2008 JBMO Shortlist G1

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Source: 2008 JBMO Shortlist G1

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#1 h Oct 10, 2017, 5:03 PM • 2 Y

Y by Adventure10, Mango247

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 $\angle KTC = \angle KNL$ .

This post has been edited 1 time. Last edited by parmenides51, Jun 21, 2022, 1:32 AM

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#2 h Dec 16, 2017, 4:23 AM • 1 Y

Y by Adventure10

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 $\angle KTC = \angle ADC = 180^{\circ} - \angle AMC = \angle KNL$

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#3 h Jan 16, 2024, 11:46 AM

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#4 h Jan 21, 2024, 6:53 PM

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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.

This post has been edited 1 time. Last edited by scarface, Jan 21, 2024, 7:00 PM
Reason: typo

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#5 h Feb 7, 2025, 9:38 AM • 1 Y

Y by nurlan2024

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.

This post has been edited 1 time. Last edited by mathbetter, Feb 7, 2025, 9:39 AM

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