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 10B Problems/Problem 3"

(→‎Solution)
(→‎Solution)
Line 6: Line 6:
 
<math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{25}{169}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{5}\qquad\textbf{(E) }\frac{9}{25}</math>
 
<math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{25}{169}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{5}\qquad\textbf{(E) }\frac{9}{25}</math>
  
−
==Solution==
+
==Solution 1==
 
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>.  
 
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>.  
  
Line 15: Line 15:
  
 
~Mintylemon66
 
~Mintylemon66
 +
 +
==Solution 2==
 +
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 <math>\frac{5^2}{13^2} =\boxed{\textbf{(B) }\frac{25}{169}}.</math>
 +
 +
~vsinghminhas

Revision as of 16:12, 15 November 2023

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 $B$?


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

Solution 1

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 $=\boxed{\textbf{(B) }\frac{25}{169}}.$


~Mintylemon66

Solution 2

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 $\frac{5^2}{13^2} =\boxed{\textbf{(B) }\frac{25}{169}}.$

~vsinghminhas

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10B_Problems/Problem_3&oldid=203134"