Prove that the lines AM,DN,XY are concurrent

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orl

#1 h Nov 9, 2005, 9:16 PM • 11 Y

Y by Davi-8191, nguyendangkhoa17112003, samrocksnature, Adventure10, HWenslawski, jhu08, megarnie, Mango247, Rounak_iitr, Tastymooncake2, and 1 other user

Let $A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$ . The line $XY$ meets $BC$ at $Z$ . Let $P$ be a point on the line $XY$ other than $Z$ . The line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$ , and the line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$ . Prove that the lines $AM,DN,XY$ are concurrent.

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Virgil Nicula

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Virgil Nicula

#2 h Nov 9, 2005, 11:42 PM • 6 Y

Y by Tranthanhtung, samrocksnature, Adventure10, jhu08, Mango247, Tastymooncake2

I note the intersection $R\in AM\cap DN$ . We will analize the all radical axisis between the circumcircles of the (inscribed) quadrilaterals $AMPZ,\ DNPZ$ and the circumcircles with the diameters $AC,\ BD$ a.s.o.

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Virgil Nicula

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Virgil Nicula

#3 h Nov 10, 2005, 11:51 AM • 9 Y

Y by dan23, Tranthanhtung, galav, richy, Inconsistent, samrocksnature, Adventure10, jhu08, Tastymooncake2

I note: $C(XY)$ - the circle with the diameter $XY;$
$\ R\in AM\cap DN,\ R_1\in AM\cap XY,\ R_2\in DN\cap XY$ ;
the circles $a=C(AC),\ b=C(BD),\ m=C(AP),\ n=C(DB)$ ;
the radical axises $d_1,\ d_2$ for the pairs of the circles $(a,n),\ (b,m)$ respectively.

The radical center of the circles $(a,b,m)$ is $R_1\in XY\cap AM\cap d_2\Longrightarrow R_1\in d_2$ ;
The radical center of the circles $(a,b,n)$ is $R_2\in XY\cap DN\cap d_1\Longrightarrow R_2\in d_1$ ;
The radical center of the circles $(a,m,n)$ is $R_1\in AM\cap XY\cap d_1\Longrightarrow R_1\in d_1$ ;
The radical center of the circles $(b,m,n)$ is $R_2\in DN\cap XY\cap d_2\Longrightarrow R_2\in d_2.$

Therefore, $R_1\equiv R_2\in d_1\cap d_2\cap XY\cap AM\cap DN\Longrightarrow R_1\equiv R_2\equiv R\in XY.$

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e.lopes

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e.lopes

#4 h Nov 10, 2005, 4:31 PM • 12 Y

Y by anthony97, claserken, Abdollahpour, huyrude411, phoenixfire, myh2910, samrocksnature, Adventure10, wasikgcrushedbi, Yashashwi_Singhania, Tastymooncake2, Rounak_iitr

we know that ${XY}\perp{AD}$ at $Z$ .

let ${AM}\cap{XY} = R$ .

we have $\angle{RMC} = 90$ (AC is diameter)

so, $RMCZ$ is cyclic, that implies $PR.PZ = PM.PC$ .

let ${ND}\cap{XY} = S$ .

analogous, we have $PS.PZ = PN.PB$ .

$XY$ is the radical axes of the two circles, so, $PM.PC = PN.PB \Longrightarrow PR = PS \Longrightarrow R = S$ .

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jorgejacinto9701

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jorgejacinto9701

#5 h Sep 24, 2007, 12:24 AM • 2 Y

Y by samrocksnature, Adventure10

I view the four circles have the same eje radical

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limac

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limac

#6 h Dec 5, 2010, 6:18 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Suppose $E = AM \cap XY$ and $N' = ED \cap XY.$ We have $\angle AMC = 90^\circ.$ Let $\angle MAC = \alpha, \angle EDZ = \beta, \angle AEZ = 180^\circ - \alpha = \theta_1,$ and $\angle DEZ = 180^\circ - \beta = \theta_2.$

Now if we show that $\angle PBZ = \theta_2,$ we show that $N' = N$ and we are done. Since $Z$ lies on the radical axis $XY$ of the two given circles, $(AZ)(CZ) = (BZ)(ZD).$ It can easily be seen that $\triangle EAZ \sim \triangle CPZ$ from which we get $(EZ)/(CZ) = (AZ)/(ZP).$ Thus, $(EZ)(ZP) = (AZ)(CZ) = (BZ)(ZD),$ so $(EZ)/(BZ) = (ZD)/(ZP),$ so $\triangle BPZ \sim \triangle EDZ,$ and we are done.

