Maximum area of pentagon in the quarter circle

Question

In the figure, quarter arc $AD$ of $x^2+y^2=16$ is given. Points $B,C$ are arbitrary on the arc $AD$ such that $C$ is between $D$ and $B$.

What is the maximum value of $\text{Area}(OABCD)$ ?

Notes:

1. Problem is mine. I have sent for sharing. I hope that you like it and enjoy.
2. I will post my solution(s) after yours. You can use any way for your solution.

Answer 1

Let $\measuredangle DOC=\alpha$, $\measuredangle COB=\beta$ and $\measuredangle BOA=\gamma$.

Hence, $\alpha+\beta+\gamma=\frac{\pi}{2}$ and since $\sin$ is a concave function on $\left[0,\frac{\pi}{2}\right]$, by Jensen we obtain: $$S_{OABCD}=\frac{1}{2}\cdot4^2(\sin\alpha+\sin\beta+\sin\gamma)=$$ $$=8(\sin\alpha+\sin\beta+\sin\gamma)\leq24\sin\left(\frac{\alpha+\beta+\gamma}{3}\right)=12.$$ The equality occurs for $\alpha=\beta=\gamma=30^{\circ},$ which gives the answer: $12$.

Answer 2

Points B and C should divide the circle into three equal arcs. This can be shown by fixing one of the points and letting the other point move around. You maximize the area of the triangle BCD when C divides the arc BD into two equal arcs, because at that point the height of the triangle is maximized. By applying this same argument to triangle ABC, we see that the area of both triangles is maximized when AB=BC=CD. Therefore, the maximum area of the pentagon occurs under the same conditions. This means we have three triangles with two side lengths of 4 and an angle between them of 30°. The entire area is therefore: $$3*\left(\frac12*4*4*\sin30°\right)=12$$

Answer 3

Isoperimetric inequalities are very elegant theorems in Euclidean geometry. One of them:

Isoperimetric inequality for cyclic polygons: If $A_1A_2\dots A_n$ is a cyclic $n$-gon, then area of $A_1A_2\dots A_n$ will be maximum when $n$-gon regular.

We will use this idea in our solution. Let's reflect the figure w.r.t $x$-axis and $y$-axis like following figure.

We yields a dodecagon. By isoperimetry theorem for cyclic $n$-gon, maximum area of it will be $$\dfrac{1}{2}\cdot 12\cdot 4^2\cdot \sin 30^\circ = 48$$ Then, quarter of it is $$\dfrac{48}{4}=12 $$

Answer 4

Let $C(a,\sqrt{4-a^2})$ and $B(b,\sqrt{4-b^2})$.

Hence, $$S_{OABCD}=f(a,b)=\tfrac{\left(4+\sqrt{4-a^2}\right)a}{2}+\tfrac{\left(\sqrt{4-b^2}+\sqrt{4-a^2}\right)(b-a)}{2}+\tfrac{\sqrt{4-b^2}(4-b)}{2},$$ where $0\leq a\leq b\leq4$.

We see that $f$ is continuous function on $\Phi=\{(a,b)|0\leq a\leq b\leq4\}$, which is compact.

Thus, $f$ gets a maximal value for each $(a_0,b_0)\in\Phi$.

In another hand, if we'll replace $C$ on a middle point of $\smallsmile DB$ or if we'll replace $B$ on a middle point of $\smallsmile AC$, then a value of $f$ increases, which says that if $f(a,b)=f(a_0,b_0)$ then $DC=CB=BA$, otherwise we can increase $f$ by previous way, which is contradiction.

Thus, $\max S_{OABCD}=f(a_0,b_0)=12$ (the number $12$ I took from my previous solution)

and we are done.