A piece of wooden board in the shape of an isosceles right triangle, with sides 1, 1, , is to be sawn into two pieces. Find the length and location of the shortest straight cut which divides the board into two parts of equal area.

We consider two cases:

- Cut 1, across one of the acute angles.
- Cut 2, across the right angle.

Let X lie on AB with AX = x, and Y lie on AC with AY = y. Then XY is a straight cut of length z.

Area ABC = ½ × *base* × *perpendicular height* = ½.

Area AXY, considering AX as the base, equals ½ × x × (y/) = xy / (2).

Since Area AXY = ½ × Area ABC, we have xy = 1/.

Applying the law of cosines (also known as the cosine rule) to AXY:

z^{2} | = x^{2} + y^{2} − 2xy cos A |

= x^{2} + y^{2} − 1, since cos A = 1/ | |

= (x − y)^{2} + ( − 1) |

Hence the minimum value of z^{2} (and therefore of z) occurs when x = y, so that z^{2} = − 1.

Then, since xy = 1/, x = y = 1/.

Intuitively, it seems clear that cutting across the smaller angle, as above, will yield a shorter minimal cut than cutting across the right angle. We verify this intuition below.

Let Y lie on AB with BY = y, and X lie on BC with BX = x. Then XY is a straight cut of length z.

Area BXY = ½xy.

Since Area BXY = ½ × Area ABC, we have xy = ½.

Applying Pythagoras' Theorem to BXY:

z^{2} | = x^{2} + y^{2} |

= (x − y)^{2} + 1 |

Hence the minimum value of z^{2} occurs when x = y, so that z^{2} = 1.

This is longer than the minimal length established for cut 1, above.

Therefore the minimal straight cut has length , with, in the first diagram, AX = AY = (Of course, by symmetry, there is an equivalent cut of equal length from BC to AC.)

It is natural to ask whether a shorter cut is possible if we are not restricted to using a straight line. The answer is: yes! To see why, we use symmetry.

Consider the diagram below, obtained by successive reflection of the triangle in its sides. The area of the whole square is 4; the area of the (regular) octagon is 2.

A result known as the isoperimetric theorem states that of all planar shapes with the same area the circle has the shortest perimeter. Hence a circle with the same area as the octagon will have minimal perimeter. It then follows that the minimal arc which bisects an isosceles right triangle, *while passing through one leg and the hypotenuse*, is an arc of such a circle, with its center at a 45° vertex of the triangle. See below.

(In order to prove that this is the *shortest* arc that bisects an isosceles right triangle, we need to show that no other arc, such as one passing through both legs, is shorter. This can be confirmed by a similar argument based upon symmetry.)

It is easy to show that the shortest arc, shown above, has length 0.626657, versus 0.643594 for the straight line bisector.

Source: Traditional