Difference between revisions of "1999 AHSME Problems/Problem 21"
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== Solution 2 (Alternative to realize that the triangle is Right) == | == Solution 2 (Alternative to realize that the triangle is Right) == | ||
− | Click this link for the diagram (NOT TO SCALE): <math>[https://geogebra.org/classic/psg3ugzm Sample diagram] | + | Click this link for the diagram (NOT TO SCALE): <math>[https://geogebra.org/classic/psg3ugzm Sample diagram]</math> |
− | Let < | + | Let <math>\circle{O}</math> be the circumcircle of <math>\triangle{XYZ}</math> in the problem, and let the circle have a radius of <math>r</math>. Let <math>XY=20, YZ=21, XZ=29</math>. |
− | Using the law of cosines: < | + | Using the law of cosines: <math>29^2=20^2+21^2-2*20*21*\cos{XYZ}</math>. |
− | Thus < | + | Thus <math>\cos{XYZ}=0</math>, thus the triangle is right. Thus follow as above: "Moreover, the area of the triangle is <math>\frac{20\cdot 21}{2} = 210</math>. Therefore the area of the other half of the circumcircle can be expressed as <math>A+B+210</math>. Thus the answer is <math>\boxed{\mathrm{(B)}}</math>" (I quoted the solution above to show you where to continue). |
~hastapasta | ~hastapasta |
Revision as of 10:47, 5 December 2022
Contents
[hide]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):
Let $\circle{O}$ (Error compiling LaTeX. Unknown error_msg) 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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.