Another triangle

DanishDynamite · Post by **DanishDynamite** » Wed Sep 08, 2004 7:13 pm

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

shemp · Post by **shemp** » Wed Sep 08, 2004 9:03 pm

DanishDynamite wrote:Given a right-angled triangle with the two sides of length 1, what is the shortest segment which divides this triangle into two figures with equal area?

Call the hypotenuse "z".

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.

DanishDynamite · Post by **DanishDynamite** » Thu Sep 09, 2004 6:38 pm

Sorry shemp. Though you've correctly calculated the length of the segment bisecting the right-angle, this isn't the shortest segment possible.

shemp · Post by **shemp** » Thu Sep 09, 2004 8:12 pm

DanishDynamite wrote:Sorry shemp. Though you've correctly calculated the length of the segment bisecting the right-angle, this isn't the shortest segment possible.

Yes, I realized that was possible when I woke up this morning. I knew it seemed too easy.

millirem · Post by **millirem** » Fri Sep 10, 2004 12:57 am

DanishDynamite wrote:Given a right-angled triangle with the two sides of length 1, what is the shortest segment which divides this triangle into two figures with equal area?

Length of segment is sqrt [sqrt(2) - 1], which is approximately 0.64359

(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.)

ceptimus · Post by **ceptimus** » Fri Sep 10, 2004 4:44 am

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>

DanishDynamite · Post by **DanishDynamite** » Fri Sep 10, 2004 8:16 pm

Well done, millirem and ceptimus!

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