Difference between revisions of "2023 AMC 12B Problems/Problem 3"
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==See also== | ==See also== | ||
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{{AMC12 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}} | {{MAA Notice}} |
Revision as of 22:33, 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.
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 triangle 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 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 this is a right triangle the hypotenuse of each triangle will be the diameter of the circle. This yields the expression
~vsinghminhas
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
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.