1961 AMC 12 Problem 8

Attempt Problem 8 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).

Let the two base angles of a triangle be

A

and

B,

with

B

larger than

A.

The altitude to the base divides the vertex angle

C

into two parts,

C_1

and

C_2,

with

C_2

adjacent to side

a.

Then:

C_1+C_2=A+B

C_1-C_2=B-A

C_1-C_2=A-B

C_1+C_2=B-A

C_1-C_2=A+B

The altitude creates two right triangles

Express each part of

C

as a complement of the opposite base angle

The two right triangles give

\begin{aligned} A+C_1&=90^\circ,\\ B+C_2&=90^\circ. \end{aligned}

Subtracting the second relation from the first gives

C_1-C_2=B-A.

Thus, the correct answer is B.

Problem 8 in other years