Difference between revisions of "1992 OIM Problems/Problem 3"

Revision as of 20:35, 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

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

Point $P$ coordinates is $(P_x,P_y)$ and

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

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.

@@ Line 14: / Line 14: @@
 [[File:1992_OIM_P3.png|center|400px]]
+Construct the triangle in the cartesian plane as shown above with the shown vertices coordinates.
+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>
 * 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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Difference between revisions of "1992 OIM Problems/Problem 3"

Revision as of 20:35, 14 December 2023

Problem

Solution

See also