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)
Click this link for the diagram (NOT TO SCALE): $[https://geogebra.org/classic/psg3ugzm Sample diagram]
Let$ (Error compiling LaTeX. Unknown error_msg)\circle{O}
\triangle{XYZ}
r
XY=20, YZ=21, XZ=29$.
Using the law of cosines:$ (Error compiling LaTeX. Unknown error_msg)29^2=20^2+21^2-2*20*21*\cos{XYZ}$.
Thus$ (Error compiling LaTeX. Unknown error_msg)\cos{XYZ}=0
\frac{20\cdot 21}{2} = 210
A+B+210
\boxed{\mathrm{(B)}}$" (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 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
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