Difference between revisions of "1994 AIME Problems/Problem 8"
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Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number <math>\sqrt{3}</math> can lead you down the path of using simple vectors. | Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number <math>\sqrt{3}</math> can lead you down the path of using simple vectors. | ||
− | + | First, drop a perpendicular from <math>O</math> to <math>AB</math>. Call this midpoint of <math>AB M</math>. Thus, <math>M=(\frac{a+b}{2}, 24)</math>. Now we can use perpendicularity and slope. That of <math>OM</math> is <math>\frac{48}{a+b}</math>. The vector is <math>[\frac{a+b}{2}, 24]</math>. Meanwhile from point <math>M</math> we can use a vector with <math>\frac{\sqrt{3}}{3}</math> the distance; we have to switch the <math>x</math> and <math>y</math> and our displacement is <math>[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]</math>. (Do you see why we switched <math>x</math> and <math>y</math> due to the rotation of 90 degrees?) | |
We see this displacement from <math>M</math> to <math>A</math> is <math>[\frac{a-b}{2}, 13]</math> as well. Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}</math>. And the answer is <math>\boxed{315}</math>. | We see this displacement from <math>M</math> to <math>A</math> is <math>[\frac{a-b}{2}, 13]</math> as well. Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}</math>. And the answer is <math>\boxed{315}</math>. | ||
'''Note''': This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems". | '''Note''': This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems". | ||
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== See also == | == See also == | ||
{{AIME box|year=1994|num-b=7|num-a=9}} | {{AIME box|year=1994|num-b=7|num-a=9}} |
Revision as of 17:11, 3 September 2017
Contents
Problem
The points ,
, and
are the vertices of an equilateral triangle. Find the value of
.
Solution
Consider the points on the complex plane. The point is then a rotation of
degrees of
about the origin, so:
Equating the real and imaginary parts, we have:
Solving this system, we find that . Thus, the answer is
.
Note: There is another solution where the point is a rotation of
degrees of
; however, this triangle is just a reflection of the first triangle by the
-axis, and the signs of
and
are flipped. However, the product
is unchanged.
Solution Two
Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number can lead you down the path of using simple vectors.
First, drop a perpendicular from to
. Call this midpoint of
. Thus,
. Now we can use perpendicularity and slope. That of
is
. The vector is
. Meanwhile from point
we can use a vector with
the distance; we have to switch the
and
and our displacement is
. (Do you see why we switched
and
due to the rotation of 90 degrees?)
We see this displacement from to
is
as well. Equating the two vectors, we get
and
. Therefore,
and
. And the answer is
.
Note: This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.