2009 AIME II Problems/Problem 13
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 .
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 .
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 , we have
It can be shown that = , so = = = , so the answer is
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 .
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