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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) ==
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Click this link for the diagram (NOT TO SCALE): <math>[https://geogebra.org/classic/psg3ugzm Sample diagram]
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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

Problem

A circle is circumscribed about a triangle with sides $20,21,$ and $29,$ thus dividing the interior of the circle into four regions. Let $A,B,$ and $C$ be the areas of the non-triangular regions, with $C$ be the largest. Then

$\mathrm{(A) \ }A+B=C \qquad \mathrm{(B) \ }A+B+210=C \qquad \mathrm{(C) \ }A^2+B^2=C^2 \qquad \mathrm{(D) \ }20A+21B=29C \qquad \mathrm{(E) \ } \frac 1{A^2}+\frac 1{B^2}= \frac 1{C^2}$

Solution

$20^2 + 21^2 = 841 = 29^2$. Therefore the triangle is a right triangle. But then its hypotenuse is a diameter of the circumcircle, and thus $C$ is exactly one half of the circle. Moreover, the area of the triangle is $\frac{20\cdot 21}{2} = 210$. Therefore the area of the other half of the circumcircle can be expressed as $A+B+210$. Thus the answer is $\boxed{\mathrm{(B)}}$.

To complete the solution, note that $\mathrm{(A)}$ is clearly false. As $A+B < C$, we have $A^2 + B^2 < (A+B)^2 < C^2$ and thus $\mathrm{(C)}$ is false. Similarly $20A + 21B < 21(A+B) < 21C < 29C$, thus $\mathrm{(D)}$ is false. And finally, since $0<A<C$, $\frac 1{C^2} < \frac1{A^2} < \frac 1{A^2} + \frac 1{B^2}$, thus $\mathrm{(E)}$ 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}$be the circumcircle of$\triangle{XYZ}$in the problem, and let the circle have a radius of$r$. Let$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$, thus the triangle is right. Thus follow as above: "Moreover, the area of the triangle is$\frac{20\cdot 21}{2} = 210$. Therefore the area of the other half of the circumcircle can be expressed as$A+B+210$. Thus the answer is$\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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