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Difference between revisions of "1993 AIME Problems/Problem 15"

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(the WOOT group did this problem. I was part of it. Somebody put the image up.)
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== Solution ==
 
== Solution ==
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From the [[Pythagorean Theorem]], <math>AH^2+CH^2=1994^2</math>, and <math>(1995-AH)^2+CH^2=1993^2</math>. Subtracting those two equations yields <math>AH^2-(1995-AH)^2=3987</math>. After simplification, we see that <math>2*1995AH-1995^2=3987</math>, or <math>AH=\frac{1995}{2}+\frac{3987}{2*1995}</math>. Note that <math>AH+BH=1995</math>. Therefore we have that <math>BH=\frac{1995}{2}-\frac{3987}{2*1995}</math>. Therefore <math>AH-BH=\frac{3987}{1995}</math>.
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Now note that <math>RS=|HR-HS|</math>, <math>RH=\frac{BH+CH-BC}{2}</math>, and <math>HS=\frac{CH+AH-AC}{2}</math>. Therefore we have
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<math>RS=|\frac{BH+CH-BC-CH-AH+AC}{2}|=\frac{|BH-AH-1993+1994|}{2}</math>.
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Plugging in <math>AH-BH</math> and simplifying, we have <math>RS=\frac{1992}{1995*2}=\frac{332}{665}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1993|num-b=14|after=Last question}}
 
{{AIME box|year=1993|num-b=14|after=Last question}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 21:27, 3 January 2009

Problem

Let $\overline{CH}$ be an altitude of $\triangle ABC$. Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$. If $AB = 1995\,$, $AC = 1994\,$, and $BC = 1993\,$, then $RS\,$ can be expressed as $m/n\,$, where $m\,$ and $n\,$ are relatively prime integers. Find $m + n\,$.

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

From the Pythagorean Theorem, $AH^2+CH^2=1994^2$, and $(1995-AH)^2+CH^2=1993^2$. Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$. After simplification, we see that $2*1995AH-1995^2=3987$, or $AH=\frac{1995}{2}+\frac{3987}{2*1995}$. Note that $AH+BH=1995$. Therefore we have that $BH=\frac{1995}{2}-\frac{3987}{2*1995}$. Therefore $AH-BH=\frac{3987}{1995}$.

Now note that $RS=|HR-HS|$, $RH=\frac{BH+CH-BC}{2}$, and $HS=\frac{CH+AH-AC}{2}$. Therefore we have

$RS=|\frac{BH+CH-BC-CH-AH+AC}{2}|=\frac{|BH-AH-1993+1994|}{2}$.

Plugging in $AH-BH$ and simplifying, we have $RS=\frac{1992}{1995*2}=\frac{332}{665}$.

See also

1993 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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