Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1991 AHSME Problems/Problem 29"
(→‎Problem)
 
(One intermediate revision by the same user not shown)
Line 3: Line 3:
 
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
  
−
<math>\text{(A) } \frac{8}{5}
+
<math>\text{(A) } \frac{8}{5}  
 
\text{(B) } \frac{7}{20}\sqrt{21}
 
\text{(B) } \frac{7}{20}\sqrt{21}
 
\text{(C) } \frac{1+\sqrt{5}}{2}
 
\text{(C) } \frac{1+\sqrt{5}}{2}

Latest revision as of 20:15, 20 December 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 $\fbox{(B)}$.

Video Solution by Pi Academy

https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh

~ TheSkibidiSigma9

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_29&oldid=237441"