Smallest Equilateral Triangle

Geometry Level 5

An equilateral triangle is drawn such that its vertices lie on the different sides of a $\triangle ABC$ .

Find the ratio of the area of $\triangle ABC$ to the minimum possible area of the equilateral triangle.

For $A=\frac{\pi}{7}$ , $B=\frac{2\pi}{7}$ and $C=\frac{4\pi}{7}$ , the answer is $\dfrac{k+l\sqrt{m}}{n}$ , for positive coprime integers ${k,l,m,n}$ . Submit ${k+l+m+n}$ .

The answer is 32.

1 solution

Diogo Marques
Apr 4, 2021

I was able to get the result with the help of a powerful solver, but i'm still curious to see an easier path.

I will just explain my overall approach:

With 3 angles provided in the question we can create the blue triangle with a base of any size. I think 1 is a good choice. We follow by creating 2 points that can only move on their respective sides. R and Q as an example. From those we create a line, and then a perpendicular to that line that passes through the middle point between R and Q. The intersection of that line with the base, will create a point S that is dependent on the other two points. After that by setting a distance formula equating RQ and RS, this will turn the whole structure dependent on just one point which then can be used to minimize the equilateral.

The final ratio is $\LARGE\frac{6+2\sqrt{21}}{3}$

The geometry is straight foward but algebraically it is a mess. The slopes given by $\tan \theta$ don't seem to be the source of the square root form. There has to be a big simplification somewhere and hope someone points to it.

Smallest Equilateral Triangle

The answer is 32.

1 solution

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