Two circles concur on a line

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math154

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math154

#1 h Jul 2, 2012, 3:10 AM • 7 Y

Y by Amir Hossein, abacadaea, Davi-8191, Adventure10, Mango247, buddyram, Rounak_iitr

In acute triangle $ABC$ , let $D,E,F$ denote the feet of the altitudes from $A,B,C$ , respectively, and let $\omega$ be the circumcircle of $\triangle AEF$ . Let $\omega_1$ and $\omega_2$ be the circles through $D$ tangent to $\omega$ at $E$ and $F$ , respectively. Show that $\omega_1$ and $\omega_2$ meet at a point $P$ on $BC$ other than $D$ .

Ray Li.

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Luis González

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Luis González

#2 h Jul 2, 2012, 3:56 AM • 7 Y

Y by DwyerX, zarengold, DanDumitrescu, myh2910, Adventure10, Mango247, NxD_

$M$ is the midpoint of $\overline{BC}$ and $H$ is the orthocenter of $\triangle ABC.$ Since the circle with diameter $\overline{BC}$ is orthogonal to the circle $\omega$ through $A,E,F,H,$ then $ME,MF$ are tangents of $\omega.$ Let $Q \equiv EF \cap BC.$ Since $(B,C,D,Q)=-1,$ we have $MB^2=ME^2=MF^2=MD \cdot MQ$ $\Longrightarrow$ $\odot(DEQ)$ and $\odot(DFQ)$ are tangent to $\omega$ at $E,F$ $\Longrightarrow$ $\odot(DEQ)$ and $\odot(DFQ)$ coincide with $\omega_1$ and $\omega_2$ $\Longrightarrow$ $P \equiv Q.$

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WakeUp

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WakeUp

#3 h Jul 2, 2012, 4:53 PM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the centre of $\omega$ , on which the orthocentre $H$ clearly lies diametrically opposite to $A$ . An angle chase shows that $O,E,D,F$ are concyclic: $\angle EOF=2\angle EAF=2A$ and of course $\angle FDE=180^{\circ}-2A$ .

Invert with respect to $\omega$ (we now know $D'$ lies on $EF$ ).

The problem becomes:

In the triangle $AEF$ with circumcentre $O$ and circumcircle $\omega$ , the diameter $AH$ meets $EF$ at $D'$ . The circle tangent to $\omega$ at $E$ and passing through $D'$ and the circle tangent to $\omega$ at $F$ passing through $D'$ intersect at $D'$ and $P$ . Prove that $OP\perp PD'$ .

Note that we can actually ignore points $B'$ and $C'$ , who also lie on the circle with diameter $OD'$ along with $P$ , so we've only really inverted $D,\omega_1,\omega_2$ ...

Anyway:

Consider the tangent $\ell$ to $\omega$ and $(OD'E)$ at $E$ . By considering the angle $ED'$ makes with this tangent, we get, by the alternate segment theorem that this angle is both equal to $\angle D'PE$ and $\angle FAE$ . Hence $\angle D'PE=\angle FAE$ and similarly we get that $\angle FPD'=\angle FAE$ . Therefore $\angle FPE=\angle FPD'+\angle D'PE=2\angle FAE=\angle FOE$ so $EPOF$ is cyclic. Now consider the diameter $OO'$ of $(OEF)$ . Since $OF=OE$ , $O'$ is the midpoint of the arc $EF$ not containing $O$ . Now, since $\angle FPD'=\angle EPD'$ , we have that $PD'$ is the bisector of $\angle FPE$ and so $PD'$ passes through $O'$ . Hence $\angle OPD'=\angle OPO'=90^{\circ}$ and the result follows.

