show that out of all triangles inscribed in a circle the one with maximum area is equilateral

show that out of all triangles inscribed in a circle the one with maximum area is equilateral

How do i start. I have to use function of two variables


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The straightforward way to demonstrate this is the geometric approach, outlined in three different ways by the other solutions thus far.

If you must use multi-variable calculus, however, let the three points be designated by their angular position on a circle of radius $2$ (this is for convenience; the radius of the circle does not matter). Without loss of generality, let the first point $A$ be given angle $0$; the other points $B$ and $C$ are at angles $\alpha_1$ and $\alpha_1+\alpha_2$, with both $\alpha_1, \alpha_2 > 0$ and $\alpha_1+\alpha_2 < 2\pi$.

We can then subdivide the triangle into three sub-triangles: With the center of the circle at $O$, we have $\triangle OAB, \triangle OBC, \triangle OCA$. These combine to form $\triangle ABC$. Do not worry if $\alpha_1$ or $\alpha_2 > \pi$; we will determine such a sub-triangle’s area as negative, and you can convince yourself (with a diagram) that everything will work out.

Consider the sub-triangle $\triangle OAB$. It has base length $OA = 2$, and altitude $2\sin \alpha_1$; therefore, its area is $2\sin \alpha_1$. Likewise, the other two sub-triangles have area $2\sin \alpha_2$ and $2\sin (2\pi-\alpha_1-\alpha_2) = -2\sin (\alpha_1+\alpha_2)$. We therefore want to maximize their sum

A = 2[\sin \alpha_1+\sin \alpha_2-\sin (\alpha_1+\alpha_2)]

with respect to $\alpha_1$ and $\alpha_2$ (subject to the constraints above). Can you take it from here?

I suggest you to consider a triangle, and show that if it isn’t equilateral, you can find another triangle with a greater area.
That way you are sure that the equilateral triangle is the biggest you can get.

Consider the bisectors of the sides of your triangle.

Hint 2:

If you move a vertex parallely to the opposite side, you dont change the area. So, if you move a vertex that isn’t on the bisector to the bisector, you have the same area but the vertex you moved is now strictly inside the circle.

Most geometric treatments of this problem tacitly assume that a triangle of largest area in fact exists, and argue as follows: Since non-equilateral triangles can be transformed into inscribed triangles with larger area, the maximal triangle has to be equilateral. It is a nontrivial geometric exercice to provide a proof that any non-equilateral triangle has actually smaller area than the equilateral one.

A non-equilateral triangle has an angle $\alpha<60^\circ$. After a preliminary step we may assume that this triangle ABC is isosceles with base $BC$, see the following figure. In the same figure we draw the equilateral triangle $AB’C’$.

From $\alpha<60^\circ$it follows that the point $B’$ lies strictly between the point $B$ and the point $S$ on the circle. This allows to draw the following conclusion:
$${\rm area\,}(ABC)<{\rm area\,}(AB’C)<{\rm area\,}(AB’C’)\ ,$$
because at each step we make a triangle more isosceles, keeping two of ists vertices fixed.

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I’m sure this can be done with multivariable calculus, but I would just do the following.

  • Fix (for a moment) one side of the triangle. Using that side as a base show that the height is maximized, when the other two sides have equal length. This is a necessary condition for maximal area. Remember that the height is the projection of the third vertex of the triangle on the line perpendicular to the base, passing through its midpoint.
  • Do the same for all the sides of the triangle, and rejoice.