Difference between revisions of "2014 AIME II Problems/Problem 10"
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Let <math>z</math> be a complex number with <math>|z|=2014</math>. Let <math>P</math> be the polygon in the complex plane whose vertices are <math>z</math> and every <math>w</math> such that <math>\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}</math>. Then the area enclosed by <math>P</math> can be written in the form <math>n\sqrt{3}</math>, where <math>n</math> is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>. | Let <math>z</math> be a complex number with <math>|z|=2014</math>. Let <math>P</math> be the polygon in the complex plane whose vertices are <math>z</math> and every <math>w</math> such that <math>\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}</math>. Then the area enclosed by <math>P</math> can be written in the form <math>n\sqrt{3}</math>, where <math>n</math> is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>. | ||
− | ==Solution 1 (long but non-bashy)== | + | ==Solution 1 (long but non-bashy)== |
Note that the given equality reduces to | Note that the given equality reduces to | ||
Line 33: | Line 33: | ||
and we are done. <math>\blacksquare</math> | and we are done. <math>\blacksquare</math> | ||
− | ==Solution 2 (short | + | ==Solution 2 (short)== |
Assume <math>z = 2014</math>. Then | Assume <math>z = 2014</math>. Then | ||
<cmath>\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}</cmath> | <cmath>\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}</cmath> | ||
Line 47: | Line 47: | ||
Thus <math>P</math> is an isosceles triangle with area <math>\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}</math> and <math>n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.</math> | Thus <math>P</math> is an isosceles triangle with area <math>\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}</math> and <math>n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.</math> | ||
− | ==Solution 3== | + | ==Solution 3 (Roots of Unity) == |
− | + | Notice that <cmath>\frac1{w+z} = \frac{w+z}{wz} \implies 0 = w^2 + wz + z^2 = \frac{w^3-z^3}{w-z}.</cmath> | |
− | <cmath>\frac{ | + | Hence, <math>w=ze^{2\pi i/3},ze^{4\pi i/3}</math>, and <math>P</math> is an equilateral triangle with circumradius <math>2014</math>. Then, <cmath>[P]=\frac{3}{2}\cdot 2014^2\cdot\sin\frac{\pi}3=3\cdot 1007^2\sqrt3,</cmath> |
− | <cmath> | + | and the answer is <math>3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}</math>. |
− | <cmath>w | + | |
− | We | + | ==Solution 4 (Slick)== |
− | + | I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation: | |
+ | <cmath>\frac1{z+w}=\frac1z+\frac1w</cmath> | ||
+ | <cmath>\frac w{z+w}=\frac wz+1</cmath> | ||
+ | <cmath>\frac w{z+w}=\frac{w+z}z</cmath> | ||
+ | <cmath>\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0}</cmath> | ||
+ | We can interpret the difference of two complex numbers as a vector from one to the other, and we can interpret the quotient as a vector with an angle equal to the angle between the two vectors. Therefore, after labeling the complex numbers with <math>W</math> (<math>w</math>), <math>V</math> (<math>-z</math>), and <math>O</math> (<math>0</math>), we can interpret the above equation to mean that the <math>\angle OWV=\angle OVW</math>, and hence triangle <math>OWV</math> is isosceles, so length <math>OW</math> = <math>OV</math>. Rearranging the equation | ||
+ | <cmath>\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0},</cmath> | ||
+ | we find that | ||
+ | <cmath>(w-0)((-z)-0)=-(w-(-z))^2.</cmath> | ||
+ | Taking the magnitude of both sides and using the fact that <math>OW=OV\implies |w-0|=|(-z)-0|</math>, we find that | ||
+ | <cmath>|w-0|^2=|w-(-z)|^2,</cmath> | ||
+ | so length <math>OW=VW</math> and triangle <math>OWV</math> is equilateral. There are only two possible <math>W</math>'s for which <math>OWV</math> is equilateral, lying on either side of <math>OV</math>. After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that <math>z</math> and the two <math>w</math>'s form an equilateral triangle (this can be verified by simple angle chasing). Drawing a perpendicular bisector of one of the sides and using 30-60-90 triangles shows that the side length of this triangle is <math>2014\sqrt3</math> and hence its area is <math>\frac{\sqrt3(2014\sqrt3)^2}4=\boxed{147}\sqrt3+1000k\sqrt3,</math> for some integer <math>k</math>. | ||
+ | |||
+ | ~SymbolicPermutation | ||
− | + | ==Notes== | |
+ | This problem is killed by polar coordinates (complex numbers) . Solve using cosine-i-sine. | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=II|num-b=9|num-a=11}} | {{AIME box|year=2014|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:08, 21 January 2024
Contents
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)
Assume . Then
Thus is an isosceles triangle with area
and
Solution 3 (Roots of Unity)
Notice that
Hence,
, and
is an equilateral triangle with circumradius
. Then,
and the answer is
.
Solution 4 (Slick)
I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation:
We can interpret the difference of two complex numbers as a vector from one to the other, and we can interpret the quotient as a vector with an angle equal to the angle between the two vectors. Therefore, after labeling the complex numbers with
(
),
(
), and
(
), we can interpret the above equation to mean that the
, and hence triangle
is isosceles, so length
=
. Rearranging the equation
we find that
Taking the magnitude of both sides and using the fact that
, we find that
so length
and triangle
is equilateral. There are only two possible
's for which
is equilateral, lying on either side of
. After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that
and the two
's form an equilateral triangle (this can be verified by simple angle chasing). Drawing a perpendicular bisector of one of the sides and using 30-60-90 triangles shows that the side length of this triangle is
and hence its area is
for some integer
.
~SymbolicPermutation
Notes
This problem is killed by polar coordinates (complex numbers) . Solve using cosine-i-sine.
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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