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Triangle In Square

Problem

A line segment is placed on top of a unit square so as to form a triangle region.


Given the length of the line segment, L, find the maximum area of the triangle.


Solution

Clearly for L greater than or equal radical2, the maximum area will be ½, when the segment is placed along the diagonal of the square.

For L less than radical2, consider the following diagram.


b = radical(L2 minus a2).

So area of triangle, A = ½ab = ½aradical(L2 minus a2) = ½a(L2 minus a2)½.

We may proceed from here via a calculus or non-calculus approach:

Calculus Method

dA/da = ½(L2 minus a2)½ + ½a½(L2 minus a2)(-2a)
  = ½(L2 minus a2)½ minus ½a2(L2 minus a2)

At turning point, dA/da = 0.

Therefore, ½(L2 minus a2)½ = ½a2(L2 minus a2)

radical(L2 minus a2) = a2/radical(L2 minus a2)

Hence, L2 minus a2 = a2, leading to, a = L/radical2.

As area of triangle, A = ½aradical(L2 minus a2),

Amax = ½(L/radical2)radical(L2 minus L2/2)
  = ½(L/radical2)radical(L2/2)
  = ½(L/radical2)(L/radical2)
  = L2/4

Non-calculus Method

As the area of the triangle, A = ½aradical(L2 minus a2) is defined for positive values, the value of a for which it maximises is the same for A2.

Therefore 4A2 = a2(L2 minus a2) = a2L2 minus a4 = L2/4 minusL minus a2)2.

Hence, 4A2 will maximise when (½L minus a2)2 = 0 implies a = L/radical2.

Proceeding as in previous method we deduce that Amax = L2/4.

Problem ID: 147 (Jan 2004)     Difficulty: 3 Star

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