Difference between revisions of "1979 AHSME Problems/Problem 3"

(Solution)
Line 28: Line 28:
 
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>.
  
 +
== 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. Apply the law of cosines then the law of sines, we find that </math>\angle{AED}=15^\circ<math>. Select </math>\boxed{C}$.
 +
 +
~hastapasta
 
== Video Solution ==
 
== Video Solution ==
 
https://youtu.be/FDgcLW4frg8?t=1341
 
https://youtu.be/FDgcLW4frg8?t=1341

Revision as of 14:07, 3 May 2022

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]

In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$. What is the measure of $\measuredangle AED$ in degrees?

$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution

Solution by e_power_pi_times_i

Notice that $\measuredangle DAE = 90^\circ+60^\circ = 150^\circ$ and that $AD = AE$. Then triangle $ADE$ is isosceles, so $\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{\textbf{(C) } 15}$.

Solution 2 (Obnoxiously tedious)

WLOG, let the side length of the square and the equilateral triangle be $1$. $\angle{DAE}=90^\circ+60^\circ=150^\circ. Apply the law of cosines then the law of sines, we find that$\angle{AED}=15^\circ$. Select$\boxed{C}$.

~hastapasta

Video Solution

https://youtu.be/FDgcLW4frg8?t=1341

~ pi_is_3.14

See also

1979 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png