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.
Contents
Problem
A right triangle is inscribed in circle , and a right triangle is inscribed in circle . What is the ratio of the area of circle to the area of circle ?
Solution 1
Because the triangles are right triangles, we know the hypotenuses are diameters of circles and . Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply to get and as the areas of the circles. Multiply 4 on both numbers to get and . Cancel out the , and lastly, divide, to get your answer
~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 which means the area ratio is just so,
-Mismatchedcubing/Andrew_Lu
Solution 2
Since the arc angle of the diameter of a circle is degrees, the hypotenuse of each these two triangles is respectively the diameter of circles and .
Therefore the ratio of the areas equals the radius of circle squared : the radius of circle squared the diameter of circle , squared : the diameter of circle , squared the diameter of circle , squared: the diameter of circle , squared
~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
~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!)
~Education, the Study of Everything
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
Video Solution
~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.