Difference between revisions of "2014 AIME II Problems/Problem 10"
m (→Solution 2 (short but a little bashy)) |
m (→Solution 1 (long but non-bashy)) |
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<cmath>x = R \sqrt{3}</cmath> | <cmath>x = R \sqrt{3}</cmath> | ||
<cmath>\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}</cmath> | <cmath>\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}</cmath> | ||
+ | |||
+ | Since we're concerned with the non-radical part of this expression and <math>R = 2014</math>, | ||
+ | |||
<cmath>\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}</cmath> | <cmath>\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}</cmath> | ||
Revision as of 17:56, 29 March 2014
Problem
Let be a complex number with
. Let
be the polygon in the complex plane whose vertices are
and every
such that
. Then the area enclosed by
can be written in the form
, where
is an integer. Find the remainder when
is divided by
.
Solution 1 (long but non-bashy)
Note that the given equality reduces to
Now, let and likewise for
. Consider circle
with the origin as the center and radius 2014 on the complex plane. It is clear that
must be one of the points on this circle, as
.
By DeMoivre's Theorem, the complex modulus of is cubed when
is cubed. Thus
must lie on
, since its the cube of its modulus, and thus its modulus, must be equal to
's modulus.
Again, by DeMoivre's Theorem, is tripled when
is cubed and likewise for
. For
,
, and the origin to lie on the same line,
must be some multiple of 360 degrees apart from
, so
must differ from
by some multiple of 120 degrees.
Now, without loss of generality, assume that is on the real axis. (The circle can be rotated to put
in any other location.) Then there are precisely two possible distinct locations for
; one is obtained by going 120 degrees clockwise from
about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.
Let the two possible locations for be
and
and the location of
be point
. Note that by symmetry,
is equilateral, say, with side length
. We know that the circumradius of this equilateral triangle is
, so using the formula
and that the area of an equilateral triangle with side length
is
, so we have
Since we're concerned with the non-radical part of this expression and ,
and we are done.
Solution 2 (short but a little bashy)
Without loss of generality, let . Then
Thus is an isosceles triangle with area
and