Difference between revisions of "2007 Alabama ARML TST Problems/Problem 15"
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[[File:2007AlabamaARMLTST15.png]] | [[File:2007AlabamaARMLTST15.png]] | ||
− | Let <math>D</math>, <math>E</math>, and <math>F</math> be the reflections of <math>P</math> over sides <math>BC</math>, <math>CA</math>, and <math>AB</math>, respectively. We then have that <math>[AEC]=[APC]</math>, <math>[FAB]=[PAB]</math>, and <math>[BCD]=[BCP]</math>. This shows that <math>[AECDBF]=2[ABC]</math>. I shall now proceed to find <math>[AECDBF]</math>. | + | Let <math>D</math>, <math>E</math>, and <math>F</math> be the reflections of <math>P</math> over sides <math>BC</math>, <math>CA</math>, and <math>AB</math>, respectively. We then have that <math>[AEC]=[APC]</math>, <math>[FAB]=[PAB]</math>, and <math>[BCD]=[BCP]</math>. This shows that <math>[AECDBF]=2[ABC]</math>. I shall now proceed to find <math>[AECDBF]</math>. This is equal to |
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<cmath>[AECDBF]=[AEF]+[BDF]+[DEF]</cmath> | <cmath>[AECDBF]=[AEF]+[BDF]+[DEF]</cmath> | ||
− | Now note that <math>AE=AF=AP=5</math> and <math>BD=BF=BP=13</math>. Therefore <math>[AEF]=\frac{25}{2}</math> and <math>[BDF]=\frac{169}{2}</math>. Also note that <math>EF=5\sqrt{2}</math> and <math>DF=13\sqrt{2}</math>. We also know that <math>D</math>, <math>C</math>, and <math>E</math> are collinear, so <math>DE=EC+CD=2CP=12\sqrt{2}</math>. This shows that <math>DEF</math> is a 5-12-13 right triangle, so it has area <math>\frac{EF\cdot DE}{2}=60</math>, so | + | Note that <math>\angle CAE=\angle CAP</math> and <math>\angle BAP=\angle BAF</math>, so <math>\angle EAF=2\angle CAB=90^{\circ}</math>. Similarly, <math>\angle FBD=90^{\circ}</math> and <math>\angle DCE=180^{\circ}</math>. Now note that <math>AE=AF=AP=5</math> and <math>BD=BF=BP=13</math>. Therefore <math>[AEF]=\frac{25}{2}</math> and <math>[BDF]=\frac{169}{2}</math>. Also note that <math>EF=5\sqrt{2}</math> and <math>DF=13\sqrt{2}</math>. We also know that <math>D</math>, <math>C</math>, and <math>E</math> are collinear, so <math>DE=EC+CD=2CP=12\sqrt{2}</math>. This shows that <math>DEF</math> is a 5-12-13 right triangle, so it has area <math>\frac{EF\cdot DE}{2}=60</math>, so |
<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> |
Revision as of 12:59, 30 September 2011
Problem
Let be a point inside isosceles right triangle such that , , , and . Find the area of .
Solution
Let , , and be the reflections of over sides , , and , respectively. We then have that , , and . This shows that . I shall now proceed to find . This is equal to
Note that and , so . Similarly, and . Now note that and . Therefore and . Also note that and . We also know that , , and are collinear, so . This shows that is a 5-12-13 right triangle, so it has area , so
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 |