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Difference between revisions of "2007 Alabama ARML TST Problems/Problem 15"

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==Solution==
 
==Solution==
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=== Solution 1 ===
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[[File:2007AlabamaARMLTST15.png]]
 
[[File:2007AlabamaARMLTST15.png]]
  
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<cmath>[ABC]=\frac{[AECDBF]}{2}=\frac{\frac{25}{2}+\frac{169}{2}+60}{2}=\boxed{\frac{157}{2}}</cmath>
 
<cmath>[ABC]=\frac{[AECDBF]}{2}=\frac{\frac{25}{2}+\frac{169}{2}+60}{2}=\boxed{\frac{157}{2}}</cmath>
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=== Solution 2 ===
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Rotate the diagram by 90 degrees about <math>C</math> so that <math>B</math> goes to <math>A</math>, <math>A</math> goes to a point <math>D</math>, and <math>P</math> goes to <math>P'</math>. Since the image of <math>BP</math> under this rotation is <math>AP'</math>, <math>AP' = 13</math>. Since <math>\triangle PCP'</math> is a 45-45-90 right triangle, <math>PP' = 12</math>. Thus, <math>APP'</math> is a 5-12-13 right triangle, with <math>\angle APP' = 90^\circ</math>. Note that <math>\angle APC = \angle APP' + PP'C = 90^\circ + 45^\circ = 135^\circ</math>, so by Law of Cosines on triangle <math>APC</math>, <math>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</math>, so <math>[ABC] = \frac{AC^2}{2} = \boxed{\frac{157}{2}}</math>.
  
 
==See also==
 
==See also==
 
{{ARML box|year=2007|state=Alabama|num-b=14|after=Last Problem}}
 
{{ARML box|year=2007|state=Alabama|num-b=14|after=Last Problem}}

Revision as of 17:11, 21 March 2018

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

2007AlabamaARMLTST15.png

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

See also

2007 Alabama ARML TST (Problems)
Preceded by:
Problem 14 		Followed by:
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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