Problem 2, Olympic Revenge 2013

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hvaz

148 posts

hvaz

#1 h Jan 26, 2013, 10:17 PM • 4 Y

Y by mathuz, Davi-8191, jhu08, Adventure10

Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$ , with $K$ closer to $B$ than $L$ . Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$ , with $X$ closer to $C$ than $Y$ . Prove that the intersection of $XL$ and $KY$ lies on $BC$ .

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pi37

2079 posts

pi37

#2 h Jan 27, 2013, 3:23 AM • 5 Y

Y by myh2910, Adventure10, Mango247, poirasss, and 1 other user

Let $H$ be the orthocenter of $ABC$ , and let $D,E,F$ be its projections onto $BC,AC,AB$ respectively. Note that the circle with diameter $AB$ goes through $D,E$ , and the circle with diameter $AC$ goes through $D,F$ . Now by PoP
$HY\cdot HX=HD\cdot HA=HK\cdot HL$
So $KLXY$ is cyclic. Additionally, since $AF,AE$ are perpendicular bisectors of $XY,KL$ , $A$ is the center of this circle. Now since $KL,XY$ intersect at $H$ , by Brokard's theorem it suffices to show that the polar of $H$ is $BC$ with respect to $(KLXY)$ , or equivalently that $H,D$ are inverses with respect to this circle.
Now since $\angle BXA=\angle BYA=\frac{\pi}{2}$ , and $F$ is the midpoint of $XY$ , we know $F$ and $B$ are mutually inverse with respect to this circle. Similarly, $E$ and $C$ are mutually inverse. Now since $\angle AFH=\angle ADB=\frac{\pi}{2}$ and $\angle AED=\angle ADC=\frac{\pi}{2}$ , $D$ is mapped to $H$ in this inversion, so the problem is solved.

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apluscactus

73 posts

apluscactus

#3 h Jan 29, 2013, 8:20 PM • 1 Y

Y by Adventure10

Cross ratio let us show that XK and LY also intersect on BC...
Let's denote by H the orthocenter of ABC.Then CXHY and BKHL are both harmonic.
Since they've got a common point H, the results follow immediately.

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mathocean97

606 posts

mathocean97

#4 h Feb 3, 2013, 7:09 AM • 4 Y

Y by AllanTian, Adventure10, Mango247, and 1 other user

An alternative finish after proving that $KLXY$ is cyclic.

Let $E, F$ be the projections from $B$ onto $AC$ , $C$ onto $AB$ . Clearly, $\angle BXA = 90^\circ$ , so $\triangle BXA \sim \triangle BFX$ , so

$BX^2 = BF \cdot BA = BK \cdot BL$ , so $BK$ is tangent to the circumcircle of $KLXY$ . Similarly, since $BY = BX$ , $BY^2 = BK \cdot BL$ , so $BY$ is also tangent.

Similarly, $CL$ and $CK$ are tangent to the circumcircle of $KLXY$ .

So using Pascal's Theorem on $XXLLKY$ gives $XX\cap LK$ , $LL\cap XY$ , and $XL\cap KY$ are collinear, which is the same as $B$ , $C$ and the intersection of $XL$ and $KY$ are collinear, which is equivalent to the problem statement.

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v_Enhance

6865 posts

v_Enhance

#5 h Jul 30, 2013, 5:24 AM • 3 Y

Y by Davi-8191, Adventure10, ItsBesi

Let $ABC$ be an acute triangle. Circle $\omega_1$ , with diameter $AC$ , intersects side $BC$ at $F$ (other than $C$ ). Circle $\omega_2$ , with diameter $BC$ , intersects side $AC$ at $E$ (other than $C$ ). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$ . Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$ . Prove that lines $AB$ , $ML$ , $NK$ are concurrent.

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v_Enhance

6865 posts

v_Enhance

#6 h Jul 30, 2013, 5:30 AM • 2 Y

Y by Adventure10, Mango247

Sorry for posting just a reworded version of the problem, but I wanted the original wording for this.

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mathuz

1510 posts

mathuz

#7 h Oct 7, 2013, 11:59 AM • 1 Y

Y by Adventure10

very nice problem!

