Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 6"
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== Solution == | == Solution == | ||
− | {{ | + | 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 and are on the same side of line in the plane. is units away from is units away from . The distance between and is . For all points on what is the smallest value of the sum of the distances from to and from to ?
Solution
Construct , the reflection of across , and , the foot of the perpendicular from to the segment connecting to . By Pythagorean Theorem, . The minimum possible value of is equal to the minimum value of , which is equal to the length of . By Pythagorean Theorem, .
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 |