1961 AMC 12 Problem 25

Attempt Problem 25 of the 1961 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 1961 AMC 12 solutions, or check the answer key.

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

25.

Triangle

ABC

is isosceles with base

AC.

Points

P

and

Q

are respectively in

CB

and

AB

and such that

AC=AP=PQ=QB.

The number of degrees in angle

B

is:

25\dfrac57

26\dfrac13

30

40

not determined by the information given

Let

\angle B=m

and use each pair of equal segments to create isosceles triangles

Chase the angles through triangles

BQP,

AQP,

and

APC

Let

\angle B=m.

Since

PQ=QB,

triangle

BQP

gives

\angle QPB=m

and

\angle AQP=2m.

Since

AP=PQ,

triangle

AQP

gives

\angle QAP=2m,

\angle QPA=180^\circ-4m.

B,P,C

are collinear,

\angle APC=3m.

Now

AP=AC

gives

\angle ACP=3m.

Finally

AB=BC

makes both base angles of

ABC

equal to

3m,

and

m+3m+3m=180^\circ.

Therefore

m=\frac{180^\circ}{7}=25\dfrac57^\circ.

Thus, the correct answer is A.

Problem 25 in other years