If $r$ is the radius of the circle, $\Delta{ABC}$ a triangle inscribed in it and $x,y,z$ are the angles at $A,B,C$ then how to find the triangle that has the maximum area.
In my book it's mentioned that the area$=\frac12ab \sin z$ where $a,b$ are the sides opposite to $A,B$, also $a=2r\sin z, b=2r\sin z .$ Can anyone tell me how to find expression for $a,b$ and the area mentioned above?
On
The maximum area will be for an equilateral triangle. Draw a radial line from each vertex to the circle's centre, to make three isosceles triangles. The central angle $θ$ is 120 degrees or 2π/3 radians. The area of each triangle is $\frac {r^2\sinθ}{2}$ and the whole area will be 3 times that.
Edit after a comment from @miracle173.
Why is it an equilateral triangle? Consider such an inscribed triangle with one vertex centrally at the top and a horizontal base line. The area of a triangle is height * base / 2. Now, if you move the top vertex either way, the height decreases and the base remains the same, so the area decreases.
The side lengths $a$ and $b$ can also be found from simple trigonometry
$a=b=2r\sin(θ/2)$
On
If $x$, $y$ and $z$ are given then $$S_{\Delta ABC}=2r^2\sin{x}\sin{y}\sin{z}$$ if it's not so,
we'll get much interesting problem.
Let $a$, $b$ and $c$ be sides-lengths of the triangle.
For $a=b=c$ we have $r=\frac{a}{\sqrt3}$ and $$S_{\Delta ABC}=\frac{a^2\sqrt3}{4}=\frac{3\sqrt3}{4}r^2.$$
We'll prove that $$S_{\Delta ABC}\leq\frac{3\sqrt3}{4}r^2.$$ Indeed, since $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=\frac{1}{4}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}$$ and $$S_{\Delta ABC}=\frac{abc}{4r},$$ we need to prove that $$\frac{1}{4}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}\leq\frac{3\sqrt3}{4}\frac{a^2b^2c^2}{16\left(\frac{1}{4}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}\right)^2}$$ or $$\left(\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}\right)^3\leq3\sqrt3a^2b^2c^2$$ or $$\sum\limits_{cyc}(2a^2b^2-a^4)\leq3\sqrt[3]{a^4b^4c^4}$$ or $$\sum_{cyc}(a^4+\sqrt[3]{a^4b^4c^4}-2a^2b^2)\geq0.$$ Now, let $a^2=x^3$, $b^2=y^3$ and $c^2=z^3$.
Hence, we need to prove that $$\sum_{cyc}(x^6+x^2y^2z^2-2x^3y^3)\geq0,$$ which follows from Schur: $\sum_{cyc}(x^6-x^4y^2-x^4z^2+x^2y^2z^2)\geq0.$
Indeed, by Schur we obtain: $$\sum_{cyc}(x^6+x^2y^2x^2-2x^3y^3)\geq\sum_{cyc}(x^4y^2+x^4z^2-2x^3y^3)=\sum_{cyc}x^2y^2(x-y)^2\geq0$$ and we are done!
On
HINT
First, I suppose only $r$ is given, otherwise it doesn't make much sense.
Since $S_{\Delta ABC}=2r^2\sin{x}\sin{y}\sin{z}$, we'll have to find the maximum of $X = \sin{x}\sin{y}\sin{z}$. Now, let's fix $z$. Then $X=\frac {\cos(x - y) - \cos(x + y)}{2}\sin z$. Since $z$ it's fixed, $x + y$ is fixed and $X$ will reach a maximum when $\cos(x - y)$ reaches a maximum, that is when $x=y$
I would think of it without trigonometry and calculus: If the triangle has maximum area, then we can't make the area larger by moving a single corner.
Pick one side of the triangle as base (rotate the circle so that the base is horizontal for clearer view). Then the largest area you can get is by putting the point opposite the base as far away from from the base as possible. This happens to be halfway along the longest arc between the end points of the base. In other words, the opposite point must be exactly in the middle of the two end points of the base for the area to be maximal.
Now, if the triangle is not equilateral, there is at least one point which is not in the middle of the two others, so we can increase the area of the triangle. An equilateral triangle is the only one which cannot be increased this way, so it must be maximal.
To find the area of an equilateral triangle inscribed in a circle, the height of the triangle is $\frac32r$. This is because in an equilateral triangle, heights, medians and side bisectors coincide, and we know that
Now you have an equilateral triangle where you know the height is $\frac32r$, so you can work out that the side length is $\sqrt3r$, which gives the triangle an area of $\frac{3\sqrt3}2r^2$.