Difference between revisions of "2009 AIME II Problems/Problem 13"
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=== Solution 1 === | === Solution 1 === | ||
− | Let the radius be 1 instead. All lengths will be halved so we will multiply by <math>2^{12}</math> 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 <math>C_1,\ldots, C_6</math> are 6 of the 14th roots of unity. Let <math>\omega=\text{cis}\frac{360^{\circ}}{14}</math>; then <math>C_1,\ldots, C_6</math> correspond to <math>\omega,\ldots, \omega^6</math>. Let <math>C_1',\ldots, C_6'</math> be their reflections across the diameter. These points correspond to <math>\omega^8\ldots, \omega^{13}</math>. Then the lengths of the segments are <math>|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|</math>. | + | Let the radius be 1 instead. All lengths will be halved so we will multiply by <math>2^{12}</math> 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 <math>C_1,\ldots, C_6</math> are 6 of the 14th roots of unity. Let <math>\omega=\text{cis}\frac{360^{\circ}}{14}</math>; then <math>C_1,\ldots, C_6</math> correspond to <math>\omega,\ldots, \omega^6</math>. Let <math>C_1',\ldots, C_6'</math> be their reflections across the diameter. These points correspond to <math>\omega^8\ldots, \omega^{13}</math>. Then the lengths of the segments are <math>|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|</math>. Noting that <math>B</math> represents 1 in the complex plane, the desired product is |
<cmath> | <cmath> | ||
+ | BC_1\cdots BC_6 \cdot AC_1\cdots AC_6= | ||
+ | BC_1\cdots BC_6 \cdot BC_1'\cdots BC_6 | ||
|(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| | |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| | ||
</cmath> | </cmath> |
Revision as of 20:32, 11 February 2011
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
= .
= - , 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 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 |