Solution
One Point per Side
Show the problem again
Triangle ABC has area 1. Points P, Q, R are chosen independently and uniformly at random on sides BC, CA and AB respectively (uniform with respect to length). What is the expected area of triangle PQR? Give a reduced fraction.
Worked solution
The answer is 1/4. Write AR = r·AB, AQ = (1 − q)·AC, and so on, with each parameter uniform on [0, 1] and independent. Triangle PQR is ABC minus three corner triangles. The corner triangle at A has area (AR/AB)(AQ/AC) times the area of ABC, a product of two independent uniforms (one from each of the two sides meeting at A), so its expected area is (1/2)(1/2) = 1/4. The same holds at B and at C. By linearity of expectation, E[area PQR] = 1 − 3·(1/4) = 1/4. The answer is affine-invariant, so it does not matter what shape the triangle is. Choosing every point at a midpoint also gives 1/4, but randomness does not change the average.
Source: Classic geometric-probability exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.