Problem 5: Balkan MO 1996, Problem 3

by henderson, Feb 14, 2016, 4:56 PM

$\bf\color{red}Problem \ 5 \$ In a convex pentagon $ABCDE$ , the points $M$ , $N$ , $P$ , $Q$ , $R$ are the midpoints of the sides $AB$ , $BC$ , $CD$ , $DE$ , $EA$ , respectively. If the segments $AP$ , $BQ$ , $CR$ and $DM$ pass through a single point, prove that $EN$ contains that point as well.
$\bf\color{red}Solution \ 1 \$ Let the four lines concur at $X$ . Let $EX$ meet $BC$ at $N'$ . Let $\angle BXM = \angle DXQ = m$ , $\angle MXA = \angle PXD = n$ , $\angle AXR = \angle CXP = p$ , $\angle RXE = \angle N'XC = q$ , $\angle EXQ = \angle BXN' = r$ .
Let $[ABC]$ denote the area of $\triangle ABC$ . Then,
$[AXM] = [MXB]$ (since they have the same base and height) $\implies \frac {1}{2} BX \cdot MX \cdot \sin m = \frac {1}{2}MX \cdot \sin n$ , or
$BX \sin m = \ AX \sin n$ . Similarly,
$AX \sin p = EX \sin q$
$EX \sin r = DX \sin m$
$DX \sin n = CX \sin p$
Multiplying these,
$BX \sin r = CX \sin q$
Thus, $[BXN'] = [CXN']$ , so $N' = N$ is the midpoint of $BC$ .

So the problem generalizes to any convex $n$ -gon.

$\bf\color{red}Solution \ 2 \$ Here is my solution of the problem. It uses a generalization of the Ceva theorem indeed (by the way, I have never seen it being called Poncelet theorem, but who knows):

Theorem 1. Let n be a positive integer, and let $A_1A_2...A_{2n+1}$ be any (2n + 1)-gon. We work with indices cyclic modulo 2n + 1, so that $A_{2n+1+i}=A_i$ for any i. Let P be a point in the plane; for every index i, let the line $A_iP$ meet the line $A_{n+i}A_{n+1+i}$ at a point $X_i$ . Then,

$\prod_{i=1}^{2n+1}\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=1$ .

Hereby, we work with directed segments.

Proof of Theorem 1. Let's use not only directed segments, but also directed areas, and let's denote the directed area of a triangle $P_1P_2P_3$ by $\left[P_1P_2P_3\right]$ .

Let S and T be the orthogonal projections of the points $A_{n+i}$ and $A_{n+1+i}$ on the line $A_iP$ . Then, the two lines $A_{n+i}S$ and $A_{n+1+i}T$ are parallel to each other (since they are both perpendicular to $A_iP$ ); if we direct these two lines in the same way, then Thales yields $\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\frac{A_{n+i}S}{A_{n+1+i}T}$ . On the other hand, the segments $A_{n+i}S$ and $A_{n+1+i}T$ are the altitudes of the triangles $A_iPA_{n+i}$ and $A_iPA_{n+1+i}$ to the common side $A_iP$ ; since the area of a triangle equals $\pm\frac12\cdot\text{ side }\cdot\text{ corresponding altitude}$ , where the $\pm$ sign is + or - depending on the orientation of the triangle, it follows that $\left[A_iPA_{n+i}\right]=\pm\frac12\cdot A_iP\cdot A_{n+i}S$ and $\left[A_iPA_{n+1+i}\right]=\pm\frac12\cdot A_iP\cdot A_{n+1+i}T$ ; note that the two $\pm$ signs are either both + or both - (since the lines $A_{n+i}S$ and $A_{n+1+i}T$ are directed in the same way). Hence, $\frac{\left[A_iPA_{n+i}\right]}{\left[A_iPA_{n+1+i}\right]}=\frac{A_{n+i}S}{A_{n+1+i}T}$ . Comparing this with $\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\frac{A_{n+i}S}{A_{n+1+i}T}$ , we obtain $\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\frac{\left[A_iPA_{n+i}\right]}{\left[A_iPA_{n+1+i}\right]}$ . Thus,

