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Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 9"

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== Problem ==
 
== Problem ==
  
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Let <math>P</math> be a point inside the rectangle <math>ABCD</math>. If <math>AP=5</math> , <math>BP=11</math> and <math>CP=10</math>, find the length of <math>DP</math>.  
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Let <math>P</math> be a point inside the rectangle <math>ABCD</math>. If <math>AP=5</math> , <math>BP=10</math> and <math>CP=11</math>, find the length of <math>DP</math>.  
 
(Hint: draw helpful vertical and horizontal lines.)
 
(Hint: draw helpful vertical and horizontal lines.)
  
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== Solution ==
 
== Solution ==
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By the British Flag Theorem, we have <math>AP^2</math>+<math>CP^2</math>=<math>BP^2</math>+<math>DP^2</math>. Substituting in, we have 25+121=100+<math>DP^2</math>. We find <math>DP</math> to be <math>\sqrt{46}</math>.
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By the [[British Flag Theorem]], we have <math>AP^2+CP^2</math>=<math>BP^2+DP^2</math>. Substituting in, we have <math>25+121=100+DP^2</math>. We find <math>DP</math> to be <math>\boxed{\sqrt{46}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 00:07, 20 January 2023

Problem

Let $P$ be a point inside the rectangle $ABCD$. If $AP=5$ , $BP=10$ and $CP=11$, find the length of $DP$. (Hint: draw helpful vertical and horizontal lines.)

[asy] pair P=(2,2); draw((0,0)--(0,5)--(10,5)--(10,0)--cycle,dot); draw((0,0)--P,black); draw((0,5)--P,black); draw((10,5)--P,black); draw((10,0)--P,black); dot(P); MP("P",(1,1),N); MP("5",(1.5,2.9),N); MP("10",(6.5,2.8),N); MP("11",(6.5,.9),N); MP("A",(0,5),NW); MP("B",(10,5),NE); MP("C",(10,0),SE); MP("D",(0,0),SW); [/asy]


Solution

By the British Flag Theorem, we have $AP^2+CP^2$=$BP^2+DP^2$. Substituting in, we have $25+121=100+DP^2$. We find $DP$ to be $\boxed{\sqrt{46}}$.

See also

1993 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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