Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 1.5 (Way FASTER! Use in the contest))
 
(18 intermediate revisions by 11 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2023 AMC 10B Problems/Problem 3|2023 AMC 10B #3]] and [[2023 AMC 12B Problems/Problem 3|2023 AMC 12B #3]]}}
  
 
==Problem==
 
==Problem==
−
A 3-4-5 right triangle is inscribed circle <math>A</math>, and a 5-12-13 right triangle is inscribed in circle <math>B</math>. What is the ratio of the area of circle <math>A</math> to circle <math>B</math>?
+
A <math>3-4-5</math> right triangle is inscribed in circle <math>A</math>, and a <math>5-12-13</math> right triangle is inscribed in circle <math>B</math>. What is the ratio of the area of circle <math>A</math> to the area of circle <math>B</math>?
  
 
<math>\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}</math>
 
<math>\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}</math>
  
 +
==Solution 1==
 +
Because the triangles 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>
  
−
==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>. Lastly, divide, to get your answer: <math>\frac{25}{169}</math> = <math>\boxed{\textbf{(D) \frac{25}{169}}}</math>.
 
  
 
~Failure.net
 
~Failure.net
 +
 +
==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 <math>\frac{5}{13}</math>  which means the  area ratio is just <math>\frac{5^2}{13^2}</math>  so,  <math>\boxed{\textbf{(D) }\frac{25}{169}}.</math>
 +
 +
-Mismatchedcubing/Andrew_Lu
 +
 +
==Solution 2==
 +
Since the arc angle of the diameter of a circle is <math>90</math> degrees, the hypotenuse of each these two triangles is respectively the diameter of circles <math>A</math> and <math>B</math>.
 +
 +
Therefore the ratio of the areas equals the radius of circle <math>A</math> squared : the radius of circle <math>B</math> squared
 +
<math>=</math> <math>0.5\times</math> the diameter of circle <math>A</math>, squared : <math>0.5\times</math> the diameter of circle <math>B</math>, squared
 +
<math>=</math> the diameter of circle <math>A</math>, squared: the diameter of circle <math>B</math>, squared <math>=\boxed{\textbf{(D) }\frac{25}{169}}.</math>
 +
 +
 +
~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 <math>\frac{5^2}{13^2} =\boxed{\textbf{(D) }\frac{25}{169}}.</math>
 +
 +
~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
 +
 +
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=2|num-a=4}}
 +
{{AMC12 box|year=2023|ab=B|num-b=2|num-a=4}}
 +
{{MAA Notice}}

Latest revision as of 20:13, 29 January 2024

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 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.net

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_Lu

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

See also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_3&oldid=213774"