+586 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)
+128399 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