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vslmat

154 posts

vslmat

#8 h Oct 7, 2013, 12:43 PM • 2 Y

Y by Adventure10, Mango247

Let the two circles with diameters $AC$ and $AB$ cut at another point $E$ and let $H\equiv BL\cap CY$ , than $H$ is the orthocenter of $\Delta ABC$ and $A, H, E$ are collinear. As $YH. HX = AH. HE = KH.HL$ , quadrilateral $YKXL$ is cyclic, as $AY^{2} = AF. AB = AX^{2} = AH. AE = AD. AC = AL^{2}$ , $A$ it the center of $(YKXL)$ . Now notice that because $AH. AE = AY^{2}$ and $AE\perp BC$ , $BC$ is in fact the polar of $H$ wrt. $(YKXL)$ . But the intersection $G\equiv YK\cap XL$ also lies on the polar of $H$ , thus $G$ lies on $BC$ .

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JuanOrtiz

366 posts

JuanOrtiz

#9 h Jan 15, 2014, 5:46 PM • 1 Y

Y by Adventure10

Notice by power of point at orthocenter that XLKY is cyclic. Then notice that the orthocenter and the foot of the altitude from A to BC are inverse w.r.t this circle. This can be done by inverting lines XY and KL. Therefore polar line of the orthocenter is BC. But XL and KY also must intersect on the polar line of the orthocenter. So we are done.

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jayme

9767 posts

jayme

#10 h Jan 16, 2014, 4:03 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,according with the cocyclicity (with the three chords theorem) of vslmat , we can finish with the Pascal's theorem.
Sincerely
Jean-Louis

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thecmd999

2860 posts

thecmd999

#11 h Mar 15, 2014, 6:59 AM • 7 Y

Y by AE-TheRocket, Durjoy1729, richy, myh2910, Adventure10, Mango247, Stuffybear

Solution

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shinny98NT

62 posts

shinny98NT

#12 h Mar 15, 2014, 9:08 PM • 2 Y

Y by Adventure10, Mango247

Another solution to this interesting problem, which just uses elementary knowledge (angle chasing and similar triangles):

Assume $AB \le AC$ . Let $H$ is the orthocenter. $D$ is the pedal of $H$ on $BC$ . $(AB), (AC)$ are the circle of diameter $AB$ and $AC$ . Observe that $(AB)$ and $(AC)$ meet at $D$ .

$H$ is a point in $(AB)$ , so $HA.HD=HX.HY$ . Similarly, $HA.HD=HK.HL$ . Therefore, $X, Y, L, K$ are on the same circle. Also, $A$ is the intersection of the bisectors of $KL$ and $XY$ . So, $A$ is the center of this circle.

We need to prove: $\angle KYD = \angle XLD$ .
Having, $\angle KYD = \angle HYD - \angle HYK$ , $\angle XLD = \angle HLX - \angle HLD$ .
Now we need to show that: $\angle HYK + \angle HLK = \angle HYD + \angle HLD$ .
It's equivalent to: $\angle KAX = \angle HYD + \angle HLD$ , which is true because $\angle KAX = \angle KAD + \angle XAD$ .

We need orther equal angles: $\angle DYK = \angle DLX$ .
Or $\angle XYD - \angle XYK = \angle XLK - \angle DLK$ .
It's equivalent to: $\angle KAX = \angle XLK + \angle XYK$ . That's true because A is the center of $(XYLK)$ .

From two pairs of equal angles, we've shown that $\Delta DYK$ is similar to $\Delta DLX$ .
As a result, $\Delta DYL$ is similar to $\Delta DKX$ .

Finishing the proof: Let $KY$ and $XL$ meet at $N$ . It's easy to see that $N$ is the intersection of $(DKX)$ and $(DYL)$ .
Assume $(DKX)$ meets $BC$ at $N'$ . We have $L, X, N'$ are collinear. So $\angle LN'C = \angle DKX = \angle DYL$ .
This means $DYLN'$ is a cyclic quadrilateral. Thus, $N \equiv N'$ . We are done.

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sayantanchakraborty

505 posts

sayantanchakraborty

#13 h Dec 6, 2014, 6:04 AM • 2 Y

Y by Adventure10, Mango247

Nice,but a bit too trivial for TST.Of course you have $KH \cdot HM=CH \cdot HD=NH \cdot HL$ so $KLMN$ is cyclic. $C$ lies on the perpendicular bisector of $KM$ as well as $FN$ so it is the center of $KLMN$ . $CH \cdot CD=CE \cdot CA=CN^2$ so $D$ is the inverse point of $H$ w.r.t this circle and consequently $AB$ is the polar line of $H$ .Now the result is just a consequence of Brokard's theorem.