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xeroxia

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xeroxia

#7 h Mar 19, 2011, 10:06 PM • 6 Y

Y by iarnab_kundu, john10, phoenixfire, samrocksnature, Elyson, Adventure10

Let $R = AM \cap DN$ . $P$ is a point on radical axis of the circlec with diameter $AC$ and $BC$ .
So we have $PC\cdot PM = BP \cdot PN$ which implies $B,C,N,M$ are concyclic.
That is $\angle NMC = \alpha = \angle NBD \Rightarrow \angle BDN = 90^{\circ}-\alpha$ and $\angle AMN=90^{\circ}+\alpha$ which also implies $A,M,N,D$ are concyclic.
Thus $RM\cdot RA = RN \cdot RD$ which means $R$ is a point on the radical axis of those circles. So $R,X,P,Y$ are concurrent.

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SCP

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SCP

#8 h Jun 23, 2011, 6:10 PM • 2 Y

Y by samrocksnature, Adventure10

This message might beput away

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Learner94

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Learner94

#9 h Nov 28, 2011, 4:14 PM • 6 Y

Y by dan23, anthony97, phoenixfire, samrocksnature, Adventure10, and 1 other user

Solution using Menelaus' theorem:

Let $AM \cap XY = T$ . Equivalently we have to prove that $T,N,D$ are collinear. By Menelaus' theorem $T,N,D$ will be collinear if and only if $\frac{ZT}{PT}\cdot \frac{PN}{BN} \cdot \frac{BD}{DZ} = -1$

Observe that $\triangle AMC \sim \triangle PZC, \implies \frac{CA}{CM}= \frac{PC}{ZC}$ and $\triangle BND \sim \triangle PZB \implies \frac{PB}{ZB}= \frac{BD}{BN}$ . By intersecting chord theorem $PB \cdot PN = PM\cdot PC$ and $ZB \cdot ZD = ZA\cdot ZC$ .

Note that $\frac{PM}{ZA}\cdot \frac{CA}{CM} = \frac{PM\cdot PC}{ZC \cdot ZA} = \frac{PM\cdot PC}{ZB\cdot ZA} = \frac{PB\cdot PN}{ZB \cdot ZD}= \frac{BD\cdot PN}{ZD \cdot BN }$ .

Since $T,M,A$ are collinear $\frac{ZT}{PT}\cdot \frac{PM}{CM} \cdot \frac{CA}{AZ} = \frac{ZT}{PT}\cdot \frac{PN}{BN} \cdot \frac{BD}{DZ} =-1$ . $Q.E.D$

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dgrozev

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dgrozev

#10 h Mar 11, 2012, 9:06 AM • 4 Y

Y by huricane, phoenixfire, samrocksnature, Adventure10

Let first establish that:

(1) $\frac{ZA}{ZB}=\frac{ZD}{ZC}$ .

From the right triangle $AXC$ we get $ZA.ZC=ZX^2$ . Similarly, from the right triangle $BXD$ it follows: $ZB.ZD = ZX^2$ . Hence $ZA.ZC=ZB.ZD$ , which implies (1).

Now the homothety with center $Z$ and ratio $k=\frac{ZA}{ZB}$ will send the altitude through $B$ to $PC$ , to the line $AM$ ; the altitude through $C$ to $BP$ will go to the line $DN$ , and the line $ZP$ (which is the altitude through $P$ to $BC$ ) will be sent to itself.

This means that $AM, DN, ZP$ are concurent and its common point is the image of the orthocenter of $\Delta BCP$ by this homothety.

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HAPPYNEWYEAR1729

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HAPPYNEWYEAR1729

#11 h Apr 22, 2013, 3:56 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Let $AD \cap XY=L$
Notice that $ALOM$ and $NDLO$ are cyclic .So $\angle OMY=\angle MOY$ and $\angle YON=\angle YNO$ .
So O is the circumcentre of $\triangle OMN$ .and thus by angle chasing we get $AMND$ cyclic.THus the lines $AM$ and $DN$ are meeting on the radical axis.

