How can we find the area of the triangle which covers a finite point set in $\mathbb{R}^2$ by using the interior triangles with specified area? Suppose we have given a finite point set $X \subset \mathbb{R}^2$ in a way that any triangle made by vertices of $X$ has area at most 1. How can we prove that there is a triangle of area 4 which is covering the whole point set $X$?
And why we cannot take any other number less than 4 for the area of the covering triangle?
Hint: Suppose one of the triangles with vertices in $X$ and use the hyper planes which have intersections with mentioned triangle just in its vertices and find the bigger triangle.
 A: Let $PQ$ be the segment with vertices in $X$ and let $R$ be another point of $X$ such that $PQR$ has the maximum area (of all possible triangles with $P$ and $Q$ being two of their vertices). Now let $l_R$ be the line passes through the point $R$ and parallel to the edge $PQ$, and let $H_R$ be the half plane through $l_R$ which does not contain the edge $PQ$. Since $PQR$ has the maximal area between the triangles with edge $PQ$, so any other point of $X$ does not belong to $H_R$, because if does? we will have a triangle with an area bigger than $PQR$; that is contradiction. So whole $X$ will stay in the other side of $H_R$. And if we repeat this process for other vertices $P$ and $Q$, we will find that $X$ will be inside the intersection of outers of the half plans $H_R$, $H_P$ and $H_Q$ which is the triangle $ABC$ with similarity ratio 2 as shown in the picture, which will give us:
$S_{ABC} = \frac{AC \times h_{ABC}}{2} = \frac{2 PR \times 2 h_{PQR}}{2} = 2 PR \times h_{PQR} = 4 S_{PQR}$.
It shows that area of $ABC$ is at most 4.

And for the next part of question, we need to prove that the triangle with maximum size inscribed in the circumference of a circle of radios $r$ is a regular (equilateral) triangle.
$\textbf{Solution:}$
$>$  suppose we have the following isosceles triangle inscribed in a circle with radios $r$ and want to maximize its area.

We first notice that the vertical line creates a right angle with the base of the triangle and bisects the
triangle since it is isosceles. Next, we write the equations 
\begin{equation}
x^2 + y^2 = r^2
\end{equation}
from the Pythagorean theorem with the small triangles. Then, we write the equation
\begin{equation}
A = \frac{1}{2} (2x)(r + y)
\end{equation}
for the area of the triangle. Now, solve (0.0.1) for $x$, we obtain 
$$x = \sqrt{r^2 - y^2}$$
by Substituting this into (0.0.2), we have
$$A(y) = \sqrt{r^2 - y^2} (y + r) $$
Taking the derivative by applying the product rule, we obtain
$$\frac{\partial A(y)}{\partial y} = \sqrt{r^2 - y^2} + (y + r) \frac{- 2 y}{2 \sqrt{r^2 - y^2}}$$
Simplifying this expression, we have
$$\frac{\partial A(y)}{\partial y} = \frac{r^2 - ry - 2 y^2}{\sqrt{r^2 - y^2}}$$
Now, to find the maximum, we set $\frac{\partial A(y)}{\partial y} = 0$ and solve:
$$\frac{r^2 - ry - 2 y^2}{\sqrt{r^2 - y^2}} = 0$$
Since $y = r$ and $y = −r$ give areas of $0$, we are not interested in the points where the derivative is not
defined, so we may assume that $r^2 - y^2 \neq 0$ . Therefore, we must solve 
$$ r^2 - ry - 2 y^2 = 0 \Rightarrow y = \frac{r \pm \sqrt{9 r^2}}{-4}$$
Therefore, we have 
$$y = −r or y = \frac{r}{2}$$
Clearly $y \neq −r$ since $A(−r) = 0$. Therefore, we must check to see that $y = \frac{r}{2}$ is a maximum. To do
this, we determine that 
$$\frac{\partial A(y)}{\partial y} > 0 for y < \frac{r}{2} and \frac{\partial A(y)}{\partial y} < 0 for y > \frac{r}{2}$$
Therefore,$\frac{r}{2}$
is a maximum. 
Now, we solve for the maximizing dimensions of the triangle. Using (0.0.1), we obtain 
$$x^2 + (\frac{r}{2})^2= r^2 \Rightarrow x^2 = \frac{3}{4} r^2 \Rightarrow x = \frac{\sqrt{3}}{2} r$$
Now, writing the pythagorean theorem for the large triangles, we see that
$$x^2 + (r+ y)^2 = s^2 \Rightarrow \frac{3}{4} r^2 + (r + \frac{r}{2})^2 = \frac{12}{4} r^2 = s^2$$
Thus, we have $s = \frac{2\sqrt{3}}{2} r. $ Finally, we compute the area of the triangle and obtain
$$A(\frac{r}{2})= x(y + r) = \frac{\sqrt{3}}{2} r (\frac{r}{2} + r) = \frac{3\sqrt{3}}{4} r^2$$
Thus, the triangle with maximum area has dimensions
$$s = 2x = \sqrt{3} r, ~~~ h = r + y = \frac{3}{2}r.$$
Which shows that is a regular triangle.
Now we need to prove that the triangle with minimum area size circumscribed around the circumference of a circle of radios $r$ is a regular (equilateral) triangle.
$\textbf{Solution:}$
$>$  The area of the triangle is $A = \frac{1}{2} h b.$ By similar triangles (see figure)