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TelvCohl

2311 posts

TelvCohl

#14 h Dec 6, 2014, 6:20 AM • 4 Y

Y by Atpar, myh2910, Adventure10, and 1 other user

My solution:

Let $H$ be the orthocenter of $\triangle ABC$ .

Since $B$ lie on the polar of $H$ WRT $(AC)$ ,
so we get $Y(B,H;K,L)=-1$ . ... $(1)$
Since $C$ lie on the polar of $H$ WRT $(AB)$ ,
so we get $L(H,C;X,Y)=-1$ . ... $(2)$

From $(1)$ and $(2)$ we get $LX \cap KY \in BC$ .

Q.E.D

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EulerMacaroni

851 posts

EulerMacaroni

#15 h Aug 21, 2016, 3:03 AM • 2 Y

Y by Adventure10, Mango247

Define $H\equiv BK\cap CX$ , $F\equiv AH\cap BC$ ; it is easy to see that an inversion centered at $A$ with power $r^2=AH\cdot AF$ fixes each of $\{Y, K, X, L\}$ , so it follows that $AY=AK=AX=AL$ and thus quadrilateral $YKXL$ is cyclic. Then $BY, BX, CL, CK$ are tangent to $\odot(YKXL)$ , and so $B$ and $C$ lie on the polar of $H$ and we're done by Brokard's theorem.

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EpicBird08

1740 posts

EpicBird08

#61 h Jul 11, 2023, 4:32 PM

Y by

A problem I came across in EGMO

Let $D$ be the second intersection of $\omega_1$ and $\omega_2;$ clearly $D,E,F$ are the feet of the altitudes from $C,B,A$ to $BA,AC,CB,$ respectively.

We will now prove that $\measuredangle CHM = \measuredangle DMC.$ But this is easy, we have $\measuredangle CHM = \measuredangle CHF = 90^\circ - \measuredangle FCH = 90^\circ - (90^\circ - \measuredangle DBC) = \measuredangle DBC = \measuredangle DMC.$ Hence we know that $\triangle CHM \sim \triangle CMD.$ Consequently, $CH \cdot CD = CM^2.$ Similarly, we can deduce that $CH \cdot CD = CM^2 = CN^2 = CK^2 = CL^2.$ Therefore, not only do we know that $K,L,M,N$ are concyclic, but we also know that the polar of $H$ is just the line through $D$ perpendicular to $DH,$ which is precisely $AB.$ On top of that, if $X = ML \cap NK$ and $Y = KL \cap MN,$ then by Brocard the polar of $H$ is $XY.$ Therefore, the lines $XY$ and $AB$ are equivalent to each other. In particular, $X,$ which is the intersection of $ML$ and $NK,$ lies on $AB,$ as desired.

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john0512

4170 posts

john0512

#62 h Sep 3, 2023, 10:17 PM

Y by

Clearly, $E$ and $F$ are just the feet of the altitudes. Let $P$ be the feet of the altitude from $C$ to $AB$ (so that $P$ lies on both $\omega_1$ and $\omega_2$ ). Let $H$ denote the orthocenter of $\triangle ABC$ .

Claim: $MKNL$ is cyclic. By Power of a Point, $HN\cdot HL=HP\cdot HC=HK\cdot HM,$ hence shown.

Furthermore, note that the perpendicular bisector of $NL$ is side $AC$ , and the perpendicular bisector of $KM$ is side $BC$ . Hence, since $MKNL$ is cyclic, its circumcenter is the intersection of these two perpendicular bisectors, which is $C$ .

Note that $\angle ALC=\angle ANC=90$ , so $AN$ and $AL$ are tangent to $(MNKL)$ . Hence, $MNKL$ is a harmonic quadrilateral. Let $NK$ intersect $AB$ at $T_1$ , and let $ML$ intersect $AB$ at $T_2.$ Finally, let $KL$ intersect $AB$ at $X$ .

