I solved a), but solution is a little computational. Your remark $a \Rightarrow b$ is true, if we are allowed to coincide hexagon's vertices, i.e. instead of hexagon we can find pentagon, for example.

Hmm... No comments Here is my solution. Suppose $AB$ is the largest diagonal of $P$ (see Figure 1). Note that it can be side of $P$. Construct two perpedicular lines to $AB$ via p.$A$ and p.$B$. Then $P$ lies strictly between these two lines. We will prove that thre is chord $CD||AB$ s.t. $S_{ABCD}\geq S_{P_r}$, where $P_r$ is the right part of $P$ with respect to $AB$. It is clear that problem will be solved after that. I state that chord $CD$ with $x=\frac{3}{4}$ will be appropriate choice. To prove it I will modificate problem. Left $f(x)$ be length of $CD$, $x$ is absciss of $C$. Problem reduces to the following one: Given concave function $f:[0,1]\to[0,1]$, $f(0)=1$, $f(1)=1$. Prove that there is $b\in [0,1]]$ s.t. $S_{OATb}\geq \frac{3}{4}\int_0^1f(t)dt=S$. Indeed, since $f$ is concave, $f(0)=1$ and $f(1)$, we know that $f$ lies under broken line $AEF$ (actually, $EF$ is tangent line to graph $f$ in point $T$). So $S\leq S_{OAEFC}$ (see Figure 2) Thus we have come to computational step of my solution. We need to show $S_{OATb}\geq \frac{3}{4}S_{OAEFC}$ (remember, $b=3/4$). I leave this pure algebraic fact to you (just denote some segments and express areas)! I checked it, but I believe there is geometrical approach.

Attachments:

Wow it's an impresive solution . However I am more interested in what you called Myth wrote: I believe there is geometrical approach. which is what I tried but with no success.

This is a more geometric approach. Take any two parallel lines of support $s,t$ touching $\mathcal{P}$ at $S,T$. The segment $ST$ partitions $\mathcal{P }$ into two convex polygons $\mathcal{P}^{\prime}$ and $\mathcal{P} ^{\prime\prime}$. Let $K,L$ be points on the boundary of $\mathcal{P}$ such that the segment $KL$ is parallel to $ST$ and has length $ST/2$. We claim that the trapezoid $SKLT$ has area at least $3/4$ the area of $\mathcal{P} ^{\prime}$. This yields the result because the same argument, applied to $\mathcal{P}^{\prime\prime}$, produces another trapezoid, of area at least $3/4$ the area of $\mathcal{P}^{\prime\prime}$. Combining the two trapezoids, we obtain a hexagon as needed. Draw the lines of support of $\mathcal{P}$ through $K$ and $L$, and let them intersect at $Z$. We assume for definiteness that $K$ is closer to the line $s$ than $L$. Let $ZK$ meet $s$ at $X$, let $ZL$ meet $t$ at $Y$, and let the line through $Z$, parallel to $ST$, meet $s$ and $t$ at $A$ and $B$, respectively. Clearly $\mathcal{P}^{\prime }$ is contained in the pentagon $SXZYT$, (which may reduce to a triangle or a quadrilateral), so it suffices to show that \[ \lbrack SKLT]\geq \frac{3}{4}[SXZYT].\ \ \ \ \ \mathbf{(1)} \] Denote ${AZ=a}$, ${BZ=b}$, and let the distances from the points $K$, $X$, $Y$, $S$ to the line $AB$ be $k$, $x$, $y$, $s$. Taking $U$, $V$ on $AB$ so that ${s\parallel KU\parallel LV\parallel t}$, we obtain \[ \frac{a+b}{2}=KL=UZ+ZV=\frac{ak}{x}+\frac{bk}{y}.\ \ \ \ \ \mathbf{(2)} \] The areas we are about to compare are expressed by \[ \lbrack SKLT]=\frac{3}{4}(a+b)(s-k), \] \[ \quad \lbrack SXZYT]=[SABT]-[AXZ]-[BYZ]=(a+b)s-\frac{ax}{2}-\frac{by}{2}, \] so the claimed inequality (1) comes down to \[ ax+by-2(a+b)k\geq 0.\ \ \ \ \ \mathbf{(3)} \] On computing $k$ from (2) and plugging into (3), a short calculation shows that the left-hand side of (3) is equal to ${ab(x-y)^{2}(ay+bx)^{-1}}$, which is nonnegative. [Moderator edit: This is the second proposed solution of this problem, avaliable at http://www.mathlinks.ro/Forum/viewtopic.php?t=15622 .]

