Difference between revisions of "1992 OIM Problems/Problem 6"
(→Solution) |
|||
| Line 11: | Line 11: | ||
== Solution == | == Solution == | ||
| − | * 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 on this one. I don't remember what I did. | + | For this problem were going to denote <math>H</math> and <math>T</math> as the area of hexagon <math>H</math> and triangle <math>T</math> respectively. |
| − | {{solution}} | + | |
| + | We know that the area of an isosceles triangle is given by <math>\frac{a^2sin(\theta)}{2}</math> where <math>a</math> is the length of the sides that are equal and <math>\theta</math> 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 <math>H</math> by adding all of those triangles and then subtracting 2 times the area of the triangle: | ||
| + | |||
| + | <math>H=\frac{a^2sin(A)}{2}+\frac{b^2sin(B)}{2}</math>+\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 solution}} | ||
== 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.
