In triangle $ABC$ let $AD,BE$ and $CF$ be the altitudes. Let the points $P,Q,R$ and $S$ fulfil the following requirements: i) $P$ is the circumcentre of triangle $ABC$. ii) All the segments $PQ,QR$ and $RS$ are equal to the circumradius of triangle $ABC$. iii) The oriented segment $PQ$ has the same direction as the oriented segment $AD$. Similarly, $QR$ has the same direction as $BE$, and $Rs$ has the same direction as $CF$. Prove that $S$ is the incentre of triangle $ABC$.
Problem
Source: Baltic Way 2007
Tags: geometry, incenter, circumcircle, trigonometry, geometry proposed
jgnr
01.12.2010 11:42
Clearly $Q$ is the midpoint of arc $BC$. Let $I$ be the incenter of $\triangle ABC$, and $T$ is the intersection of the line through $S$ perpendicular to $AB$ and the line through $Q$ perpendicular to $AC$. We have $\angle TQI=90^{\circ}-\frac{A}2=\angle QIT$, so $QT=IT$. Also note that $QT=\frac{IQ}{2\cos\angle TQI}=\frac{QC}{2\sin\frac{A}2}$ which equals the circumradius of $ABC$. So $T=R$ and $S=I$, as desired.
KRIS17
04.08.2019 23:41
$Q$ lies midway on the arc $BC$ of circumcircle $C$ of $\triangle ABC$ and lies on the internal bisector through $A$. Let $C1$ be the circle with center $Q$ passing through $P$. $C1$ has the same radius as $C$. Let the perpendicular bisector of $AC$ passing through $P$ intersect circumcircle $C$ at $T$. Clearly $PT$ = $QR$ = circumradius of $\triangle ABC$ and they are both parallel. Since $PQ$ = $PT$ = $QR$, $PQRT$ is a rhombus. Thus $RT$ = $PQ ---> (1)$ Let $C2$ be the circle with center $R$ passing through $Q$. From (1) above $C2$ also passes through $T$. Finally $S$ lies on $C2$ as well since $RS$ = $QR$. We have that $BT$ is the internal bisector through $B$ and since $BQTA$ is cyclic, $\angle BTQ$ = $\angle BAQ$ = $\angle A/2 ---> (2)$ In circle $C2$, $\angle STQ$ = $\angle SRQ/2$ = $\angle A/2$ = $\angle BTQ$ from (2) above. Thus $S$ lies on the internal bisector though $B ---> (3)$ Since $PQ$ is parallel to altitude from $A$, by alternate angles $\angle AQP$ = $\angle A/2 - (90 - \angle B)$ = $(\angle B - \angle C)/2 ---> (4)$ In circle $C2$, $\angle SQP$ = $\angle SQR - \angle PQR$ = $(90 - \angle A/2)$ - $\angle C$ = $(\angle B - \angle C)/2$ = $\angle AQP$ from (4) above. Thus $S$ lies on the internal bisector though $A ---> (5)$ Thus $S$ is the incenter of $\triangle ABC$ from (3) and (5) above.
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