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sayantanchakraborty

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sayantanchakraborty

#12 h Apr 28, 2014, 4:19 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

This problem is easy but tasty....

Notations: $AM \cap DN=K,\angle{BNM}=\theta,\odot{AMC}=\Omega_1,\odot{BND}=\Omega_2$ .

ProofSince $P$ lies on the radical axis $XY$ of $\Omega_1$ and $\Omega_2$ ,we have $BP*PN=CP*PM \Rightarrow B,C,N,M$ concyclic.So $\angle{MCB}=\angle{MNB}=\theta$ .Hence $\angle{MAD}=90^{\circ}-\theta \Rightarrow M,A,D,N$ concyclic.Hence $KM*KA=KN*KD \Rightarrow K$ lies on the radical axis $XY$ as desired.

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infiniteturtle

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infiniteturtle

#13 h May 19, 2014, 1:21 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Let $Q=AM\cap XY$ and let $QD$ intersect the circle with diameter $BD$ at $D, N'$ . We wish to show $N'\equiv N$ .
Note that
$\angle AMC =\angle QMP =\angle BND =\angle BN'D =\angle AZQ =\angle QN'P=\frac{\pi}{2}.$
Then $\angle QMP +\angle QN'P = \pi$ , so $QMPN'$ is cyclic. Now $\angle MPN' =\angle BPC$ and
$\angle MN'P=\angle MQP =\frac{\pi}{2}-\angle QAC =\angle BCP,$
so $\triangle MN'P\sim \triangle BPC$ . Hence,
$\frac{PM}{PB}=\frac{PN'}{PC}\implies PM\cdot PC=PB\cdot PN'.$
But since $P$ is on the radical axis,
$PM\cdot PC=PB\cdot PN\implies PB\cdot PN=PB\cdot PN'\implies PN=PN'.$
Hence $N\equiv N'$ and $AM,XY,DN$ concur as desired. $\blacksquare$

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utkarshgupta

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utkarshgupta

#14 h May 19, 2014, 9:35 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

A very nice proof is also given here:
https://www.math.ust.hk/excalibur/v5_n3.pdf

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DaChickenInc

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DaChickenInc

#15 h Jun 22, 2014, 10:25 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

This is from the book! There's no radical axes or bashing.
Thought Process
Solution
The second paragraph of the solution is well-known.

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blueberryfaygo_55

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blueberryfaygo_55

#63 h Mar 25, 2024, 9:29 PM

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Let $T = AM \cap DN$ . Since $XY$ is the radical axis of $(AC)$ and $(BD)$ , it suffices to show that $T$ also lies on the radical axis of $(AC)$ and $(BD)$ .

Claim: $AMND$ is cyclic.
Proof. Since $P \in XY$ , $P$ lies on the radical axis of $(AC)$ and $(BD)$ . It follows that $\text{Pow}_{(AC)} P = \text{Pow}_{(BD)} P$ , or $PM \cdot PC = PN \cdot PB$ Thus, $BMNC$ is cyclic. Further, since $AC$ is the diameter of $(AC)$ and $M$ lies on $(AC)$ , we have $\angle AMC = 90^{\circ}$ . By angle chasing, we have $\begin{align*} \angle MAD &= 90^{\circ} - \angle MCA \\ &= 90^{\circ} - \angle MNB \\ \end{align*}$ However, we also have $\angle MND = 90^{\circ} + \angle MNB$ , which leads to $\angle MND + \angle MAD = 180^{\circ}$ , and the conclusion follows. $\blacksquare$

Then, by Power of a Point, we have $TM \cdot TA = TN \cdot TD$ , which is equivalent to $T$ lying on the radical axis of $(AC)$ and $(BD)$ , as desired. $\blacksquare$

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joshualiu315

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joshualiu315

#64 h Apr 24, 2024, 5:50 PM

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Note that $\operatorname{Pow}_{(AC)} (P) = \operatorname{Pow}_{(BD)} (P)$ , so $BNMC$ is cyclic. Hence,

$\angle MAD = 90^\circ - \angle MCB = \angle MNB - 90^\circ = \angle MND,$
which implies $ANMD$ is cyclic.

Then, note that $Z$ lies on $(AP)$ and $(DP)$ , and using the radical center of $(AP)$ , $(DP)$ , and $(ANMD)$ concludes.

