AMC 10 Step by Step

2023 AMC 10B Problem 3

P1-10~1 minPrint

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 ?

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Key insight

When a right triangle sits inside a circle with all three corners on it, its longest side is a diameter. So the diameters are 5 and 13.

Solution

To compare the circles, we need their sizes. The triangles give us those sizes through one key fact.

1. The key fact. If a right triangle has all three corners on a circle, its longest side (the hypotenuse) goes straight through the center. In other words, the hypotenuse is a diameter. (The right angle "looks at" half the circle, and half a circle is cut off by a diameter.)
2. Read off the diameters. The 33-44-55 triangle has hypotenuse 55, so circle AA has diameter 55 and radius 52\frac{5}{2}. The 55-1212-1313 triangle has hypotenuse 1313, so circle BB has diameter 1313 and radius 132\frac{13}{2}.
3. Compare the areas. Area is πr2\pi r^2:
π(52)2π(132)2=2541694=25169.\frac{\pi \left(\frac{5}{2}\right)^2}{\pi \left(\frac{13}{2}\right)^2} = \frac{\frac{25}{4}}{\frac{169}{4}} = \frac{25}{169}.
The π\pi and the 14\frac{1}{4} cancel, so you could simply square 513\frac{5}{13}.

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

Why this works

"A right triangle inside a circle" should make you think "the hypotenuse is a diameter" right away; the two shorter sides do not matter after that. And when one circle is a scaled copy of another, the area ratio is the square of the length ratio, so you never need to multiply out π\pi.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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}}.

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.