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Square Lattice Triangles

Problem

A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.


Find the probability that three randomly chosen points of a 5 by 5 lattice will form a triangle.


Solution

There are 25C3 = 2300 ways of picking three points from twenty-five. However, some of the sets will be collinear, and this can happen in a number of ways.

Along each vertical, horizontal, and main diagonal:
125C3 = 12times10 = 120

Along the other diagonals:
4(4C3+3C3)=4(4+1)= 20


Then, we have twelve more:


2300 minus (120+20+12) = 2148, hence the probability of three random points forming a triangle will be 2148/2300 = 537/575 approximately 0.934.

What about a 6 by 6 lattice?
Although there is no general formula, investigate other sized lattices to find an algorithm.

Problem ID: 162 (Mar 2004)     Difficulty: 4 Star

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