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MagicalToaster53

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MagicalToaster53

#65 h Sep 15, 2024, 12:59 AM

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Let the point $Q$ be the intersection of $AM$ and $DN$ . We use directed angles $\measuredangle$ modulo $\pi$ . It suffices to then show that $MNDA$ is cyclic, as then $QM \cdot QA = QN \cdot QD,$ which would imply $Q$ has the same respective power to both circles and consequently $Q, X, Y$ are collinear.

Make the first observation that $QMPN$ is cyclic as $\measuredangle QMP = \measuredangle QMC = 90^{\circ} = \measuredangle QNB = \measuredangle QNP.$ We subsequently discover
$\begin{align*} \measuredangle QMN &= 90^{\circ} - \measuredangle NMC \\ &= 90^{\circ} - \measuredangle NBC \\ &= 90^{\circ} - \measuredangle NBD \\ &= \measuredangle BDN, \end{align*}$ so that indeed $MNDA$ is cyclic, as was to be shown. $\blacksquare$

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DEKT

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DEKT

#66 h Oct 7, 2024, 11:49 AM

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Can cord-bash work??

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maths_enthusiast_0001

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maths_enthusiast_0001

#67 h Oct 14, 2024, 1:41 PM

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Claim: Denote $AM \cap DN =K$ then, $K$ lies on the line $XY$ .
Proof: Denote the circumcircles of $\Delta AMC$ and $\Delta BND$ as $\Omega_{1}$ and $\Omega_{2}$ . Note that $XY$ is the radical axis of $\Omega_{1}$ and $\Omega_{2}$ . Thus by Power of Point, $BP.PN=CP.PM$ which implies that points $B,M,N,C$ are concyclic. As $AC$ is the diameter of $\Omega_{1}$ and $BD$ is the diameter of $\Omega_{2}$ , we have $\angle AMC=\angle BND=90^{\circ}$ . Thus, $90^{\circ}-\angle MAC=\angle MCB=\angle MNB \implies \angle MND= \angle MNB+\angle BND=180^{\circ}-\angle MAD$ . Hence, points $A,M,N,D$ are concylic and as $K=AM \cap DN$ which implies, $KM.KA=KN.KD$ or the power of $K$ with respect to $\Omega_{1}$ and $\Omega_{2}$ is same $\implies K \in XY$ as desired. $\blacksquare$

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Sadigly

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Sadigly

#68 h Nov 18, 2024, 7:38 PM

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Since $AC$ and $BD$ are diameters, $\measuredangle AMC=90$ and $\measuredangle BND=90$
Claim: $MCBN$ is cyclic
Proof:
$XP\times PY=BP\times PN$
$XP\times PY=MP\times PC=BP\times PN\Rightarrow MCBN$ cyclic

$\measuredangle MND=\measuredangle MNB+\measuredangle BND=90+\measuredangle MNB=90+\measuredangle MCB=90+\measuredangle MCA$
$\measuredangle MCA+\measuredangle AMC+\measuredangle CAM=180\Rightarrow \measuredangle CAM=90-\measuredangle MCA\Rightarrow \measuredangle MAC=\measuredangle MCA-90\Rightarrow \measuredangle MAD=\measuredangle MCA-90=\measuredangle MCA+90$

$\measuredangle MAD=\measuredangle MCA+90=\measuredangle MND\Rightarrow MAND$ cyclic.

Circle $MAND$ and $AMCY$ intersects at $A;M\Rightarrow AM$ is the radical axis of these 2 circles.
Circle $MAND$ and $BNDY$ intersects at $B;N\Rightarrow BN$ is the radical axis of these 2 circles.
Circle $AMCY$ and $BNDY$ intersects at $X;Y\Rightarrow XY$ is the radical axis of these 2 circles.
SInce $P\neq Z$ ,center of these 3 circles aren't colinear,so their radical axes( $AM;BN;XY$ ) concur.

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DensSv

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#69 h Nov 18, 2024, 8:11 PM

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Vedoral

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#70 h Dec 20, 2024, 10:48 PM

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MR.1

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MR.1

#71 h Jan 14, 2025, 10:44 AM • 1 Y

Y by MIC38

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ehuseyinyigit

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ehuseyinyigit

#72 h Jan 16, 2025, 8:53 PM

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$XP\cdot PY=BP\cdot PN=CP\cdot PM$ gives $BCMN$ concyclic. Recall that $AC$ and $BD$ was diameters. Hence, $\angle MAC=90^{\circ}-\angle MCA=90^{\circ}-\angle MNB$ implies $ADMN$ concyclic. Then, $AM$ , $DN$ and $XY$ are concurrent as desired.

