Suppose two convex quadrangles in the plane $P$ and $P'$, share a point $O$ such that, for every line $l$ trough $O$, the segment along which $l$ and $P$ meet is longer then the segment along which $l$ and $P'$ meet. Is it possible that the ratio of the area of $P'$ to the area of $P$ is greater then $1.9$?
Problem
Source:
Tags: ratio, geometry, geometric transformation, reflection, integration, function, homothety
dinoboy
06.03.2013 19:46
dr_Civot wrote: Suppose two convex quadrangles in the plane $P$ and $P'$, share a point $O$ such that, for every line $l$ trough $O$, the segment along which $l$ and $P$ meet is longer then the segment along which $l$ and $P'$ meet. Is it possible that the ratio of the area of $P'$ to the area of $P$ be greater then $1.9$?
It can be shown that ratio of the area of $P'$ to the area of $P$ is always less then $2$.
First, note that it suffices to find a polygon $P'$ whose area is double of a quadrilateral $P$ which contains the point $O$ and that the segment statement things in the problem are equal (OK it doesn't exactly suffice, but to find one of these is the driving intuition). Let $P = ABCD$ and set $O = AC \cap BD$. The intuition behind setting this as $O$ is because its one of the few actually useful points for arbitrary quadrilaterals (or maybe I just suck at combinatorial geometry), and it makes measuring the segments a lot easier since we can do some sort of reflection. Reflect $O$ about $A,B,C$ to get $A',B',C'$. Then $[A'B'C'] = 4 [ABC]$. So if $[ABC] = [CDA]$ we have $[A'B'C'] = 2[P]$ which is good. The motivation for defining these reflections is that it gives a natural way to have the segments equal. Now note that if $P$ is symmetric about $AC$ (as in reflected across $AC$ and then $AC$'s perpendicular bisector) we have then the equality condition of the lengths hold (note I could have just set $P$ as a square, but I feel like it would be more instructive to walk through my intuition and work). Now to finish the problem just take $A'DOE$ where $D$ lies on $A'B'$ and $E$ lies on $A'C'$ that $D$ is very close to $B'$ and $E$ very close to $A'C'$ to get the desired result. since as $A'DOE$ is strictly smaller than $A'B'C'$ the segment length condition holds.
JuanOrtiz
18.01.2015 02:20
Here is my solution together with how I reached it. Let's rename $Q=P'$. Reasoning:
First, we have to relate the area of the polygon to the lengths of the segments that pass through $O$. Let's work more generally. Take any region $R$ and any point $O$ on its boundary. As $\alpha$ ranges from $0$ to $\pi$, let $l_{\alpha}$ be the ray that passes through $O$ with angle $\alpha$ w.r.t the x-axis, and let $f(\alpha)$ be the length of the segment $OX$, where $X$ is the intersection point of the boundary of $R$ with $l_{\alpha}$. This sounds like integration. Indeed, if we choose $0 = \alpha_0 < \ldots < \alpha_n = \pi$ then we will have that
$(R) \sim \displaystyle\sum_{i=1}^n \left( f(\alpha_i)^2 \right) \left( \displaystyle\frac{\alpha_i - \alpha_{i-1}}{2} \right) $,
which approximates the area of $R$ as a sum of sectors of disks. Therefore
$(R) = \left( \displaystyle\int_0^{\pi} f(\alpha)^2 d\alpha \right) \left( \displaystyle\frac{1}{2} \right)$.
Therefore, say $P = P_1 \cup P_2$ and $Q=Q_1 \cup Q_2$ are partitions of both polygons using the same line $l$, which passes through $O$ and is parallel to the x-axis. Let $f,g,h,k$ be the functions for $R=P_1, P_2, Q_1, Q_2$, respectively. Then
$(P) = (P_1)+(P_2) = \left( \displaystyle\int_0^{\pi} f(\alpha)^2 d\alpha \right) \left( \displaystyle\frac{1}{2} \right) + \left( \displaystyle\int_0^{\pi} g(\alpha)^2 d\alpha \right) \left( \displaystyle\frac{1}{2} \right) = \left( \displaystyle\int_0^{\pi} f(\alpha)^2+g(\alpha)^2 d\alpha \right) \left( \displaystyle\frac{1}{2} \right)$.
Similarly,
$(Q) = \left( \displaystyle\int_0^{\pi} h(\alpha)^2+k(\alpha)^2 d\alpha \right) \left( \displaystyle\frac{1}{2} \right)$.
Moreover, we have $f+g < h+k$, and we want $(P) \sim 2(Q)$.
Now, given $a+b=c$, the minimum $a^2+b^2$ is achieved when $a=b=c/2$ and the maximum (which equals two times the minimum) is achieved when $a=c$, $b=0$. Since we want $(P) \sim 2(Q)$, we want $f(\alpha)^2+g(\alpha)^2 \sim 2(h(\alpha)^2+k(\alpha)^2)$ because of our formula. Taking the previous result into account, we want $f(\alpha) \sim c$, $g(\alpha)=0$, $h(\alpha)=k(\alpha)=c/2$.
This would give that $P$ is a homothety (with radius 2) of the upper part of $Q$. With this in mind, we choose the easiest $O$, namely $(0,0)$, the easiest $Q$, namely $(0,1)(1,0)(0,-1)(-1,0)$, and $P$ will be the triangle $(-2,0)(2,0)(0,2)$. Slightly modifying this, we get the result.
Result: Choose $\epsilon_1, \epsilon_2$ sufficiently small but positive numbers.
$O=(0,0)$
$Q=(0,1)(1,0)(0,-1)(-1,0)$
$P=(0,0)(2-\epsilon_1,0)(0, 2-\epsilon_1)(-2+\epsilon_1+\epsilon_2, \epsilon_2)$
rkm0959
17.02.2016 17:38
Yes. In fact, we can make one for all $2-\epsilon$, where $\epsilon > 0$. Let $P$ be a square, and let $O$ be the center of $P$. Extend half of $P$ by ratio of $2-\frac{\epsilon}{2}$ to make $P'$. Then we have $|P'|=\frac{1}{2} \cdot (2-\frac{\epsilon}{2})^2 \cdot |P| = (2-\epsilon+\frac{\epsilon^2}{8}) \cdot |P| > (2-\epsilon) \cdot |P|$, as desired.
IAmTheHazard
29.01.2024 04:58
Yes. Let $P$ be a square $ABCD$, and $O$ be its center. Pick points $A' \in \overline{BA}$ (the segment) and $C' \in \overline{BC}$ with $AA'=CC'=\varepsilon_1$. Then make $P'$ the image of $OA'BC'$ under a homothety of scale factor $2-\varepsilon_2$. For small enough $\varepsilon_1,\varepsilon_2>0$ this makes $[P']>1.9[P]$. Any choice of $\ell$ intersects $P$ at two symmetric points $T_1$ and $T_1'$ such that $T_1$ lies on polyline $A'BC'$, and intersects $P'$ at $O$ and $T_2$. Then $\frac{T_1T_1'}{OT_2}=\frac{2OT_1}{OT_2}=\frac{2}{2-\varepsilon_2}>1$. $\blacksquare$