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2023 AMC 10B Problem 3

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2023 AMC 10B Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Circles · difficulty 1 of 5 · about 1 min · position P1-10

A 3−4−53-4-5 right triangle is inscribed in circle AA , and a 5−12−135-12-13 right triangle is inscribed in circle BB . What is the ratio of the area of circle AA to the area of circle BB ?

  1. A)

    925\frac{9}{25}

  2. B)

    19\frac{1}{9}

  3. C)

    15\frac{1 }{5}

  4. D)

    25169\frac{25}{169}

  5. E)

    425\frac{4}{25}

Answer

D

Key insight

The hypotenuse of an inscribed right triangle is a diameter, so the radii are 5/2 and 13/2 and areas scale as (5/13)^2.

Solution

An inscribed angle of 90∘90^\circ subtends a semicircle, so the hypotenuse of each right triangle is a diameter of its circle. Circle AA has diameter 55 and circle BB has diameter 1313.

Area is proportional to the square of the radius (or of the diameter), so
area of Aarea of B=(513)2=25169.\frac{\text{area of } A}{\text{area of } B} = \left(\frac{5}{13}\right)^2 = \frac{25}{169}.

The answer is (D) 25169\boxed{\textbf{(D)}\ \frac{25}{169}}.

Why this works

"Right triangle inscribed in a circle" is a signal that the hypotenuse is the diameter; the legs are irrelevant once you know that. Ratios of areas of similar figures are squares of ratios of corresponding lengths, so no π\pi and no explicit area computation are needed.

The trap

Squaring the ratio of the shorter legs (3/5) instead of the hypotenuses, giving 9/25.

Common mistakes

  • Squaring the ratio of the shorter legs (3/5) instead of the hypotenuses, giving 9/25.
  • Reporting the ratio of radii 5/135/13 without squaring, or using the hypotenuse as the radius rather than the diameter (which cancels in the ratio, but only by luck).

Techniques

Set up the equation/formula and compute; no special trick needed