Difference between revisions of "1996 AIME Problems/Problem 15"
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== Problem == | == Problem == | ||
− | 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 | + | 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>. |
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and the double and triple angle (<math>\cos 3x = 4\cos^3 x - 3\cos x</math>) formulas further simplify this to | 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 \theta + 3 = (4\cos^2 2\theta - 3)(\cos 2\theta - 1) = 0</cmath> | + | <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>. | 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>. |
Latest revision as of 21:04, 14 September 2020
Problem
In parallelogram , let
be the intersection of diagonals
and
. Angles
and
are each twice as large as angle
, and angle
is
times as large as angle
. Find
.
Contents
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]](http://latex.artofproblemsolving.com/d/0/9/d09e1f91df6f1a2f004955349478e82b4cf67dad.png)
Let . Then
,
, and
. Since
is a parallelogram, it follows that
. By the Law of Sines on
,
![$\frac{\sin \angle CBO}{OC} = \frac{\sin \angle ACB}{OB} \quad \text{and} \quad \frac{\sin \angle DBA}{OC} = \frac{\sin \angle BAC}{OB}.$](http://latex.artofproblemsolving.com/c/4/5/c45cc618c84f220b8d2b555f9a4961f1716b91e5.png)
Dividing the two equalities yields
Pythagorean and product-to-sum identities yield
and the double and triple angle () formulas further simplify this to
The only value of that fits in this context comes from
. The answer is
.
Solution 2 (trigonometry)
Define as above. Since
, it follows that
, and so
. The Law of Sines on
yields that
Expanding using the sine double and triple angle formulas, we have
By the quadratic formula, we have , so
(as the other roots are too large to make sense in context). The answer follows as above.
Solution 3
We will focus on . Let
, so
. Draw the perpendicular from
intersecting
at
. Without loss of generality, let
. Then
, since
is the circumcenter of
. Then
.
By the Exterior Angle Theorem, and
. That implies that
. That makes
. Then since by AA (
and reflexive on
),
.
![$\frac {CO}{BC} = \frac {BC}{AC} \implies 2 = BC^2 = \implies BC = \sqrt {2}.$](http://latex.artofproblemsolving.com/8/b/6/8b64d0330d044a95b98d2f79590db3e51384ef77.png)
Then by the Pythagorean Theorem, . That makes
equilateral. Then
. 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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