Another triangle

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Another triangle
Given a rightangled triangle with the two sides of length 1, what is the shortest segment which divides this triangle into two figures with equal area?

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Re: Another triangle
Call the hypotenuse "z".DanishDynamite wrote:Given a rightangled triangle with the two sides of length 1, what is the shortest segment which divides this triangle into two figures with equal area?
1^2 + 1^2 = z^2
2 = z^2
(sq rt)2 = z
Call the segment "a".
a^2 + (z/2)^2 = 1
a^2 + ([(sq rt)2]/2)^2 = 1
a^2 + (2/4) = 1
a^2 = 1  .5
a^2 = .5
a = (sq rt).5
So the shortest segment is the square root of 1/2.

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Re: Another triangle
Length of segment is sqrt [sqrt(2)  1], which is approximately 0.64359DanishDynamite wrote:Given a rightangled triangle with the two sides of length 1, what is the shortest segment which divides this triangle into two figures with equal area?
(Let the original triangle be ABC, with A = 45 degrees and B = 90 degrees. The searched segment cuts off a triangle AB'C'; a function with this segment B'C' and another side AB' as variables can be constructed, and then the minimum of the segment is determined, using derivatives.)

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I worked out the same solution as millirem. I might as well post it, now that I've made a diagram.
The area of the original triangle is 1/2. In the diagram, I've added a line of length b, making both of the angles marked with a dot 67.5 degrees.
<img src="http://www.mround.pwp.blueyonder.co.uk/ ... angle3.gif" width="215" height="235" alt="triangle diagram" title="triangle diagram">
To divide the big triangle into two equal areas, the length of the line is:
b = sqrt(tan(45 / 2)) = 0.64359
I didn't use any calculus. I just figured that the isosceles triangle was the way to go, and used a bit of trigonometry.
<BR>
The area of the original triangle is 1/2. In the diagram, I've added a line of length b, making both of the angles marked with a dot 67.5 degrees.
<img src="http://www.mround.pwp.blueyonder.co.uk/ ... angle3.gif" width="215" height="235" alt="triangle diagram" title="triangle diagram">
To divide the big triangle into two equal areas, the length of the line is:
b = sqrt(tan(45 / 2)) = 0.64359
I didn't use any calculus. I just figured that the isosceles triangle was the way to go, and used a bit of trigonometry.
<BR>

 Posts: 2608
 Joined: Mon Jun 07, 2004 4:58 pm
 Location: Copenhagen