1996 AIME Problems/Problem 15

Problem

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 $1000r$ .

Solution

Solution 1

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Let $\theta = \angle DBA$ . Then $\angle CAB = \angle DBC = 2\theta$ , $\angle AOB = 180 - 3\theta$ , and $\angle ACB = 180 - 5\theta$ . Since $ABCD$ is a parallelogram, it follows that $OA = OC$ . By the Law of Sines on $\triangle ABO,\, \triangle BCO$ ,

$\frac{\sin \angle CBO}{OC} = \frac{\sin \angle ACB}{OB} \quad \text{and} \quad \frac{\sin \angle DBA}{OC} = \frac{\sin \angle BAC}{OB}.$

Dividing the two equalities yields

$\frac{\sin 2\theta}{\sin \theta} = \frac{\sin (180 - 5\theta)}{\sin 2\theta} \Longrightarrow \sin^2 2\theta = \sin 5\theta \sin \theta.$

Pythagorean and product-to-sum identities yield

$1 - \cos^2 2 \theta = \cos 4\theta - \cos 6 \theta,$

and the double and triple angle ( $\cos 3x = 4\cos^3 x - 3\cos x$ ) formulas further simplify this to

$4\cos^3 2\theta - 4\cos^2 2\theta - 3\cos \theta + 3 = (4\cos^2 2\theta - 3)(\cos 2\theta - 1) = 0$

The only value of $\theta$ that fits in this context comes from $4 \cos^2 2\theta - 3 = 0 \Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}$ . The answer is $\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \frac{7000}{9} \right \rfloor = \boxed{777}$ .

Solution 2

We will focus on $\triangle ABC$ . Let $\angle ABO = x$ , so $\angle BAO = \angle OBC = 2x$ . Draw the perpendicular from $C$ intersecting $AB$ at $H$ . Without loss of generality, let $AO = CO = 1$ . Then $HO = 1$ , since $O$ is the circumcenter of $\triangle AHC$ . Then $\angle OHA = 2x$ .

By the Exterior Angle Theorem, $\angle COB = 3x$ and $\angle COH = 4x$ . That implies that $\angle HOB = x$ . That makes $HO = HB = 1$ . Then since by AA ( $\angle HBC = \angle HOC = 3x$ and reflexive on $\angle OCB$ ), $\triangle OCB \sim \triangle BCA$ .

$\frac {CO}{BC} = \frac {BC}{AC} \implies 2 = BC^2 = \implies BC = \sqrt {2}.$

Then by the Pythagorean Theorem, $1^2 + HC^2 = \left(\sqrt {2}\right)^2\implies HC = 1$ . That makes $\triangle HOC$ equilateral. Then $\angle HOC = 4x = 60 \implies x = 15$ . Then $\angle AOB = 180 - 3 \times 15 = 135^{\circ}$ and $\angle ACB = 180 - 5 \times 15 = 105^{\circ}$ .

Then $\frac {\angle ACB}{\angle AOB} = r = \frac {7}{9}$ . Then it follows that $\left\lfloor \frac {7000}{9}\right\rfloor = \boxed{777}$ .

1996 AIME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Final Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1996 AIME Problems/Problem 15

Problem

Contents

Solution

Solution 1

Solution 2

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