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Difference between revisions of "1996 AIME Problems/Problem 15"

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The only value of <math>\theta</math> that fits in this context comes from <math>4 \cos^2 2\theta - 3 = 0 \Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\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}</math>.
 
The only value of <math>\theta</math> that fits in this context comes from <math>4 \cos^2 2\theta - 3 = 0 \Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\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}</math>.
  
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=== Solution 2 (trignometry) ===
+
=== Solution 2 (trigonometry) ===
 
Define <math>\theta</math> as above. Since <math>\angle CAB = \angle CBO</math>, it follows that <math>\triangle COB \sim \triangle CBA</math>, and so <math>\frac{CO}{BC} = \frac{BC}{AC} \Longrightarrow BC^2 = AC \cdot CO = 2CO^2 \Longrightarrow BC = CO\sqrt{2}</math>. The [[Law of Sines]] on <math>\triangle BOC</math> yields that  
 
Define <math>\theta</math> as above. Since <math>\angle CAB = \angle CBO</math>, it follows that <math>\triangle COB \sim \triangle CBA</math>, and so <math>\frac{CO}{BC} = \frac{BC}{AC} \Longrightarrow BC^2 = AC \cdot CO = 2CO^2 \Longrightarrow BC = CO\sqrt{2}</math>. The [[Law of Sines]] on <math>\triangle BOC</math> yields that  
  

Revision as of 20:51, 22 December 2015

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 (trignometry)

[asy]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--C,B--D); D(MP("A",A,N)--MP("B",B)--MP("C",C)--MP("D",D,N)--cycle); D(B--D); D(A--C); D(MP("O",O,SE)); D(anglemark(D,B,A,4));D(anglemark(B,A,C,3.5));D(anglemark(B,A,C,4.5));D(anglemark(C,B,D,3.5));D(anglemark(C,B,D,4.5)); [/asy]

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 (trigonometry)

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

\[2\sqrt {2} \sin \theta \cos \theta = \sin\theta( - 4\sin ^2 \theta + 3) \Longrightarrow \sin \theta\left(4\cos^2\theta - 2\sqrt {2} \cos \theta - 1\right) = 0.\]

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 • 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
All AIME Problems and Solutions

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