Problem 1237

I. Solution: Let the intersection of $A M$ and $N D$ be $F$. According to the note attached to the problem - taking into account the direction of the segments - the value of the ratio to be examined, regardless of whether $F$ is on the segment $A L$ or outside it, is:

$$\frac{A L}{L M} = \frac{A F}{L M} + \frac{F L}{L M}$$

The ratio in the second term on the right-hand side, conventionally denoted by $(F M L) \sqrt{1} - D N$ and the parallelism of $B C$, can be replaced by the equal ratio $(D E L)$, and similarly, by the parallelism of $N E \| A M$, by the ratio $(D N F)$, and finally, by the parallelism of $D N \| B C$, by the ratio $(B C M)$, whose value is given:

$$(F M L) = \frac{F L}{L M} = \frac{D L}{L E} = \frac{D F}{F N} = \frac{B M}{M C} = (B C M) = m$$

(any two consecutive triples of points have their first, second, and third points corresponding to each other).
!

After multiplying the numerator and the denominator of the first term by $F M$, the product is equal to the ratio $(A M F) = (A C N) = n$ (again by the parallelism of $D N \| B C$) and the ratio $\frac{F M}{L M}$, and the latter factor is equal to $F M = F L + L M$, and by (2) to $m + 1$:

$$\frac{A F}{L M} = \frac{A F}{F M} \cdot \frac{F M}{L M} = \frac{A N}{N C} \cdot \frac{F L + L M}{L M} = n \left( \frac{F L}{L M} + 1 \right) = n(m + 1)$$

Now, according to (1)-(3), the required equality indeed holds:

$$\frac{A L}{L M} = n(m + 1) + m = m n + m + n$$

Müller Miksa (Makó, József A. g. II. o. t.)

II. Solution: Recognizing the three terms of the product $(m + 1)(n + 1)$ on the right-hand side of the equality to be proven, we add the missing fourth term, 1, to both sides and try to express the ratio

$$\frac{A L}{L M} + 1 = \frac{A L + L M}{L M} = \frac{A M}{L M}$$

as the product of the ratios $m + 1 = \frac{B M}{M C} + 1 = \frac{B C}{M C}$ and $n + 1 = \frac{A N}{N C} + 1 = \frac{A C}{N C}$. (These can also be considered as division ratios according to the transformations $\frac{A M}{L M} = -\frac{A M}{M L}$, $\frac{B C}{M C} = -\frac{B C}{C M}$, and $\frac{A C}{N C} = -\frac{A C}{C N}$, although it is unusual that in $-(A L M)$, $-(B M C)$, and $-(A N C)$, the "later appearing" points $L$, $M$, and $N$ serve as base points.) Indeed, using the transformations from the first solution or similar ones:

$$\begin{gathered} \frac{A M}{L M} = \frac{A F + F M}{L M} = \frac{\frac{A F}{F M} + 1}{\frac{L M}{F M}} = \frac{\frac{A N}{N C} + 1}{\frac{L M}{F L + L M}} = \\ = (n + 1) \cdot \frac{F L + L M}{L M} = (n + 1) \left( \frac{F L}{L M} + 1 \right) = (n + 1)(m + 1) \end{gathered}$$

This completes our proof.

Remarks. 1. In both solutions, we were somewhat superficial, not considering whether all the ratios, points we talked about, exist and are uniquely determined for any choice of $M$ and $N$, and whether the transformations we applied can always be carried out. Now we will make up for these omissions.

It is quite simple to ensure that the given division ratios (divisions) make sense: $M C$ and $N C$ cannot be 0, i.e., $M$ and $N$ cannot be chosen at $C$. The more complex question is whether the point $L$ constructed in several steps is determined. This would only not be the case if the lines $A M$ and $D E$ coincided or were parallel.

Their coincidence would only occur if the common point $A$ of $A B$ and the common point $D$ of $A$ and the common point $M$ of $B C$ and the common point $E$ of $B C$ coincided. Already in the interpretation of $D$, $D \equiv A$ would only hold if $N \equiv A$; in this case, from the interpretation of $E$, $E \equiv M$ would also follow, so the coincidence of $N \equiv A$, or rather the case $n = 0$, must also be excluded. Conversely, from $E \equiv M$, $N \equiv A$ also follows; thus, the necessary and sufficient condition for the coincidence of $N \equiv A$ is the coincidence of $E \equiv M$.

The parallelism of $D E$ with $A M$ would mean, through the interpretation of $E$, that $D E$ coincides with $N E$, so either $D \equiv N$ (then they coincide with $A$, which case we have already excluded), or $D N \| A M$. However, by definition, $D N \| B C$, so in this case $A M \| B C$, which contradicts the definition of $M$. Therefore, no further case needs to be excluded here.

For the ratio $\frac{A L}{L M}$ to make sense, we cannot place $M$ and $N$ such that the constructed $L$ coincides with $M$. This could only be caused by the coincidence of $E \equiv M$, which we have already excluded.

The last such question: was it permissible to "extend" or "simplify" the ratio with $F M$, is there no "danger" of $F M = 0$? - No, because $F$ and $M$ are points on the parallel lines and $N \not \equiv C$, which are distinct from each other.

According to these, the proven equality holds if $M$ does not fall on $C$ and $N$ does not fall on either $C$ or $A$.

2. It is easy to show that the potentially "dangerous" coincidence $M \equiv B$ does not need to be excluded. In this case, for any allowed position of $N$, $L \equiv D$ and the left-hand side of our equality

$$\frac{A L}{L M} = \frac{A D}{D B} = \frac{A N}{N B} = n$$

matches the value of the right-hand side with $m = 0$.

3. Our diagrams differ only in the placement of $M$ and $N$, essentially one would be enough, or even none. The direction of the segments did not complicate but rather simplified and unified the examination of the various possibilities. Nevertheless, we recommend to our readers to follow the given line of thought on even more diagrams. For each of $M$ and $N$, there are already three possibilities: on the side segment (here $m, n > 0$), or on one of the extensions (here $-1 < m, n < 0$ or $m, n < -1$).

[^0]: ${ }^{1}$ See, for example, Kárteszi Ferenc: The Theorems of Menelaus and Ceva. KML. Vol. XI, pp. 67-75, November 1955.