Difference between revisions of "1997 AIME Problems/Problem 15"
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Consider points on the [[complex plane]] <math>A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)</math>. Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one [[vertex]] of the triangle at <math>A</math>, and the other two points <math>E</math> and <math>F</math> on <math>BC</math> and <math>CD</math>, respectively. Let <math>E (11,a)</math> and <math>F (b, 10)</math>. Since it's equilateral, then <math>E\cdot\text{cis}60^{\circ} = F</math>, so <math>(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i</math>, and expanding we get <math>\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i</math>. | Consider points on the [[complex plane]] <math>A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)</math>. Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one [[vertex]] of the triangle at <math>A</math>, and the other two points <math>E</math> and <math>F</math> on <math>BC</math> and <math>CD</math>, respectively. Let <math>E (11,a)</math> and <math>F (b, 10)</math>. Since it's equilateral, then <math>E\cdot\text{cis}60^{\circ} = F</math>, so <math>(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i</math>, and expanding we get <math>\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i</math>. |
Revision as of 15:51, 25 June 2016
Problem
The sides of rectangle have lengths
and
. An equilateral triangle is drawn so that no point of the triangle lies outside
. The maximum possible area of such a triangle can be written in the form
, where
,
, and
are positive integers, and
is not divisible by the square of any prime number. Find
.
Solution 1
Consider points on the complex plane . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at
, and the other two points
and
on
and
, respectively. Let
and
. Since it's equilateral, then
, so
, and expanding we get
.
We can then set the real and imaginary parts equal, and solve for . Hence a side
of the equilateral triangle can be found by
. Using the area formula
, the area of the equilateral triangle is
. Thus
.
Solution 2
This is a trigonometric re-statement of the above. Let ; by alternate interior angles,
. Let
and the side of the equilateral triangle be
, so
by the Pythagorean Theorem. Now
. This reduces to
.
Thus, the area of the triangle is , which yields the same answer as above.
Solution 3
Since and
, it follows that
. Rotate triangle
degrees clockwise. Note that the image of
is
. Let the image of
be
. Since angles are preserved under rotation,
. It follows that
. Since
, it follows that quadrilateral
is cyclic with circumdiameter
and thus circumradius
. Let
be its circumcenter. By Inscribed Angles,
. By the definition of circle,
. It follows that triangle
is equilateral. Therefore,
. Applying the Law of Cosines to triangle
,
. Squaring and multiplying by
yields
-Solution by thecmd999
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.