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r1234

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r1234

#4 h Jul 4, 2012, 11:05 AM • 1 Y

Y by Adventure10

First inverting the figure wrt $H$ with power $HA.HD$ we get the following problem

In $\triangle ABC$ , two circles $\omega_1, \omega_2$ pass through $A$ and touch the line $BC$ at $B,C$ respectively. Show that the intersection point of these two circles lies on $\odot AEF$ . Now inverting this figure again wrt $A$ with power $AH.AD$ we get the following problem:

In $\triangle ABC$ , two tangents are drawn at the points $E,F$ to the circle $\odot AEF,$ then these two tangent lines intersect on $BC,$ and this is obvious. They meet on at the midpoint of $BC$ .

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Bigwood

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Bigwood

#5 h Jul 7, 2012, 1:18 AM • 2 Y

Y by Asynchrone, Adventure10

Let $H$ be the orthocenter and $M$ be midpoint of $BC$ , and second intersection point of $\omega_1,\omega_2$ be $Q$ . $\angle{AEH}=\angle{AFH}=\frac{\pi}{2}$
implies $H$ is on $\omega$ . Note that $B,F,E,C$ lie on the circle with diameter $BC$ .
Then $\angle{HEM}=\angle{BEM}=\angle{EBM}=\angle{EAD}=\angle{EAH}$ , which means $ME$ tangents $\omega$ . Similarly so does $MF$ . Then we can conclude that radical axis of $(\omega,\omega_1),(\omega,\omega_2),(\omega_1,\omega_2)$ is $EM,FM,DQ$ . thus $M,D,Q$ is collinear, which is equivalent to the claim.

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polya78

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polya78

#6 h Apr 11, 2013, 5:46 PM • 2 Y

Y by Adventure10, Mango247

The radical axes of the three circles are $DQ$ and the tangents to $w$ at $E,F$ . But those two tangents meet at $M$ , the midpoint of $BC$ , because $M$ and the center of $w$ are antipodal on the nine-point circle.

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yugrey

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yugrey

#7 h Apr 14, 2013, 10:03 PM • 2 Y

Y by Adventure10, Mango247

My solution is more or less the same as the last couple; if you let $PB=PC$ on $BC$ , showing that $P$ is the intersection of the tangents to the circle at $E$ and $F$ is straightforward, and then $PD$ , the same line as $BC$ , must be the radical axis so if the circles meet again they must meet on $BC$ .

This has one slight flaw, but that flaw is actually inherent in the problem statement. If $P=D$ , then $AB=AC$ and the two circles $w_1$ and $w_2$ are tangent, so they don't meet on $BC$ . If the problem were reworded to "If $w_1$ and $w_2$ meet at a point $P$ other than $D$ , show that point is on $BC$ " or if it was stated that the triangle was not isosceles, this would have been avoided.

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leader

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leader

#8 h Apr 21, 2013, 1:58 PM • 2 Y

Y by Adventure10, Mango247

let $AD$ be the diameter of circle $ABC$ and let $DH$ meet circle $ABC$ again at $R$ . now let $AR$ meet $BC$ at $K$ .since $\angle HRA=90=\angle HFA=\angle HEA$ $R$ is on $\omega$ . since $\angle CRH=\angle CAD=90\angle CDA=90-\angle CBA=\angle BCH$ circle $CHR$ touches $BC$ . Similarly circle $BHR$ touches $BC$ . Consider inversion of pole $A$ and power $AE*AC$ . it pictures $\omega$ into $BC$ so $R$ into $K$ . it pictures $H,E,F$ into $D,C,B$ so $\omega$ touches circles $EDK$ and $FDK$ therefore $K=P$ so $P$ is on $BC$ .

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giratina150

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giratina150

#9 h Apr 24, 2013, 4:18 AM • 2 Y

Y by Adventure10, Mango247

solution

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thecmd999

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thecmd999

#10 h May 6, 2013, 9:56 PM • 3 Y

Y by Math1331Math, Adventure10, Mango247

Nice and quick

Solution

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anantmudgal09

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anantmudgal09

#11 h Jun 4, 2015, 9:02 PM • 2 Y

Y by sa2001, Adventure10

Let $EF$ meet $BC$ at $X$ . Now it is some easy angle chase that $X$ is the desired point.

