A convex hexagon is given. Let $ s$ be the sum of the lengths of the three segments connecting the midpoints of its opposite sides. Prove that there is a point in the hexagon such that the sum of its distances to the lines containing the sides of the hexagon does not exceed $ s.$ Author: N. Sedrakyan
Problem
Source: Tuymaada 2008, Senior League, Second Day, Problem 8.
Tags: analytic geometry, inequalities, geometry unsolved, geometry
18.07.2008 15:35
If the lines connecting the midpoints of the opposite side-segments of the given hexagon, are concurrent at one point, then their concurrency point has the property as the problem states. A configuration of this particular case, may be constructed by three similar isosceles triangles, as the Kyriazis problem states, in the topic A difficult concurrency. In the general case of three distinct points $ X,\ Y,\ Z,$ of intersection of the above three midlines, taken per two of them, these points have the property as the problem states too. In the schema t=215641, we can see a ''constructive'' solution without words, which may be used also in the particular case. It is easy to show that all the points inwardly to the triangle $ \bigtriangleup XYZ,$ including the points of its side-segments, have the property as the problem states, by the argument of $ (d_{3} + d_{6}) < M'Q' < MQ,$ where $ MQ'\parallel MQ.$ Kostas Vittas.
Attachments:
t=215641.pdf (9kb)
24.07.2008 10:35
1 point
17.01.2014 21:23
I've got to say that I am not entirely satisfied with the following solution; at the end, I resort to some rather messy computations to solve a fact that seems solvable by purely geometric means (I know a solution to the problem which asks to prove that if a hexagon has parallel opposite sides then the lines that connect midpoints of opposite sides are concurrent, which is a particular case of the more general result I proved). I would be really interested if anyone could point a geometric approach to the final part. Till then, here is my solution: We first work with the line $MQ$; the exact same considerations can be then carried out for the other similar lines. Suppose that $AB$ and $ED$ intersect at $X$ (possibly an infinite point) and let $H_{MQ}$ be the semiplane determine by $MQ$ that contains $X$ (if $X$ is an infinite point we choose the semiplane arbitrarely). Let $L$ be a point that is both inside of the hexagon and of $H_{MQ}$. We assert that the sum of distances from $L$ to the lines $AB$ and $ED$ is at most equal to $MQ$: let $L_1$ and $L_2$ be the intersections of the parallels from $L$ to the lines $ED$ and $AB$, respectively (see figure 1.1). Because of the semiplane in which $L$ lies (and the fact that it lies inside the hexagon), the order of the points on line $MQ$ is $M-L_2-L_1-Q$. This means that $MQ\geq ML_2+QL_1$. But $QL_1$ is at not less than the distance from $L$ to $ED$ (because this distance is equal to the distance from $L_1$ to $ED$ and in a right triangle, the longest side is the hypotenuse). Similarly, we have that the distance from $L$ to $AB$ is a t most equal to $ML_2$ and we have thus proved that $MQ$ is at least equal to the sum of distances from $L$ to the sides $AB$ and $ED$. As we will use them further, we will define (in a similar way to the one we defined $H_{MQ}$) $H_{NR}$ and $H_{PS}$. It is rather easy to see that each of the semiplanes defined contains three vertices. We may suppose that no vertex belongs to all three semiplanes, because otherwise the problem would immediately result from the fact proved in the preceeding paragraph. We may therefore suppose that $A,F,E\in H_{MQ}$, $E,D,C\in H_{NR}$ and $A,B,C\in H_{PS}$ (the other possible case, $B,C,D\in H_{MQ}$ and so on is proved similarly; in any other case we may find a vertex that belongs to all three semiplanes). We will prove that the three semiplanes have a point in common, from which the problem results by invoking the fact proved earlier. First, we will prove that we may consider that at least two pairs from $(AB,DE)$, $(BC,EF)$ and $(CD,FA)$ consist of parallel lines. Suppose that this is not the case, and let's suppose that neither $AB$ and $DE$, nor $BC$ and $FE$ are parallel. To prove that the three semiplanes have a point in common, it is enough to prove that if $MQ\cap NR=\{X\}$ and $MQ\cap PS=\{Y\}$ then the order of the points on $MQ$ is $M-X-Y-Q$ (we have used the fact that $A,F,E\in H_{MQ}$). Let $B'$ be in the interior of the heaxagon so that $BB'||MQ$ and at least one of the pairs of sides $(AB',ED)$ and $(CB',FE)$ consists of parallel sides, while the semiplane defined by the other still "points" in the same direction as the semiplane defined by the corresponding line in $ABCDEF$ (or, in other words, $H_{M'Q'}$, $H_{N'R'}$ and $H_{P'S'}$ have no vertex in common; see figure 1.2). If $B'$ happens to be in the interior of the penatgon $ACDEF$, we replace it with a point $B'$ so that $BB'||MQ$ and $B'\in AC$. We shall study this case, in which $B'\in AC$, first. We have that $A,F,E\in H_{M'Q'}$ and $C,D,E,\in H_{N'R'}$. Let $\{Z\}=ED\cap AB$. Then $A$ lies between $Z$ and $C$. Because $ACDEF$ is convex, $A$ also lies between $C$ and $U$, the intersection of $EF$ and $AC$ (see figure 1.3). Because $B'\in AC$, this means that $A,F,B'\in H_{N'R}$, a contradiction. We may therefore suppose that $B'$ does not lie in the interior of $ACDEF$ and so $AB'CDEF$ is a convex hexagon. Because of the definition of $B'$ we have that $M'\in MQ$ and therefore $Y$ remains unchanged. Because $\overrightarrow{NN'}=1/2\overrightarrow{BB'}$ and $\overrightarrow{BB'}$ and $\overrightarrow{MQ}$ have the same \textit{direction}, $\overrightarrow{NN'}$ and $\overrightarrow{MQ}$ also have the same direction. Therefore, as both $X$ and $X'$ are in the interior of $MQ$, we have that $X\in (MX')$. This means that if $Y\in{MX'}$, then $Y\in {MX}$. As such, if we manage to prove the problem for $AB'CDEF$ the problem for $ABCDEF$ immediately follows. We may therefore suppose that two of the pairs $(AB,DE)$, $(BC,EF)$ and $(CD,FA)$ consist of parallel lines. Without loss of generality, let $(AB,ED)$ and $(BC,EF)$ be those. Although now we may theoretically choose $H_{MQ}$ as we wish, the transformations we made in the previous paragraphs force us to take it as the one containing $A,F$ and $E$ (similarly for $H_{NR}$). Let $\{T\}=AB\cap CD$, $\{V\}=CD\cap FE$ and $\{W\}=AB\cap EF$ (see figure 1.4). To make computations easier, we shall make an affine transformation that sends $\Delta TWV$ into an isosceles right triangle with sidelenght $1$ and right angle at $W$ (we are allowed to perform such a tranformation maintains parallelism and and midpoints). Let now the coordinates of $W$ be $(0,0)$, of $V=(1,0)$, $T=(0,1)$, $A=(0,x)$, $F=(x_1,0)$, $B=(0,x_2)$ and $E=(x_3,0)$ (see figure 1.5). Because of the parallelisms we have (and the fact that $A,B,C\in H_{PS}$) we have that $C=(1-x_2,x_2)$, $D=(x_3,1-x_3)$ and $x\geq x_1$. As before, let $\{X\}=MQ\cap NR$ and $\{Y\}=PS\cap NR$. We shall now find the coordinates of $X$ and $Y$. Let the coordinates of $X$ be $a(\frac{x_1+x_3}{2},0)+(1-a)(\frac{1-x_2}{2},x_2)=b(0,\frac{x+x+2}{2})+(1-b)(x_3,\frac{1-x_3}{2})$ and those of $Y$ be $c(\frac{x_1+x_3}{2},0)+(1-c)(\frac{1-x_2}{2},x_2)=d(\frac{x_1}{2},\frac{x}{2})+(1-d)(\frac{1+x_3-x_2}{2},\frac{1+x_2-x_3}{2})$. All we have to prove is that $c\geq a$. By solving the systems of equations we get that \[c\left(\frac{x_1+x_3+x_2-1}{2}\cdot\frac{1+x_2-x-x_1}{2}+x_2\cdot\frac{x_3+1-x_1-x_2}{2}\right)=\] \[\frac{x_3}{2}\cdot\frac{1+x_2-x_3-x}{2}+\frac{x_3+x_2-1}{2}\cdot\frac{1+x_3-x_1-x_2}{2}\] and \[a\left(x_2x_3-\frac{x_1+x_2+x_3-1}{2}\cdot\frac{x+x_2+x_3-1}{2}\right)=\] \[\frac{1-x_2}{2}\cdot\frac{x+x_2+x_3-1}{2}+x_3\cdot\frac{x_2-x}{2}.\] By using the fact that the hexagon is convex we get the following inequalities: $x_2\geq x(\geq x_1)$, $x_3 \geq x_1$ and $x_1+x_2\geq 1$. From these inequalities it results that all the brackets written above are positive (actually, nonnegative; however, we may suppose that in the inequalities written earlier, equality does not hold, as this case follows by continuity from the other). As \[\frac{x_1+x_2+x_3-1}{2}\cdot\frac{1+x_2-x-x_3}{2}+x_2\cdot\frac{x_3+1-x_1-x_2}{2}=\] \[x_2x_3-\frac{x_1+x_2+x_3-1}{2}\cdot\frac{x+x_2+x_3-1}{2}\] all we have to prove is that \[\frac{x_3}{2}\cdot\frac{1+x_2-x-x_3}{2}+\frac{x_3+x_2-1}{2}\cdot\frac{1+x_3-x_1-x_2}{2}\geq\] \[\frac{1-x_2}{2}\cdot\frac{x+x_2+x_3-1}{2}+x_3\cdot\frac{x_2-x}{2}\] which is actually equivalent to $(x-x_1)(x_2+x_3-1)\geq 0$ which follows from the inequalities written at the beginning of this paragraph. The problem is now solved, as we have proved that the three semiplanes $H_{MQ}$, $H_{NR}$ and $H_{PS}$ have a point in common, which then has the property requested in the problem.