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

(solutions, (2) by Altheman)
m (→‎Solution 1: forgot <asy>)
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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<center><asy>
+
<center><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></center>
 
</asy></center>
  
−
Let <math>\theta = \angle DBA</math>. Then <math>\angle CAB = \angle DBD = 2\theta</math>, <math>\angle AOB = 180 - 3\theta</math>, and <math>\angle ACB = 180 - 5\theta</math>. Since <math>ABCD</math> is a parallelogram, it follows that <math>OA = OC</math>. By the [[Law of Sines]] on <math>\triangle ABO,\, \triangle BCO</math>,
+
Let <math>\theta = \angle DBA</math>. Then <math>\angle CAB = \angle DBC = 2\theta</math>, <math>\angle AOB = 180 - 3\theta</math>, and <math>\angle ACB = 180 - 5\theta</math>. Since <math>ABCD</math> is a parallelogram, it follows that <math>OA = OC</math>. By the [[Law of Sines]] on <math>\triangle ABO,\, \triangle BCO</math>,
  
 
<center><math>\frac{\sin \angle CBO}{OC} = \frac{\sin \angle ACB}{OB} \quad \text{and} \quad \frac{\sin \angle DBA}{OC} = \frac{\sin \angle BAC}{OB}.</math></center>
 
<center><math>\frac{\sin \angle CBO}{OC} = \frac{\sin \angle ACB}{OB} \quad \text{and} \quad \frac{\sin \angle DBA}{OC} = \frac{\sin \angle BAC}{OB}.</math></center>

Revision as of 22:15, 13 August 2008

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

[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

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 $\left\lfloor \frac {7000}{9}\right\rfloor = \boxed{777}$.

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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