Difference between revisions of "2024 AMC 10A Problems/Problem 14"
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− | Draw radii to the tangency points, the arc is 60 degrees because <math>\angle ACB</math> is 60, and since <math>\angle DCE</math> is supplementary, it's <math>120^{\circ}</math>. The sum of the angles in a quadrilateral is 360, which means <math>\angle COD</math> is <math>60^{\circ}</math> | + | Draw radii to the tangency points, the arc is 60 degrees because <math>\angle ACB</math> is <math>60</math>, and since <math>\angle DCE</math> is supplementary, it's <math>120^{\circ}</math>. The sum of the angles in a quadrilateral is <math>360</math>, which means <math>\angle COD</math> is <math>60^{\circ}</math> |
− | Triangle ODC is 30-60-90 triangle so CD is <math>4\sqrt{3}</math>. | + | Triangle ODC is <math>30</math>-<math>60</math>-<math>90</math> triangle so CD is <math>4\sqrt{3}</math>. |
− | Since we have 2 congruent triangles (<math>\Delta ODC</math> and <math>\Delta OEC</math>), the combined area of both is <math>48\sqrt{3}</math>. | + | Since we have <math>2</math> congruent triangles (<math>\Delta ODC</math> and <math>\Delta OEC</math>), the combined area of both is <math>48\sqrt{3}</math>. |
The area of the arc is <math>144*60/360*\pi</math> which is <math>24\pi</math>, so the answer is <math>48\sqrt{3}-24\pi</math> | The area of the arc is <math>144*60/360*\pi</math> which is <math>24\pi</math>, so the answer is <math>48\sqrt{3}-24\pi</math> | ||
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~ASPALAPATI75 | ~ASPALAPATI75 | ||
+ | ~andliu766 (latex) | ||
edits by 9897 | edits by 9897 |
Revision as of 19:41, 8 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
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 60 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
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.