Difference between revisions of "2009 AIME II Problems/Problem 13"
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Let <math>O</math> be the midpoint of <math>A</math> and <math>B</math>. Assume <math>C_1</math> is closer to <math>A</math> instead of <math>B</math>. <math>\angle AOC_1</math> = <math>\frac {\pi}{7}</math>. Using the [[Law of Cosines]], | Let <math>O</math> be the midpoint of <math>A</math> and <math>B</math>. Assume <math>C_1</math> is closer to <math>A</math> instead of <math>B</math>. <math>\angle AOC_1</math> = <math>\frac {\pi}{7}</math>. Using the [[Law of Cosines]], | ||
− | <math>\overline {AC_1}</math> = <math>8 - 8 cos \frac {\pi}{7}</math> | + | <math>\overline {AC_1}^2</math> = <math>8 - 8 cos \frac {\pi}{7}</math> |
− | <math>\overline {AC_2}</math> = <math>8 - 8 cos \frac {2\pi}{7}</math> | + | <math>\overline {AC_2}^2</math> = <math>8 - 8 cos \frac {2\pi}{7}</math> |
. | . | ||
. | . | ||
. | . | ||
− | <math>\overline {AC_6}</math> = <math>8 - 8 cos \frac {6\pi}{7}</math> | + | <math>\overline {AC_6}^2</math> = <math>8 - 8 cos \frac {6\pi}{7}</math> |
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 |
Revision as of 22:37, 17 April 2009
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
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 sin sin
sin
=
, so
=
=
=
, so the answer is