1962 AMC 12 Problem 7

Attempt Problem 7 of the 1962 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 1962 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 bisectors of the exterior angles at

B

and

C

of triangle

ABC

meet at

D.

Then, if all measurements are in degrees, angle

BDC

equals:

\dfrac12(90-A)

90-A

\dfrac12(180-A)

180-A

180-2A

Each exterior-angle bisector makes an angle

90^\circ-\frac{B}{2}

90^\circ-\frac{C}{2}

with a side

Apply the angle sum in triangle

BDC

and use

B+C=180^\circ-A

The angles of triangle

BDC

B

and

C

are

90^\circ-\frac{B}{2}

and

90^\circ-\frac{C}{2}.

Thus

\begin{aligned} \angle BDC &=180^\circ-\left(90^\circ-\frac B2\right) \\ &\quad-\left(90^\circ-\frac C2\right)\\ &=\frac{B+C}{2} \\ &=\frac{180^\circ-A}{2}. \end{aligned}

Therefore, the correct answer is C.

Problem 7 in other years