Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 10"

Revision as of 18:16, 14 March 2009

Problem

$ABCDEFG$ is a regular heptagon inscribed in a unit circle centered at $O$ . $l$ is the line tangent to the circumcircle of $ABCDEFG$ at $A$ , and $P$ is a point on $l$ such that triangle $AOP$ is isosceles. Let $p$ denote the value of $AP \cdot BP \cdot CP \cdot DP \cdot EP \cdot FP \cdot GP$ . Determine the value of $p^2$ .

Solution

Let in the complex plane $P = 0+i0$ , $A = 0+i$ , $O = -1+i$ . Then the vertices of our hexagon are at points $z_1, \cdots , z_7$ , where $z_k - (-1+i)$ are the 7th roots of unity, ie. the complex roots of $f(z) = (z-(-1+i))^7=1$ . If $z_k = a_k+ib_k$ , then what we want is $p^2 = \prod_{i=1}^7 (a_k^2+ib_k^2) = \prod_{i=1}^7 (a_k+b_k)(a_k-b_k) = \prod_{i=1}^7 z_k\cdot\bar{z_k}$ . Notice that $\bar{z_k}-(-1+-i)$ are also the 7th roots of unity, ie. the complex roots of $g(z) = (z_k -(-1-i))^7=1$ . From Vieta, $\prod_{i=1}^7 z_k$ is the constant term of $f(z)$ , or $f(0) = 7+i8$ and $\prod_{i=1}^7 \bar{z_k}$ is $g(0) = 7-i8$ . Thus, $p^2 = (7-i8)(7+i8) = 7^2+8^2 = 49+64 = \boxed{113}$ .

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 10"

Revision as of 18:16, 14 March 2009

Problem

Solution

See also