1979 AMC 12 Problem 24

Question

A polygon is called “simple” if it is not self-intersecting. Sides

AB,

BC,

and

CD

of simple quadrilateral

ABCD

have lengths

4,

5,

and

20,

respectively. If vertex angles

B

and

C

are obtuse and

\sin C=-\cos B=\frac35,

then side

AD

has length

24

24.5

24.6

24.8

25

Place

B=(0,0)

and

C=(5,0)

, then use the signs forced by the obtuse angles

Determine coordinates for

A

and

D

from the

3

-

4

-

5

ratios

Put

B=(0,0)

and

C=(5,0).

Since

B

is obtuse and

-\cos B=\frac{3}{5},

we may take

\begin{aligned} A&=4(-\frac{3}{5},\frac{4}{5})\\ &=(-\frac{12}{5},\frac{16}{5}). \end{aligned}

The simple configuration and the obtuse angle at

C

give

\begin{aligned} D&=C+20(\frac{4}{5},\frac{3}{5})\\ &=(21,12). \end{aligned}

Thus

\begin{aligned} AD&=\sqrt{\begin{gathered} (21+\frac{12}{5})^2\\ {}+(12-\frac{16}{5})^2 \end{gathered}}\\ &=\sqrt{(\frac{117}{5})^2+(\frac{44}{5})^2}\\ &=25. \end{aligned}

Therefore, the correct answer is E.

Problem 24 in other years