2017 AMC 10A Problems/Problem 22
Problem
Sides and of equilateral triangle are tangent to a circle at points and respectively. What fraction of the area of lies outside the circle?
Solution 1
Let the radius of the circle be , and let its center be . Since and are tangent to circle , then , so . Therefore, since and are equal to , then (pick your favorite method) . The area of the equilateral triangle is , and the area of the sector we are subtracting from it is . The area outside of the circle is . Therefore, the answer is
Note
The sector angle is because and are both 90 degrees meaning , so is cyclic. Thus, the angle is
~mathboy282
Solution 2
(same diagram as Solution 1)
Without the Loss of Generality, let the side length of the triangle be .
Then, the area of the triangle is . We are looking for the area of the portion inside the triangle but outside the circle divided by the area of the triangle. Since , and , we know , and . Drop an angle bisector of onto , call the point of intersection . By SAS congruence, , by CPCTC (Congruent Parts of Congruent Triangles are Congruent) and they both measure . By 30-60-90 triangle, . The area of the sector bounded by arc BC is one-third the area of circle O, whose area is . Therefore, the area of the sector bounded by arc BC is .
We are nearly there. By 30-60-90 triangle, we know , so the area of is . The area of the region inside both the triangle and circle is the area of the sector bounded by arc BC minus the area of : . The area of the region outside of the circle but inside the triangle is and the ratio is .
~JH. L
Video Solution
https://www.youtube.com/watch?v=GnJDNtjd57k&feature=youtu.be
https://youtu.be/ADDAOhNAsjQ -Video Solution by Richard Rusczyk
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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