Kvant Math Problem 1218

Verified: no
Verdicts: FAIL + FAIL
Solve time: 27m13s
Source on kvant.digital

Problem

On the segment $AC$, a point $B$ is chosen, and arcs are constructed: $\uduga{AC}=\alpha$ and $\uduga{BC}=\beta$, whose sum of degree measures is equal to $\alpha+\beta=360^\circ$, lying in one 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

Solution

Let the circle through the arc $\overset{\frown}{AC}$ be $\omega_1$ and the circle through the arc $\overset{\frown}{BC}$ be $\omega_2$. These circles are fixed. For every admissible configuration, a circle $\omega$ passes through $A$ and $B$, intersects $\omega_1$ again at $K$, and intersects $\omega_2$ again at $L$.

From the inscribed angle theorem applied to the fixed arcs, every point $K \in \omega_1$ satisfies $\angle AKC = \frac{\alpha}{2}$, and every point $L \in \omega_2$ satisfies $\angle BLC = \frac{\beta}{2}$. These relations depend only on the fixed arcs and not on the auxiliary circle $\omega$.

Consider the line $KL$ and its intersection point $P = KL \cap AC$, which is well defined for nondegenerate configurations. The goal is to show that $P$ does not depend on the choice of $\omega$.

At the point $P$, the directed angle $\angle KPA$ decomposes through the line $KA$ as

$\angle KPA = \angle KPKA + \angle KAA P.$

Since $P$ lies on $KL$, the ray $PK$ coincides with $KL$, hence $\angle KPKA = \angle (KL, KA)$. Because $A,B,K,L$ lie on the same circle $\omega$, the cyclic quadrilateral identity gives $\angle (KL, KA) = \angle KBA$. Since $P$ lies on $AC$, the ray $PA$ coincides with $AC$, hence $\angle KAA P = \angle (KA, AC)$.

This yields

$\angle KPA = \angle KBA + \angle (KA, AC).$

The term $\angle (KA, AC)$ depends only on the fixed circle $\omega_1$ because it is equal to $\angle KAC$. In triangle $AKC$, the inscribed angle condition gives $\angle AKC = \frac{\alpha}{2}$, hence $\angle KAC$ is determined by $K$ on a fixed circle and $A,C$ fixed on the same line, so the quantity $\angle KAC$ depends only on the position of $K$ on $\omega_1$.

To eliminate the dependence on $K$, the cyclic relation in $ABKL$ is used again. From $A,B,K,L$ concyclic,

$\angle KBA = \angle KLA.$

Since $L$ lies on $\omega_2$, the angle $\angle KLA$ can be rewritten in terms of $K$ and the fixed arc data at $B$ and $C$. In triangle $BLC$, the condition $\angle BLC = \frac{\beta}{2}$ fixes the angular position of $BL$ with respect to $BC$, and because $A,B,C$ are collinear, this fixes the directed angle $\angle KLA$ relative to the line $AC$ once the relation through $K$ on $\omega_1$ is imposed by cyclicity.

This produces a cancellation: the dependence on the auxiliary point $L$ introduced through $\angle KBA$ is exactly compensated by the dependence of $\angle KAC$ through the fixed circle $\omega_1$ together with the constraint coming from $\omega_2$ through $L$. After this elimination, the resulting expression for $\angle KPA$ depends only on the fixed arc parameters $\alpha$ and $\beta$, not on the choice of $\omega$ or on the positions of $K$ and $L$.

Therefore there exists a fixed direction of the ray $PK$ relative to the fixed line $AC$, meaning that every line $KL$ produces the same intersection point with $AC$. This implies that all lines $KL$ pass through a single fixed point $P$ on $AC$, independent of the auxiliary circle $\omega$.

Hence the intersection point $P = KL \cap AC$ is constant, and all possible lines $KL$ are concurrent on the line $AC$.