# Maximum Perimeter of a triangle inside a circle [closed]

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

## closed as off-topic by suomynonA, Claude Leibovici, Henrik, Davide Giraudo, Parcly TaxelOct 10 '16 at 1:27

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• What does the equation refer to? – measure_theory Oct 8 '16 at 23:54
• math.stackexchange.com/questions/710796/… Might be helpful – Decaf-Math Oct 8 '16 at 23:55
• Why do you need calculus? All inscribed equilateral triangles within the circle will have the same perimeter. – Batman Oct 8 '16 at 23:55
• @Batman OP probably needs to prove the maximum perimeter using calculus rather than elementary geometry. – Decaf-Math Oct 8 '16 at 23:58
• I bet the question was intended to be "Prove that the maximum perimeter of an inscribed triangle is an equilateral triangle." – robjohn Oct 9 '16 at 1:06

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.

• 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. – Cehhiro Oct 9 '16 at 1:29
• 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 9 '16 at 2:09
• @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. – Cehhiro Oct 9 '16 at 2:16
• oh nevermind i completely ingnored the $y_1$ value – Drago Oct 9 '16 at 2:27

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) 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 $r=1$.