Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 10"
Brut3Forc3 (talk | contribs) m (fixed box to say between 9 and 11 instead of 8 and 10) |
Monkeythyme (talk | contribs) m (→Fixed a vocab error, hexagon to heptagon.) |
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== Solution == | == Solution == | ||
− | Let in the complex plane <math>P = 0+i0</math>, <math>A = 0+i</math>, <math>O = -1+i</math>. Then the vertices of our | + | Let in the complex plane <math>P = 0+i0</math>, <math>A = 0+i</math>, <math>O = -1+i</math>. Then the vertices of our heptagon are at points <math>z_1, \cdots , z_7</math>, where <math>z_k - (-1+i)</math> are the 7th roots of unity, ie. the complex roots of <math>f(z) = (z-(-1+i))^7=1</math>. If <math>z_k = a_k+ib_k</math>, then what we want is <math>p^2 = \prod_{i=1}^7 (a_k^2+ib_k^2) = \prod_{i=1}^7 (a_k+b_k)(a_k-b_k) = \prod_{i=1}^7 z_k\cdot\bar{z_k}</math>. Notice that <math>\bar{z_k}-(-1+-i)</math> are also the 7th roots of unity, ie. the complex roots of <math>g(z) = (z_k -(-1-i))^7=1</math>. From Vieta, <math>\prod_{i=1}^7 z_k</math> is the constant term of <math>f(z)</math>, or <math>f(0) = 7+i8</math> and <math>\prod_{i=1}^7 \bar{z_k}</math> is <math>g(0) = 7-i8</math>. Thus, <math>p^2 = (7-i8)(7+i8) = 7^2+8^2 = 49+64 = \boxed{113}</math>. |
== See also == | == See also == |
Revision as of 11:58, 6 March 2010
Problem
is a regular heptagon inscribed in a unit circle centered at . is the line tangent to the circumcircle of at , and is a point on such that triangle is isosceles. Let denote the value of . Determine the value of .
Solution
Let in the complex plane , , . Then the vertices of our heptagon are at points , where are the 7th roots of unity, ie. the complex roots of . If , then what we want is . Notice that are also the 7th roots of unity, ie. the complex roots of . From Vieta, is the constant term of , or and is . Thus, .
See also
Mock AIME 1 Pre 2005 (Problems, Source) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |