1976 AMC 12 Problem 3

Attempt Problem 3 of the 1976 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 1976 AMC 12 solutions, or check the answer key.

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

The sum of the distances from one vertex of a square with sides of length two to the midpoints of each of the sides of the square is

2\sqrt5

2+\sqrt3

2+2\sqrt3

2+\sqrt5

2+2\sqrt5

Place the chosen vertex at

(0,0)

and the other vertices at

(2,0),

(2,2),

and

(0,2)

Two midpoints are one unit away, while each opposite-side midpoint forms legs

2

and

1

The two midpoints on sides meeting the chosen vertex are each at distance

1.

The other two midpoints have coordinate differences

2

and

1,

so each is at distance

\sqrt{2^2+1^2}=\sqrt5.

The sum is

2+2\sqrt5.

Therefore, the correct answer is E.

Problem 3 in other years