Can You Design a Quadrangular Logo?¶

Fiddler¶

Inspired by Delta Airlines, you and a graphic designer colleague are crafting a new logo for Fiddler Airlines. Your colleague starts with an equilateral triangle with a side length of 1 unit. Next, you pick a random point inside the triangle, giving you four points: the three corners of the triangle and this fourth point inside. There are several (simple) quadrilaterals that can be drawn whose four vertices are these four points. Your colleague picks one of these quadrilaterals at random, with each having an equal probability. This quadrilateral will be the new logo.

On average, what can you expect the area of the quadrilateral to be?

Solution¶

Given a random point, there are three quadrilaterals that could be drawn.

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The sum of the gray triangles is the original equilateral triangle.

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Answer¶

The sum of every three quadrilaterals for a given random point is two of the original triangle. Thus, the average of the quadrilaterals is $\dfrac{2}{3}$ the original triangle.
$$\dfrac{2}{3} \times \dfrac{\sqrt{3}}{4} = \boxed{\dfrac{1}{2\sqrt{3}} \approx 0.289}$$

Extra Credit¶

While picking random points inside the equilateral triangle with side length 1, you and your colleague notice that some points are “special.” When one of these special points is connected with the triangle’s three vertices, it creates three smaller triangles whose areas form an arithmetic sequence.

After further investigation, you realize these special points all belong to a locus, the total length of which is L. What is this length L?

Solution¶

As shown above, the sum of the triangles is the original triangle. That means the middle value of the arithmetic sequence must be $\dfrac{1}{3}$ the original triangle.

As all the bases are 1, the height of one of the triangles must always be
$$\dfrac{1}{3} \times \dfrac{\sqrt{3}}{2} = \boxed{\dfrac{1}{2\sqrt{3}}}$$
This only happens along these dotted line segments.

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Answer¶

Each of those segments is of length $\dfrac{2}{3}$.

The answer is $\boxed{2}$.

Animation¶

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Rohan Lewis¶

2026.10.26¶

Code can be found here.