Difference between revisions of "2009 AIME II Problems/Problem 13"
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So <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {2\pi}{7})\dots(1 - cos \frac{6\pi}{7})</math>. It can be rearranged to form | So <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {2\pi}{7})\dots(1 - cos \frac{6\pi}{7})</math>. It can be rearranged to form | ||
− | <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {6\pi}{7})\dots(1 - cos \frac {3\pi}{7})(1 - cos \frac { | + | <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {6\pi}{7})\dots(1 - cos \frac {3\pi}{7})(1 - cos \frac {4\pi}{7})</math>. |
<math>cos a</math> = - <math>cos (\pi - a)</math>, so we have | <math>cos a</math> = - <math>cos (\pi - a)</math>, so we have |
Revision as of 21:49, 11 June 2009
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 be the midpoint of
and
. Assume
is closer to
instead of
.
=
. Using the Law of Cosines,
=
,
=
,
.
.
.
=
So =
. It can be rearranged to form
=
.
= -
, so we have
=
=
=
It can be shown that =
, so
=
=
=
, so the answer is
Solution 2
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 1.
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
. However, since sin
= sin
, then
would be the square root of
, or
, which is what we needed to find.
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 |