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

ebaines · 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:

$<br /> A = (2L\cos \theta - 2x \cos \theta)(x \sin \theta)<br />$

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 · 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:

$<br /> A = (2L\cos \theta - 2x \cos \theta)(x \sin \theta)<br />$

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?

EB, you answered a more difficult question than the OP asked.

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 · 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 · 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.

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

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