2007 Alabama ARML TST Problems/Problem 15

Problem

Let $P$ be a point inside isosceles right triangle $ABC$ such that $\angle C = 90^{\circ}$ , $AP = 5$ , $BP = 13$ , and $CP = 6\sqrt{2}$ . Find the area of $ABC$ .

Solution

Solution 1

Let $D$ , $E$ , and $F$ be the reflections of $P$ over sides $BC$ , $CA$ , and $AB$ , respectively. We then have that $[AEC]=[APC]$ , $[FAB]=[PAB]$ , and $[BCD]=[BCP]$ . This shows that $[AECDBF]=2[ABC]$ . I shall now proceed to find $[AECDBF]$ . This is equal to

$[AECDBF]=[AEF]+[BDF]+[DEF]$

Note that $\angle CAE=\angle CAP$ and $\angle BAP=\angle BAF$ , so $\angle EAF=2\angle CAB=90^{\circ}$ . Similarly, $\angle FBD=90^{\circ}$ and $\angle DCE=180^{\circ}$ . Now note that $AE=AF=AP=5$ and $BD=BF=BP=13$ . Therefore $[AEF]=\frac{25}{2}$ and $[BDF]=\frac{169}{2}$ . Also note that $EF=5\sqrt{2}$ and $DF=13\sqrt{2}$ . We also know that $D$ , $C$ , and $E$ are collinear, so $DE=EC+CD=2CP=12\sqrt{2}$ . This shows that $DEF$ is a 5-12-13 right triangle, so it has area $\frac{EF\cdot DE}{2}=60$ , so

$[ABC]=\frac{[AECDBF]}{2}=\frac{\frac{25}{2}+\frac{169}{2}+60}{2}=\boxed{\frac{157}{2}}$

Solution 2

Rotate the diagram by 90 degrees about $C$ so that $B$ goes to $A$ , $A$ goes to a point $D$ , and $P$ goes to $P'$ . Since the image of $BP$ under this rotation is $AP'$ , $AP' = 13$ . Since $\triangle PCP'$ is a 45-45-90 right triangle, $PP' = 12$ . Thus, $APP'$ is a 5-12-13 right triangle, with $\angle APP' = 90^\circ$ . Note that $\angle APC = \angle APP' + PP'C = 90^\circ + 45^\circ = 135^\circ$ , so by Law of Cosines on triangle $APC$ ,

$AC^2 = AP'^2 + P'C^2 - 2(AP')(P'C)\cos 135^\circ = 25 + 72 - 2(-\frac{1}{\sqrt{2}}) (6\sqrt{2})(5) = 157,$

so $[ABC] = \frac{AC^2}{2} = \boxed{\frac{157}{2}}$ .

2007 Alabama ARML TST (Problems)
Preceded by: Problem 14	Followed by: Last Problem
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2007 Alabama ARML TST Problems/Problem 15

Contents

Problem

Solution

Solution 1

Solution 2

See also