Let $\{R_i\}_{1\leq i\leq n}$ be a family of disjoint closed rectangular surfaces with total area 4 such that their projections of the $Ox$ axis is an interval. Prove that there exist a triangle with vertices in $\displaystyle \bigcup_{i=1}^n R_i$ which has an area of at least 1. [Thanks Grobber for the correction]
Problem
Source: Romanian ROM TST 2004, problem 2, created by Dan Ismailescu
Tags: geometry, rectangle, inequalities, analytic geometry, geometry proposed
psykyk
02.05.2004 16:35
hi! just a question about problem 2, Can the rectangles have points in common? and in that case are the common areas considered once for the total area? psykyk p.s. sorry for posting in the wrong place the last occasion
grobber
02.05.2004 20:12
If I remember correctly, the closed rectangular surfaces had to be disjoint. I think this was mentioned in the text.
psykyk
08.05.2004 05:27
another question my friend, and thx for clearing my previous doubts. First the rectangles must be disjoint, and can their projections onto the Ox axis be not disjoint? Thank you! psykyk
psykyk
08.05.2004 05:52
i am deadly confused with this problem !!! It seems that the interval condition is not at all needed... after being puzzled trying to figure out how could i use this condition i found the following: sinc U Rn (the union)is closed, there is among all triangles with vertices in this region one with maximum area.... then all points are within a triangle that has the first one as median traingle, and it's area is at least 4, so the original one has area at least 1, i'm not sure but with this argument even a strict inequality can be proved.
Hawk Tiger
20.05.2006 10:41
psykyk wrote: i am deadly confused with this problem !!! It seems that the interval condition is not at all needed... after being puzzled trying to figure out how could i use this condition i found the following: sinc U Rn (the union)is closed, there is among all triangles with vertices in this region one with maximum area.... then all points are within a triangle that has the first one as median traingle, and it's area is at least 4, so the original one has area at least 1, i'm not sure but with this argument even a strict inequality can be proved. i really agree with you .i think if there exists a set of points with area 4,then exist three points $A,B,C$ in the set such that the area of triangle $ABC$ is at least 1
enescu
20.05.2006 23:31
The corect statement of the problem ends with: Prove that there exist a triangle with vertices in $\displaystyle \bigcup_{i=1}^n R_i$ which has the area equal to 1.
SurrealisticStranger
27.08.2013 11:57
Let $[a, b]$ be the projection of $ \bigcup_{i=1}^n R_i $ onto $Ox$. For $x \in [a, b],$ define $m(x) = \text{min}\{y|(x, y) \in \bigcup_{i=1}^n R_i\}$ and $M(x) = \text{max}\{y|(x, y) \in \bigcup_{i=1}^n R_i\}.$ Since $\bigcup_{i=1}^n R_i$ is included in $\{(x, y)|x \in [a, b], m(x) \le y \le M(x)\}$ and has area of at least $4$, it follows that there exists $x_0 \in [a, b]$ such that \[(b-a)(M(x_0)-m(x_0)) \ge 4.\] Consider the points $A$ and $B$ having coordinates $(x_0, m(x_0))$ and $(x_0, M(x_0))$ respectively, and suppose, for instance, $x_0 \le {{a+b}\over2}.$ Then, for \[x_1 = x_0 + {2\over{M(x_0) - m(x_0)}}\] we have $x_0 < x_1 \le x_0 + {{b-a}\over2} \le b,$ therefore we can find a point $C \in \bigcup_{i=1}^n R_i$ having coordinates $(x_1, y_1)$ such that $\text{area}(ABC) = 1$.