Problem
ALG-B1-M04-P023 Three Concurrent Lines
Let \(P,Q\) be quadratic polynomials. For each positive integer \(n\), draw the line \(L_n: y=P(n)x+Q(n)\). If three distinct lines \(L_k,L_m,L_s\) pass through one point, prove that all lines \(L_n\) pass through one point.
The abscissa of the intersection of \(L_k\) and \(L_m\) can be expressed through \(k+m\).
Let \(P(n)=an^2+bn+c\), \(Q(n)=un^2+vn+w\). If \(P(k)\ne P(m)\), the abscissa of the intersection of \(L_k,L_m\) is
\[x_{km}=\frac{Q(m)-Q(k)}{P(k)-P(m)}=-\frac{u(k+m)+v}{a(k+m)+b}.\]
If three lines \(L_k,L_m,L_s\) are concurrent, then \(x_{km}=x_{ks}\). Since \(m\ne s\), equality of the fractions implies \(ub-av=0\), so the expression for \(x_{km}\) does not depend on \(k,m\). Thus all pairs meet at the same abscissa \(x_0\).
Substituting \(x_0\), the values \(P(n)x_0+Q(n)\) are equal for three different \(n\). This is a quadratic polynomial in \(n\); after subtracting the common value, it has three roots, so it is identically zero. Therefore all \(L_n\) pass through one point.
Preserves the idea: intersection is expressed as a rational function of the sum of indices, then identity of a quadratic polynomial is used.