TITLE: What is the probability that a point chosen randomly from inside an equilateral triangle is closer to the center than to any of the edges?
QUESTION [128 upvotes]: My friend gave me this puzzle:

What is the probability that a point chosen at random from the interior of an equilateral triangle is closer to the center than any of its edges? 


I tried to draw the picture and I drew a smaller (concentric) equilateral triangle with half the side length. Since area is proportional to the square of side length, this would mean that the smaller triangle had $1/4$ the area of the bigger one. My friend tells me this is wrong. He says I am allowed to use calculus but I don't understand how geometry would need calculus. Thanks for help.

REPLY [14 votes]: The problem can be solved without calculus by using Archimedes Quadrature of the Parabola. To prove this Archimedes did, however, use techniques that are closely related to modern calculus.

According to Archimedes Quadrature of the Parabola the area between the parabola and the edge EF is $4\over3$ the area of triangle EFG.
Let's assume that EK has length x, then due to similarity DI has length ${3\over2}x$ and DH has length ${1\over2}x$. Because point G is on the parabola we have that the length of DG is half the length of DI, hence the length of DG is ${3\over4}x$. Now the length of HG is ${3\over4}x-{1\over2}x={1\over4}x$. So the length of DH is twice the length of HG
From this it follows that the area of triangle EFG is half the area of triangle EFD.
The area between EF and the parabola is ${4\over3}\cdot{1\over2}={2\over3}$ times the area of triangle EFD.
As the area of triangle EFD is equal to ${1\over27}$ the area of triangle ABC we now find that the area of points closer to the middle then to the edges is equal to $3\cdot1{2\over3}\cdot{1\over27}={5\over27}$ the area of triangle ABC.
Hence P(point is closer to the center)=${5\over27}$.