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steppewolf

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steppewolf

#12 h Jan 28, 2016, 9:02 PM • 1 Y

Y by Adventure10

First of all, it's easy to see that $AEHF$ is cyclic.
Let $M$ be the pole of $EF$ wrt the circumcircle of $\triangle AEF$ . Let $S$ be the point of intersection of the lines $EF$ and $AH$ .
Because $AEHF$ is cyclic, the polar of $S$ is the line $BC$ , so $M$ lies on $BC$ because $S$ lies on the polar of $M$ . Now if circle $\omega_1$ intersects $BC$ at $P$ , and circle $\omega_2$ intersects $BC$ at $Q$ , by power of point we have $MD\cdot MP = MF^{2} = ME^{2} = MD\cdot MQ$ So $MP = MQ$ , which means that $P = Q$ ;

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Kayak

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Kayak

#13 h Dec 17, 2017, 9:27 AM • 2 Y

Y by Adventure10, Mango247

After a inversion centered at $D$ with power (or radius or whatever it is called) $\sqrt{DB*DC}$ , and reflecting the inverted figure in line $DA$ , we see $\omega$ maps to itself, $F$ maps to $E$ , $E$ maps to $F$ , and the question becomes equivalent to proving whether the tangents to $\omega$ at $E$ and $F$ meet at $BC$ or not.

But by a very well known property of nine-point circle, they meet at the midpoint of $BC$ , hence QED.

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khanhnx

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khanhnx

#14 h Mar 10, 2019, 6:56 AM • 5 Y

Y by Imayormaynotknowcalculus, Mathasocean, tiendung2006, Adventure10, Mango247

Let $M$ be midpoint of $BC$ then $ME$ , $MF$ are common tangents of ( $\omega$ ; $\omega_1$ ), ( $\omega$ ; $\omega_2$ ) then $M$ is radical center of $\omega$ , $\omega_1$ , $\omega_2$
But: $\omega_1$ , $\omega_2$ have common point $D$ so $DM$ is radical axis of $\omega_1$ , $\omega_2$
Hence: $\omega_1$ , $\omega_2$ intersect again at a point on $BC$

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e_plus_pi

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e_plus_pi

#15 h Jul 27, 2019, 1:28 PM • 2 Y

Y by Adventure10, Mango247

Good problem although easy.

math154 wrote:

In acute triangle $ABC$ , let $D,E,F$ denote the feet of the altitudes from $A,B,C$ , respectively, and let $\omega$ be the circumcircle of $\triangle AEF$ . Let $\omega_1$ and $\omega_2$ be the circles through $D$ tangent to $\omega$ at $E$ and $F$ , respectively. Show that $\omega_1$ and $\omega_2$ meet at a point $P$ on $BC$ other than $D$ .

Ray Li.

Let $X$ be the $A -$ EX point of $\triangle ABC$ . We claim that $X \equiv P$ .
$\text{[math]}$
Denote by $\Omega$ the circumcircle of $\triangle XFD$ . Let $l$ be the tangent line to $\Omega$ at $F$ . Then note that $\angle (l , \overline{FH}) = \angle HFD - (\angle B - \angle C) = 90^{\circ} - \angle B = \angle FAH$ .
So $l$ is also tangent to $\odot(AFHE) \implies {\Omega, \omega}$ are tangent. Therefore $\omega_2 = \Omega \implies X \in \omega_2$ . Similarly $X \in \omega_1 \iff X \equiv P$ .

