Convex pentagon

[MathIsEverything]

Hello I'm struggling with this problem:
In a convex pentagon ABCDE, each diagonal cuts a triangle with an area of 1. Calculate the area of the pentagon.
My main question is whether this pentagon is regular and if it is how to prove that. When you prove that this problem becomes very easy, and i solved that, for regular pentagon. I should be trying to prove that it is a convex pentagon or I should be trying to solve that without knowing It is regular. Thanks for helping

 

[blamocur]

Hello I'm struggling with this problem:
In a convex pentagon ABCDE, each diagonal cuts a triangle with an area of 1. Calculate the area of the pentagon.
My main question is whether this pentagon is regular and if it is how to prove that. When you prove that this problem becomes very easy, and i solved that, for regular pentagon. I should be trying to prove that it is a convex pentagon or I should be trying to solve that without knowing It is regular. Thanks for helping

No, the pentagon does not have to be regular. I believe that the pentagon with the coordinates specified in the table below has triangles with an area of 1/2 but it is not regular. Multiplying all coordinates by [imath]\sqrt{2}[/imath] will bring all triangle areas to 1:

[math]\begin{array}{|l|l|} \hline X & Y \\\hline\hline 0 & 0\\\hline 1 & 0\\\hline \frac{1}{2}\left(\sqrt{5}+1 \right) & 1 \\\hline 1 & \frac{1}{2}\left(\sqrt{5}+1 \right) \\\hline 0 & 1\\\hline \end{array}[/math]

 

[blamocur]

While I know the answer I don't know how to prove it for an arbitrary pentagon. But I can prove that each diagonal is parallel to one of the sides. Given that we get lots of similar triangles and can probably work out some equations leading to the answer.
Good luck

 

[LCKurtz]

I must be misunderstanding this question. Why can't you pick a vertex, draw diagonals from it to the two opposite vertices, which divides the pentagon into three disjoint triangles, each of area 1, giving a total area of 3?

 

[Dr.Peterson]

I must be misunderstanding this question. Why can't you pick a vertex, draw diagonals from it to the two opposite vertices, which divides the pentagon into three disjoint triangles, each of area 1, giving a total area of 3?

The middle triangle doesn't have to have area 1; it isn't "cut by [one] diagonal".