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[Solutions] Iranian Geometry Olympiad 2014

Junior

  1. $ABC$ is a triangle with $\angle A=90^\circ$ and $\angle C=30^\circ$. Let $M$ be the midpoint of $BC$. Let $W$ be a circle passing through $A$ tangent in $M$ to $BC$. Let $P$ be the circumcircle of $ABC$. $W$ is intersecting $AC$ in $N$ and $P$ in $M$. Prove that $MN$ is perpendicular to $BC$. 
  2. The inscribed circle of $\triangle ABC$ touches $BC$, $AC$ and $AB$ at $D$, $E$ and $F$ respectively. Denote the perpendicular foots from $F$, $E$ to $BC$ by $K$, $L$ respectively. Let the second intersection of these perpendiculars with the incircle be $M$, $N$ respectively. Show that $\dfrac{{{S}_{\triangle BMD}}}{{{S}_{\triangle CND}}}=\dfrac{DK}{DL}$
  3. Each of Mahdi and Morteza has drawn an inscribed $93$-gon. Denote the first one by $A_1A_2…A_{93}$ and the second by $B_1B_2…B_{93}$. It is known that $A_iA_{i+1} // B_iB_{i+1}$ for $1 \le i \le 93$ ($A_{93} = A_1$, $B_{93} = B_1$). Show that $\dfrac{A_iA_{i+1} }{ B_iB_{i+1}}$ is a constant number independent of $i$.
  4. In a triangle ABC we have $\angle C = \angle A + 90^\circ$. The point $D$ on the continuation of $BC$ is given such that $AC = AD$. A point $E$ in the side of $BC$ in which $A$ doesn’t lie is chosen such that $\angle EBC = \angle A$, $\angle EDC = \frac{1}{2} \angle A$ . Prove that $\angle CED = \angle ABC$.
  5. Two points $X$, $Y$ lie on the arc $BC$ of the circumcircle of $\triangle ABC$ (this arc does not contain $A$) such that $\angle BAX = \angle CAY$. Let $M$ denotes the midpoint of the chord $AX$. Show that $$BM +CM > AY.$$

Senior

  1. $ABC$ is a triangle with $\angle BAC=90^{\circ}$ and $\angle ACB=30^{\circ}$. Let $M_1$ be the midpoint of $BC$. Let $W$ be a circle passing through $A$ tangent in $M_1$ to $BC$. Let $P$ be the circumcircle of $ABC$. $W$ is intersecting $AC$ in $N$ and $P$ in $M$. Prove that $MN$ is perpendicular to $BC$.
  2. In the quadrilateral $ABCD$ we have $ \measuredangle B=\measuredangle D = 60^\circ $. $M$ is midpoint of side $AD$. The line through $M$ parallel to $CD$ meets $BC$ at $P$. Point $X$ lying on $CD$ such that $BX=MX$. Prove that $AB=BP$ if and only if $\measuredangle MXB=60^\circ$.
  3. Let $ABC$ be an acute triangle. A circle with diameter $BC$ meets $AB$ and $AC$ at $E$ and $F$, respectively. $M$ is midpoint of $BC$ and $P$ is point of intersection $AM$ with $EF$. $X$ is an arbitary point on arc $EF$ and $Y$ is the second intersection of $XP$ with a circle with diameter $BC$. Prove that $ \measuredangle XAY=\measuredangle XYM $.
  4. A tangent line to circumcircle of acute triangle $ABC$ ($AC>AB$) at $A$ intersects with the extension of $BC$ at $P$. $O$ is the circumcenter of triangle $ABC$. Point $X$ lying on $OP$ such that $\measuredangle AXP=90^\circ$. Points $E$ and $F$ lying on $AB$ and $AC$, respectively, and they are in one side of line $OP$ such that $ \measuredangle EXP=\measuredangle ACX $ and $\measuredangle FXO=\measuredangle ABX $. $K$, $L$ are points of intersection $EF$ with circumcircle of triangle $ABC$. Prove that $OP$ is tangent to circumcircle of triangle $KLX$.
  5. Two points $P$ and $Q$ lying on side $BC$ of triangle $ABC$ and their distance from the midpoint of $BC$ are equal.The perpendiculars from $P$ and $Q$ to $BC$ intersect $AC$ and $AB$ at $E$ and $F$,respectively. $M$ is point of intersection $PF$ and $EQ$. If $H_1$ and $H_2$ be the orthocenters of triangles $BFP$ and $CEQ$, respectively, prove that $ AM\perp H_1H_2 $.

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MOlympiad: [Solutions] Iranian Geometry Olympiad 2014
[Solutions] Iranian Geometry Olympiad 2014
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