S must be the incentre of triangle ABC

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S must be the incentre of triangle ABC

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Source: Baltic Way 2007

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WakeUp

#1 h Nov 30, 2010, 5:09 PM • 2 Y

Y by Adventure10, Mango247

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 $ABC$ .

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jgnr

#2 h Dec 1, 2010, 9:42 AM • 2 Y

Y by Adventure10, Mango247

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.

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KRIS17

#3 h Aug 4, 2019, 8:41 PM • 2 Y

Y by Adventure10, Mango247

$Q$ 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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This post has been edited 7 times. Last edited by KRIS17, Aug 4, 2019, 9:01 PM

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#4 h Apr 6, 2025, 6:19 AM

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Nice and easy one. For the convenience let's solve the following problem.
Let $I$ and $O$ be the incenter and circumcircle of a $\triangle ABC$ . $M$ be the midpoint of minor arc $BC$ . And incircle touches $AB$ at $F$ . Let $R$ be the intersection of $FI$ and perpendicular bisector of $IM$ . Then prove that $IR=R$ .
Proof: Let $G$ be the midpoint of $IM$ . Then $AFGR$ is cyclic. So by the Thales theorem we have that:
$IR=\frac{AI*IG}{IF}=\frac{AI*IM}{2*IF}=\frac{R^2-OI^2}{2r}=R$ , and we are done. We used Euler's theorem, which states that $OI^2=R(R-2r)$ .

Result follows from this problem.

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