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Right Triangle Maximization Problem

  1. Aug 9, 2012 #1
    This is a fairly standard maximization problem in calculus, but I was wondering if anybody could help me come up with a nice geometric solution. It seems like it should be possible to make an argument based on symmetry, but I haven't quite been able to work it out yet. Note, I have already solved this problem using calculus and/or trigonometry, but I'm interested in a geometric solution. Help is greatly appreciated.

    The Right Triangle Problem
    The sum of the lengths of the two legs of a right triangle is greater
    than the length of its hypotenuse. [Let’s agree that the three sided
    of a triangle must be genuine line segments, not single points.]
    Considering only right triangles with hypotenuse of length 1, what is
    the largest the difference between the sum of the lengths of the two
    sides and the length of the hypotenuse can be?
     
  2. jcsd
  3. Aug 9, 2012 #2

    pwsnafu

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    If x is the base and y is the height then by the generalized mean equality [tex]\frac{x+y}{2} \leq \sqrt{\frac{x^2+y^2}{2}} = \frac{1}{\sqrt2}[/tex]
    and we attain equality when x = y. So max is when they are the same.
     
  4. Aug 11, 2012 #3
    You can represent all the possible triangles by a circle of radius 1. It is clear that when one side of the triangle decreases, the other must decrease. The direction in which the vertex of the triangle moves is a tangent to the circle. When you are moving at a less than 45 degree angle with respect to the x axis, you get less change in y than change in x. When you start at point (0,1) going around the circle counterclockwise, x is decreasing and y in increasing. Since at first you are going almost vertically, you get a lot of change in y for a change in x, and since y is increasing, it must be that their sum is also increasing. Y is changing faster all the way up to 45 degrees. After 45 degrees, x is changing faster, that means the sum is decreasing, because x is decreasing. Because the sum is increasing all the way up to 45 degrees, it means that it must be the biggest at 45 degrees.
     
  5. Aug 14, 2012 #4
    Thanks for the responses.

    @pwsnafu: That's interesting - I wasn't familiar with the Generalized Mean Equality and that is a slick solution. Do you have any suggestions on how to make it more accessible and easy to think about for people with less math experience?

    @chingel: This was the way that I thought about the problem without using calculus. However, the difficulty I encountered was arguing in a rigorous fashion that "Since at first you are going almost vertically, you get a lot of change in y for a change in x, and since y is increasing, it must be that their sum is also increasing." Is there a simple and accessible way to show this?
     
  6. Aug 14, 2012 #5
    I'm not sure about how rigorous it is but basically it's just that a right triangle has the shorter leg opposite the smaller angle.

    Consider a point A on the circle in the first quadrant making a 30 degree angle with the x axis moving counterclockwise on the circle. You can approximately for small movement consider the point moving on the line that is tangent to the circle at point A. The point is moving left and up at an 60 degree angle. The length of the hypotenuse represents how much the point A has moved, and it's legs represent how much x and y has changed. Since in a right triangle the leg opposite the 60 degree angle is longer than the leg opposite the 30 degree angle, this means y has changed more than x. And since you are moving up and left, x is decreasing and y is increasing. Since the change in y is bigger, the sum of x and y is also increasing.
     
  7. Aug 15, 2012 #6
    Gotcha. It's essentially a calculus-like argument when it comes down to it. Thanks for your thoughts. :-)
     
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