MHB The Smartest Triangles: Equilateral Triangles

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The discussion establishes that the IQ of a triangle, defined as the ratio of its area to the square of its perimeter, is maximized by equilateral triangles. Using Heron's formula and Lagrange multipliers, it is shown that the optimal configuration occurs when all side lengths are equal, leading to the conclusion that for a given perimeter, the triangle with the highest IQ is equilateral. The conversation also touches on concerns about originality in problem-solving, with some participants suggesting that similar problems may lead to accusations of academic dishonesty. Overall, the mathematical proof reinforces the unique properties of equilateral triangles in maximizing this specific ratio. The findings highlight the significance of equilateral triangles in geometric optimization.
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Let the IQ of a triangle be the ratio $$\frac{\text{area of the triangle}}{(\text{perimeter of the triangle})^2}$$.

This is a dimensionless number. Show that the smartest triangles are equilateral triangles.
 
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Using Heron's formula, one notices by symmetry that interchanging any two of the variables representing the side lengths results in the same area and perimeter functions, hence Lagrange multipliers used with:

The objective function $$A(x,y,z)=\frac{\sqrt{s(s-x)(s-y)(s-z)}}{4s^2}$$ where $$s=\frac{x+y+z}{2}$$

Subject to the constraint $$g(x,y,z)=x+y+z-p=0$$

will necessarily lead to the implication:

$$x=y=z=\frac{p}{3}$$

Thus, the triangle having the greatest IQ is equilateral.
 
Hey MarkFL, this is the fastest solution that I have ever received so far!

View attachment 952

Great job!:cool:;)
 

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Let's just say this problem is nearly identical to a recent university level POTW...copy/paste/edit/done. (Rofl)
 
Looks to me these copy/paste/edit things are signs of you're on the verge of cheating...(Rofl)
 
anemone said:
Looks to me these copy/paste/edit things are signs of you're on the verge of cheating...(Rofl)

It's only cheating if I copy/paste the work of someone else...which I would never do of course. (Angel)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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