# Problem A. 806. (October 2021)

**A. 806.** Four distinct lines are given in the plane, which are not concurrent and no three of which are parallel. Prove that it is possible to find four points in the plane, \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) and \(\displaystyle D\) with the following properties:

\(\displaystyle (i)\) \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) and \(\displaystyle D\) are collinear in this order;

\(\displaystyle (ii)\) \(\displaystyle AB=BC=CD\);

\(\displaystyle (iii)\) with an appropriate order of the four given lines \(\displaystyle A\) is on the first, \(\displaystyle B\) is on the second, \(\displaystyle C\) is on the third and \(\displaystyle D\) is on the fourth line.

Proposed by *Kada Williams,* Cambridge

(7 pont)

**Deadline expired on November 10, 2021.**

**Solution.** Note that after an affine transformation conditions (i) and (ii) are still satisfied by points \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) and \(\displaystyle D\).

If three of the lines form a triangle, then it is well known that after applying an appropriate affine transformation it can be achieved that the lines become \(\displaystyle y=0\), \(\displaystyle x=0\) and \(\displaystyle x+y=1\). Let's find triplets of points \(\displaystyle P\), \(\displaystyle Q\) and \(\displaystyle R\) such that \(\displaystyle P\) is on \(\displaystyle x=0\), \(\displaystyle Q\) is on \(\displaystyle x+y=1\), \(\displaystyle R\) is on \(\displaystyle y=0\), and \(\displaystyle Q\) is the midpoint of line segment \(\displaystyle PR\).

Let \(\displaystyle Q\) be \(\displaystyle (t,1-t)\) (where \(\displaystyle t\) takes all the real values). Let's reflect \(\displaystyle y=0\) across this point to get the coordiantes of \(\displaystyle P\): \(\displaystyle (0,2-2t)\). Now it is easy to get the coordinates of \(\displaystyle R\): \(\displaystyle (2t,0)\).

Let's define \(\displaystyle A=P\), \(\displaystyle B=Q\) and \(\displaystyle C=R\), and let's find point \(\displaystyle D\) such that conditions (i) and (ii) are satisfied. Since \(\displaystyle C\) is the midpoint of line segment \(\displaystyle BD\), the coordinates of \(\displaystyle D\) are \(\displaystyle (3t,t-1)\). This means that if \(\displaystyle t\) suns through all real values (i.e. \(\displaystyle Q=B\) runs through line \(\displaystyle x+y=1\)), then point \(\displaystyle D\) will run through line \(\displaystyle x-3y=3\).

Similarly let \(\displaystyle A=R\), \(\displaystyle B=Q\) and \(\displaystyle C=P\). A similar argument shows that the coordinates of \(\displaystyle D\) will be \(\displaystyle (-t,3-3t)\), thus in this case \(\displaystyle D\) will run through line \(\displaystyle y-3x=3\).

Since lines \(\displaystyle x-3y=3\) and \(\displaystyle y-3x=3\) are not parallel to each other, one of them must intersect the fourth line, and choosing \(\displaystyle D\) to be the point of intersection we will get a solution to the problem.

If three among the four given lines pass through the same point, then the fourth won't pass through this point, and it can be parallel only with one of these three lines, so it will form a triangle with the other two lines, which makes the argument above work.

If no three lines pass through the same point, and we cannot find three lines that form a triangle, then the four lines can have only two directions. And since there are no three parallel among the four lines, thus we have to have two lines parallel to each direction, so the four lines must form a parallelogram. This case is really easy to solve: if the parallelogram is \(\displaystyle XYZW\), then let \(\displaystyle A\) be reflection of \(\displaystyle W\) across \(\displaystyle X\), \(\displaystyle B\) be the point of line segment \(\displaystyle XY\) such that \(\displaystyle XB:BX=1:2\), \(\displaystyle C\) be the point of line segment \(\displaystyle ZW\) such that \(\displaystyle ZC:CW=1:2\) and \(\displaystyle D\) be the reflection of \(\displaystyle Y\) across \(\displaystyle Z\).

### Statistics:

13 students sent a solution. 7 points: Molnár-Szabó Vilmos, Simon László Bence, Sztranyák Gabriella, Varga Boldizsár. 6 points: Bán-Szabó Áron, Diaconescu Tashi. 5 points: 1 student. 3 points: 2 students. 2 points: 2 students. 1 point: 1 student. 0 point: 1 student.

Problems in Mathematics of KöMaL, October 2021