Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 9"
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== Problem == | == Problem == | ||
| − | Let <math>P</math> be a point inside the rectangle <math>ABCD</math>. If <math>AP=5</math> , <math>BP= | + | 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.) | ||
| Line 24: | Line 24: | ||
== 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 <math>\sqrt{46}</math>. | + | 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 
be a point inside the rectangle 
. If 
, 
and 
, find the length of 
.
(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]](http://latex.artofproblemsolving.com/c/e/5/ce5c11133ade1b8f6d288e91b6d710637a5ee630.png)
Solution
By the British Flag Theorem, we have 
=
. Substituting in, we have 
. We find to be 
.
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 | ||
