Difference between revisions of "1991 AHSME Problems/Problem 29"
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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>\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}</math> | ||
== Solution == | == Solution == | ||
Revision as of 21:39, 31 July 2016
Problem
Equilateral triangle 
has 
on 
and 
on 
. The triangle is folded along 
so that vertex 
now rests at 
on side 
. If 
and 
then the length of the crease is
Solution
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Geometry