$$\frac{\frac{b}{2}}{h} = \frac{r}{\sqrt{h^2 - 2rh}} \Rightarrow b = \frac{2rh}{\sqrt{h^2 - 2rh}}$$
So $A = \frac{rh^2}{\sqrt{h^2 - 2rh}}$ for $h > 2r,$ and then by taking the derivative, we have 
$$\frac{\partial A(h)}{\partial h} = \frac{rh^2 (h - 3r)}{(h^2 - 2rh)^{\frac{3}{2}}}$$
Then $\frac{\partial A(h)}{\partial h} = 0$ for $h > 2r$ when $h = 3r$ and by the first derivative test $A$ is minimum when $h = 3r$.
If $h = 3r$ then $b = 2 \sqrt{3} r$, shows that the triangle is regular (equilateral).
Now the natural question which come to the mind is that " what is the  value for the radios of the circle for to have inscribe triangles to have maximum area 1?"
$\textbf{Solution:}$ 
$>$ For to have this result we should have
$$x(y + r) \leq 1 \Rightarrow \frac{\sqrt{3} r}{2} \frac{3r}{2} \leq 1 \Rightarrow r \leq \frac{2}{3^{\frac{3}{4}}}$$
The another question will be the opposite one  "what is the  value for the radios of the circle for to have circumscribed triangle around the circumference circle  to have minimum area 4?"
$\textbf{Solution:}$ 
$>$ the same as previous for to have the result we should have
$$x . 3 r \geq 4 \Rightarrow 3 \sqrt{3 } r^2  \geq 4 \Rightarrow r \geq \frac{2}{3^{\frac{3}{4}}}$$
And it will give the answer that $r$ should be equal to $ \frac{2}{3^{\frac{3}{4}}}$ .
A: Lemma: let $ABC$ a triangle and $P,Q,R$ the middle points, then the area of $PQR$ is $\frac{1}{4}$ of $ABC$.
Proof: note that $ABC$ was dividing in 4 congruent triangles.
Now, for the problem: let $PQ$ the longest segment with vertices in $X$ and let $R$ other point of $X$ such that $PQR$ has the max area (of all possible triangles with P and Q). If you construct the triangle ABC of lemma, the problem is solved.
The way to prove that is: if exist a point $D\in X$ but not into $ABC$, we have area $PQD$ $>$ area $PRQ$, absurd.
