Difference between revisions of "2024 AMC 10A Problems/Problem 14"
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+ | == Video Solution by Pi Academy == | ||
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+ | https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM | ||
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== Video Solution 1 by Power Solve == | == Video Solution 1 by Power Solve == |
Revision as of 10:07, 9 November 2024
Contents
[hide]Problem
One side of an equilateral triangle of height lies on line . A circle of radius is tangent to line and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line can be written as , where , , and are positive integers and is not divisible by the square of any prime. What is ?
Diagram
~MRENTHUSIASM
Solution 1
Call the bottom vertices and (the one closer to the circle is ) and the top vertice . The tangency point between the circle and the side of triangle is , and the tangency point on line , and the center of the circle is
Draw radii to the tangency points, the arc is degrees because is , and since is supplementary, it's . The sum of the angles in a quadrilateral is , which means is
Triangle ODC is -- triangle so CD is .
Since we have congruent triangles ( and ), the combined area of both is .
The area of the arc is which is , so the answer is
is which is
~ASPALAPATI75
~andliu766 (latex)
edits by 9897
Note
There were two possible configurations from this problem; the one described in the solution above and the configuration in which the circle is tangent to the bottom of line and the base of the equilateral triangle. However, since the area in this configuration is simply we can infer that the problem is talking about the configuration in Solution 1.
~dbnl
Video Solution by Pi Academy
https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
Video Solution 1 by Power Solve
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.