1987 AMC 12 Problem 2

Attempt Problem 2 of the 1987 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 1987 AMC 12 solutions, or check the answer key.

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

As shown in the figure, a triangular corner with side lengths

DB=EB=1

is cut from equilateral triangle

ABC

of side length

3.

The perimeter of the remaining quadrilateral

ADEC

6

6\frac12

7

7\frac12

8

Determine the angle between

BD

and

BE

Start from the original perimeter, remove

BD

and

BE,

and add

DE

Since

\angle DBE=60^\circ

and

DB=EB=1,

triangle

DBE

is equilateral, so

DE=1.

The original perimeter is

9.

Cutting off the corner removes two unit segments and adds one:

9-1-1+1=8.

Thus the correct answer is E.

Problem 2 in other years