+1671 Let PQR be an equilateral triangle, centered at O. A point X is chosen at random inside the triangle. Find the probability that X is closer to O than to any of the sides. (In other words, find the probability that XO is shorter than XA, XB, and XC.)
+618 To solve this problem, we can first consider the perpendicular bisector of any side of the triangle. This line will divide the triangle into two smaller triangles, each of which is similar to the original triangle. Furthermore, the point O will be the incenter of each of the smaller triangles.
[asy] unitsize(2 cm);
pair A, B, C, D, O;
A = (1,0); B = (0,1); C = intersectionpoint(arc(A,1,0,180),arc(B,1,180,0)); D = (C + A)/2; O = incenter(A,B,C);
draw(A--B--C--cycle); draw(O--D);
label("A", A, SW); label("B", B, SE); label("C", C, NE); label("D", D, S); label("O", O, N); [/asy]
Now, consider a point X inside the triangle. If X is closer to O than to any of the sides, then it must lie inside the smaller triangle that is centered at O. Therefore, the probability that X is closer to O than to any of the sides is equal to the ratio of the area of the smaller triangle to the area of the original triangle.
Since the smaller triangle is similar to the original triangle, its area is one-fourth the area of the original triangle. Therefore, the probability that X is closer to O than to any of the sides is 1/4.
