1993 UNCO Math Contest II Problems/Problem 9

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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 (ProblemsAnswer KeyResources)
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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