Two perpendicular chords of a circle, $AM, BN$ , which intersect at point $K$, define on the circle four arcs with pairwise different length, with $AB$ being the smallest of them. We draw the chords $AD, BC$ with $AD // BC$ and $C, D$ different from $N, M$ . If $L$ is the intersection point of $DN, M C$ and $T$ the intersection point of $DC, KL,$ prove that $\angle KTC = \angle KNL$.
Problem
Source: 2008 JBMO Shortlist G1
Tags: JBMO, geometry
ACGNmath
16.12.2017 07:23
We use directed angles modulo $180^{\circ}$. First we prove $KMLN$ cyclic. $\measuredangle KML + \measuredangle KNL = \measuredangle AMC + \measuredangle BND = \measuredangle ABC + \measuredangle BND = \measuredangle BCD + \measuredangle BND = 0^{\circ}$. Thus $KMLN$ is cyclic. Now $\angle KTC = \angle ADC = 180^{\circ} - \angle AMC = \angle KNL$
scarface
21.01.2024 21:53
ACGNmath wrote: We use directed angles modulo $180^{\circ}$. First we prove $KMLN$ cyclic. $\measuredangle KML + \measuredangle KNL = \measuredangle AMC + \measuredangle BND = \measuredangle ABC + \measuredangle BND = \measuredangle BCD + \measuredangle BND = 0^{\circ}$. Thus $KMLN$ is cyclic. Now $\angle KTC = \angle ADC = 180^{\circ} - \angle AMC = \angle KNL$ Hi, why $\angle KTC = \angle ADC$? You have not proven that it is $ KL // AD$. I also don't see you using the given perpendicularity. Your solution seems incomplete.
mathbetter
07.02.2025 12:38
From the cyclic quadrilaterals \( \text{ADCM} \) and \( \text{DNBC} \), we have: \[ \angle DCL = \angle DAM \quad \text{and} \quad \angle CDL = \angle CBN. \]Thus, we obtain \[ \angle DCL + \angle CDL = \angle DAM + \angle CBN. \]Since \( AD \parallel BC \), if \( Z \) is the point of intersection of \( AM \) and \( BC \), then \( \angle DAM = \angle BZA \), and we have \[ \angle DCL + \angle CDL = \angle BZA + \angle CBN = 90^\circ, \]so \[ \angle DLC = 90^\circ. \] Now, let \( P \) be the point of intersection of \( KL \) and \( AC \). Then \( NP \perp AC \), because the line \( KPL \) is a Simson line of the point \( N \) with respect to the triangle \( ACM \). From the cyclic quadrilaterals \( NPCL \) and \( ANDC \), we obtain: \[ \angle CPL = \angle CNL \quad \text{and} \quad \angle CNL = \angle CAD, \]so \[ \angle CPL = \angle CAD. \]Since \( KL \parallel AD \parallel BC \), we conclude that \[ \angle KTC = \angle ADC \quad \text{(1)}. \] Since \( ANDC \) is cyclic, we have \( \angle ADC = \angle ANC = \angle ANK + \angle KNC = \angle CNL + \angle KNC = \angle KNL \) (\( \angle ANK = \angle CNL \) because \( AD \parallel BC \), so the arcs \( AB \) and \( DC \) are equal). Therefore, \[ \angle ADC = \angle KNL \quad \text{(2)}. \] From (1) and (2), we obtain the result.