Let $ ABC$ be the triangle among the vertices of $ P$ with the largest area, and let $ A'B'C'$ be the triangle with $ ABC$ as its medial triangle (oriented canonically so that $ \triangle A'B'C' \sim \triangle ABC$). Note that if a vertex $ V$ of $ P$ lay on the side of $ A'B'$ opposite of $ AB$, then we would have $ [ABV] > [ABC]$, which contradicts the maximality of $ [ABC]$. Applying symmetric arguments, we conclude that $ P$ lies within $ A'B'C'$. Considering only the portion of $ P$ lying within $ A'BC$, let $ V_A$ be the vertex farthest away from $ BC$, and set $ [V_ABC]/[A'BC] = 1 - a$. Defining $ V_B$, $ V_C$, $ b$, and $ c$ similarly, we claim that $ V_ABV_CAV_BC$ works as our desired hexagon. To simplify calculations, we will assume henceforth that $ [ABC] = 1$. Note that the area of our hexagon is $ 1 + 1 - a + 1 - b + 1 - c = 4 - a - b - c$. If the line through $ V_A$ parallel to $ BC$ intersects $ A'B$ and $ A'C$ at $ X$ and $ Y$ respectively, then we know that no part of $ P$ can lie within $ A'XY$, which has area $ a^2 [A'BC] = a^2$. Since the entirety of $ A'B'C'$ has area $ 4$, we then know that $ P$ has area at most $ 4 - a^2 - b^2 - c^2$. It now suffices to show $ 4 - \sum_{cyc} a \ge \frac{3}{4} \left( 4 - \sum_{cyc} a^2 \right)$ $ \iff 1 + \frac{3}{4} \sum_{cyc} a^2 \ge \sum_{cyc} a$, but this is easy to see from AM-GM: $ \frac{1}{3} + \frac{3}{4} \cdot x^2 \ge x$.

Is this correct? For simplicity assume $P$ is a convex figure (like a circle or an oval). It is wellknown result that there exists a triangle $ABC$ inscribed in $P$ of at least half its area. Take the inscribed triangle with maximal area. Now take this triangle, and the tangent lines to $P$ that pass through its vertices.These tangents form a triangle $XYZ$. Now, if $P$ is inscribed inside of $XYZ$, then we can take a point $A'$ on "arc" $BC$ that doesn't contain $A$ such that the area of triangle $A'BC$ is at least half of the area of the region determined by line $BC$ (on the other side than $A$) and $P$. This is because, since the tangents intersect on the other side of $BC$ than $A$, we can take the point $A'$ on $P$ that is furthest apart from $BC$, and if we take the tangent to $P$ through $A'$ (it obviously will be parallel to $BC$) then it forms a quadrilateral together with line $BC$ and the tangents to $P$ through $B$ and $C$. It has parallel sides and completely covers the part of $P$ that is on this side of line $BC$. Since $BC$ is longer than the other parallel side, the existence of $A'$ is guaranteed. Similarly we can take $B'$ and $C'$ and it is easy to see we are finished taking $AC'BA'CB'$ as our hexagon. Otherwise, if triangle $XYZ$ does not cover $P$, then, if we take $B'$ and $C'$ on the other side of $BC$ than $A$, such that $B'C' || BC$ and they are really close to $B$ and $C$, then, because there are points on $P$ arbitrarily close to the tangents through $B$ and through $C$, we can assume $BC < B'C'$ and so triangle $AB'C'$ has a greater area than triangle $ABC$. Contradiction.

