Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
 
(9 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
In parallelogram <math>ABCD</math>, let <math>O</math> be the intersection of diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>. Angles <math>CAB</math> and <math>DBC</math> are each twice as large as angle <math>DBA</math>, and angle <math>ACB</math> is <math>r</math> times as large as angle <math>AOB</math>. Find the greatest integer that does not exceed <math>1000r</math>.
+
In [[parallelogram]] <math>ABCD</math>, let <math>O</math> be the intersection of [[diagonal]]s <math>\overline{AC}</math> and <math>\overline{BD}</math>. Angles <math>CAB</math> and <math>DBC</math> are each twice as large as angle <math>DBA</math>, and angle <math>ACB</math> is <math>r</math> times as large as angle <math>AOB</math>. Find <math>\lfloor 1000r \rfloor</math>.
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
=== Solution 1 (trigonometry) ===
 +
<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>
 +
 
 +
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>
 +
 
 +
Dividing the two equalities yields
 +
 
 +
<cmath>\frac{\sin 2\theta}{\sin \theta} = \frac{\sin (180 - 5\theta)}{\sin 2\theta} \Longrightarrow \sin^2 2\theta = \sin 5\theta \sin \theta.</cmath>
 +
 
 +
Pythagorean and product-to-sum identities yield
 +
 
 +
<cmath>1 - \cos^2 2 \theta = \frac{\cos 4\theta - \cos 6 \theta}{2},</cmath>
 +
 
 +
and the double and triple angle (<math>\cos 3x = 4\cos^3 x - 3\cos x</math>) formulas further simplify this to
 +
 
 +
<cmath>4\cos^3 2\theta - 4\cos^2 2\theta - 3\cos 2\theta + 3 = (4\cos^2 2\theta - 3)(\cos 2\theta - 1) = 0</cmath>
 +
 
 +
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>.
 +
 
 +
=== 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
 +
 
 +
<cmath>\frac{BC}{CO} = \frac{\sin (180-3\theta)}{\sin 2\theta} = \frac{\sin 3\theta}{\sin 2\theta} = \sqrt{2}</cmath>
 +
 
 +
Expanding using the sine double and triple angle formulas, we have
 +
 
 +
<cmath>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.</cmath>
 +
 
 +
By the [[quadratic formula]], we have <math>\cos \theta = \frac {2\sqrt {2} \pm \sqrt {8 + 4 \cdot 1 \cdot 4}}{8} = \frac {\sqrt {6} \pm \sqrt {2}}{4}</math>, so <math>\theta = 15^{\circ}</math> (as the other roots are too large to make sense in context). The answer follows as above.
 +
 
 +
=== Solution 3 === <!-- I would put this solution first, but needs a bit of formatting for clarity - azjps -->
 +
We will focus on <math>\triangle ABC</math>. Let <math>\angle ABO = x</math>, so <math>\angle BAO = \angle OBC = 2x</math>. Draw the [[perpendicular]] from <math>C</math> intersecting <math>AB</math> at <math>H</math>. [[Without loss of generality]], let <math>AO = CO = 1</math>. Then <math>HO = 1</math>, since <math>O</math> is the [[circumcenter]] of <math>\triangle AHC</math>. Then <math>\angle OHA = 2x</math>.
 +
 
 +
By the Exterior Angle Theorem, <math>\angle COB = 3x</math> and <math>\angle COH = 4x</math>. That implies that <math>\angle HOB = x</math>. That makes <math>HO = HB = 1</math>. Then since by AA (<math>\angle HBC = \angle HOC = 3x</math> and reflexive on <math>\angle OCB</math>), <math>\triangle OCB \sim \triangle BCA</math>.
 +
 
 +
<center><math>\frac {CO}{BC} = \frac {BC}{AC} \implies 2 = BC^2 = \implies BC = \sqrt {2}. </math></center>
 +
 
 +
Then by the [[Pythagorean Theorem]], <math>1^2 + HC^2 = \left(\sqrt {2}\right)^2\implies HC = 1</math>. That makes <math>\triangle HOC</math> equilateral. Then <math>\angle HOC = 4x = 60 \implies x = 15</math>. The answer follows as above.
  
 
== See also ==
 
== See also ==
−
*[[1996 AIME Problems]]
+
{{AIME box|year=1996|num-b=14|after=Final Problem}}
  
−
{{AIME box|year=1996|num-b=14|after=Final Problem}}
+
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:04, 14 September 2020

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 $\lfloor 1000r \rfloor$.

Solution

Solution 1 (trigonometry)

[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

\[1 - \cos^2 2 \theta = \frac{\cos 4\theta - \cos 6 \theta}{2},\]

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_15&oldid=133658"