Bisecting a triangle

[bonanova]

A triangle has sides of length {1, 1, sqrt(2)}.

What is the length of the shortest cut that divides the triangle into two pieces that have equal areas?

[Pickett]

I'll take a quick stab at it:

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^sqrt(2π) / ₄ = ~0.62666

The "obvious" first answer is ^sqrt(2) / ₂ (bisect the 90 degree angle perpendicular to the hypotenuse)...But if we step back and look at it a different way: The current triangle has an area of ¹/₂...

So, we simply need to make a cut that leaves two areas of ¹/₄ (not necessarily similar shapes). The shortest STRAIGHT cut would be the ^sqrt(2) / ₂, but if we decide to make a curved cut, we can get it shorter. I'll try to describe it (can't add pictures right now)

Choose one of the 45 degree angle points and center a circle there with radius r. Mark the points along the hypotenuse and adjacent leg at points to create a segment of a circle with a 45 degree angle and radius r. We know, then, that is exactly ¹/₈ of the area of the full circle. Since we are trying to get an area of ¹/₄, and we have ¹/8 of a circle, the full area of the circle would be 2. so we can figure out the length of r by doing πr²=2. Leaving us with r = ^sqrt(2π) / _π

Given that, we need to find the length of just that arc of the circle...which is ¹/₈ of the perimeter. So we know P=πd...which leaves us with P=2*sqrt(2π). Divide it by 8, and you get ^sqrt(2π) / ₄ as the length of the arc connecting your points...leaving 2 pieces of the triangle with areas of ¹/₄ each.

 

 

Edited February 28, 2017 by Pickett
It duplicated my response for some reason....

[bonanova]

Nice stab!

I made a picture of it.

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