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 October 24th, 2013, 12:07 PM #1 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Largest possible inscribed triangle in a circle Problem: "Show that the, in area, largest triangle of the in a circle inscribed triangles is an equilateral triangle. Guidance: First show that the equilateral triangle is the largest of the isosceles triangles." My first thought was to begin with defining the area of an inscribed isosceles triangle as a function of theta, and then setting its derivative to zero. Hopefully you could from there solve for theta, which then should be 60Â° since the largest one is supposed to be an equilateral triangle. But I couldn't solve the equation and I don't know how I would have continued anyway. This was meant to be the hardest problem in the Finnish Matriculation Exam for high schoolers in 97 If the translation is unclear, what you're supposed to show is that the largest possible triangle you can inscribe in a circle is an equilateral one. Last edited by skipjack; December 11th, 2014 at 01:06 PM.
 October 24th, 2013, 12:29 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Largest possible inscribed triangle in a circle I would assume that the inscribed triangle must contain the center of the circle, and then divide the triangle into 3 triangles. Without loss of generality, assume the circle is a unit circle. Then our objective function becomes: $A$$\theta_1,\theta_2,\theta_3$$=\frac{1}{2}$$\sin\ (\theta_1$$+\sin$$\theta_2$$+\sin$$\theta_3$$\)$ subject to the constraint: $\theta_1+\theta_2+\theta_3-2\pi=0$ We see that the 3 variables are cyclically symmetric, which means the maximum occurs for: $\theta_1=\theta_2=\theta_3=\frac{2\pi}{3}$ This leads to the inscribed triangle being equilateral.
 October 24th, 2013, 10:37 PM #3 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Re: Largest possible inscribed triangle in a circle What does it mean that they're cyclically symmetric? That they're interchangeable? And how do you then know when the maximum occurs? From a theorem? Last edited by skipjack; December 11th, 2014 at 01:07 PM.
 October 24th, 2013, 11:08 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Largest possible inscribed triangle in a circle Yes, cyclic symmetry means any two of the variables may be interchanged with no change in the objective and constraint functions. That the critical point comes from the equation of the variables to one another in such a case can be shown use Lagrange multipliers.
 October 24th, 2013, 11:21 PM #5 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Re: Largest possible inscribed triangle in a circle Ok, thanks. Multivariable calculus isn't part of the curriculum here but I do know how to use Lagrange Multipliers. The only method I can think of that is taught, for finding a maximum is setting the derivative to zero. I believe this problem is "supposed" to be done that way, since it was under that chapter in the book. Would it be possible to substitute the constraint into the area formula and optimize from there?
 October 24th, 2013, 11:54 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Largest possible inscribed triangle in a circle You would still have a function of two variables. I would suggest another approach.
 October 25th, 2013, 03:49 AM #7 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: Largest possible inscribed triangle in a circle $A=\frac{abc}{4R}=2R^2 \sin \alpha \sin \beta \sin \gamma$ We need to find maximum of $\sin \alpha \sin \beta \sin \gamma$. From A-G inequality $\sin \alpha \sin \beta \sin \gamma \leq (\sin \alpha+\sin \beta +\sin \gamma)^3$ where equality occurs when $\sin \alpha=\sin \beta = \sin \gamma$ ($\alpha=\beta=\gamma=\frac{\pi}{3}$) [1] If you want to find the maximum value of $\sin \alpha+\sin \beta +\sin \gamma$ $f(x)=\sin x$ is concave on interval $<0,\pi>$. Applying Jensen inequality $\frac{\sin \alpha+\sin \beta +\sin \gamma}{3} \leq sin(\frac{\alpha+\beta+\gamma}{3})=\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ $\sin \alpha+\sin \beta +\sin \gamma \leq \frac{3 \sqrt{3}}{2}$
October 25th, 2013, 08:51 AM   #8
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Re: Largest possible inscribed triangle in a circle

Quote:
 Originally Posted by MarkFL You would still have a function of two variables. I would suggest another approach.
It's definitely ugly, but I would assume that's why they put the "clue", to prove it for isosceles triangles first, since you then have one variable. Don't know how you'd go from there though.
Yours and Crom's methods are much cleaner, but sadly slightly above my current level.

 October 25th, 2013, 08:59 AM #9 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Re: Largest possible inscribed triangle in a circle Since we're on the topic of optimization, how do you determine the largest value the expression (3x+4y) can take, when you have the constraint x^2+y^2=14x+6y+6 for the coordinates (x,y)? The answer is 73, I can get that using substitution and then the derivative or Lagrange Multipliers. But how do you determine that without any calculus at all? Is there some trick since the constraint is a circle? It's a problem from a while back, when we hadn't done any calculus.
October 25th, 2013, 11:40 AM   #10
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Re: Largest possible inscribed triangle in a circle

Quote:
Originally Posted by Daltohn
Quote:
 Originally Posted by MarkFL You would still have a function of two variables. I would suggest another approach.
It's definitely ugly, but I would assume that's why they put the "clue", to prove it for isosceles triangles first, since you then have one variable. Don't know how you'd go from there though.
Yours and Crom's methods are much cleaner, but sadly slightly above my current level.
If we assume an isosceles triangle, then consider the following diagram:

[attachment=0:30uox1gp]daltohn.jpg[/attachment:30uox1gp]

We have bisected the circle and inscribed isosceles triangle. Optimizing the area of the resulting right triangle, will also do the same for the isosceles triangle from which it comes.

The area of the triangle is:

$A=\frac{1}{2}bh$

Using the Law of Cosines, we find:

$1^2=1^2+k^2-2k\cos(\theta)$

$k=2\cos(\theta)$

Now, we also have:

$b=k\cos(\theta)=2\cos^2(\theta)=1+\cos(2\theta)$

$h=k\sin(\theta)=2\sin(\theta)\cos(\theta)=\sin(2 \theta)$

Hence, the area of the right triangle as a function of $\theta$ is:

$A(\theta)=\frac{1}{2}$$1+\cos(2\theta)$$$$\sin(2\t heta)$$=\frac{1}{2}\sin(2\theta)+\frac{1}{4}\sin(4 \theta)$

Differentiating and equating to zero, we find:

$A'(\theta)=\cos(2\theta)+\cos(4\theta)=2\cos^2(2 \theta)+\cos(2\theta)-1=(2\cos(2\theta)-1)(\cos(2\theta)+1)=0$

The second factor give the minimal area of zero, and so the first factor gives:

$\cos(2\theta)=\frac{1}{2}$

$\theta=\frac{\pi}{6}$

This means the right triangle is a 30Â°-60Â°-90Â° triangle, and so the entire triangle must be equilateral.
Attached Images
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Last edited by skipjack; December 11th, 2014 at 01:24 PM.

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