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Difference between revisions of "2013 OIM Problems/Problem 2"

(Created page with "== Problem == Let <math>X</math>, <math>Y</math> be the ends of a diameter of a circle <math>\Gamma</math> and <math>N</math> the the midpoint of one of the <math>XY</math> ar...")
 
 
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== Solution ==
 
== Solution ==
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{{solution}}
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This proof won't use the fact that XY is a diameter of <math>\Gamma</math> and will prove it for every chord XY.
  
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== See also ==
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Let <math>M'</math> be the midpoint of <math>CD</math> and <math>S = AB \cap NM'</math>.
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[[OIM Problems and Solutions]]
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We observe that <math>NM' \equiv NS</math> is the median and <math>NM</math> is the symmedian of <math>\triangle NCD</math>, hence <math>\angle CNM \equiv \angle ANM = \angle DNM' \equiv \angle BNS</math>.
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Therefore, it suffices to show that <math>NS</math> is symmedian of <math>\triangle ABN</math>, which is equivalent to <math>AB</math> and <math>CD</math> being antiparallel, in other words, we only need to prove that <math>ACDB</math> is cyclic:
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<math>\angle NCD = \frac {\overarc (DN)}{2} = \frac {\overarc (DY + YN)}{2} = \frac {\overarc (DY + NX)}{2} = \angle XYN + \angle DNY
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\equiv \angle BYN + \angle YNB = \angle ABN</math>, where <math>\overarc PQ</math> stands for the arc <math>PQ</math>, which ends the problem
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-zuat.e

Latest revision as of 11:19, 5 May 2024

Problem

Let $X$, $Y$ be the ends of a diameter of a circle $\Gamma$ and $N$ the the midpoint of one of the $XY$ arcs of $\Gamma$. Let $A$ and $B$ be two points on the segment $XY$. The straight lines $NA$ and $NB$ cut $\Gamma$ again at points $C$ and $D$, respectively. The tangents to $\Gamma$ in $C$ and $D$ intersect at $P$. Let $M$ be the point of intersection of segment $XY$ with segment $NP$. Show that $M$ is the midpoint of segment $AB$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

This proof won't use the fact that XY is a diameter of $\Gamma$ and will prove it for every chord XY.

Let $M'$ be the midpoint of $CD$ and $S = AB \cap NM'$. We observe that $NM' \equiv NS$ is the median and $NM$ is the symmedian of $\triangle NCD$, hence $\angle CNM \equiv \angle ANM = \angle DNM' \equiv \angle BNS$.

Therefore, it suffices to show that $NS$ is symmedian of $\triangle ABN$, which is equivalent to $AB$ and $CD$ being antiparallel, in other words, we only need to prove that $ACDB$ is cyclic:

$\angle NCD = \frac {\overarc (DN)}{2} = \frac {\overarc (DY + YN)}{2} = \frac {\overarc (DY + NX)}{2} = \angle XYN + \angle DNY \equiv \angle BYN + \angle YNB = \angle ABN$, where $\overarc PQ$ stands for the arc $PQ$, which ends the problem


-zuat.e

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