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Difference between revisions of "1991 AHSME Problems/Problem 29"
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== Problem ==
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Equilateral triangle <math>ABC</math> has <math>P</math> on <math>AB</math> and <math>Q</math> on <math>AC</math>. The triangle is folded along <math>PQ</math> so that vertex <math>A</math> now rests at <math>A'</math> on side <math>BC</math>. If <math>BA'=1</math> and <math>A'C=2</math> then the length of the crease <math>PQ</math> is
 
Equilateral triangle <math>ABC</math> has <math>P</math> on <math>AB</math> and <math>Q</math> on <math>AC</math>. The triangle is folded along <math>PQ</math> so that vertex <math>A</math> now rests at <math>A'</math> on side <math>BC</math>. If <math>BA'=1</math> and <math>A'C=2</math> then the length of the crease <math>PQ</math> is
  
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(A) <math>\frac{8}{5}</math> (B) <math>\frac{7}{20}\sqrt{21}</math> (C) <math>\frac{1+\sqrt{5}}{2}</math> (D) <math>\frac{13}{8}</math> (E) <math>\sqrt{3}</math>
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<math>\text{(A) } \frac{8}{5}
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{{MAA Notice}}
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\text{(B) } \frac{7}{20}\sqrt{21}
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\text{(C) } \frac{1+\sqrt{5}}{2}
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\text{(D) } \frac{13}{8}
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\text{(E) } \sqrt{3}</math>
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== Solution ==
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<math>ABC</math> has side length <math>3</math>. Let <math>AP=A'P=x</math> and <math>AQ=A'Q=y</math>. Thus, <math>BP=3-x</math> and <math>CQ=3-y</math>. Applying Law of Cosines on triangles <math>BPA'</math> and <math>CQA'</math> using the <math>60^{\circ}</math> angles gives <math>x=\frac{7}{5}</math> and <math>y=\frac{7}{4}</math>. Applying Law of Cosines once again on triangle <math>APQ</math> using the <math>60^{\circ}</math> angle gives <cmath>PQ^2=\frac{(21)(49)}{400}</cmath> so <cmath>PQ=\frac{7}{20}\sqrt{21}</cmath> The answer is <math>\fbox{(B)}</math>.
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== See also ==
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{{AHSME box|year=1991|num-b=28|num-a=30}} 
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}Geometry

Revision as of 22:51, 31 July 2016

Problem

Equilateral triangle $ABC$ has $P$ on $AB$ and $Q$ on $AC$. The triangle is folded along $PQ$ so that vertex $A$ now rests at $A'$ on side $BC$. If $BA'=1$ and $A'C=2$ then the length of the crease $PQ$ is

$\text{(A) } \frac{8}{5} \text{(B) } \frac{7}{20}\sqrt{21} \text{(C) } \frac{1+\sqrt{5}}{2} \text{(D) } \frac{13}{8} \text{(E) } \sqrt{3}$

Solution

$ABC$ has side length $3$. Let $AP=A'P=x$ and $AQ=A'Q=y$. Thus, $BP=3-x$ and $CQ=3-y$. Applying Law of Cosines on triangles $BPA'$ and $CQA'$ using the $60^{\circ}$ angles gives $x=\frac{7}{5}$ and $y=\frac{7}{4}$. Applying Law of Cosines once again on triangle $APQ$ using the $60^{\circ}$ angle gives \[PQ^2=\frac{(21)(49)}{400}\] so \[PQ=\frac{7}{20}\sqrt{21}\] The answer is $\fbox{(B)}$.

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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Geometry

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