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Difference between revisions of "1992 OIM Problems/Problem 3"
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Point <math>P</math> coordinates is <math>(P_x,P_y)</math> and <math>P_x^2+P_y^2=r^2=\left( \frac{1}{\sqrt{3}} \right)^2=\frac{1}{3}</math>
 
Point <math>P</math> coordinates is <math>(P_x,P_y)</math> and <math>P_x^2+P_y^2=r^2=\left( \frac{1}{\sqrt{3}} \right)^2=\frac{1}{3}</math>
  
−
Let <math>a, b, c</math> be the distances from the vertices to point <math>P</math> and <math>S</math> the sum of the squares of those distances.
+
Let <math>a, b, c</math> be the distances from the vertices to point <math>P</math>.
  
 
Part a.
 
Part a.
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<math>a^2=-2P_x+\frac{2}{\sqrt{3}}P_y+\frac{5}{3}</math>
 
<math>a^2=-2P_x+\frac{2}{\sqrt{3}}P_y+\frac{5}{3}</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.
+
<math>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}</math>
  
−
{{solution}}
+
<math>b^2=2P_x+\frac{2}{\sqrt{3}}P_y+\frac{5}{3}</math>
 +
 
 +
<math>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}</math>
 +
 
 +
<math>c^2=-\frac{4}{\sqrt{3}}P_y+\frac{5}{3}</math>
 +
 
 +
<math>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}</math>
 +
 
 +
<math>P_x</math> and <math>P_y</math> cancels in the above equation. So,
 +
 
 +
<math>a^2+b^2+c^2=\frac{15}{3}=5</math> Proving proves part a.
 +
 
 +
Part b.
 +
 
 +
Using Heron's formula:
 +
 
 +
<math>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)}</math>
 +
 
 +
<math>4A=\sqrt{\left( a+b+c \right)\left( a+b-c \right)\left( a-b+c \right)\left( -a+b+c \right)}</math>
 +
 
 +
<math>4A=\sqrt{\left( \left( a+b \right)^2-c^2 \right)\left(c^2- \left( a-b \right)^2 \right)}</math>
 +
 
 +
<math>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)}</math>
 +
 
 +
Substitute what we proved in part a we get:
 +
 
 +
<math>4A=\sqrt{\left( 5-2c^2+2ab \right)\left( 2c^2-5+2ab \right)}</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>
 +
 
 +
<math>a^2b^2= \frac{4}{3}P_y^2+\frac{20}{3\sqrt{3}}P_y-4P_x^2+\frac{25}{9}</math>
 +
 
 +
Now let's find <math>-c^4</math>
 +
 
 +
<math>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}</math>
 +
 
 +
<math>-c^4=-\frac{16}{3}P_y^2+\frac{40}{3\sqrt{3}}P_y-\frac{25}{9}</math>
 +
 
 +
Now we find <math>5c^2</math>:
 +
 
 +
<math>5c^2=-\frac{20}{\sqrt{3}}P_y+\frac{25}{3}</math>
 +
 
 +
Now we find <math>a^2b^2-c^4+5c^2</math>
 +
 
 +
<math>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}</math>
 +
 
 +
The <math>P_y</math> terms cancel and we get:
 +
 
 +
<math>a^2b^2-c^4+5c^2=-4P_y^2-4P_x^2+\frac{25}{3}</math>
 +
 
 +
Since <math>P_x^2+P_y^2=\frac{1}{3}</math>,
 +
 
 +
<math>a^2b^2-c^4+5c^2=-\frac{4}{3}+\frac{25}{3}=7</math>
 +
 
 +
We substitute <math>a^2b^2-c^4+5c^2=7</math> into the equation for <math>4A</math> to get:
 +
 
 +
<math>4A=\sqrt{4\left(7 \right)-25}=\sqrt{28-25}=\sqrt{3}</math>
 +
 
 +
Therefore, <math>A=\frac{\sqrt{3}}{4}</math>
 +
 
 +
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 <math>\frac{\sqrt{3}}{4}</math>
 +
 
 +
~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}}
  
 
== See also ==
 
== See also ==
 +
[[OIM Problems and Solutions]]
 +
 
https://www.oma.org.ar/enunciados/ibe7.htm
 
https://www.oma.org.ar/enunciados/ibe7.htm

Latest revision as of 08:42, 23 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.

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

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

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.

See also

OIM Problems and Solutions

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

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