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shendrew7

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shendrew7

#46 h Feb 10, 2024, 11:35 PM

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WLOG assume $AB < AC$ . Let $H$ be the orthocenter and $X = BC \cap EF$ . Then

$EB$ bisects $\angle DEX$ , so the $AE$ passes through midpoint of arc $DX$ not containing $E$ . Hence there exists a homothety at $E$ sending $\omega$ to $(DEX)$ , meaning the two are tangent, so $(DEX) = \omega_1$ .
$FA$ is the external angle bisector $\angle DFX$ , so we use the same steps to get $(DFX) = \omega_2$ . Hence $X$ is our desired point on $BC$ . $\blacksquare$

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Pyramix

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Pyramix

#47 h Mar 1, 2024, 1:26 PM

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Let $L$ be the mid-point of $\overline{AH}$ and $M$ be the mid-point of $\overline{BC}$ . Then, $M,D,E,F,L$ lie on the nine-point circle. Moreover, $\angle MDL=90^\circ=\angle MEL=\angle MFL$ . Note that $L$ is the center of circle $(AEF)$ as $H\in(AEF)$ and $\angle AEH=\angle AFH=90^\circ$ .
It follows that $ME$ and $MF$ are tangents to circle $(AEF)$ . Since $\omega_1,\omega_2$ are tangent to $(AEF)$ as well, which means that $ME,MF$ is the radical axis of $(AEF),\omega_1$ and $(AEF),\omega_2$ , respectively. Hence, $M$ is the radical centre of $(AEF),\omega_1,\omega_2$ . So, $M$ lies on the radical axis of $\omega_1,\omega_2$ . So, $MD$ is the radical axis of the two circles. Since $P\in MD$ , and $M,D,B,C$ are collinear, it follows that $P\in BC$ .

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Reason: Forgot the diagram!

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HamstPan38825

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HamstPan38825

#48 h Mar 2, 2024, 12:13 AM

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Note that $M = \overline{EE} \cap \overline{FF}$ , the midpoint of $\overline{BC}$ by three tangent lemma, is the radical center of $\omega_1, \omega_2$ , and $(AEF)$ . Hence $\overline{DM} \equiv \overline{BC}$ is the radical axis of $\omega_1$ and $\omega_2$ , and the result follows.

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ihatemath123

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ihatemath123

#49 h Apr 30, 2024, 3:55 AM

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They meet at the $A$ -ex point; if we let this point be $X$ , let $H$ be the orthocenter and let $N$ be the midpoint of $\overline{AH}$ , then $\angle FXD = \angle DNM$ (it's well known that $\overline{NM} \perp \overline{FE}$ ). Because of the nine point circle, $\angle DNM = \angle DFM$ , so $(XFD)$ is tangent to $\overline{FM}$ . The three tangents lemma gives us that $\overline{MF}$ is tangent to $(AFE)$ , so indeed $(XFD)$ is tangent to $(AFE)$ . Likewise, $(XED)$ is tangent to $(AFE)$ .

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EpicBird08

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EpicBird08

#50 h May 18, 2024, 12:44 AM

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We claim that the point $P$ is just the intersection of $EF$ and $BC.$ To do so, we let $X = EF \cap BC,$ then we will show that $(DEX)$ and $\omega$ are tangent; this will imply by symmetry that $(DFX)$ and $\omega$ are tangent, which would finish.

Let $M$ be the midpoint of $BC.$ By the Three Tangents Lemma, $EM$ is tangent to $\omega.$ We wish to show that $ME$ is also tangent to $(DEX).$ Note that $(X,D;B,C) = -1,$ so by the midpoint of harmonic bundles lemma, we see that $MB^2 = ME^2 = MD \cdot MX,$ which finishes.

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BestAOPS

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BestAOPS

#51 h Jun 29, 2024, 2:57 PM

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Let $M$ be the midpoint of $BC$ . It is well known that $B$ , $F$ , $E$ , and $C$ lie on a circle centered at $M$ and lines $ME$ and $MF$ are tangent to $\omega$ . Thus, line $ME$ is tangent to $w_1$ and line $MF$ is tangent to $w_2$ , so $M$ has the same power with respect to $w_1$ and $w_2$ . $D$ also has the same power $0$ with respect to both circles, so $\overline{DM} = \overline{BC}$ must be the radical axis of the two circles, making the result obvious.

