# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – suomynonA, Claude Leibovici, Henrik, Davide Giraudo, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $r=1$.