Four circles

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April

1270 posts

April

#1 h Jul 26, 2007, 8:53 AM • 8 Y

Y by Davi-8191, bobcats4thewin, Adventure10, HWenslawski, ImSh95, and 3 other users

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.

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vittasko

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vittasko

#2 h Jul 27, 2007, 12:07 AM • 10 Y

Y by john10, bobcats4thewin, richy, Adventure10, wuyisince9271, ImSh95, and 4 other users

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.

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orl

3647 posts

orl

#3 h Oct 1, 2008, 11:29 PM • 5 Y

Y by jam10307, Adventure10, Mango247, and 2 other users

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.

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nickthegreek

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nickthegreek

#4 h Jan 4, 2012, 2:59 PM • 6 Y

Y by shmm, Flash_Sloth, dili96, Adventure10, Mango247, Math_.only.

Let me give another,quicker approach to this beautiful problem...

For the first part, it is easy to observe that all of the circles must pass through the incenter ( to see this just drop the perpendiculars from $D,E,F$ to the sides of $BC,CA,AB$ respectively). In fact, it is not even necessary to use the incenter. We could just use the well-known Miquel point of the triangle with respect to the points $D,E,F$ .

Now, as far as the second part is concerned, it is easy to prove (using nothing more than angle-chasing) that the three quadrilaterals $PEDQ,QFER,PFDR$
are cyclic. Hence , the three lines $PD,QE,RF$ are the radical axes of the circles taken two at a time. Finally, the three lines concur at the radical center of the three circles!

Best,
Nick

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sayantanchakraborty

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sayantanchakraborty

#5 h Jan 27, 2014, 4:34 PM • 3 Y

Y by Adventure10, Mango247, Math_.only.

Excellent,excellent....This was the solution I looked for...Thanks a lot,nick.I think even before approaching you thought of radical center...But how can you say that $PEDQ$ is cyclic?

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utkarshgupta

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utkarshgupta

#6 h Jan 4, 2015, 6:45 AM • 3 Y

Y by aayush-srivastava, Adventure10, Mango247

Consider the spiral symmetry that takes $E \to F$ and $B \to C$
It is well known that $P$ is the centre of this spiral symmetry.
Thus we have $\triangle PEB \sim \triangle PFC$
$\implies \frac{PB}{PC}=\frac{BE}{CF}=\frac{BD}{CD}$
$\implies PD$ is the angle bisector of $\angle BPC$

Let $X$ be the midpoint of arc $BC$ of circumcircle of $\triangle ABC$

We have $PDX$ collinear.

Define $Y,Z$ similarly.

Thus we are left to prove $XD,YE,ZF$ are collinear which is easy

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utkarshgupta

2280 posts

utkarshgupta

#7 h Jan 4, 2015, 6:53 AM • 3 Y

Y by Adventure10, Mango247, Math_.only.

nickthegreek wrote:

Let me give another,quicker approach to this beautiful problem...

For the first part, it is easy to observe that all of the circles must pass through the incenter ( to see this just drop the perpendiculars from $D,E,F$ to the sides of $BC,CA,AB$ respectively). In fact, it is not even necessary to use the incenter. We could just use the well-known Miquel point of the triangle with respect to the points $D,E,F$ .

Now, as far as the second part is concerned, it is easy to prove (using nothing more than angle-chasing) that the three quadrilaterals $PEDQ,QFER,PFDR$
are cyclic. Hence , the three lines $PD,QE,RF$ are the radical axes of the circles taken two at a time. Finally, the three lines concur at the radical center of the three circles!

Best,
Nick

How is $PEDQ$ cyclic ?

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61plus

252 posts

61plus

#8 h Jan 6, 2015, 9:06 AM • 2 Y

Y by Adventure10, Mango247

U can show that $DE$ is parallel to the line joining midpoint of arcs $BC,AC$ , then from $P,D,X$ collinear etc. we can deduce that those $4$ points are concyclic.

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john10

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john10

#9 h Aug 26, 2015, 4:29 PM • 1 Y

Y by Adventure10

61plus wrote:

U can show that $DE$ is parallel to the line joining midpoint of arcs $BC,AC$ , then from $P,D,X$ collinear etc. we can deduce that those $4$ points are concyclic.

I don't understand you proof @61plus , caan you explain more ?

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va2010

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va2010

#11 h Jan 20, 2016, 3:33 AM • 2 Y

Y by Adventure10, Mango247

utkarshgupta wrote:

Thus we are left to prove $XD,YE,ZF$ are collinear which is easy

This is tricky. We use the Cevian Nest Theorem. Observe that $AX$ , $YE$ , and $ZF$ are concurrent at the incenter $I$ , and that $AD$ , $BE$ , and $CF$ are concurrent at the Gergonne point. Hence, by the Cevian Nest theorem, we are done.