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dolphinday

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dolphinday

#52 h Jun 29, 2024, 3:24 PM

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It is well known that the tangents at $E$ and $F$ at $(AEF)$ concur at the midpoint of $BC$ , which we will call $M$ . So it follows that $M$ is the radical center of $\omega_1$ , $\omega_2$ and $(AEF)$ , and since $\omega_1$ and $\omega_2$ pass through $D$ , the radical axis of the two circles is $\overline{DM} = \overline{BC}$ , so the two circles intersect again on $BC$ .

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AshAuktober

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AshAuktober

#53 h Jun 29, 2024, 4:07 PM • 1 Y

Y by GeoKing

Let $M$ be the midpoint of $BC$ .
Observe that $AB \ne AC$ , else $\omega_1, \omega_2$ cannot exist. Therefore $D \ne M$ .
Claim 1: $\omega_1$ and $\omega_2$ cannot be tangent to each other.
Proof sketch: Note that from the three tangents lemma, $MD = ME = MF$ . But the key idea is at least one of $\angle MDE, \angle MDF$ is $\ge 90^\circ$ . Therefore, $M$ cannot be the circumcentre of $\Delta DEF$ . (in the case that $AB = AC$ , we get a contradiction anyways as $M = D$ .)

Therefore, $\omega_1$ and $\omega_2$ meet at $P \ne D$ .

Claim: The radical axis of the above two circles is line $BC$ .
(Note that this implies the statement)

Proof: Again from the Three Tangents Lemma, $M$ is the radical centre of $\omega, \omega_1, \omega_2$ , so it lies on the radical axis of these circles. But so does $D$ , so the radical axis is line $MD$ which is also line $BC$ . $\square$

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Kappa_Beta_725

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Kappa_Beta_725

#54 h Jul 31, 2024, 2:18 PM

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Inversion with radius $\sqrt{AF.AB}$ switches $(F,B),(H,D),(E,C),(\omega, BC)$ . Hence $\omega_1^*$ and $\omega_2^*$ intersect at $H$ humpty point of $HBC$ which lies on $(EHF)$ (well known). QED

I am only capable of inverting and finding 2 line solutions. Dying of skill issue.

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Eka01

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Eka01

#55 h Aug 4, 2024, 4:52 AM • 1 Y

Y by Sammy27

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MagicalToaster53

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MagicalToaster53

#56 h Sep 3, 2024, 1:29 AM

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Observe that the radical center of $\omega, \omega_1, \omega_2$ is $M$ , where $M$ is the midpoint of $\overline{BC}$ (by three tangents lemma). Now as $\omega_1, \omega_2$ meet at $D$ on $\overline{BC}$ , it follows that the radical axis of $\omega_1, \omega_2$ is $\overline{BC}$ , so that their second point of intersection (not equal to $D$ ) likewise lies on $\overline{BC}$ , as desired.

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Mathandski

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Mathandski

#57 h Oct 8, 2024, 9:15 PM

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$\text{[math]}$

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cj13609517288

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cj13609517288

#58 h Dec 4, 2024, 8:10 PM

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Let $M$ be the midpoint of $BC$ . It is well known that $ME,MF$ are tangent to $\omega$ and $DMEF$ is cyclic.

We claim that $P=EF\cap BC$ . It suffices to prove that $P$ lies on $\omega_1$ , so it suffices to prove that $(MP)(MD)=ME^2$ . But $ME^2=(MB)(MC)$ so this turns into the well known harmonic config since $(BC;DP)=-1$ .

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Saucepan_man02

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#59 h Feb 8, 2025, 1:11 AM

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Due to Three Tangents Lemma, $M$ is the radical center of $\omega, \omega_1, \omega_2$ .
Thus, $P=\omega_1 \cap \omega_2$ ( $P \neq D$ ) lies on $MD$ (which is $BC$ ).

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Mathandski

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#60 h Mar 17, 2025, 8:41 PM

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Mathandski wrote:

$\text{[math]}$

Non-inversion solution
Subjective Rating (MOHs) Please contact westskigamer@gmail.com if there is an error with any of my solution for cash bounties by 3/18/2025.

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