Projecting through $K$ , we have $(MK;NL)=^K(AB;T_1X)=-1.$ Similarly, $(MK;NL)=^L(T_2X;BA)=-1.$ Hence, $T_1=T_2$ , as desired.

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signifance

140 posts

signifance

#63 h Sep 23, 2023, 6:15 AM

Y by

Relatively easier problem

Since H is on the radax of the two circles (it's on CD), HL\cdot HN=HK\cdot HM so KLMN is cyclic; in particular, since the perp. bisectors of MK,NL are BC,AB, respectively (due to them passing through the center and perp.), C is the center of (KLMN). It suffices to prove that AB is the polar of H, which means it suffices to prove CH\cdot CD=CM^2, since it would finish by Brocard's. Indeed, CMB\sim CFM\implies CM^2=CF\cdot CB=CH\cdot CD, as desired.

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IAmTheHazard

5000 posts

IAmTheHazard

#64 h Nov 17, 2023, 2:56 PM

Y by

IAmTheHazard wrote:

We use the actual wording that Evan wrote

Evidently, $E$ and $F$ are the feet of the obvious altitudes, so add $D$ , the foot of the $C$ -altitude, so $ADFC$ and $BDEC$ are cyclic. By radical center on these two circles, it follows that $KLMN$ is cyclic as well. Furthermore, because $C$ lies on the perpendicular bisectors of $\overline{LN}$ and $\overline{KM}$ , $C$ is the center of $(KLMN)$ .
Now, note that since $\angle ALC=\angle ANC=90^\circ$ , $A$ is the intersection of the tangents at $L$ and $N$ to $(KLMN)$ , hence $A$ and $E$ are polar and $CE\cdot CA=r^2$ where $r$ is the radius of $(KLMN)$ . But since $AEHD$ is also cyclic, we have $CH \cdot CD=r^2$ as well, hence $\overline{AB}$ is the polar of $H$ . By Brocard, it thus follows that we should have $\overline{KN} \cap \overline{LM} \in \overline{AB}$ , as desired. $\blacksquare$

ive done this problem before its actually over

We first prove the following lemma.

Lemma: Let $\ell$ be a line and $A$ be a point on it. Suppose lines $\ell,\ell_1,\ell_2$ are concurrent at a point $X$ . Let $P_1,P_2 \in \ell_1$ and $Q_1,Q_2 \in \ell_2$ such that $\ell$ bisects $\angle P_1AP_2$ and $\angle Q_1AQ_2$ . Then $\overline{P_1Q_1} \cap \overline{P_2Q_2}$ lies on the perpendicular to $\ell$ through $A$ .
Proof: Let the perpendicular to $\ell$ through $A$ intersect $\ell_1$ and $\ell_2$ at $T_1$ and $T_2$ respectively. Then $(P_1,P_2;X,T_1)=(Q_1,Q_2;X,T_2)=-1$ , so by Prism Lemma $\overline{P_1Q_1} \cap \overline{P_2Q_2}$ lies on $\overline{T_1T_2}$ as desired.

Use Evan's wording, and let $D$ be the foot of the $C$ -altitude. Clearly $\overline{CD}$ bisects $\angle LDN$ and $\angle KDM$ , so we are done by the lemma. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Nov 17, 2023, 2:57 PM

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shendrew7

792 posts

shendrew7

#65 h Nov 22, 2023, 11:15 PM

Y by

v_Enhance wrote:

Let $ABC$ be an acute triangle. Circle $\omega_1$ , with diameter $AC$ , intersects side $BC$ at $F$ (other than $C$ ). Circle $\omega_2$ , with diameter $BC$ , intersects side $AC$ at $E$ (other than $C$ ). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$ . Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$ . Prove that lines $AB$ , $ML$ , $NK$ are concurrent.

We first note $\omega_1$ and $\omega_2$ will pass through the feet of the altitudes in $\triangle ABC$ . We see $KLMN$ is cyclic, as
$HK \cdot HM = HB \cdot HE = HC \cdot HG = HL \cdot HN,$
where $G$ is the foot from $C$ to $AB$ . In particular, the center of $(KLMN)$ is the intersection of the perpendicular bisectors of $KM$ and $LN$ , or $C$ .

