Difference between revisions of "1992 OIM Problems/Problem 6"

Revision as of 11:39, 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:

$area(H) \ge 13.area(T)$

~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:

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

$2H=(a^2+b^2+c^2)(sin(A)+sin(B)+sin(C))+(2ab.sin(C)+2ac.sin(B)+2bc.sin(A))-4T$

Using the extended law of sines:

$sin(A)=\frac{a}{2R}$ ; $sin(B)=\frac{b}{2R}$ ; $sin(C)=\frac{c}{2R}$

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.

@@ Line 20: / Line 20: @@
 <math>2H=(a^2+b^2+c^2)(sin(A)+sin(B)+sin(C))+(2ab.sin(C)+2ac.sin(B)+2bc.sin(A))-4T</math>
+Using the extended law of sines:
+<math>sin(A)=\frac{a}{2R}</math>; <math>sin(B)=\frac{b}{2R}</math>; <math>sin(C)=\frac{c}{2R}</math>

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Difference between revisions of "1992 OIM Problems/Problem 6"

Revision as of 11:39, 20 December 2023

Problem

Solution

See also