Kvant Math Problem 1218
The three arcs are parts of three circles.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m49s
Source on kvant.digital
Problem
On the segment $AC$, a point $B$ is chosen and the arcs $\uduga{AC}=\alpha$ and $\uduga{BC}=\beta$ are constructed, the sum of whose angular measures is $\alpha+\beta=360^\circ$, lying in the same half-plane with respect to the line $AC$. An arbitrary arc $AB$ intersects them at points $K$ and $L$ (Fig. 1). Prove that all possible lines $KL$ intersect the line $AC$ at the same point.
Fig. 1
B. Mikhailov
Exploration
The three arcs are parts of three circles. Let $\omega_1$ be the circle containing the arc $\overset{\frown}{AC}$, let $\omega_2$ be the circle containing the arc $\overset{\frown}{BC}$, and let $\omega$ be the circle containing the arbitrary arc $\overset{\frown}{AB}$.
The condition $\alpha+\beta=360^\circ$ is the essential geometric datum. If $\alpha=2t$, then $\beta=360^\circ-2t$, hence
$$\sin \frac{\beta}{2} =\sin(180^\circ-t) =\sin t.$$
Consequently the radii of $\omega_1$ and $\omega_2$ are expressed through the same factor $\sin t$.
A natural coordinate choice is to place $A,B,C$ on the $x$-axis. Then the equations of $\omega_1$ and $\omega_2$ become particularly simple. The arbitrary circle $\omega$ through $A$ and $B$ depends on one parameter. After computing the intersection points $K$ and $L$, one can calculate the intersection of the line $KL$ with the $x$-axis. The crucial question is whether the parameter disappears.
The most likely place for an error is the computation of the second intersection point of a line with a circle. Since both $K$ and $L$ are obtained in that way, every sign must be checked carefully.
The calculation eventually produces an $x$-inter