1967 AMC 12 Problem 32

Attempt Problem 32 of the 1967 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 1967 AMC 12 solutions, or check the answer key.

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32.

In quadrilateral

ABCD

with diagonals

AC

and

BD

intersecting at

O,

BO=4,

OD=6,

AO=8,

OC=3,

and

AB=6.

The length of

AD

is:

9

10

6\sqrt3

8\sqrt2

\sqrt{166}

Answer: E

Concepts:law of cosines angle chasing

Difficulty rating: 1690

Small hint:

Use triangle

AOB

to find

\cos\angle AOB

Big hint:

Angles

AOB

and

AOD

are supplementary

Solution:

In triangle

AOB,

\begin{aligned} \cos\angle AOB &=\frac{8^2+4^2-6^2}{2\cdot8\cdot4}\\ &=\frac{11}{16}. \end{aligned}

Therefore

\cos\angle AOD=-\frac{11}{16}.

Applying the law of cosines in triangle

AOD

gives

\begin{aligned} AD^2 &=8^2+6^2\\ &\quad+2(8)(6)\frac{11}{16}\\ &=166. \end{aligned}

Hence

AD=\sqrt{166}.

Therefore, the correct answer is E.