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

(Solution)
(Solution)
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== Solution ==
 
== 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 /squt{46}.
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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 \sqrt{46}.
  
 
== See also ==
 
== See also ==

Revision as of 13:42, 30 December 2014

Problem

Let $P$ be a point inside the rectangle $ABCD$. If $AP=5$ , $BP=11$ and $CP=10$, 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 \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