2023 AMC 12B Problems/Problem 3

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

1 Problem
2 Solution 1
3 Solution 1.5 (Way FASTER! Use in the contest)
4 Note for Solution 1
5 Solution 2
6 Solution 3
7 Video Solution by Math-X (First understand the problem!!!)
8 Video Solution (Quick and Easy!)
9 Video Solution 1 by SpreadTheMathLove
10 Video Solution
11 Video Solution by Interstigation
12 See also

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 $B$ ?

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

Solution 1

Because the triangles 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

Solution 1.5 (Way FASTER! Use in the contest)

With right triangles inscribed in circles, the hypotenuse must be the diameter. Therefore, the ratio of radii is $\frac{5}{13}$ which means the area ratio is just $\frac{5^2}{13^2}$ so, $\boxed{\textbf{(D) }\frac{25}{169}}.$

-Mismatchedcubing/Andrew_

Note for Solution 1

The reason why the right triangle has its hypotenuse on the diameter is because of its properties with the circle and the central angle theorem. The central angle theorem states that an angle inscribed on a circle is just arc length in degrees. The diameter of a circle splits the circle into two halves with arc of $ 180^\circ $, and therefore the angle inscribed in that arc is just $ 180/2 = 90 $ degrees. The longest side of a triangle is opposite to the longest angle, and therefore the length of the hypotenuse is just the diameter of the circle.

~Pinotation

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 they are similar figures). 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

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/EuLkw8HFdk4?si=R0r3rh5MCqQXLG9G&t=587

~Math-X

Video Solution (Quick and Easy!)

https://youtu.be/Chw1TTyPiZE

~Education, the Study of Everything

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=SUnhwbA5_So

Video Solution

https://youtu.be/JeokECZJQko

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

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

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