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

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m (→‎Solution)
 
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Edit by GameMaster402:  
 
Edit by GameMaster402:  
  
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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>.
+
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>.
  
 
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>
 
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>
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~minor edit by [[User: Yiyj1|Yiyj1]]
 
~minor edit by [[User: Yiyj1|Yiyj1]]
  
−
Note: We can also just right it as <math>RS=\frac{|b-a|(a+b-c)}{2c}</math> since <math>a+b-c \leq 0</math> by the triangle inequality. ~[[User: Yiyj1|Yiyj1]]
+
Note: We can also just right it as <math>RS=\frac{|b-a|(a+b-c)}{2c}</math> since <math>a+b-c \geq 0</math> by the triangle inequality. ~[[User: Yiyj1|Yiyj1]]
  
 
== See also ==
 
== See also ==

Latest revision as of 20:39, 21 September 2023

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 $\frac{332}{665}$

~minor edit by Yiyj1

Edit by phoenixfire:

It can further be shown for any triangle with sides $a=BC, b=CA, c=AB$ that \[RS=\dfrac{|b-a|}{2c}|a+b-c|\] Over here $a=1993, b=1994, c=1995$, so using the formula gives \[RS = \dfrac{|1994 - 1993|}{2 \cdot 1995}|1993 + 1994 - 1995| = \dfrac{1 \cdot 1992}{2(1995)} = \frac{332}{665}.\]

~minor edit by Yiyj1

Note: We can also just right it as $RS=\frac{|b-a|(a+b-c)}{2c}$ since $a+b-c \geq 0$ by the triangle inequality. ~Yiyj1

See also

1993 AIME (Problems • Answer Key • Resources)
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