Geometry: 102

Minimal straight cut 

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.

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Solution to puzzle 77: Minimal straight cut

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.

Cut 1

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 to AXY:

z2	= x2 + y2 - 2xy cos A
	= x2 + y2 - 1,  since cos A = 1/
	= (x - y)2 + ( - 1)

Hence the minimum value of z² (and therefore of z) occurs when x = y, so that z² =  - 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.

Cut 2

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:

z2	= x2 + y2
	= (x - y)2 + 1

Hence the minimum value of z² occurs when x = y, so that z² = 1.
This is longer than the minimal length established for cut 1, above.

Minimal cut

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.)

Remarks

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.