Next we find that the inverse of $H$ with respect to $(KLMN)$ is $G$ , as
$CH \cdot CG = CF \cdot CB = CK^2.$
Thus $AB$ is the polar of $H$ , and Brocard tells us $ML \cap NK$ lies on $AB$ . $\blacksquare$

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asdf334

7586 posts

asdf334

#66 h Nov 23, 2023, 3:17 AM

Y by

short.

Step 1: Consider $(BC)$ ; the power of $A$ with respect to this circle is $AX^2=AY^2=AK^2=AL^2$ , and now I claim $H=XY\cap KL$ has polar $BC$ .

Step 2: Previous claim is trivial after noting that inverting $B$ and $C$ around $(YKXL)$ maps them to the feet of the $C$ and $B$ -altitudes.

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dolphinday

1318 posts

dolphinday

#68 h Feb 13, 2024, 4:13 PM

Y by

Let $D$ be the second intersection point. Note that $AF$ , $BE$ , and $CD$ concur at the orthocenter $H$ of $\triangle ABC$ . Then from Radical Axis, we have $KLMN$ cyclic since $\overline{KM} \cap \overline{LN} \cap \overline{CD} = H$ . Then let $\overline{KL} \cap \overline{MN} = X$ and $\overline{KN} \cap \overline{LM} = Y$ . By Brokard's, $H$ is the pole of $XY$ wrt $(KLMN)$ . We wish to show that $AB$ is also the pole of $H$ which would imply that $X \in AB$ . Note that $\angle MLC = \angle MNC = 90^{\circ} \implies LA, LN$ tangents to $(MLKN)$ so $MLKN$ is a harmonic quadrilateral. We also have $(A, H; K, M) = -1$ and similarly $(B, H; L, N) - 1$ . Note that $(A, H; K, M) \overset{X}= (B, H; XK \cap BN, L) = -1$ which implies that $XK \cap BN = N$ which finishes by symmetry.

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Pyramix

419 posts

Pyramix

#69 h Mar 9, 2024, 7:40 PM

Y by

We rephrase the problem because $C-$ centered problems are... annoying. Great problem though

Problem:
Let $ABC$ be an acute triangle. Circle $\omega_1$ , with diameter $\overline{AC}$ , intersects side $\overline{AB}$ at $F$ (other than $A$ ). Circle $\omega_2$ , with diameter $\overline{AB}$ , intersects side $\overline{AC}$ at $E$ (other than $A$ ). Ray $CF$ intersects $\omega_2$ at $K$ and $M$ with $CK < CM$ . Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$ . Prove that lines $BC, ML, NK$ are concurrent.

Solution:
We know that $BE,CF$ are altitudes. Let $AD$ be the third altitude and $H$ be the orthocenter. Note that since $\overline{AC}$ is the diameter of $\omega_1$ , and $BN\perp AC$ . Hence, $ALNC$ is a cyclic kite, which means that $(A,C;L,N)=-1$ . Projecting onto line $BN$ with perspective at $D$ gives $(B,H;L,N)=-1$ . Similarly, we get $(C,H;K,M)=-1$ .

Let $T=LK\cap BC$ , $X_1=NK\cap BC$ and $X_2=ML\cap BC$ .
Projecting $(B,H;L,N)$ onto line $BC$ with perspective at $K$ gives $(B,C;T,X_1)=(B,H;L,N)=-1$ , while projecting $(C,H;K,M)$ onto line $BC$ with perspective at $L$ gives $(B,C;T,X_2)=(C,H;K,M)=-1$ . Hence, $(B,C;T,X_1)=(B,C;T,X_2)\Longrightarrow X_1=X_2$ . Hence, $ML, NK, BC$ concur at a point. $\blacksquare$

Attachments:

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eibc

596 posts

eibc

#70 h Mar 11, 2024, 10:43 PM • 1 Y

Y by Pyramix

Swap the definitions of $A$ and $C$ as @above does, and let $H$ be the orthocenter of $\triangle ABC$ . By symmetry, we know that $E$ is the midpoint of $\overline{LN}$ , and since
$BL \cdot BN = BD \cdot BC = BH \cdot BE,$ by some famous harmonic bundle lemma we have $(B, H; L, N) = -1$ . So,
$-1 = (B, H; L, N) \overset{M}{=} (B, C; \overline{ML} \cap \overline{BC}, \overline{MN} \cap \overline{BC}).$ Analogously, we find that $(B, C; \overline{NK} \cap \overline{BC}, \overline{MN} \cap \overline{BC}) = -1$ , so $ML$ and $NK$ intersect $BC$ at the same point, and we are done.

