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

m (fixed box to say between 9 and 11 instead of 8 and 10)
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 hexagon 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>.
+
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

$ABCDEFG$ 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 heptagon 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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