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,

$a^2+b^2+c^2=\frac{15}{3}=5$ Proving proves part a.

Part b.

Using Heron's formula:

$A=\sqrt{\left( \frac{a+b+c}{2} \right)\left( \frac{a+b+c}{2} -a\right)\left( \frac{a+b+c}{2} -b\right)\left( \frac{a+b+c}{2} -c\right)}$

$4A=\sqrt{\left( a+b+c \right)\left( a+b-c \right)\left( a-b+c \right)\left( -a+b+c \right)}$

$4A=\sqrt{\left( \left( a+b \right)^2-c^2 \right)\left(c^2- \left( a-b \right)^2 \right)}$

$4A=\sqrt{\left( \left( a^2+b^2+c^2 \right)-2c^2+2ab \right)\left( 2c^2-\left( a^2+b^2+c^2 \right)+2ab \right)}$

Substitute what we proved in part a we get:

$4A=\sqrt{\left( 5-2c^2+2ab \right)\left( 2c^2-5+2ab \right)}$

$4A=\sqrt{4a^2b^2-\left( 2c^2-5 \right)^2}=\sqrt{4\left( a^2b^2-c^4+5c^2 \right)-25}$

Now let's find $a^2b^2$ and $c^4$ separately.

$a^2b^2=\left( \frac{2}{\sqrt{3}}P_y+\frac{5}{3}-2P_x \right)\left( \frac{2}{\sqrt{3}}P_y+\frac{5}{3}+2P_x \right)=\left( \frac{2}{\sqrt{3}}P_y+\frac{5}{3}\right)^2-4P_x^2$

$a^2b^2= \frac{4}{3}P_y^2+\frac{20}{3\sqrt{3}}P_y-4P_x^2+\frac{25}{9}$

Now let's find $-c^4$

$c^4=\left(-\frac{4}{\sqrt{3}}P_y+\frac{5}{3}\right)^2=\frac{16}{3}P_y^2-\frac{40}{3\sqrt{3}}P_y+\frac{25}{9}$

$-c^4=-\frac{16}{3}P_y^2+\frac{40}{3\sqrt{3}}P_y-\frac{25}{9}$

Now we find $5c^2$ :

$5c^2=-\frac{20}{\sqrt{3}}P_y+\frac{25}{3}$

Now we find $a^2b^2-c^4+5c^2$

$a^2b^2-c^4+5c^2=\frac{4}{3}P_y^2+\frac{20}{3\sqrt{3}}P_y-4P_x^2+\frac{25}{9}-\frac{16}{3}P_y^2+\frac{40}{3\sqrt{3}}P_y-\frac{25}{9}-\frac{20}{\sqrt{3}}P_y+\frac{25}{3}$

The $P_y$ terms cancel and we get:

$a^2b^2-c^4+5c^2=-4P_y^2-4P_x^2+\frac{25}{3}$

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

$a^2b^2-c^4+5c^2=-\frac{4}{3}+\frac{25}{3}=7$

We substitute $a^2b^2-c^4+5c^2=7$ into the equation for $4A$ to get:

$4A=\sqrt{4\left(7 \right)-25}=\sqrt{28-25}=\sqrt{3}$

Therefore, $A=\frac{\sqrt{3}}{4}$

And since we use Heron's formula, if the triangle was not possible it would have given us imaginary numbers. Therefore it is now proven that it is possible to construct that triangle and the area is $\frac{\sqrt{3}}{4}$

~Tomas Diaz. orders@tomasdiaz.com

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 but partial points for part b. because I messed up the algebra on the Heron's formula and couldn't finish the proof. So, I think I got 6 or 7 points out of 10.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1992 OIM Problems/Problem 3

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