Difference between revisions of "2023 AMC 12B Problems/Problem 3"

Revision as of 20:47, 15 November 2023

The following problem is from both the 2023 AMC 10B #3 and 2023 AMC 12B #3, so both problems redirect to this page.

Problem

A 3-4-5 right triangle is inscribed 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 circle $B$ ?

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

Solution 1

Because the triangle are right triangles, we know the hypotenuses are diameters of circles $A$ and $B$ . Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply $\pi$ to get $6.25\pi$ and $42.25\pi$ as the areas of the circles. Multiply 4 on both numbers to get $25\pi$ and $169\pi$ . Cancel out the $\pi$ , and lastly, divide, to get your answer $=\boxed{\textbf{(D) }\frac{25}{169}}.$

~Failure.net

Solution 2

Since the arc angle of the diameter of a circle is $90$ degrees, the hypotenuse of each these two triangles is respectively the diameter of circles $A$ and $B$ .

Therefore the ratio of the areas equals the radius of circle $A$ squared : the radius of circle $B$ squared $=$ $0.5\times$ the diameter of circle $A$ , squared : $0.5\times$ the diameter of circle $B$ , squared $=$ the diameter of circle $A$ , squared: the diameter of circle $B$ , squared $=\boxed{\textbf{(D) }\frac{25}{169}}.$

~Mintylemon66

Solution 3

The ratio of areas of circles is the same as the ratios of the diameters squared. Since this is a right triangle the hypotenuse of each triangle will be the diameter of the circle. This yields the expression $\frac{5^2}{13^2} =\boxed{\textbf{(D) }\frac{25}{169}}.$

~vsinghminhas

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 20:19, 15 November 2023 (view source) Eevee9406 (talk \| contribs) (redirect) ← Older edit		Revision as of 20:47, 15 November 2023 (view source) Failure.net (talk \| contribs) (→‎Solution 1) Newer edit →
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	==Solution 1==		==Solution 1==
−	Because the triangle are right triangles, we know the hypotenuses are diameters of circles <math>A</math> and <math>B</math>. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply <math>\pi</math> to get <math>6.25\pi</math> and <math>42.25\pi</math> as the areas of the circles. Multiply 4 on both numbers to get <math>25\pi</math> and <math>169\pi</math>. Cancel out the <math>\pi</math>, and lastly, divide, to get your answer <math>\boxed{\textbf(D)\frac{25}{169}}</math>.	+	Because the triangle are right triangles, we know the hypotenuses are diameters of circles <math>A</math> and <math>B</math>. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply <math>\pi</math> to get <math>6.25\pi</math> and <math>42.25\pi</math> as the areas of the circles. Multiply 4 on both numbers to get <math>25\pi</math> and <math>169\pi</math>. Cancel out the <math>\pi</math>, and lastly, divide, to get your answer <math>=\boxed{\textbf{(D) }\frac{25}{169}}.</math>

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