IV Lusophon Mathematical Olympiad 2014 - Problem 4

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IV Lusophon Mathematical Olympiad 2014 - Problem 4

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#1 h Dec 28, 2014, 10:31 PM • 1 Y

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From a point $K$ of a circle, a chord $KA$ (arc $AK$ is greather than $90^{o}$ ) and a tangent $l$ are drawn. The line that passes through the center of the circle and that is perpendicular to the radius $OA$ , intersects $KA$ at $B$ and $l$ at $C$ . Show that $KC = BC$ .

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#2 h May 21, 2024, 2:55 PM

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What a very nice and simple problem.

Let $D$ be a point in the circumference such that $DA$ is diameter of the circle and $E$ the point on the perpendicular of $OA$ that lies on the circle.
Since $DA$ is a diameter we have $\angle DKA=90^\circ$ and notice $EO\perp DA$ $\implies$ $\angle DOB=90^\circ$ so $DKOB$ is cyclic $\implies$ $\angle ODK= \angle KBC$

Now since $l$ is tangent at $K$ we must have $\angle ADK= \angle AKC$ because they're looking at the same arc (small arc $KA$ ). So we have $\angle CKB= \angle CBK$ $\implies$ $\triangle CBK$ is isosceles with $KC=BC$ . $Q.E.D$ $\square$ .

This post has been edited 1 time. Last edited by iMqther_, May 21, 2024, 2:56 PM
Reason: typo

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