Difference between revisions of "1992 OIM Problems/Problem 3"
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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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <math>b^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. | * 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 21:43, 14 December 2023
Problem
In an equilateral triangle whose side has length 2, the circle
is inscribed.
a. Show that for every point of
, the sum of the squares of its distances to the vertices
,
and
is 5.
b. Show that for every point in
it is possible to construct a triangle whose sides have the lengths of the segments
,
and
, and that its area is:
~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 coordinates is
and
Let be the distances from the vertices to point
and
the sum of the squares of those distances.
Part a.
Since ,
- 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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