2009 AIME II Problems/Problem 13
Contents
Problem
Let and
be the endpoints of a semicircular arc of radius
. The arc is divided into seven congruent arcs by six equally spaced points
,
,
,
. All chords of the form
or
are drawn. Let
be the product of the lengths of these twelve chords. Find the remainder when
is divided by
.
Solution
Solution 1
Let the radius be 1 instead. All lengths will be halved so we will multiply by at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then
are 6 of the 14th roots of unity. Let
; then
correspond to
. Let
be their reflections across the diameter. These points correspond to
. Then the lengths of the segments are
. Noting that
represents 1 in the complex plane, the desired product is
for .
However, the polynomial
has as its zeros all 14th roots of unity except for
and
. Hence
Thus the product is
(
) when the radius is 1, and the product is
. Thus the answer is
.
Solution 2
Let be the midpoint of
and
. Assume
is closer to
instead of
.
=
. Using the Law of Cosines,
=
,
=
,
.
.
.
=
So =
. It can be rearranged to form
=
.
Since , so we have
=
=
=
It can be shown that =
, so
=
=
=
, so the answer is
Solution 3
Note that for each the triangle
is a right triangle. Hence the product
is twice the area of the triangle
. Knowing that
, the area of
can also be expressed as
, where
is the length of the altitude from
onto
. Hence we have
.
By the definition of we obviously have
.
From these two observations we get that the product we should compute is equal to , which is the same identity as in Solution 2.
Computing the product of sines
In this section we show one way how to evaluate the product .
Let . The numbers
are the
-th complex roots of unity. In other words, these are the roots of the polynomial
. Then the numbers
are the roots of the polynomial
.
We just proved the identity .
Substitute
. The right hand side is obviously equal to
. Let's now examine the left hand side.
We have:
Therefore the size of the left hand side in our equation is . As the right hand side is
, we get that
.
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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