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Kezer

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Kezer

#12 h Apr 2, 2016, 10:23 PM • 2 Y

Y by Adventure10, Mango247

Revival, as the answer I'm looking for, wasn't answered very well in here.
How do we show that $PEDQ$ is a cyclic quadrilateral?

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v_Enhance

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v_Enhance

#13 h Apr 24, 2016, 4:47 PM • 21 Y

Y by Kezer, jam10307, vsathiam, claserken, QWERTYphysics, TheUltimate123, mshang, scimaths, DPS, Jafarly8097, kirillnaval, a_n, HamstPan38825, AllanTian, anonman, SSaad, alexgsi, Adventure10, Mango247, vrondoS, Math_.only.

Kezer wrote:

How do we show that $PEDQ$ is a cyclic quadrilateral?

$\begin{align*} \measuredangle QPE &= \measuredangle QPA + \measuredangle APE \\ &= \measuredangle QBA + \measuredangle AIE \\ &= \measuredangle QBA + \measuredangle ABI + \measuredangle IDE \\ &= \measuredangle QBI + \measuredangle IDE \\ &= \measuredangle QDI + \measuredangle IDE \\ &= \measuredangle QDE. \end{align*}$ All angles directed. With this, of course, the problem falls to radical axis.

A shorter but more advanced way to show the quadrilateral is cyclic is to note that $PQDE$ becomes the circle with diameter $DE$ after inverting around the incircle. (The circumcircle of $ABC$ maps to the nine-point circle of $DEF$ .)

(A little addendum: for anyone that's reading this because of my book, the above angle chase is actually a lot harder to find than it looks, or at least it took me half an hour to work out again just now. I wish now that I had included another hint or two afterwards, but too late now.)

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Reason: addendum

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anantmudgal09

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anantmudgal09

#14 h Apr 24, 2016, 11:56 PM • 2 Y

Y by Adventure10, Mango247

We notice that $PD$ passes through the midpoint af arc $BC$ not containing $A$ since $\angle API=90$ . (This follows since $P,D$ are in verses under inversion about $M$ with radius $MB=MC=MI$ ) Now, this concurrency point has to be centre of positive Homothety sending the in touch triangle $DEF$ to the triangle formed by the midpoint of arcs $MKL$ . Thus, we are done.

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trumpeter

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trumpeter

#16 h May 18, 2017, 9:37 PM • 3 Y

Y by kirillnaval, Adventure10, Mango247

Solution

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bobthesmartypants

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bobthesmartypants

#17 h May 18, 2017, 10:43 PM • 1 Y

Y by Adventure10

Let me post an unenlightening solution since trumpeter already took the time to bump this

solution

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minusonetwelth

225 posts

minusonetwelth

#49 h Oct 28, 2022, 5:50 AM

Y by

a) Trivially the incircle.
b) This common point is the exsimilicenter by Miquel's incircle configuration.

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Taco12

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Taco12

#50 h Dec 23, 2022, 4:01 AM

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Invert about the incircle.

Canadian MO 2007 Problem 5, Inverted wrote:

Let $\triangle ABC$ be a triangle with circumcenter $O$ and orthic triangle $\triangle DEF$ . Show that the circumcircles of $\triangle ADO, \triangle BEO, \triangle CFO$ are coaxial.

Note that the orthocenter of $\triangle ABC$ has equal power to all three circles, which finishes. $\blacksquare$

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HoRI_DA_GRe8

597 posts

HoRI_DA_GRe8

#52 h May 24, 2023, 4:24 PM

Y by

In todays world of config geo it is a shame to use angle chasing to prove that $PQDE$ and other variations cyclic.

Dude $P,Q$ are the $A$ and $B \text{ Sharkydevil point}$ .If $M_A,M_B$ are the midpoints of minor arcs $\widehat{BC}'\widehat{CA}$ respectively, it is well known that, $\overline{P-D-M_A}$ and $\overline{Q-E-M_B}$ .
Now as it is well known that $M_AM_B \perp CI ,DE \perp CI$ we have $M_AM_B \parallel DE$ .Since $PQM_AM_B$ is cyclic, by converse of Reim's Theorem $PQDE$ is also cyclic , thats all $\blacksquare$

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MagicalToaster53

159 posts

MagicalToaster53

#53 h Jun 26, 2023, 3:40 PM

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Solution (a): This part of the problem is trivialized by Miquel's theorem.

Solution (b): We claim that $PDEQ$ is cyclic, effectively proving that both $EQRF$ and $RFPD$ are cyclic. Thus by the existence of the radical center, we prove the desired concurrency as was necessary in the problem.

