Four circles

April

Let the incircle of triangle $ABC$ touch sides $BC,\, CA$ and $AB$ at $D,\, E$ and $F,$ respectively. Let $\omega,\,\omega_{1},\,\omega_{2}$ and $\omega_{3}$ denote the circumcircles of triangle $ABC,\, AEF,\, BDF$ and $CDE$ respectively.

Let $\omega$ and $\omega_{1}$ intersect at $A$ and $P,\,\omega$ and $\omega_{2}$ intersect at $B$ and $Q,\,\omega$ and $\omega_{3}$ intersect at $C$ and $R.$

$a.$ Prove that $\omega_{1},\,\omega_{2}$ and $\omega_{3}$ intersect in a common point.

$b.$ Show that $PD,\, QE$ and $RF$ are concurrent.

vittasko

From cyclic quadrilateral $APFE,$ we have that $\angle EPF = \angle A$ $,(1)$

Froma cyclic quadrilateral $APBC,$ we have that $\angle BPC = \angle A$ $,(2)$ and $\angle ABP = \angle ACP$ $,(3)$

From $(1),$ $(2)$ $\Longrightarrow$ $\angle EPC = \angle FPB$ $,(4)$

From $(3),$ $(4)$ we conclude the similarity of the triangles $\bigtriangleup EPC,$ $\bigtriangleup FPB$ and so, we have $\frac{PB}{PC}= \frac{BF}{CE}= \frac{BD}{CD}$ $,(5)$

From $(5)$ we conclude that the segment line $PD,$ is the angle bisector of $\angle BPC$ and so, it passes through the point $A',$ as the intersection point of $(\omega)$ from $AI,$ where $I$ is the incenter of $\bigtriangleup ABC.$

Similarly the segment line $QE,$ passes through the point $B'\equiv (\omega)\cap BI.$

$\bullet$ We denote the points $M\equiv OI\cap PA',$ $M'\equiv OI\cap QB'$ and we will prove that $M'\equiv M.$

Because of $ID\parallel OA'$ $\Longrightarrow$ $\frac{MI}{MO}= \frac{ID}{OA'}$ $,(6)$ and similarly $\frac{M'I}{M'O}= \frac{IE}{OB'}$ $,(6)$

From $(6),$ $(7)$ $\Longrightarrow$ $\frac{MI}{MO}= \frac{M'I}{M'O}$ $,(8)$ $($ because of $ID = IE$ and $OA' = OB'$ $).$

From $(8)$ $\Longrightarrow$ $M'\equiv M$ $,(9)$

From $(9)$ we conclude that the intersection point of the segment lines $PA'\equiv PD,$ $QB'\equiv QE,$ lies the constant segment line $OI.$

By the same way we can prove that the intersection point of the segment lines $PD,$ $RF,$ lies also on $OI.$

Hence, the segment lines $PD,$ $QE,$ $RF,$ are concurrent at one point and the proof is completed.

Kostas Vittas.

PS. I don’t understand well the part (a) of the problem. If by ''comment point'', you mean ''common point'', then it is clear that the concurrency point of $(\omega_{1}),$ $(\omega_{2}),$ $(\omega_{3}),$ is the incenter $I$ of $\bigtriangleup ABC.$

Sorry, I will post here later a figure, because now I am very tired.

**Posted:** Thu Jul 26, 2007 4:07 pm

orl

Problem statement posted by tobeno_1:

$\odot I$ and $\odot O$ are the incircle and circumcircle of $\triangle ABC$ respectively. $D,E,F$ are the tangent points.

$\odot O_1$ is the circumcirlce of $\triangle AEF$ , $\odot O_2$ is the circumcirlce of $\triangle BDF$ , $\odot O_3$ is the circumcirlce of $\triangle CDE$ .

$\odot O_1 \cap \odot O = P(P\neq A)$ , similarly define $Q,R$ .

Prove: $PD,QE,RF$ are concurrent. (see image below)

Approach by QuattoMaster 6000:

We consider the circle, $T_A$ that is tangent to the incircle at $D$ and tangent to $(O)$ at a point, say $P'$ . Notice that the homothety that maps $T_A$ to $(O)$ maps $D$ to the midpoint of arc $BC$ , say $K$ . Thus, $P'DK$ is a line. We also have that $\angle PBF = \angle PCE$ and $\angle PFB = 180 - \angle PFA = 180 - \angle PEA = \angle PEC$ , so $\triangle PBF\sim \triangle PCE$ . This gives us that $\frac {PB}{PC} = \frac {FB}{EC} = \frac {BD}{DC}$ , so $PD$ bisects $\angle BPC$ , which implies that $PD$ passes through the midpoint of arc $BC$ . Hence, $PDK$ is a line, which means that $P = P'$ . Now, $P$ is the center of homothety between $(O)$ and $T_A$ and $D$ is the center of homothety between $(I)$ and $T_A$ . This implies that the center of homothety between $(O)$ and $(I)$ lies on $PD$ by Monge's Theorem. Similarly, this center of homothety lies on $FR$ and $\text{[math]}$ , which implies that $PD$ , $FR$ , and $\text{[math]}$ concur, as desired.

Approach by pohoatza:

Consider the inversion $\Psi$ with respect to the incircle of triangle $ABC$ . The vertices $A$ , $B$ , $C$ go into the midpoints $A'$ , $B'$ , $C'$ of the segments $EF$ , $FD$ , $DE$ , respectively, and the circumcircle of $ABC$ goes into the circumcircle of $A'B'C'$ . Since the circumcircle of $AEF$ passes through the pole of inversion $I$ , its image is the line $EF$ , and therefore the image $P'$ of the second intersection $P$ of the circles $AEF$ and $ABC$ is the intersection of $EF$ and the nine-point circle $(A'B'C')$ of $DEF$ (the one different from $A'$ ). This means that $P'$ is the foot of the perpendicular from $D$ to $EF$ . Similarly, we get that the images $Q'$ , $R'$ of $Q$ , $R$ under $\Psi$ are the feet of the perpendiculars from $E$ , $F$ to $FD$ , $DE$ , respectively, and thus $P'Q'R'$ is the orthic triangle of $DEF$ . Now, since the orthocenter of $DEF$ has the same power with respect to the circles $(DIP')$ , $(EIQ')$ and $(FIR')$ , it is easy to see that they are coaxal, and thus they have another common point, different then $I$ . This means that the lines $DP$ , $\text{[math]}$ , $FR$ are concurrent.

**Posted:** Wed Oct 01, 2008 3:29 pm

_________________
Math is like love. A simple idea but it can get complicated.