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Difference between revisions of "1992 OIM Problems/Problem 6"
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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:
 
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$
+
<math>H=\frac{a^2sin(A)}{2}+\frac{b^2sin(B)}{2}+\frac{c^2sin(C)}{2}+\frac{(a+b)^2sin(C)}{2}+\frac{(a+c)^2sin(B)}{2}+\frac{(b+c)^2sin(A)}{2}-2T</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 11:05, 20 December 2023

Problem

From the triangle $T$ with vertices $A$, $B$ and $C$, the hexagon $H$ with vertices $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ is constructed as shown in the figure.

Show that:

\[area(H) \ge 13.area(T)\]

Ibe7 2.gif

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

For this problem were going to denote $H$ and $T$ as the area of hexagon $H$ and triangle $T$ respectively.

We know that the area of an isosceles triangle is given by $\frac{a^2sin(\theta)}{2}$ where $a$ is the length of the sides that are equal and $\theta$ 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 $H$ by adding all of those triangles and then subtracting 2 times the area of the triangle:

$H=\frac{a^2sin(A)}{2}+\frac{b^2sin(B)}{2}+\frac{c^2sin(C)}{2}+\frac{(a+b)^2sin(C)}{2}+\frac{(a+c)^2sin(B)}{2}+\frac{(b+c)^2sin(A)}{2}-2T$

  • 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.

See also

https://www.oma.org.ar/enunciados/ibe7.htm

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