AIME 1993 · 第 15 题

Solution

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 $AH-BH=\frac{3987}{1995}$.

$\textbf{Note: }$An easier method is to use both facts together:

First, by the Pythagorean Theorem, we find that $1993^2-BH^2=1994^2-AH^2$, so $AH^2-BH^2=1994^2-1993^2=3987$. We know that $AH+BH=1995$, so by difference of squares and dividing we find that $AH-BH=\frac{3987}{1995}=\frac{1329}{665}$. ~eevee9406

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 $RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}$.

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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

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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 write it as $RS=\frac{|b-a|(a+b-c)}{2c}$ since $a+b-c \geq 0$ by the triangle inequality. ~Yiyj1