Given a convex figure in the Cartesian plane that is symmetric with respect of both axis, we construct a rectangle $A$ inside it with maximum area (over all posible rectangles). Then we enlarge it with center in the center of the rectangle and ratio lamda such that is covers the convex figure. Find the smallest lamda such that it works for all convex figures.
Problem
Source: China TST 1987, problem 4
Tags: geometry, rectangle, ratio, analytic geometry, symmetry, conics, hyperbola
freemind
07.04.2007 16:22
I guess $\lambda=2$ works. It is easy to prove that the maximal rectangle has center $O$ - origin of Coordinate Axes. Indeed, suppose not, then the coordinate axes divide $A$ into 4 rectangles (possibly less than $4$). If $O$ is not the center of rectangle, then these 4 smaller rectangles are not all equal. Take one with largest area , and consider the rectangle $R$ with center in $O$ having this greatest rectangle as one quarter of it. Then $R$ will fit in the figure, but it's area will be greater than the area of $A$. Contradiction. So $O$ is the center of $A$ Now, due to the symmetry of the figure wrt $Ox$ and $Oy$ we can work only in the first quarter of the plane. It is clear that the part of the figure lying in the first quarter must be as well convex. Let $M=(a,b)$ be the point such the rectangle with one vertex $O$ and the opposite one - $(a,b)$ has maximal area $k$. Take any other point $N=(x,y)$ of the figure. Let $l$ be the tangent through $M$ to the hyperbola $Y=\frac{k}{X}$. Suppose $l$ interesects $Ox$ and $Oy$ at $P$ and $Q$ respectively. Then $N$ must lie inside $\triangle OPQ$. Otherwise $NM$ will intersect once again (except $M$) the hyperbola. And since the figure is convex, part of it lies "above" $MN$, that is some point of it will be strictly inside the hyperbola, generating a rectangle of area greater than $k$. But $P$ and $Q$ have coordinates $(2a,0)$ and $(0,2b)$. So $\lambda$ must be at most $2$. Because for a rhombus centered at $O$, $\lambda$ will be exactly $2$ (that is, the border of the figure is exactly $l$ - the tangent to the hyperbola), we have the smallest $\lambda$ is $2$.
srulikbd
01.07.2007 11:38
ns problem! there is some hole in your solution, you need to show that the rectangle with max area have sides parabell to the axes(isn't so difficult). I did in a similar way- because it convex we need to consider this figure: by the way, the china TST is the best one I think.
P-H-David-Clarence
14.10.2016 04:18
srulikbd wrote: ns problem! there is some hole in your solution, you need to show that the rectangle with max area have sides parabell to the axes(isn't so difficult). I did in a similar way- because it convex we need to consider this figure: by the way, the china TST is the best one I think. It's not true! The maximum rectangle don't have to have sides parallel to axis (take a rhombus)