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1996 AIME Problems/Problem 15

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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 $1000r$ .

Contents

1 Problem
2 Solution
3 See also

Solution

Solution 1 (trignometry)

size(180); pathpen = black+linewidth(0.7);pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A-...

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 $\triangle ABO,\, \triangle BCO$ ,

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

Dividing the two equalities yields

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

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 $\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \frac{7000}...$ .

Solution 2 (trignometry)

Define $\theta$ as above. Since $\angle CAB = \angle CBO$ , it follows that $\triangle COB \sim \triangle CBA$ , and so $\frac{CO}{BC} = \frac{BC}{AC} \Longrightarrow BC^2 = AC \cdot CO = 2CO^2 \Longrightarrow BC = CO\sqrt{2}$ . The Law of Sines on $\triangle BOC$ yields that

$\frac{BC}{CO} = \frac{\sin (180-3\theta)}{\sin 2\theta} = \frac{\sin 3\theta}{\sin 2\theta} = \sqrt{2}$

Expanding using the sine double and triple angle formulas, we have

$2\sqrt {2} \sin \theta \cos \theta = \sin\theta( - 4\sin ^2 \theta + 3) \Longrightarrow \sin \left\theta (4\cos^2\theta - 2\s...$

By the quadratic formula, we have $\cos \theta = \frac {2\sqrt {2} \pm \sqrt {8 + 4 \cdot 1 \cdot 4}}{8} = \frac {\sqrt {6} \pm \sqrt {2}}{4}$ , so $\theta = 15^{\circ}$ (as the other roots are too large to make sense in context). The answer follows as above.

Solution 3

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 $\angle HOC = 4x = 60 \implies x = 15$ . The answer follows as above.

See also

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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1996_AIME_Problems/Problem_15"

Category: Intermediate Geometry Problems

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