Instructions

Dumy solution

Your first thought will be : "This is an equilateral triangle"

Now, let's prove it. We've to find some relations. To do that, let's use this drawing :

We have the following relations :
[display]\alpha+\beta+\gamma=360[/display]
The area of each inner triangle is bound to inner angles.

Each of the inner triangle is isocele

Now, we will find how to calculate small triangle area with help of the following diagram :

You can see a rectangle triangle (in black). His Hypothenuse is the radius.

You can quickly calculate the length of the sides using basic trigonometry ! [display]a_1=e_1 \sin(\frac{\gamma}{2})[/display] [display]c_1=e_1 \cos(\frac{\gamma}{2})[/display] So, the area of the [math]\gamma[/math] triangle is : [display]T_\gamma=(e_1)^2 [\sin(\frac{\gamma}{2}).\cos(\frac{\gamma}{2})][/display] We can deduce the area of the two others triangles : [display]T_\alpha=(e_1)^2 [\sin(\frac{\alpha}{2}).\cos(\frac{\alpha}{2})][/display] [display]T_\beta=(e_1)^2 [\sin(\frac{\beta}{2}).\cos(\frac{\beta}{2})][/display] Let's take a radius of 1 to get all simplified : [display]T_\alpha= [\sin(\frac{\alpha}{2}).\cos(\frac{\alpha}{2})][/display] [display]T_\beta= [\sin(\frac{\beta}{2}).\cos(\frac{\beta}{2})][/display] [display]T_\gamma= [\sin(\frac{\gamma}{2}).\cos(\frac{\gamma}{2})][/display] With help of that identity : [display]\sin(2a) = 2 \sin(a)\cos(a) [/display] [display]\sin(a)\cos(a)=\frac{\sin(2a)}{2} [/display] We obtain the following relations : [display]T_\alpha=\frac{\sin(\alpha)}{2}[/display] [display]T_\beta=\frac{\sin(\beta)}{2}[/display] [display]T_\gamma=\frac{\sin(\gamma)}{2}[/display] The area is given by : [display]\frac{\sin(\alpha)+\sin(\beta)+\sin(\gamma)}{2}[/display] You can swap any pair of symbol in the expression upside. That means that they're equals ! [display]\alpha=\beta=\gamma=120[/display]

Now, let's prove it. We've to find some relations. To do that, let's use this drawing :

Each of the inner triangle is isocele

Now, we will find how to calculate small triangle area with help of the following diagram :

You can quickly calculate the length of the sides using basic trigonometry ! [display]a_1=e_1 \sin(\frac{\gamma}{2})[/display] [display]c_1=e_1 \cos(\frac{\gamma}{2})[/display] So, the area of the [math]\gamma[/math] triangle is : [display]T_\gamma=(e_1)^2 [\sin(\frac{\gamma}{2}).\cos(\frac{\gamma}{2})][/display] We can deduce the area of the two others triangles : [display]T_\alpha=(e_1)^2 [\sin(\frac{\alpha}{2}).\cos(\frac{\alpha}{2})][/display] [display]T_\beta=(e_1)^2 [\sin(\frac{\beta}{2}).\cos(\frac{\beta}{2})][/display] Let's take a radius of 1 to get all simplified : [display]T_\alpha= [\sin(\frac{\alpha}{2}).\cos(\frac{\alpha}{2})][/display] [display]T_\beta= [\sin(\frac{\beta}{2}).\cos(\frac{\beta}{2})][/display] [display]T_\gamma= [\sin(\frac{\gamma}{2}).\cos(\frac{\gamma}{2})][/display] With help of that identity : [display]\sin(2a) = 2 \sin(a)\cos(a) [/display] [display]\sin(a)\cos(a)=\frac{\sin(2a)}{2} [/display] We obtain the following relations : [display]T_\alpha=\frac{\sin(\alpha)}{2}[/display] [display]T_\beta=\frac{\sin(\beta)}{2}[/display] [display]T_\gamma=\frac{\sin(\gamma)}{2}[/display] The area is given by : [display]\frac{\sin(\alpha)+\sin(\beta)+\sin(\gamma)}{2}[/display] You can swap any pair of symbol in the expression upside. That means that they're equals ! [display]\alpha=\beta=\gamma=120[/display]

Smart solution (click to open)

No calculus is needed !

Let's fixate B and C point.

The area of the triangle is given by : [display]\frac{|BC|\times h}{2}[/display] If we remove the constants, formula became : [display]h[/display] That means that you only have to maximize the height !

And you can find where to put D, on the intersection of the circle and BC mediatrice !

Now, we could have fixed any two of the tree points

The triangle must have any of his vertices on his sides bissectors. The only one having this property is equilateral.

Let's fixate B and C point.

The area of the triangle is given by : [display]\frac{|BC|\times h}{2}[/display] If we remove the constants, formula became : [display]h[/display] That means that you only have to maximize the height !

And you can find where to put D, on the intersection of the circle and BC mediatrice !

Now, we could have fixed any two of the tree points

The triangle must have any of his vertices on his sides bissectors. The only one having this property is equilateral.