A point P is chosen at random inside equilateral triangle ABC Find the probability that P is closer to the center of triangle than to any of the vertices of the triangle.

(In other words, let O be the center of the triangle. Find the probability that OP is shorter than all of AP, BP and CP)

xXxTenTacion Oct 4, 2019

#1**+3 **

See the image here :

Let ABC be an equilateral triangle of side 6

The area of this triangle is (√3/4) *side^2 = (√3/4)*6^2 = √3 (36/4) = 9√3

Any point falling into [ ILMNKJ] will be nearer to the center O than to any vertex

Looking at triangle ICL....

IL = 2 and this triangle is similar to triangle ABC

So.....the scale factor of ICL to ABC = (2/6) = (1/3)

So....the area of triangle ICL = area of triangle ABC * (scale factor)^2 = 9√3 * (1/3)^2 = √3

And triangles AJK and BNM are congruent to ICL so they have the same areas

So [ ILMNKJ] has an area of [ABC] - [ ICL] - [ AJK] - [ BNM] =

9√3 - √3 - √3 - √3 =

9√3 - 3√3 =

6√3

So.....the probability that P is closer to the center O than to any vertex =

[ ILMNKJ] 6√3 6 2

________ = _____ = ___ = ___

[ ABC ] 9√3 9 3

CPhill Oct 4, 2019