Difference between revisions of "1979 AHSME Problems/Problem 3"
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Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
| − | Notice that <math>\measuredangle DAE = 90\circ+60\circ = 150\circ</math> and that <math>AD = AE</math>. Then triangle <math>ADE</math> is isosceles, so <math>\measuredangle AED = \dfrac{180\circ-150\circ}{2} = \boxed{\textbf{(C) } 15}</math>. | + | Notice that <math>\measuredangle DAE = 90^\circ+60^\circ = 150^\circ</math> and that <math>AD = AE</math>. Then triangle <math>ADE</math> is isosceles, so <math>\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{\textbf{(C) } 15}</math>. |
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| + | == Solution 2 (Obnoxiously tedious) == | ||
| + | |||
| + | WLOG, let the side length of the square and the equilateral triangle be <math>1</math>. <math>\angle{DAE}=90^\circ+60^\circ=150^\circ</math>. Apply the law of cosines then the law of sines, we find that <math>\angle{AED}=15^\circ</math>. Select <math>\boxed{C}</math>. | ||
| + | |||
| + | Please don't try this solution. This is just obnoxiously tedious and impractical. | ||
| + | |||
| + | ~hastapasta | ||
| + | |||
| + | == Video Solution by OmegaLearn== | ||
| + | https://youtu.be/FDgcLW4frg8?t=1341 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1979| | + | {{AHSME box|year=1979|num-b=2|num-a=4}} |
| − | [[Category:Introductory | + | [[Category:Introductory Geometry Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:42, 31 December 2022
Contents
Problem 3
![[asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]](http://latex.artofproblemsolving.com/4/9/7/4979f48c39324925c22031079f03410ffff35395.png)
In the adjoining figure, 
is a square, 
is an equilateral triangle and point 
is outside square .
What is the measure of 
in degrees?
Solution
Solution by e_power_pi_times_i
Notice that 
and that 
. Then triangle 
is isosceles, so 
.
Solution 2 (Obnoxiously tedious)
WLOG, let the side length of the square and the equilateral triangle be 
. 
. Apply the law of cosines then the law of sines, we find that 
. Select 
.
Please don't try this solution. This is just obnoxiously tedious and impractical.
~hastapasta
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=1341
~ pi_is_3.14
See also
| 1979 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
