle in circle $B$: right triangle with sides $5, 12, 13$; Each triangle is *inscribed* — all three vertices lie on its circle; Answer choices: (A) $\frac{9}{25}$, (B) $\frac{1}{9}$, (C) $\frac{1}{5}$, (D) $\frac{25}{169}$, (E) $\frac{4}{25}$

Unknowns: The ratio of the two circle areas

Understand

Restated: A $3$-$4$-$5$ right triangle is inscribed in circle $A$, and a $5$-$12$-$13$ right triangle is inscribed in circle $B$. Find $\frac{\text{area of } A}{\text{area of } B}$.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

A quick sketch (Tool #1) of each circle with its right triangle makes the key fact pop out: the $90^\circ$ 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.

Execute — Answer: D

#1 Draw a Diagram 7.G.B.4 Step 1

Sketch circle $A$ with the $3$-$4$-$5$ triangle inscribed.
The right angle is opposite the longest side.
A standard inscribed-angle fact says any inscribed right angle's opposite side is a diameter — so the hypotenuse of length $5$ is the diameter of circle $A$.

$$d_A = 5$$

💡 Drawing the triangle on the circle shows the hypotenuse stretches from one side of the circle to the other — that's the diameter.

#1 Draw a Diagram 7.G.B.4 Step 2

Same picture for circle $B$ with the $5$-$12$-$13$ triangle.
The hypotenuse $13$ is again the diameter, by the same inscribed-right-angle reasoning.

$$d_B = 13$$

💡 Same drawing argument, different right triangle — the hypotenuse becomes the diameter again.

#7 Identify Subproblems 7.G.B.4 Step 3

Write each area in terms of its diameter.
Since $r = d/2$, the area $\pi r^2$ equals $\pi (d/2)^2 = \frac{\pi d^2}{4}$.
Now form the ratio: $\pi$ and the $4$ cancel, leaving the square of the diameter ratio.

$$\dfrac{\text{Area}_A}{\text{Area}_B} = \dfrac{\pi d_A^2/4}{\pi d_B^2/4} = \left(\dfrac{d_A}{d_B}\right)^2$$

💡 When two circles' areas are compared, only the diameter ratio matters — the $\pi$ and the $4$ are the same on both.

#7 Identify Subproblems 6.EE.A.1 Step 4

Substitute the two diameters and square.

$$\left(\dfrac{5}{13}\right)^2 = \dfrac{25}{169} \;\Rightarrow\; \textbf{(D)}$$

💡 Squaring $\tfrac{5}{13}$ squares the top and bottom separately — a Grade 6 exponent rule.

[1] #1 7.G.B.4 Sketch circle $A$ with the $3$-$4$-$5$ triangle inscribed. The right angle is op

[2] #1 7.G.B.4 Same picture for circle $B$ with the $5$-$12$-$13$ triangle. The hypotenuse $13$

[3] #7 7.G.B.4 Write each area in terms of its diameter. Since $r = d/2$, the area $\pi r^2$ eq

[4] #7 6.EE.A.1 Substitute the two diameters and square.

Review

Reasonableness: 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 $\frac{25}{169} \approx 0.148$ fits. Among the choices, the only ones below $\frac{1}{4}$ are (B) $\frac{1}{9} \approx 0.111$, (C) $\frac{1}{5} = 0.2$, (D) $\frac{25}{169} \approx 0.148$, (E) $\frac{4}{25} = 0.16$. Only (D) matches the exact squared diameter ratio.

Alternative: Tool #13 (Convert to Algebra) computing radii directly: $r_A = 5/2$, $r_B = 13/2$. Areas: $\pi (5/2)^2 = \frac{25\pi}{4}$ and $\pi (13/2)^2 = \frac{169\pi}{4}$. Ratio: $\frac{25\pi/4}{169\pi/4} = \frac{25}{169}$ — same answer with a touch more arithmetic.

CCSS standards used (min grade 7)

6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Squaring the diameter ratio $\left(\frac{5}{13}\right)^2 = \frac{25}{169}$.)
7.G.B.4 Know the formulas for area and circumference of a circle (Writing each area as $\frac{\pi d^2}{4}$ and forming the area ratio.)

⭐ 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 $\left(\tfrac{5}{13}\right)^2$.