Difference between revisions of "2009 AIME II Problems/Problem 13"
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− | (1-\omega_k)(1-\omega_{k | + | (1-\omega_k)(1-\omega_{7-k})=|1-\omega_k|^2 |
& = \left( 1-\cos \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 | & = \left( 1-\cos \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 | ||
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<b>~Solution by sml1809</b> | <b>~Solution by sml1809</b> | ||
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+ | == Video Solution == | ||
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+ | https://youtu.be/TrKxzgR7V8U?si=FFOBCJxjGrg9sWGC | ||
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+ | ~MathProblemSolvingSkills.com | ||
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== See Also == | == See Also == |
Latest revision as of 21:32, 3 October 2024
Contents
[hide]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 , 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 .
Solution 4 (Product of Sines)
Lemma 1: A chord of a circle with center and radius has length .
Proof: Denote as the projection from to line . Then, by definition, . Thus, , which concludes the proof.
Lemma 2:
Proof: Let . Thus, Since, are just the th roots of unity excluding , by Vieta's, , thus completing the proof.
By Lemma 1, the length and similar lengths apply for . Now, the problem asks for . This can be rewritten, due to , as By Lemma 2, this furtherly boils down to
~Solution by sml1809
Video Solution
https://youtu.be/TrKxzgR7V8U?si=FFOBCJxjGrg9sWGC
~MathProblemSolvingSkills.com
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.