1999 AHSME Problems/Problem 21
Contents
Problem
A circle is circumscribed about a triangle with sides and thus dividing the interior of the circle into four regions. Let and be the areas of the non-triangular regions, with be the largest. Then
Solution
. Therefore the triangle is a right triangle. But then its hypotenuse is a diameter of the circumcircle, and thus is exactly one half of the circle. Moreover, the area of the triangle is . Therefore the area of the other half of the circumcircle can be expressed as . Thus the answer is .
To complete the solution, note that is clearly false. As , we have and thus is false. Similarly , thus is false. And finally, since , , thus is false as well.
Solution 2 (Alternative to realize that the triangle is Right)
Copy the link into the browser to see a diagram (I don't know how to link it and I tried using the brackets, but it didn't work)://geogebra.org/classic/psg3ugzm
Let be the circumcircle of in the problem, and let the circle have a radius of . Let .
Using the law of cosines: .
Thus , thus the triangle is right. Thus follow as above: "Moreover, the area of the triangle is . Therefore the area of the other half of the circumcircle can be expressed as . Thus the answer is " (I quoted the solution above to show you where to continue).
~hastapasta
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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