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Difference between revisions of "1992 OIM Problems/Problem 3"
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<math>4A=\sqrt{4a^2b^2-\left( 2c^2-5 \right)^2}=\sqrt{4\left( a^2b^2-c^4+5c^2 \right)-25}</math>
 
<math>4A=\sqrt{4a^2b^2-\left( 2c^2-5 \right)^2}=\sqrt{4\left( a^2b^2-c^4+5c^2 \right)-25}</math>
 +
 +
Now let's find <math>a^2b^2</math> and <math>c^4</math> separately.
 +
 +
<math>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</math>
  
 
* 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.
 
* 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.

Revision as of 21:02, 14 December 2023

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

1992 OIM P3.png

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.

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

  • 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.

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

https://www.oma.org.ar/enunciados/ibe7.htm

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