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

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</asy>
 
</asy>
  
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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>.
+
From the [[Pythagorean Theorem]], <math>AH^2+CH^2=1994^2</math>, and <math>(1995-AH)^2+CH^2=1993^2</math>.  
  
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Now note that <math>RS=|HR-HS|</math>, <math>RH=\frac{AH+CH-AC}{2}</math>, and <math>HS=\frac{CH+BH-BC}{2}</math>. Therefore we have
+
Subtracting those two equations yields <math>AH^2-(1995-AH)^2=3987</math>.  
  
−
<math>RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}</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>.
 +
 
 +
 
 +
 
 +
Now note that <math>RS=|HR-HS|</math>, <math>RH=\frac{AH+CH-AC}{2}</math>, and <math>HS=\frac{CH+BH-BC}{2}</math>.
 +
 
 +
Therefore we have <math>RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}</math>.
  
 
Plugging in <math>AH-BH</math> and simplifying, we have <math>RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}</math>.
 
Plugging in <math>AH-BH</math> and simplifying, we have <math>RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}</math>.
 +
------------------------
 +
Edit by GameMaster402:
 +
 +
It can be shown that in any triangle with side lengths <math>n-1, n, n+1</math>, if you draw an altitude from the vertex to the side of <math>n+1</math>, and draw the incircles of the two right triangles, the distance between the two tangency points is simply <math>\frac{n-2}{2n+2)}=\frac{n-2}{2(n+1)}</math>.
  
−
Edit by GameMaster402: It can be shown that in any triangle with side lengths <math>n-1, n, n+1</math>, if you draw an altitude from the vertex to the side of <math>n+1</math>, and draw the incircles of the two right triangles, the distance between the two tangency points is simply <math>\frac{n-2}{2n+2}</math> Plugging in <math>n=1994</math> yields that the answer is <math>1992/1995*2</math>, which simplifies to <math>332/665</math>
+
Plugging in <math>n=1994</math> yields that the answer is <math>\frac{1992}{2(1995)}</math>, which simplifies to <math>\frac{332}{665}</math>
  
 
== See also ==
 
== See also ==

Revision as of 10:58, 23 August 2020

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

[asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label("$R$",y[0],NE); label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S); [/asy]

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 $RS=|HR-HS|$, $RH=\frac{AH+CH-AC}{2}$, and $HS=\frac{CH+BH-BC}{2}$.

Therefore we have $RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}$.

Plugging in $AH-BH$ and simplifying, we have $RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}$.

Edit by GameMaster402:

It can be shown that in any triangle with side lengths $n-1, n, n+1$, if you draw an altitude from the vertex to the side of $n+1$, and draw the incircles of the two right triangles, the distance between the two tangency points is simply $\frac{n-2}{2n+2)}=\frac{n-2}{2(n+1)}$.

Plugging in $n=1994$ yields that the answer is $\frac{1992}{2(1995)}$, which simplifies to $\frac{332}{665}$

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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