The side lengths tell us that the triangle has a right angle at B (check the side lengths in Pythagoras’ Theorem). We’ll let the midpoint of AC be P and the midpoint of AB be Q.

To land closer to A than to C, Freddy needs to be on the “A” side of the perpendicular bisector of the hypotenuse AC.

To land closer to B than to A, Freddy must land on the “B” side of the perpendicular bisector of AB.

This means he lands in a trapezium-shaped area of the original triangle CBA, with vertices B, Q, P and R, where R is the point of intersection of BC and the perpendicular bisector of AC.

To help follow the rest of the solution, draw yourself a quick sketch of all this.

As P and Q are midpoints of AB and AC respectively, we see that QA = 2.5m and that QP = 6m. Therefore, the right-angled triangle AQP has area 0.5 x 2.5 x 6 m² = 15/2 m².

The triangle PRC is also right-angled, at P – because PR is part of the perpendicular bisector of AC.

PRC is similar to triangle CBA (there is a shared angle at C, and both are right-angled); the length scale factor PRC: CBA s found to be 6.5/12 (by comparing the sides CP and CR), therefore the area scale factor is (13/24)².

This means that as the area of triangle CBA is 0.5 x 12 x 5 = 30m², the area of triangle PRC is

30 x (13/24)² = 5 x 169/96 = 845/96 m².

Therefore, the area of the trapezium where Freddy should land is

(30 – 15/2 – 845/96)m² = (2880 – 720 – 845)/96 m²= 1315/96 m².

So the probability that this is where he lands is 1315/96 ÷ 30 = 263/576

Challenge 279: Proximity Probability