2008 UNCO Math Contest II Problems/Problem 6

Problem

Points $A$ and $B$ are on the same side of line $L$ in the plane. $A$ is $5$ units away from $L, B$ is $9$ units away from $L$ . The distance between $A$ and $B$ is $12$ . For all points $P$ on $L$ what is the smallest value of the sum $AP + PB$ of the distances from $A$ to $P$ and from $P$ to $B$ ?

$[asy] draw((-1,0)--(16,0),arrow=Arrow()); draw((16,0)--(-1,0),arrow=Arrow()); draw((2,0)--(2,5)--(2+sqrt(128),9)--(2+sqrt(128),0),black); draw((2,5)--(8,0)--(2+sqrt(128),9),dashed); dot((2,5));dot((8,0));dot((2+sqrt(128),9)); MP("A",(2,5),W);MP("P",(8,0),S);MP("B",(2+sqrt(128),9),E);MP("L",(14,0),S); MP("5",(2,2.5),W);MP("12",(2+sqrt(128)/2,7),N);MP("9",(2+sqrt(128),4.5),E); [/asy]$

Solution

Construct $B'$ , the reflection of $B$ across $L$ , and $Q$ , the foot of the perpendicular from $A$ to the segment connecting $B$ to $L$ . By Pythagorean Theorem, $AQ^2=AB^2-QB^2=12^2-4^2=128$ . The minimum possible value of $AP+BP$ is equal to the minimum value of $AP+B'P$ , which is equal to the length of $AB'$ . By Pythagorean Theorem, $AB'=\sqrt{AQ^2+B'Q^2}=\sqrt{128+14^2}=\boxed{18}$ .

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2008 UNCO Math Contest II Problems/Problem 6

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