# 1953 AHSME Problems/Problem 49

## Problem

The coordinates of $A,B$ and $C$ are $(5,5),(2,1)$ and $(0,k)$ respectively. The value of $k$ that makes $\overline{AC}+\overline{BC}$ as small as possible is: $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4\frac{1}{2} \qquad \textbf{(C)}\ 3\frac{6}{7} \qquad \textbf{(D)}\ 4\frac{5}{6}\qquad \textbf{(E)}\ 2\frac{1}{7}$

## Solution $k$ will be between $1$ and $5$ for $AC+BC$ to be the smallest. If we mirror point $A$ across the y-axis to $A'$, with coordinates $(-5,5),$ the distance $A'C+BC$ will be same as $AC+BC$. The minimum of $A'C+BC$ will occur when $C$ is on the straight line connecting $A'$ and $B$ (i.e., $C$ lies on the line $A'B$). Therefore, $k$ is the y-intercept of the line that passes through $A'$ and $B$.

The slope of the line is $\frac{1-5}{2-(-5)}=-\frac 47$. Using point-slope form, the equation of the line is $y-5=-\frac 47(x+5)$. Letting $x=0$ gives $y-5=-\frac{20}{7},$ so $y=\frac{15}{7}$. Therefore, $k = 2\frac17 \Rightarrow \boxed{\textbf{(E) } 2\frac17}.$

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