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
  
 
(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>
 
(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>
{{MAA Notice}}
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== Solution ==
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<math>\fbox{}</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 02:55, 28 September 2014

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

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

Solution

$\fbox{}$

See also

1991 AHSME (ProblemsAnswer KeyResources)
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