## Stick Broken Into Three Pieces(Trilinear Coordinates)

Assume a stick is broken at random into three pieces. What is the probability that the pieces can form a triangle?

The answer to the problem depends on the manner in which the stick is being broken. I shall consider several possibilities:

In any event, the diagram presented by the applet is helpful in finding the answers.

From a point $P$ inside (or on a side) of an equilateral $\Delta ABC$ drop perpendiculars $PP_{a},$ $PP_{b},$ $PP_{c}$ to its sides. By Viviani's theorem, the sum

$PP_{a} + PP_{b} + PP_{c}$

is independent of $P$ and is equal to any of the triangle's altitudes. Let's take a triangle whose altitude exactly measures the length of the stick. Connect the midpoints of the sides to split $\Delta ABC$ into four equal triangles, with the medial triangle in the middle. Then it is easy to surmise that the three pieces $PP_{a},$ $PP_{b},$ $PP_{c}$ form a triangle iff $P$ lies inside the medial triangle. (Otherwise, one of the segments will be longer than half of an altitude, therefore making the sum of the other two shorter than half of an altitude. Thus it is clear that if $P$ does not belong to the medial triangle one of the three triangle inequalities for $PP_{a},$ $PP_{b},$ $PP_{c}$ will fail. And the argument is reversible.)

### References

1. M. Gardner, The Colossal Book of Mathematics, W. W. Norton & Co, 2001, Ch. 21: Probability and Ambiguity
2. J. Whittaker, Random Triangles, Am Math Month 97, n 3 (Mar., 1990) pp. 228-230
3. A. M. Yaglom, I. M. Yaglom , Challenging Mathematical Problems With Elementary Solutions, Holden Day; Rev. edition (June 1, 1964)