1963 AMC 12 Problem 18

Attempt Problem 18 of the 1963 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 1963 AMC 12 solutions, or check the answer key.

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

18.

Chord

EF

is the perpendicular bisector of chord

BC,

intersecting it in

M.

Between

B

and

M

point

U

is taken, and

EU

extended meets the circle in

A.

Then, for any selection of

U,

as described, triangle

EUM

is similar to triangle:

EFA

EFC

ABM

ABU

FMC

Because the perpendicular bisector of a chord passes through the center,

EF

is a diameter

Compare the right angles and the shared angle at

E

Since

EF

is a diameter,

\angle EAF=90^\circ.

Also

EF\perp BC,

\angle EMU=90^\circ.

Because

E,U,A

are collinear, the two triangles share the same acute angle at

E.

Hence

\triangle EUM\sim\triangle EFA.

Therefore, the correct answer is A.

Problem 18 in other years