1979 AMC 12 Problem 3

Attempt Problem 3 of the 1979 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1979 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

\angle AED

in degrees?

10

12.5

15

20

25

Compare

DA,

AB,

and

AE

Find

\angle DAE

, then use the base angles of isosceles

\triangle DAE

The square and equilateral triangle give

DA=AB=AE.

Also

\begin{aligned} \angle DAE &=\angle DAB+\angle BAE\\ &=90^\circ+60^\circ\\ &=150^\circ. \end{aligned}

Thus

\triangle DAE

is isosceles, and each base angle is

\frac{180^\circ-150^\circ}{2}=15^\circ.

Therefore, the correct answer is C.

Problem 3 in other years