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 == | ||
− | <math>\fbox{}</math> | + | <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 correct answer is <math>\fbox{(B)}</math>. |
== See also == | == See also == | ||
{{AHSME box|year=1991|num-b=28|num-a=30}} | {{AHSME box|year=1991|num-b=28|num-a=30}} | ||
− | [[Category: Intermediate | + | [[Category: Intermediate Trigonometry Problems]] |
− | {{MAA Notice}} | + | {{MAA Notice}} |
Latest revision as of 22:05, 20 April 2024
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
has side length
. Let
and
. Thus,
and
. Applying Law of Cosines on triangles
and
using the
angles gives
and
. Applying Law of Cosines once again on triangle
using the
angle gives
so
The correct answer is
.
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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