Difference between revisions of "1992 OIM Problems/Problem 6"
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* 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. | ||
| − | {{alternate | + | {{alternate solutions}} |
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe7.htm | https://www.oma.org.ar/enunciados/ibe7.htm | ||
Revision as of 10:04, 20 December 2023
Problem
From the triangle 
with vertices 
, 
and 
, the hexagon 
with vertices 
, 
, 
, 
, 
, 
is constructed as shown in the figure.
Show that:
![\[area(H) \ge 13.area(T)\]](http://latex.artofproblemsolving.com/f/c/4/fc4e404f0e7cc3851c1dfea4d938a092dbdf78e5.png)
~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:
+\frac{c^2sin(C)}{2}+\frac{(a+b)^2sin(C)}{2}+\frac{(a+c)^2sin(B)}{2}+\frac{(b+c)^2sin(A)}{2}-2T$
- 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.