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shendrew7

792 posts

shendrew7

#71 h Jun 6, 2024, 11:22 PM

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v_Enhance wrote:

Let $ABC$ be an acute triangle. Circle $\omega_1$ , with diameter $AC$ , intersects side $BC$ at $F$ (other than $C$ ). Circle $\omega_2$ , with diameter $BC$ , intersects side $AC$ at $E$ (other than $C$ ). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$ . Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$ . Prove that lines $AB$ , $ML$ , $NK$ are concurrent.

Notice $KLMN$ is cyclic with center $C$ , as
$CK = CM = \sqrt{CF \cdot CB} = \sqrt{CE \cdot CA} = CL = CN.$
By Brocard, we have that $ML \cap NK$ lies on the polar of $MK \cap NL = H$ , which is simply $AB$ . $\blacksquare$

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Ywgh1

136 posts

Ywgh1

#72 h Aug 13, 2024, 8:11 AM

Y by

USA 2013 TST
Let the tangent at $A$ intersect $\omega_2$ at $X$ and $Y$ , then we have the claim.

Claim: $X-H-Y$ are collinear.
Pf: Radical axis.

Then we have that $(AH;KM)=-1$ , and $(B,H;N,L)=(B,H;L,N)=-1$ , applying prism Lemma finishes the problem.

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Eka01

204 posts

Eka01

#73 h Aug 17, 2024, 7:04 AM • 1 Y

Y by AaruPhyMath

Similar as the above posted but doing it anyway.
It is obvious that $H$ lies on the radical axis of the mentioned circles which implies $HK.HM=HL.HN$ giving us that $(KLMN)$ is cyclic.
Since $AC$ is a diamter in a circle having $KM$ as a chord perpendicular to it, this implies it bisects it. Similarly $C$ also lies on the $\perp$ bisector of $LN$ which implies $C$ is the center of $(KLMN)$ . Now it is obvious that $CL \perp AL$ so $AL$ and similarly $AN$ are tangent to $(KLMN)$ so $A$ is pole of $LN$ . Similarly $B$ is pole of $KM$ , so $AB$ is the polar of $KM \cap LN$ which is $H$ , but by brokard's, $KN \cap LM$ must lie on this polar which is what we needed to show.

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SimplisticFormulas

81 posts

SimplisticFormulas

#74 h Dec 15, 2024, 1:30 PM

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who else is here from egmo?
(Using @v-Enhance’s statement)

Observe that
a)The two given circles meet in $D$ , the foot of perpendicular from $C$ on $AB$ .
b) $CD,BE,AF$ concur in $H$ , the orthocentre of $\triangle ABC$
c) $MLKN$ is cyclic since $NH \cdot HL=DH \cdot HC=KH \cdot HM$
d) $C$ is the centre of $\odot (MLKN)$ since the perpendicular bisectors $LN,KM$ of $AC,BC$ meet in $C$
e) $AB$ is the polar of $H$ since $AB \perp CD$ and $CE \cdot CA=CN^2=CH \cdot CD$ , so inversion around $\odot(MLKN)$ sends $H$ to $D$
f) $ML,KN$ meet in the polar of $H$ , that is, $AB$ $\blacksquare$

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Ilikeminecraft

302 posts

Ilikeminecraft

#75 h Mar 16, 2025, 1:33 AM

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Note that $HM\cdot HN = HA\cdot HD = HK\cdot HL,$ so $LMKN$ is cyclic. Furthermore, $AM = AN$ since $AC$ is diameter and $AC\perp MN,$ so $AM = AN$ while $AL = AK.$ Thus, $A$ is the center
Furthermore, $AM^2 = AC\cdot AE.$
Next, observe that $CM^2=CE\cdot CA = CK\cdot CL,$ which implies $CM$ os tangent to $(KLMN).$
Finally, Pascal’s on $KKMLLN$ finishes.

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MonkeyLuffy

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MonkeyLuffy

#76 h Mar 18, 2025, 1:05 AM

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