a convex hexagon "inscribed" in a polygon with area 3/4

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a convex hexagon "inscribed" in a polygon with area 3/4

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Source: mock tst romania 2004

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Valentin Vornicu

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Valentin Vornicu

#1 h Dec 31, 2004, 10:33 AM • 4 Y

Y by megarnie, Adventure10, Mango247, and 1 other user

Let $P$ be a convex polygon. Prove that there exists a convex hexagon that is contained in $P$ and whose area is at least $\frac34$ of the area of the polygon $P$ .

Alternative version. Let $P$ be a convex polygon with $n\geq 6$ vertices. Prove that there exists a convex hexagon with

a) vertices on the sides of the polygon (or)
b) vertices among the vertices of the polygon

such that the area of the hexagon is at least $\frac{3}{4}$ of the area of the polygon.

Proposed by Ben Green and Edward Crane, United Kingdom

This post has been edited 1 time. Last edited by djmathman, Aug 1, 2015, 2:53 AM
Reason: removed remarks

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Myth

#2 h Dec 31, 2004, 1:12 PM • 2 Y

Y by Adventure10, Mango247

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.

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Myth

#3 h Dec 31, 2004, 9:05 PM • 2 Y

Y by Adventure10, Mango247

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.

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Valentin Vornicu

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Valentin Vornicu

#4 h Jan 1, 2005, 5:36 AM • 2 Y

Y by Adventure10, Mango247

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.

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#5 h Oct 29, 2005, 10:15 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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 .]

This post has been edited 1 time. Last edited by silouan, Oct 29, 2005, 10:35 AM

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probability1.01

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probability1.01

#6 h Jul 6, 2008, 3:19 AM • 2 Y

Y by mijail, Adventure10

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: $\frac{1}{3} + \frac{3}{4} \cdot x^2 \ge x$ .

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#7 h Jun 15, 2014, 6:13 PM • 1 Y

Y by Adventure10

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.

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#8 h Aug 7, 2014, 8:28 PM • 6 Y

Y by huricane, Pluto1708, yayups, CyclicISLscelesTrapezoid, Adventure10, Mango247

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.

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#9 h Aug 7, 2014, 8:39 PM • 1 Y

Y by Adventure10

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

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#10 h Jul 3, 2020, 6:06 PM • 4 Y

Y by Imayormaynotknowcalculus, Mathematicsislovely, v4913, CyclicISLscelesTrapezoid

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] 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]$
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.

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#11 h Nov 1, 2020, 9:56 AM

Y by

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 $\frac{P}{2}+\frac{P}{4}$

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IMD2

#12 h Sep 1, 2023, 6:25 PM

Y by

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.

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awesomeming327.

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awesomeming327.

#13 h Sep 2, 2023, 12:34 AM

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:help:

why is everyone using calculus

https://cdn.aops.com/images/9/2/9/9297b284a27c2a40de3154ade4353b1d4f169990.png

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.

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