Difference between revisions of "1991 AHSME Problems/Problem 29"

Revision as of 21:05, 20 April 2024

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 correct answer is $\fbox{(B)}$ .

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.

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 {{AHSME box|year=1991|num-b=28|num-a=30}}
-[[Category: Intermediate Algebra Problems]]
+[[Category: Intermediate Trigonometry Problems]]
-{{MAA Notice}}Geometry
+{{MAA Notice}}

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Difference between revisions of "1991 AHSME Problems/Problem 29"

Revision as of 21:05, 20 April 2024

Problem

Solution

See also