Problem
GEO-B3-M02-P014 The Fourth Point from Cross-Ratio
B. New Original Problem. On a line \(l\), distinct points \(A,B,C,D\) are chosen, and on a line \(m\), distinct points \(A_1,B_1,C_1,D_1\) are chosen. It is known that there exists a projective map \(f:l\to m\) such that \(f(A)=A_1\), \(f(B)=B_1\), \(f(C)=C_1\). In addition, \((A B C D)=(A_1 B_1 C_1 D_1)\). Prove that \(f(D)=D_1\).
C. Hint 1. The image of \(D\) has the same cross-ratio with \(A_1,B_1,C_1\) as \(D\) has with \(A,B,C\).
D. Hint 2. For three fixed points on a line, the value of the cross-ratio uniquely determines the fourth point.
E. Full Solution.
Denote \(D_0=f(D)\). Since \(f\) is projective, it preserves cross-ratio:
\[ (A_1 B_1 C_1 D_0)=(A B C D). \]
By assumption, \((A B C D)=(A_1 B_1 C_1 D_1)\). Therefore,
\[ (A_1 B_1 C_1 D_0)=(A_1 B_1 C_1 D_1). \]
On the line \(m\), the points \(A_1,B_1,C_1\) are distinct. For three fixed distinct points, the value \((A_1 B_1 C_1 X)\) uniquely determines the point \(X\). Hence \(D_0=D_1\). Thus \(f(D)=D_1\).
The task is useful as a bridge between “three points determine a projectivity” and practical recovery of a fourth point.