2023 AMC 10B Problem 3

Below is the professionally curated solution for Problem 3 of the 2023 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10B solutions, or check the answer key.

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Concepts:circumcircle, circumcenter, and circumradius right triangle area ratio

Difficulty rating: 1080

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 $B?$

$\dfrac{9}{25}$

$\dfrac{1}{9}$

$\dfrac{1}{5}$

$\dfrac{25}{169}$

$\dfrac{4}{25}$

Solution:

The hypotenuse of a right triangle is a diameter of the circle around it. So circle $A$ has diameter $5$ and circle $B$ has diameter $13.$ Areas scale as the square of that, giving $\left(\frac{5}{13}\right)^2 = \frac{25}{169}.$ Thus, D is the correct answer.

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