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Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 6"

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

Latest revision as of 21:28, 18 March 2018

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, $AB'=\sqrt{AQ^2+B'Q^2}=\sqrt{128+14^2}=\boxed{18}$.

See Also

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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