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How would i find/prove the maximum perimeter of an equilateral triangle inside a circle:

$$x^2+y^2=4$$

ps. sorry for my bad english and bad question making

Drago
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  • What does the equation refer to? – measure_theory Oct 08 '16 at 23:54
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    https://math.stackexchange.com/questions/710796/maximum-perimeter-of-an-isosceles-triangle-inscribed-in-the-unit-circle?noredirect=1&lq=1 Might be helpful – Decaf-Math Oct 08 '16 at 23:55
  • Why do you need calculus? All inscribed equilateral triangles within the circle will have the same perimeter. – Batman Oct 08 '16 at 23:55
  • @Batman OP probably needs to prove the maximum perimeter using calculus rather than elementary geometry. – Decaf-Math Oct 08 '16 at 23:58
  • This may pose the same confusion some experimented with a question a few days ago: the triangle is inside the cirlce, and not the circle is circumscribing the triangle, as then as Batman already pointed out, all the equilateral triangles circumscribed in the given circle have the same perimeter: $;3\cdot2\sqrt3=6\sqrt3;$ . Yet this is the maximal possible area, and the problem assumes "small equilateral triangles" can be contained in the interior of the triangle, without even touching the circle itself. – DonAntonio Oct 09 '16 at 00:27
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    I bet the question was intended to be "Prove that the maximum perimeter of an inscribed triangle is an equilateral triangle." – robjohn Oct 09 '16 at 01:06

2 Answers2

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If you really need to use Calculus, try proposing three points $P_1,P_2,P_3$. This points have coordinates $(x_i,y_i)$, which satisfy the relation $x^2+y^2=2^2$.

You need to find the perimeter of the triangle formed by those points, so you want $Perimeter = d(P_1,P_2)+d(P_1,P_3)+d(P_3,P_3)$. Expanded is:

$$ Perimeter =\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\ + \sqrt{(x_1-x_3)^2+(y_1-y_3)^2}\ + \sqrt{(x_2-x_3)^2+(y_2-y_3)^2}. $$

You can set $P_1=(2,0)$ without loss of generality, so above is simplified to:

$$ Perimeter =\sqrt{(2-x_2)^2+y_2^2}\ + \sqrt{(2-x_3)^2+y_3^2}\ + \sqrt{(x_2-x_3)^2+(y_2-y_3)^2}. $$

You also know that both $P_2$ and $P_3$ need to satisfy the relation. Try first with $y_i^2 = 2^2-x_i^2$. Set it again in the above equation and rewrite:

$$ Perimeter =\sqrt{(2-x_2)^2+2^2-x_2^2}\ + \sqrt{(2-x_3)^2+2^2-x_3^2}\ + \sqrt{(x_2-x_3)^2+(2^2-x_2^2+2^2-x_3^2)+2\sqrt{(2^2-x_2^2)(2^2-x_3^2)}}. $$

Now to try a long-shot of intuition, set $x_2=x_3=z$. Then:

$$ Perimeter =2\sqrt{(2-z)^2+2^2-z^2}\ + \sqrt{2(2^2-z^2)+2\sqrt{(2^2-z^2)(2^2-z^2)}} $$

You might simplify further if needed. Then differentiate with respect to $z$. You should get $z=-1$. This is the easiest Calculus way I can think of.

OFRBG
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  • I tried answering what was suggested in the main comment section: "Prove that the maximum perimeter of an inscribed triangle is an equilateral triangle." There are a couple of assumptions, but I guess they're okay for the level required. – OFRBG Oct 09 '16 at 01:29
  • thanks, it was very helpful, but i was kind of lost as to why $(y_1-y_2)^2$ turned into $y_2^2$ – Drago Oct 09 '16 at 02:09
  • @Drago We decided to set $y_1$ as $0$, so after substitution, the expression is just $y_2^2$. When we did $P_1=(2,0)$ it was implicit. – OFRBG Oct 09 '16 at 02:16
  • oh nevermind i completely ingnored the $y_1$ value – Drago Oct 09 '16 at 02:27
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The equation $x^2+y^2=4$ describes a circumference of radius $\rho=2$ centered at $(0,0)$. If you have a equilateral triangle inscribed in this circle (which is unique up to rotations) then this means that $\rho=2r$, where $r$ is the inradius of your triangle; i.e. $r=1$. As in the following picture (borrowed from http://mathworld.wolfram.com/EquilateralTriangle.html) enter image description here

The length $a$ of each side happens to be $\frac{6}{\sqrt 3}r$ so your answer is $$ perimeter=3a=6\sqrt3. $$ Notice that the up-to-rotation-uniqueness implies that every equilateral triangle incribed in this circumference will have the same perimeter.

If you don't restrict yourself to the inscribed case and consider a triangle inside the circle then you can maximize the perimeter maximizing the function $p(r)=6\sqrt3r$, which will tell you that the biggest value is attained when $r$ is maximun; i.e. when $r=1$.