In fact, we can improve $ \frac{3}{4} $ to $ \frac{3\sqrt{3}}{2\pi} $, with equality only if our convex polygon was an ellipse. In fact, for any $ n $-gon we can guarantee an optimal constant of $ \frac{n}{2\pi}\sin{\frac{2\pi}{n}} $. To show this, place $ F $, the original convex polygon (which can actually be any convex body), on the Cartesian plane and assume WLOG that $ (-1, 0), (1, 0) \in \partial{F} $. Parametrize $ \partial{F} $ by: \[x(\theta) = \cos{\theta} \] \[y(\theta) = g(\theta)\sin{\theta} \] Where $ \theta $ is the angle made by the line connecting $ (x, y) $ to the origin and the positive $ x $-axis and where $ g $ is a continuous, periodic function with period $ 2\pi $. Now define $ \theta_i = \theta + \frac{2\pi(i - 1)}{n} $ for all integers $ 1 \leq i \leq n $. Define $ A(\theta) $ as the area of the $ n $-gon with vertices at coordinates $ (x(\theta_1), y(\theta_1)), (x(\theta_2), y(\theta_2)), \dots, (x(\theta_n), y(\theta_n)) $. Now let $ A[(a, b), (c, d), (e, f)] $ denote the area of the triangle whose vertices are at coordinates $ (a, b), (c, d), (e, f) $. Letting indices be taken modulo $ n $ we have that: \[ A(\theta) = \sum_{i = 1}^{n}A[(0, 0), (x(\theta_i), y(\theta_i)), (x(\theta_{i + 1}), y(\theta_{i + 1}))] = \] \[ \frac{1}{2}\sum_{i = 1}^{n}(y(\theta_{i + 1})x(\theta_i) - y(\theta_i)x(\theta_{i + 1})) = \] \[\frac{1}{2}\sum_{i = 1}^{n}y(\theta_i)(x(\theta_{i - 1}) - x(\theta_{i + 1})) = \] \[\frac{1}{2}\sum_{i = 1}^{n}g(\theta_i)\sin{\theta_i}(\cos{\theta_{i - 1} - \cos{\theta_{i + 1}) = }}\] \[\sin{\frac{2\pi}{n}}\sum_{i = 1}^{n}g(\theta_i)\sin^2{\theta_i} \] This implies that $ \frac{1}{2\pi}\int_{0}^{2\pi}A(\theta)\, d\theta = \frac{n}{2\pi}\sin{\frac{2\pi}{n}}\int_{0}^{2\pi}g(\theta)\sin^2{\theta}\, d\theta $ so since by the Pigeonhole Principle there exists a $ \phi $ such that $ A(\phi) \geq \frac{1}{2\pi}\int_{0}^{2\pi}A(\theta)\, d\theta $ it suffices to show that $ \int_{0}^{2\pi}g(\theta)\sin^2{\theta}\, d\theta $ is the area of $ F $. But by Green's Theorem this area is equal to $ -\oint_{\partial{F}}y\, dx $ and since $ dx = -\sin{\theta}\, d\theta $ we have that $ -\oint_{\partial{F}}y\, dx = \int_{0}^{2\pi}g(\theta)\sin^2{\theta}\, d\theta $ as desired.

In fact, the result of my previous post implies that to get $ \frac{3}{4} $ of the area of the original polygon, one only need inscribe an optimal pentagon, not a hexagon

Solution from Twitch Solves ISL: We are going to solve the problem when $\mathcal P$ is replaced by a differentiable convex curve. (The proof requires only trivial modifications for a polygon; in any case, polygons can approximate convex curves arbitrarily well and vice versa.) We start by committing to take the longest segment $AB$ as a major diagonal of our hexagon. Therefore, this cuts $\mathcal P$ into two halves and we deal with each half separately. Orient $AB$ on the $x$-axis with $A=(0,0)$ and $B=(1,0)$ and consider the region above the $x$-axis. Let $C$ and $D$ be the points on the curve with $x$-coordinates $\frac14$ and $\frac34$ respectively. Then the lines $x=0$ and $x=1$, together with the tangents at $C$ and $D$, determine a pentagon $AXYZB$ that encloses $\mathcal P$ (since $\mathcal P$ is convex), as shown. [asy][asy] size(8cm); pair A = (-1,0); pair B = (1,0); pair C = (-0.5,0.88); pair D = (0.5,0.95); path p = A..C..(0,1.08)..(0.3,1.1)..D..B; fill(p--cycle, rgb(0.95,1,1)); draw(p, blue+0.8); draw(A--B, deepgreen); pair X = extension(A, A+dir(90), C, C+dir(p,1)); pair Z = extension(B, B+dir(90), D, D+dir(p,4)); pair Y = extension(C,X,D,Z); for (int t=-1; t<=1; ++t) { draw( (-t,1.6)--(-t,-0.2), grey, Arrows(TeXHead)); } label(scale(0.7)*"$0$", (-1,0), dir(-70)); label(scale(0.7)*"$\frac14$", (-0.5,0), dir(-90)); label(scale(0.7)*"$\frac12$", (0,0), dir(-70)); label(scale(0.7)*"$\frac34$", (0.5,0), dir(-90)); label(scale(0.7)*"$1$", (1,0), dir(250)); draw(A--X--Y--Z--B, orange); draw(C--(C.x,0), orange+dashed); draw(D--(D.x,0), orange+dashed); draw(A--C--D--B, deepgreen); pair P = extension(Z,D,(0,0),(0,1)); pair Q = extension(X,C,(0,0),(0,1)); draw(P--Y, orange); dot("$P$", P, dir(45), grey); dot("$Q$", Q, dir(135), grey); dot("$M$", (0,0), dir(45), grey); dot("$X$", X, dir(180), orange); dot("$Y$", Y, dir(60), orange); dot("$Z$", Z, dir(0), orange); dot("$A$", A, dir(225), deepgreen); dot("$B$", B, dir(-45), deepgreen); dot("$C$", C, dir(100), orange); dot("$D$", D, dir(70), orange); label("$h_1$", (C.x,0.4), dir(0), orange); label("$h_2$", (D.x,0.4), dir(180), orange); [/asy][/asy] The claim is that $ACDB$ fits the bill: Claim: We have \[ [ACDB] \ge \frac34 [AXYZB]. \]Proof. Let $h_1$ and $h_2$ be the $y$-coordinates of $C$ and $D$. Also, as depicted, let $x=1/2$ meet $YDZ$ at $P$ and $XCY$ at $Q$, and let $M = (1/2, 0)$. Then \begin{align*} \frac43 \cdot [ACDB] &= \frac43 \left( \frac18 h_1 + \frac14(h_1+h_2) + h_2 \right) = h_1 + h_2 \\ &= [AXQM] + [BZPM] \ge [AXYZB]. \end{align*}Note that equality when $P=Q=Y$. $\blacksquare$ Repeating the same proof for the bottom half of $\mathcal P$ exhibits the desired hexagon.

