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 | ||
| − | <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} | ||
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== Solution == | == Solution == | ||
| − | <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 answer is <math>\fbox{(B)}</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>. |
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| + | == Video Solution by Pi Academy == | ||
| + | |||
| + | https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh | ||
| + | |||
| + | ~ TheSkibidiSigma9 | ||
== 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 20:15, 20 December 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 ![\[PQ^2=\frac{(21)(49)}{400}\]](http://latex.artofproblemsolving.com/e/1/e/e1e78af13b9fd1648966e93fffbf619501de8e97.png)
so ![\[PQ=\frac{7}{20}\sqrt{21}\]](http://latex.artofproblemsolving.com/1/b/f/1bfda0fe917934bce76114ed611c4c0efc45bdc0.png)
The correct answer is 
.
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. 
