1996 AIME Problem 15

Question

In parallelogram

ABCD,

let

O

be the intersection of diagonals

\overline{AC}

and

\overline{BD}.

Angles

CAB

and

DBC

are each twice as large as angle

DBA,

and angle

ACB

is

r

times as large as angle

AOB.

Find the greatest integer that does not exceed

1000r.

Answer

Answer: 777

Concepts:parallelogram law of sines trigonometric identity

Difficulty rating: 2560

Small hint:

Set

\angle DBA=\alpha

and compare triangles

ABC

and

ABD

Big hint:

Use the law of sines to obtain an equation involving

\sin5\alpha,

\sin2\alpha,

and

\sin\alpha

Solution:

Let

\angle DBA=\alpha.

Then

\angle DBC=\angle CAB=2\alpha,

so in

\triangle ABC

the angles are

2\alpha,

3\alpha,

and

180^\circ-5\alpha.

Because

AD\parallel BC,

triangle

ABD

has angles

\alpha,

2\alpha,

and

180^\circ-3\alpha.

Applying the law of sines in the two triangles and using

AD=BC

gives

\frac{AB}{BC}=\frac{\sin5\alpha}{\sin2\alpha}=\frac{\sin2\alpha}{\sin\alpha}.

With

u=\cos^2\alpha,

this becomes

16u^2-16u+1=0.

Since

5\alpha<180^\circ,

the valid root is

u=\frac{2+\sqrt3}{4}=\cos^2 15^\circ,

so

\alpha=15^\circ.

Thus

\angle ACB=105^\circ.

In

\triangle AOB,

the angles at

A

and

B

are

30^\circ

and

15^\circ,

so

\angle AOB=135^\circ.

Therefore

\begin{aligned}r&=\frac{105}{135}=\frac79,\\\left\lfloor1000r\right\rfloor&=777.\end{aligned}