Two congruent circles and a common tangent. A triangle connecting tangency points and the upper intersection point. What angle α maximises its area?

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## Solution

The triangle area is maximal if α is 60°.

First, set the radii of the circles to 1 for convenience. Now we consider one of the lower angles β and use the fact that a circle tangent is perpendicular to the corresponding radius. We find 2β in the centre.

Therefore the triangle side is easily found to be 2sin(β). This leads to a triangle base of 4sin(β)cos(β) and a height of 2sin^{2}(β). The triangle area is A(β)=4sin^{3}(β)cos(β).

Now take the derivative of this function: A'(β)=4sin^{2}(β)(3cos^{2}(β)-sin^{2}(β)). Setting this to zero to find extrema gives either sin^{2}(β)=0 or tan^{2}(β)=3. Since β lies between 45° and 90°, we find β=60°, which is indeed a maximum. Using the fact that the triangle is isosceles, it gives α=60°.

## Poem

The dots are just some points

Which only help us to join

Just a position taking its place

The Greeks knew already the truth

All figures are made of points so smooth

There are many points connected

As we are to one another

Speaking to our sister or brother