Difference between revisions of "1999 AHSME Problems/Problem 21"
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To complete the solution, note that <math>\mathrm{(A)}</math> is clearly false. As <math>A+B < C</math>, we have <math>A^2 + B^2 < (A+B)^2 < C^2</math> and thus <math>\mathrm{(C)}</math> is false. Similarly <math>20A + 21B < 21(A+B) < 21C < 29C</math>, thus <math>\mathrm{(D)}</math> is false. And finally, since <math>0<A<C</math>, <math>\frac 1{C^2} < \frac1{A^2} < \frac 1{A^2} + \frac 1{B^2}</math>, thus <math>\mathrm{(E)}</math> is false as well. | To complete the solution, note that <math>\mathrm{(A)}</math> is clearly false. As <math>A+B < C</math>, we have <math>A^2 + B^2 < (A+B)^2 < C^2</math> and thus <math>\mathrm{(C)}</math> is false. Similarly <math>20A + 21B < 21(A+B) < 21C < 29C</math>, thus <math>\mathrm{(D)}</math> is false. And finally, since <math>0<A<C</math>, <math>\frac 1{C^2} < \frac1{A^2} < \frac 1{A^2} + \frac 1{B^2}</math>, thus <math>\mathrm{(E)}</math> is false as well. | ||
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+ | == 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] | ||
+ | |||
+ | 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>. | ||
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+ | Using the law of cosines: </math>29^2=20^2+21^2-2*20*21*\cos{XYZ}<math>. | ||
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+ | 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)}}$" (I quoted the solution above to show you where to continue). | ||
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+ | ~hastapasta | ||
== See also == | == See also == | ||
{{AHSME box|year=1999|num-b=20|num-a=22}} | {{AHSME box|year=1999|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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): $[https://geogebra.org/classic/psg3ugzm Sample diagram]
Let$ (Error compiling LaTeX. Unknown error_msg)\circle{O}\triangle{XYZ}rXY=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} = 210A+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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.