MENELAOS' THEOREM: the affine version

 

          Given a triangle $ABC$ and a straight line $s$ that does not pass through any of its vertices.

The line $s$ should intersect the lines determined by the sides of the triangle at the points $M,\;N,\;P$ as in the figure. (According to the position of the line $s$ in front of

the triangle, we can have one or all three of these intersection points on the extensions of the sides of the triangle.)

We will consider some vectors associated with this configuration : $\overrightarrow{AP},\;\overrightarrow {BM},\;\overrightarrow{CN}.$

          The points $P,\;A,\;B\;$ being collinear, and so are $B,\;M,\;C\;$ as well as $A,\;C,\;N\;$, we can find some scalars $\alpha,\;\beta,\;\gamma\;$ so that (see "Barycentric coordinate on the Straight Line"):

$$\color {Red} {\overrightarrow{AP}=\gamma \cdot \overrightarrow{AB}},\;\;\overrightarrow{BP}=(\gamma-1)\overrightarrow{AB},\;\;\frac{\overline{PA}}{\overline{PB}}=\frac{\gamma}{\gamma-1} \tag{1}$$

$$\color {Red}{\overrightarrow{BM}=\alpha \cdot \overrightarrow{BC}},\;\;\overrightarrow{CM}=(\alpha-1)\overrightarrow{BC},\;\;\frac{\overline{MB}}{\overline{MC}}=\frac{\alpha}{\alpha-1} \tag{2}$$

$$\overrightarrow{CN}=\beta \cdot \overrightarrow {CA},\;\;\overrightarrow{AN}=(\beta-1)\overrightarrow{CA},\;\;\frac{\overline{NC}}{\overline{NA}}=\frac{\beta}{\beta-1} \tag{3}$$

(will continue)

My books from the ELECTRONIC LIBRARY

Scan the following code

One, two, three, four // Come in, please, and shut the door

          It is a  fun poem for children. You can learn English and Mathematics at the same time.

          The following example contains ONLY math... And some gold!

$$\arctan \frac{1}{2}+\arctan \frac{3}{4}=2\arctan \frac{1}{\varphi}$$

Enjoy!

          

Revista de Matematică a Elevilor și Profesorilor din Județul Caraș-Severin (RMCS) // Mathematics Magazine of Students and Teachers from Caraș-Severin County

          I found an interesting problem, for class 5, in RMCS_39--2012.

In translation

             "V. 241   Consider the set $M=\{1,\;8,\;15,\;22,\;29, \dots ,\;134\}$. Show that any 12-element subset of the set M contains two elements whose sum is equal to 142.

Bihor County Olympiad"

              I solve this problem with the help of the Pigeonhole principle. A post on AOPS also helped me.

             First note that since $1=7\cdot 0+1,\;8=7\cdot 1+1,\;15=7\cdot 2+1,\dots ,\;134=7\cdot 19+1$, the set M has 20 elements. Let's arrange the elements of the set M on 10 columns like this

$$\begin{array}{|r|r|}\hline 8&15&22&29&36&43&50&57&64&1 \\ \hline  134&127&120&113&106&99&92&85&78&71 \\ \hline \end{array}$$

The sum of the elements in the first nine columns is 142. Let us choose some set of 12 elements from the elements of the set M. Let's assume that, in the worst case, we also chose elements 1 and 71 (from the last column). However, 12-2=10 elements remain. With only nine columns available, two of the ten elements must be in the same column, and they will have the sum 142.

$\blacksquare$

          Remark CiP  There are sets with 11 elements that do not satisfy the requirement of the problem. For example

$$\{1,\;8,\;15,\;22,\;29,\;36,\;43,\;50,\;57,\;64,\;70\}.$$

          L.E. You can also consult the solution from RMCS here, at page 19.

                   Other issues of the RMCS can be found here.