1984 AMC 12 Problem 4

Attempt Problem 4 of the 1984 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 1984 AMC 12 solutions, or check the answer key.

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

A rectangle intersects a circle as shown:

AB=4,

BC=5

and

DE=3.

Then

EF

equals

6

7

\frac{20}{3}

8

9

The perpendicular bisectors of the two parallel chords pass through the same center

Therefore the midpoints of

BC

and

EF

have the same horizontal position

The line through the circle’s center perpendicular to the parallel chords

BC

and

EF

bisects both. Measured from the rectangle’s left side, the midpoint of

BC

4+\frac52,

while the midpoint of

EF

3+\frac{EF}{2}.

Thus

4+\frac52=3+\frac{EF}{2},

giving

EF=7.

Therefore, the correct answer is B.

Problem 4 in other years