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

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== See also ==
== See also ==
{{Mock AIME box|year=Pre 2005|n=1|num-b=8|num-a=10|source=14769}}
{{Mock AIME box|year=Pre 2005|n=1|num-b=9|num-a=11|source=14769}}
[[Category:Intermediate Complex Number Problems]]
[[Category:Intermediate Complex Number Problems]]

Revision as of 21:44, 21 February 2010


$ABCDEFG$ 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 $p^2$.


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, $p^2 = (7-i8)(7+i8) = 7^2+8^2 = 49+64 = \boxed{113}$.

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
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