https://artofproblemsolving.com/wiki/index.php?title=2006_iTest_Problems/Problem_24&feed=atom&action=history2006 iTest Problems/Problem 24 - Revision history2024-03-29T14:15:59ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2006_iTest_Problems/Problem_24&diff=99299&oldid=prevRockmanex3: Solution to Problem 24 -- Layered Rights2018-12-07T16:24:05Z<p>Solution to Problem 24 -- Layered Rights</p>
<p><b>New page</b></p><div>==Problem==<br />
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Points <math>D</math> and <math>E</math> are chosen on side <math>BC</math> of triangle <math>ABC</math> such that <math>E</math> is between <math>B</math> and <math>D</math> and <math>BE=1</math>, <math>ED=DC=3</math>. If <math>\angle BAD=\angle EAC=90^\circ</math>, the area of <math>ABC</math> can be expressed as <math>\tfrac{p\sqrt q}r</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is a positive integer not divisible by the square of any prime. Compute <math>p+q+r</math>.<br />
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<asy><br />
import olympiad;<br />
size(200);<br />
defaultpen(linewidth(0.7)+fontsize(11pt));<br />
pair D = origin, E = (3,0), C = (-3,0), B = (4,0);<br />
path circ1 = arc(D,3,0,180), circ2 = arc(B/2,2,0,180);<br />
pair A = intersectionpoint(circ1, circ2);<br />
draw(E--A--C--B--A--D);<br />
label("$A$",A,N);<br />
label("$B$",B,SE);<br />
label("$C$",C,SW);<br />
label("$D$",D,S);<br />
label("$E$",E,S);<br />
</asy><br />
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==Solution==<br />
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Note that if a circle has center <math>D</math> and radius <math>CD</math>, then <math>CE</math> is the circle's diameter because <math>CD = DE</math>. Because <math>\angle CAE</math> is a right angle, <math>A</math> is also on the circle with center <math>D</math>, so <math>DA = 3</math>. By the [[Pythagorean Theorem]], <math>AB = \sqrt{16 - 9} = \sqrt{7}</math>.<br />
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<br><br />
The area of <math>ABD</math> is <math>\tfrac12 \cdot 3 \cdot \sqrt{7} = \tfrac{3\sqrt{7}}{2}</math>. Let <math>x</math> be the length of the altitude from <math>A</math> to <math>BD</math>, so <math>\tfrac12 \cdot 4x = \tfrac{3\sqrt{7}}{2}</math>. Thus, <math>x = \tfrac{3\sqrt{7}}{4}</math>.<br />
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Since <math>BD</math> is on line <math>BC</math>, the area of <math>ABC</math> is equal to <math>\tfrac12 \cdot 7 \cdot \tfrac{3\sqrt{7}}{4} = \tfrac{21\sqrt{7}}{8}</math>, so <math>p+q+r = \boxed{36}</math>.<br />
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==See Also==<br />
{{iTest box|year=2006|num-b=23|num-a=25|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}<br />
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[[Category:Intermediate Geometry Problems]]</div>Rockmanex3