How to prove geometrically that the equilateral triangle inscribed inside a circle has the longest perimeter

Please, help me! How do I prove geometrically that the equilateral triangle inscribed inside a circle has the longest perimeter? I'm really needing it and I'm lost

• Hint: the triangle with the longest perimeter is isosceles on any base – Exodd Mar 12 '18 at 16:23
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Mar 12 '18 at 16:24
• Have a look at the related questions listed on the right – David Quinn Mar 12 '18 at 16:24

you can express the area of the triangle by the radius of the circle and the angle between the radius,like $$A=\frac{1}{2}R^2\sin(\alpha)+\frac{1}{2}R^2\sin(\beta)+\frac{1}{2}R^2\sin(\gamma)=\frac{1}{2}R^2(\sin(\alpha)+\sin(\beta)+\sin(\gamma))$$ with the condition $$\alpha+\beta+\gamma=360^{\circ}$$

Let triangle $ABC$ be isosceles on base $BC$. Choose $D$ on circle $ABC$ such that $DC>AC=AB$. Join $DB, DC$, and let $DC$ cut $AB$ at $G$. Let the circle with center $B$ and radius $BD$ cut $BA$ at $F$, and circle with center $C$ and radius $CA$ cut $CD$ at $E$.

Then since triangles $BGD$ and $CGA$ are similar,$$\frac{BG}{CG}=\frac{BD}{CA}=\frac{BF}{CE}=\frac{BF-BG}{CE-CG}=\frac{GF}{GE}$$But$$\frac{BG}{CG}=\frac{GD}{GA}$$Therefore$$\frac{GF}{GE}=\frac{GD}{GA}$$ And since $DC>AB$ and $$\frac{GC}{GA}=\frac{GB}{GD}$$then $GA>GD$ [see Euclid V, 25]. And if from $GA>GD$ we take $GF<GE$, respectively, then remainder $AF>DE$.

Therefore $$(AB-DB)>(DC-AC)$$so that $$AB+AC>DB+DC$$and the isosceles triangle has the greatest perimeter of all triangles inscribed on a given chord in a given circle.

Next we must show that, of all isosceles triangles inscribed in a circle, the equilateral triangle has the greatest perimeter.

Given equilateral triangle $ABC$, let triangle $ADE$ be isosceles, with $AD>AB$ cutting $BC$ at $K$. Join $CE$, $BD$, and $CD$ cutting $AE$ at $M$. Let the circle with center $A$ and radius $AB$ cut $AD$ and $AE$ at $F$ and $H$, let circle with center $C$ and radius $CD$ cut $BC$ at $G$, and let circle with center $D$ and radius $DE$ cut $DC$ at $L$.

Then since triangles $KAB$ and $KCD$ are similar$$\frac{AB}{CD}=\frac{KB}{KD}=\frac{AF}{CG}=\frac{AK}{CK}=\frac{AF-AK}{CG-CK}=\frac{KF}{KG}$$

And since$$\frac{AK}{CK}=\frac{BK}{DK}$$and $AD>AB=BC$, then $KB>KD$ [see Euclid V, 25]. Therefore, $KF>KG$, and subtracting $KG<KF$ from $KB>KD$, respectively, then remainder$$BG>DF$$Hence$$(BC-DC)>(AD-AB)$$ making$$AB+BC>AD+DC$$And since in triangle $CDE$, $\angle DEC$ is obtuse (arc $DEC$ is less than a semicircle), then$$DC>DE$$making$$AB+BC>AD+DE$$ Finally, the same argument that proved $BG>DF$ proves $LC>EH$, making $$(DC-DE)>(AE-AC)$$ But$$AB+BC>AD+DC$$Therefore$$AB+BC+AC>AD+DE+AE$$

and the equilateral triangle has the greatest perimeter of all isosceles triangles, and consequently of all triangles, inscribed in a given circle.