$\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=-\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=-\frac{\left[A_iPA_{n+i}\right]}{\left[A_iPA_{n+1+i}\right]}=\frac{\left[A_iPA_{n+i}\right]}{\left[A_{n+1+i}PA_i\right]}=\frac{\left[A_iPA_{n+i}\right]}{\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]}$ ,

where in the last step, we have used the fact that $A_{n+\left(n+1+i\right)}=A_{2n+1+i}=A_{i}$ (since indices are cyclic modulo 2n + 1). This identity holds for every index i; multiplying all such identities, we get

$\prod_{i=1}^{2n+1}\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=\prod_{i=1}^{2n+1}\frac{\left[A_iPA_{n+i}\right]}{\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]}=\prod_{i=1}^{2n+1}\left[A_iPA_{n+i}\right]\left/\prod_{i=1}^{2n+1}\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]\right.$ .

But actually, $\prod_{i=1}^{2n+1}\left[A_iPA_{n+i}\right]$ and $\prod_{i=1}^{2n+1}\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]$ are the same product, just with a different order of factors. Hence,

$\prod_{i=1}^{2n+1}\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=\prod_{i=1}^{2n+1}\left[A_iPA_{n+i}\right]\left/\prod_{i=1}^{2n+1}\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]\right.=1$ ,

and Theorem 1 is proven.

Now, let's reformulate the problem (note that the pentagon does not need to be convex):

Theorem 2. Let $A_1A_2A_3A_4A_5$ be an arbitrary pentagon, and let $M_1$ , $M_2$ , $M_3$ , $M_4$ , $M_5$ be the midpoints of its sides $A_3A_4$ , $A_4A_5$ , $A_5A_1$ , $A_1A_2$ , $A_2A_3$ , respectively. If the lines $A_1M_1$ , $A_2M_2$ , $A_3M_3$ and $A_4M_4$ have a common point O, then the line $A_5M_5$ also passes through this point O.

Proof of Theorem 2. Since the lines $A_1M_1$ , $A_2M_2$ , $A_3M_3$ and $A_4M_4$ pass through the point O, the points $M_1$ , $M_2$ , $M_3$ , $M_4$ are the points where the lines $A_1O$ , $A_2O$ , $A_3O$ , $A_4O$ meet the lines $A_3A_4$ , $A_4A_5$ , $A_5A_1$ , $A_1A_2$ , respectively. Let the line $A_5O$ meet the line $A_2A_3$ at a point $N_5$ . Then, by Theorem 1, we have

$\frac{A_3M_1}{M_1A_4}\cdot\frac{A_4M_2}{M_2A_5}\cdot\frac{A_5M_3}{M_3A_1}\cdot\frac{A_1M_4}{M_4A_2}\cdot\frac{A_2N_5}{N_5A_3}=1$ .

But since $M_1$ , $M_2$ , $M_3$ , $M_4$ are the midpoints of the sides $A_3A_4$ , $A_4A_5$ , $A_5A_1$ , $A_1A_2$ , we have $\frac{A_3M_1}{M_1A_4}=1$ , $\frac{A_4M_2}{M_2A_5}=1$ , $\frac{A_5M_3}{M_3A_1}=1$ , $\frac{A_1M_4}{M_4A_2}=1$ . Thus, we also have $\frac{A_2N_5}{N_5A_3}=1$ , so the point $N_5$ must be the midpoint of the side $A_2A_3$ . Thus, $N_5=M_5$ . Now, from the definition of the point $N_5$ , it is clear that the line $A_5N_5$ passes through the point O; hence, the line $A_5M_5$ passes through the point O, and Theorem 2 is proven.

Darij

This post has been edited 5 times. Last edited by henderson, Sep 12, 2016, 2:42 PM

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Problem 5: Balkan MO 1996, Problem 3

by henderson, Feb 14, 2016, 4:56 PM

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