JuanOrtiz wrote: It is wellknown result that there exists a triangle $ABC$ inscribed in $P$ of at least half its area. Assuming this is true, can we do the following: -take a triangle that is at least half the area of P - from the remaining shape take again at least half the area hence we have at least $\frac{P}{2}+\frac{P}{4}$

DanjelZone wrote: JuanOrtiz wrote: It is wellknown result that there exists a triangle $ABC$ inscribed in $P$ of at least half its area. Assuming this is true, can we do the following: -take a triangle that is at least half the area of P - from the remaining shape take again at least half the area hence we have at least $\frac{P}{2}+\frac{P}{4}$ Sadly this is false, consider a circle and the largest area of a triangle inside it is an equilateral one (area less than half that of the circle). Then approximate the circle via polygons.

why is everyone using calculus Let $ABC$ be the largest triangle whose vertices are among those of $P$, and let $A'B'C'$ be the triangle such that $A$ is midpoint of $B'C'$, $B$ is the midpoint of $C'A'$, and $C$ is the midpoint of $A'B'$. WLOG, let $[ABC]=1$. $~$ If any vertex of $P$, $D$, is outside of $\triangle A'B'C$, then assume it is on the opposite side of $A'C'$ as $AC$. Then, $[ACD]>[ACB]$, contradiction, thus, $P$ is contained in $\triangle A'B'C'$. Let $I$, $J$, $K$ be the vertices of $P$, such that $I,C,J,A,K,B$ appear in that order, and $[BIC]$, $[CJA]$, and $[AKB]$ are maximized. Thus, $d(I,BC)$, $d(J,CA)$, and $d(K,AB)$ are maximized. $~$ Then, if we let the parallel line to $BC$ through $I$ intersect $A'B$ and $A'C$ at $U$ and $V$, and define $W$, $X$, $Y$, and $Z$ similarly, then $[P]\le [UVWXYZ]$. Let \[x=\frac{d(I,BC)}{d(A',BC)}\]\[y=\frac{d(J,CA)}{d(B',CA)}\]\[z=\frac{d(K,AB)}{d(C',AB)}\]and so $[UVWXYZ]=4-[A'UV]-[B'WX]-[C'YZ]=4-(1-x)^2-(1-y)^2-(1-z)^2$. Meanwhile, $[BICJAK]=1+x+y+z$. It suffices to show that \[4-(1-x)^2-(1-y)^2-(1-z)^2\le \frac{4}{3}(1+x+y+z)\]Which rearranges to \[\left(x-\frac13\right)^2+\left(y-\frac13\right)^2+\left(z-\frac13\right)^2\ge 0\]which is obviously true.