Difference between revisions of "1992 OIM Problems/Problem 6"
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<math>H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +\frac{3abc}{2R}-2T</math> | <math>H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +\frac{3abc}{2R}-2T</math> | ||
+ | |||
+ | Also from the extended law of sines, <math>T=\frac{abc}{4R}</math> therefore, <math>\frac{3abc}{2R}=6T</math> | ||
+ | |||
+ | <math>H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +4T</math> | ||
+ | |||
* Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I think I got 1 or 2 points out of 10 on this one. I don't remember what I did. | * Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I think I got 1 or 2 points out of 10 on this one. I don't remember what I did. |
Revision as of 12:38, 20 December 2023
Problem
From the triangle with vertices
,
and
, the hexagon
with vertices
,
,
,
,
,
is constructed as shown in the figure.
Show that:
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
For this problem were going to denote and
as the area of hexagon
and triangle
respectively.
We know that the area of an isosceles triangle is given by where
is the length of the sides that are equal and
is the angle between them.
We notice that there are 6 isosceles triangles in the hexagon with three of them sharing the area of the inside triangle in common Therefore we can write an equation for the area of by adding all of those triangles and then subtracting 2 times the area of the triangle:
Using the extended law of sines:
;
;
where
is the circumradius of the triangle
Substitute into :
Also from the extended law of sines, therefore,
- Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I think I got 1 or 2 points out of 10 on this one. I don't remember what I did.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.