1992 OIM Problems/Problem 3

Problem

In an equilateral triangle $ABC$ whose side has length 2, the circle $G$ is inscribed.

a. Show that for every point $P$ of $G$ , the sum of the squares of its distances to the vertices $A$ , $B$ and $C$ is 5.

b. Show that for every point $P$ in $G$ it is possible to construct a triangle whose sides have the lengths of the segments $AP$ , $BP$ and $CP$ , and that its area is:

$\frac{\sqrt{3}}{4}$

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Construct the triangle in the cartesian plane as shown above with the shown vertices coordinates.

Point $P$ coordinates is $(P_x,P_y)$ and $P_x^2+P_y^2=r^2=\left( \frac{1}{\sqrt{3}} \right)^2=\frac{1}{3}$

Let $a, b, c$ be the distances from the vertices to point $P$ .

Part a.

$a^2=(P_x-1)^2+\left( P_y-\frac{1}{\sqrt{3})} \right)^2=P_x^2+P_y^2-2P_x+\frac{2}{\sqrt{3}}P_y+\frac{4}{3}$

Since $P_x^2+P_y^2=\frac{1}{3}$ ,

$a^2=-2P_x+\frac{2}{\sqrt{3}}P_y+\frac{5}{3}$

$b^2=(P_x-1)^2+\left( P_y+\frac{1}{\sqrt{3})} \right)^2=P_x^2+P_y^2+2P_x+\frac{2}{\sqrt{3}}P_y+\frac{4}{3}$

$b^2=2P_x+\frac{2}{\sqrt{3}}P_y+\frac{5}{3}$

$c^2=P_x^2+\left( P_y-\frac{2}{\sqrt{3}} \right)^2=P_x^2+P_y^2-\frac{4}{\sqrt{3}}P_y+\frac{4}{3}$

$c^2=-\frac{4}{\sqrt{3}}P_y+\frac{5}{3}$

$a^2+b^2+c^2=-2P_x+\frac{2}{\sqrt{3}}P_y+\frac{5}{3}+2P_x+\frac{2}{\sqrt{3}}P_y+\frac{5}{3}-\frac{4}{\sqrt{3}}P_y+\frac{5}{3}$

$P_x$ and $P_y$ cancels in the above equation. So,

$a^2+b^2+c^2=\frac{15}{3}=5$

This proves part a.

Part b.

Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I got full points for part a and partial points for part b. I don't remember what I did. I will try to write a solution for this one later.

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1992 OIM Problems/Problem 3

Problem

Solution

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