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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, root 2, 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

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.

Isosceles right triangle with sides AB=1, BC=1, AC=square root of 2. Let AX=x, AY=y, with XY=z.

Area triangleABC = ½ × base × perpendicular height = ½.

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

Since Area triangleAXY = ½ × Area triangleABC, we have xy = 1/root 2.

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

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

Hence the minimum value of z2 (and therefore of z) occurs when x = y, so that z2 = root 2 − 1.
Then, since xy = 1/root 2, x = y = 1/fourth root of 2.

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.

Isosceles right triangle with sides AB=1, BC=1, AC=square root of 2. Let BX=x, BY=y, with XY=z.

Area triangleBXY = ½xy.

Since Area triangleBXY = ½ × Area triangleABC, we have xy = ½.

Applying Pythagoras' Theorem to triangleBXY:

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 root ((root 2) - 1), with, in the first diagram, AX = AY = fourth root of 1/2   (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.

Isosceles right triangle, reflected in its sides seven times to form a square of side 2. The triangle AXY, also multiplied eight times, forms a regular octagon of area 2, centered within the square.

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.)

A circle, of area 2, centered within a square of side 2.

It is easy to show that the shortest arc, shown above, has length root (2*pi) / 4 is approximately equal to 0.626657, versus root ((root 2) - 1) is approximately equal to 0.643594 for the straight line bisector.

Further reading

  1. Bisecting Arcs

Source: Traditional

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