Claim: $PDEQ$ is cyclic.

Proof: We use directed angles modulo $\pi$ :
$\begin{align*} \measuredangle EDQ &= \measuredangle EDF + \measuredangle FDQ \\ &= \measuredangle EPA + \measuredangle FBQ \\ &= \measuredangle EPA + \measuredangle APQ \\ &= \measuredangle EPQ.\phantom{A}_{\square} \end{align*}$
Therefore by symmetry we also must have that $EQRF$ and $RFPD$ are cyclic, so that their circumcircles have a radical center determined by the concurrency of the lines $PD, QE, RF$ , as desired. $\phantom{A}_\blacksquare$

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CT17

1481 posts

CT17

#54 h Sep 2, 2023, 7:00 PM

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(a) All three circles pass through $I$ .

(b) All three lines pass through the exsimilicenter of $\omega$ and the incircle.

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peace09

5417 posts

peace09

#55 h Sep 5, 2023, 6:36 PM

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Disappointing; I thought I'd have to guess the concurrency point ( $H$ ?) or make some other clever observation, but it turns out you can just read off the concurrent lines as radical axes and prove that the resulting quadrilaterals are cyclic.

(a) Clearly, they concur at $I$ .

(b) It suffices to prove that $PDQE$ and cyclic variations are cyclic, whence $PD,QE,RF$ concur at the radical center. But
$\measuredangle PQD=\measuredangle PQI+\measuredangle IQD=\measuredangle PQI+\measuredangle IBD$ and
$\measuredangle PED=\measuredangle PEI+\measuredangle IED=\measuredangle PAI+\measuredangle ICD=\measuredangle PAB+\measuredangle FAI+\measuredangle ICD,$ which, after rearranging, reduces the condition to
$\measuredangle PAB=\measuredangle PQI+\measuredangle IBD+\measuredangle ICE+\measuredangle IAF=\measuredangle PQI+\tfrac{\pi}{2}.$ The above is true because $\measuredangle PQI+\tfrac{\pi}{2}=\measuredangle PQI+\measuredangle IQB=\measuredangle PQB$ . $\square$

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Cusofay

85 posts

Cusofay

#56 h Sep 22, 2023, 6:44 PM

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Here's my solution for the second part (the first one being obvious).

Let $M,N,P$ be the midpoints of the minor arcs $\stackrel{\frown}{BC}$ , $\stackrel{\frown}{AC}$ , $\stackrel{\frown}{AB}$ . We want to show that $(MD),(NE)$ and $(FP)$ concur at one point. This is true since the triangles $\triangle DEF$ and $\triangle MNP$ are directly similar and the sides of the triangles are parallel to each other, hence $(MD),(NE)$ and $(FP)$ concur at the center of homothety that sends $\triangle DEF$ to $\triangle MNP$ .

$\mathbb{Q.E.D.}$

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math_comb01

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math_comb01

#57 h Nov 16, 2023, 6:42 PM

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A twist to this classic,
Sketch: Extend $AP,BQ,CR$ to form a big triangle, by trig ceva and cyclic quads it is easy to see $GQ,JR,HP$ are concurrent, and obviously $JF,GE,HD$ concur at $I$ , by cevian nest we're done.

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shendrew7

794 posts

shendrew7

#58 h Feb 10, 2024, 10:35 PM

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The desired concurrency is just the incenter $I$ in part (a).

For part (b), note that we can classify $P$ , $Q$ , and $R$ as the $A$ -, $B$ -, and $C$ -Sharkydevil points, respectively. Hence $PD$ , $QE$ , and $RF$ each pass through the exsimilicenter of the incircle and circumcircle. $\blacksquare$

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HamstPan38825

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HamstPan38825

#59 h Mar 5, 2024, 2:55 PM

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For part (a), the circles just concur at the incenter. For part (b), the points $P, Q, R$ are the incenter Miquel points for each of the vertices. It follows that $\overline{PD}$ , $\overline{QE}$ , $\overline{RF}$ concur at the exsimilicenter of the incircle and circumcircle.

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Mr.Sharkman

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Mr.Sharkman

#60 h Jun 18, 2024, 6:44 PM

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Solution

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Eka01

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Eka01

#61 h Aug 15, 2024, 3:03 AM

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For part $(a)$ , the result immediately follows by Miquel's theorem and in fact the concurrency point is the incenter $I$ . For part $(b)$ , We invert around the incircle and notice that $P,Q,R$ go to feet of altitudes in $\Delta DEF$ so $(PDQE)$ and variants are cyclic. Inverting back and applying radical center theorem we are done.