Math CentralQuandaries & Queries


Question from Bjorn:

I'm putting up a shelf and thought I'd be nice and cut the sharp and pointy corners off. I want to the cut to be at a 45 degree angle, but I also want the exposed edge to be the same length as from the wall to the beginning of the cut -- so the end of shelf will look like the first two sides of an octagon.
I've managed to construct the solution, but I haven't been able to calculate it...
Please help!

Hi Bjorn.

Let D be the depth of the shelf and let x be the length from the wall to the beginning of the cut.

Thus, (D - x) is the length of the leg of the isosceles right triangle you removed. And of course, the hypotenuse of that triangle is x, so that it matches the segment connecting to the wall.

So you have a right triangle with legs (D - x) and hypotenuse of x. Pythagoras tells us:

x² = 2(D - x)²
x² = 2(D² - 2Dx + x²)
x² = 2D² - 4Dx + 2x²
-x² + 4Dx = 2D²
x² - 4Dx = - 2D²

Now we can "complete the square" to determine the value of x:
x² - 4Dx + 4D² = - 2D² + 4D²
(x - 2D)² = 2D²
x - 2D = ±D√2
x = D(2 ± √2)

But we know that x < D, so D(2 + √2) doesn't make sense. Thus the answer is this:
x = D(2 - √2)

Let's test it. Say I have a 20cm deep shelf. Where should I cut that 45 degree angle?

x = 20(2 - √2) = 20(.5858) = 11.7 cm from the wall.

Doing so, I will have cut off a piece that is 20 - 11.7 = 8.3 cm long. As a right angled isosceles triangle, the hypotenuse would be √(2[8.3]²) cm.
= √(2[68.89])
= √(137.78)
= 11.7 cm as expected.

Stephen La Rocque.

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