No Individual Distance Was Found!

Geometry Level 5

We are given a regular pentagon $A_1A_2A_3A_4A_5$ in the Cartesian plane centered at the origin with $A_1 = (1,0)$ . Let $P = (1, \sqrt{3})$ .

Denote $z = \displaystyle \prod_{i=1}^5 m(PA_i)$ , where $m(AB)$ denotes the length of segment $AB$ .
Submit your answer as $z^2$ .

The answer is 993.

1 solution

Manuel Kahayon
Aug 20, 2016

Relevant wiki: Roots of Unity

Let us compute the problem in terms of the complex plane. Replace the y-axis with the imaginary axis, and let $\large A_k = e^{\frac{2 \pi (k-1) i}{5}}$ . We then get that $P = 1 + i \sqrt{3}$ .

Now, the product we are looking for is $\large \displaystyle \prod_{k=1}^5 |P-A_k| = |\prod_{k=1}^5 P-e^{\frac{2 \pi (k-1) i}{5}}|$ . But, this is just equal to $|P^5 - 1|$ , since $\large e^{\frac{2 \pi (k-1) i}{5}}$ are just the fifth roots of unity.

$\large |(1+ i \sqrt{3})^5 -1| = |2^5(\frac{1}{2} + i \frac{\sqrt{3}}{2})^5 -1| = |32(e^\frac{ \pi i}{3})^5 - 1| = |16 - 16i \sqrt{3} - 1| = |15 - 16i \sqrt{3}|$

(Since $\frac{1}{2} + i \frac{\sqrt{3}}{2} = e^\frac{ \pi i}{3}$ ).

Therefore, $z= |15 - 16i \sqrt{3}| = \sqrt{15^2 + (16 \sqrt{3})^2} = \sqrt{993}$ .

So, our answer is $z^2 = \boxed{993}$ .

If a regular pentagon is centered at the origin, and one of the vertices is at (1,0) then all the vertices are 1 away from the origin. The point ( 1, 1.732 ) therefore is at MOST 3 units from the farthest vertex of the pentagon. Assuming they were all at that distance (an impossibility) then the maximum sum of the 5 distances is 15, which when squared is 225. My calculations show the distances to be 2.141, 1.043, 2.942, 2.771, and 1.732 which sum to 10.629. This value squared is 112.976.

Roger Erisman - 4 years, 9 months ago

which sums to 10.629

You're suppose to find their product, not the sum

Pi Han Goh - 4 years, 9 months ago

nice solution :)

A Former Brilliant Member - 2 years, 10 months ago

No Individual Distance Was Found!

The answer is 993.

1 solution

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