Problem 1450

Solution:

One easily checks that, for any nonzero real numbers $a, b, c$ chosen arbitrarily, setting $x = \frac{a + b}{c}$, $y = \frac{b + c}{a}$ and $z = \frac{a + c}{b}$, the relation to be proved becomes an algebraic identity.

To find $a, b, c$ proceed as follows: let $D$ and $E$ be the intersections of the line through $P$ parallel to $AB$ with the segments $AC$ and $BC$ respectively. Let $F$ and $G$ be the intersections of $AB$ with the lines through $P$ parallel to the segments $AC$ and $BC$ respectively, and set $AF = a$, $FG = b$, $GB = c$.

The triangles $DEC$ and $FGP$ are similar, since they have their sides pairwise parallel, and moreover the relations $DP = a$, $PE = c$ and $DA = PF$ hold, since $AFPD$ and $GBEP$ are parallelograms.

Applying Thales's theorem to the parallels $GP$ and $BA'$, it follows that $x = \frac{a + b}{c}$, and similarly it follows that $y = \frac{b + c}{a}$, by applying the same theorem to the parallels $FP$ and $AB'$. Again by Thales's theorem, we have $z = \frac{CD}{DA}$, whence $z = \frac{CD}{PF} = \frac{DE}{FG}$, using the similarity between $DEC$ and $FGP$. Hence, $z = \frac{DP + PE}{FG} = \frac{a + c}{b}$, and this concludes the proof.