# Problem: find the largest area of a rectangle inscribed in an equilateral triangle

Find the dimensions of the rectangle with the largest area that can be inscribed in an equilateral triangle of side 5.5 inches if one side of the rectangle lies on the base of the triangle.

 ebaines Posts: 10,033, Reputation: 5529 Expert #2 Feb 6, 2012, 08:07 AM
Given an equilateral triangle with legs of length L and unknown angle $\theta$ between each leg and the triangle's base, you can set up an equation that correlates the area of the reactangle to the point measured along the leg where the rectangle's corner touches - call this distance x from the base. The base of the triangle has length $2L \cos \theta$. The height of the rectangle is then $x \sin \theta$, and its width is $2L \cos \theta - 2xcos \theta$. So the area of the rectangle is:

$
A = (2L\cos \theta - 2x \cos \theta)(x \sin \theta)
$

Take the derivative of A with respect to x, and set to zero to find the value for x that gives a maximum for A. When you do this - what happens to the $\theta$ terms?
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #3 Feb 6, 2012, 12:41 PM
Quote:
 Originally Posted by ebaines Given an equilateral triangle with legs of length L and unknown angle $\theta$ between each leg and the triangle's base, you can set up an equation that correlates the area of the reactangle to the point measured along the leg where the rectangle's corner touches - call this distance x from the base. The base of the triangle has length $2L \cos \theta$. The height of the rectangle is then $x \sin \theta$, and its width is $2L \cos \theta - 2xcos \theta$. So the area of the rectangle is: $ A = (2L\cos \theta - 2x \cos \theta)(x \sin \theta)$ Take the derivative of A with respect to x, and set to zero to find the value for x that gives a maximum for A. When you do this - what happens to the $\theta$ terms?

The triangle is equilateral, not just isosceles, so you automatically DO know the angle $\theta$. Doesn't change the answer, of course; it just simplifies it a little.
 ebaines Posts: 10,033, Reputation: 5529 Expert #4 Feb 6, 2012, 02:02 PM
Quote:
 Originally Posted by jcaron2 The triangle is equilateral, not just isosceles.
You're right! My mistake comes from typing the answer before I finished reading the question... Thanks!

BTW, it turns out that for an isocoles triangle it doesn't mater what that angle theta is - the point x is always in the same place to maximize the area of the rectangle. Maybe that's beyond the scope of the homework problem, but it's an interesting phenomenon.
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #5 Feb 6, 2012, 02:24 PM
I had never thought about that either until I saw your answer. It makes sense though. If you take any isosceles triangle with the maximum rectangle inscribed, you can scale it horizontally (i.e. stretch or squish it), and the respective areas inside and outside the rectangle will scale proportionally. Hence, the maximum rectangles of all of the stretched or squished triangles will all be stretched or squished versions of the original rectangle. The height never changes.

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