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QueenArwen

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QueenArwen

#73 h Feb 14, 2025, 4:18 PM

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Let the circle with diameter $AC$ be $\omega_1$ and the circle with diameter $BD$ be $\omega_2$ . Let $Q$ be the intersection of $AM$ and $XY$ and $N'$ be the intersection of $QD$ with $\omega_2$ . Since $AC$ is the diameter of $\omega_1$ , $\angle AMC = \angle QMC = 90$ . Since $XY$ is the radical axis of $\omega_1$ and $\omega_2$ , $\angle QZD = \angle QZA = 90$ so $\triangle QMP \sim \triangle QZA$ by AAA. Hence $\frac {QM}{QZ} = \frac {QP}{QA}$ so $QM\times QA = QP\times QZ$ . From power of a point, $QM\times QA = QN'\times QD$ so $QP\times QZ = QN'\times QD$ . Hence $\frac{QP}{QD} = \frac{QN'}{QZ}$ so $\triangle QN'P \sim \triangle QZD$ , so $\angle QN'P = 90 = \angle PN'D$ as $QN'D$ is a straight line. Thus $PZDN'$ is cyclic so the radical axis of $PZDN'$ and $\omega_2$ is $DN'$ . However $PZDN$ is also cyclic (since $\angle BND = \angle PND = 90$ ) so the radical axis of $PZDN$ and $\omega _2$ is $DN$ . Since two circles cannot have more than one radical axis ( $PZDN$ and $PZDN'$ are the same circles), $N'$ must lie on $DN$ . But since $N$ and $N'$ lie on $\omega_2$ , $N=N'$ so $AM$ , $DN$ and $XY$ are concurrent

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Tony_stark0094

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Tony_stark0094

#74 h Feb 27, 2025, 8:10 AM

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let AM and DN meet at P'
we have:
$\angle AMC=90 \ and \angle DNB =90$
==> PMP'N is cyclic moreover BN and MC intersect on radial axis so BMNC is cyclic
Now:
$\angle CBP=\angle CMN = \angle PP'N = 90-\angle P'PN$
also
$\angle CBP= \angle 90-\angle BPZ=90-\angle XPB$
so
$\angle XPB= \angle P'PN$
hence X-P-P'

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ali123456

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ali123456

#75 h Mar 1, 2025, 12:07 PM

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So let's start!
$\textbf{Claim:}$ $MNCB$ is concyclic
$\textbf{Proof:}$ $PN.PB=P(P,(NDBX))=P(P,(AMC))=PM.PC$
$\textbf{Claim:}$ $MADN$ is concyclic
$\textbf{Proof:}$ $\angle{MND}=\angle{MNB}+90=\angle{MCA}+90=180-\angle{MAD}$
$\textbf{Conclusion:}$ we conclude the result by the radical center of the circles $(MAC),(NBD)$ and $(MNDA)$
:cool:

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shendrew7

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shendrew7

#76 h Mar 10, 2025, 6:46 AM

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Average projective overkill. We notice that
$\begin{align*} -1 = (A,C;X,Y) &\overset{M}{=} (AM \cap XY, P; X, Y) \\ -1 = (D,B;X,Y) &\overset{N}{=} (DN \cap XY, P; X; Y). \quad \blacksquare \end{align*}$

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golden_star_123

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golden_star_123

#77 h Mar 24, 2025, 12:02 AM

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$\textbf{Claim: BMNC is cyclic}$

$\textit{Proof.}$ Choose points $M, C$ on the circle with diameter $AC$ , and choose point $B,N$ on the circle with diameter $BD$ . As $CM$ and $BN$ both pass through point $P$ which lies on the radical axis of both circles, $XY$ , we must have $BMNC$ cyclic, as desired.

As $BMNC$ is cyclic, we know that $\angle BMC = \angle BNC$ . Also, as $AC$ and $BD$ are both diameters of their respective circles, we also know that $\angle AMC = \angle BND = 90^{\circ}$ , which leads us to the conclusion that $\angle AMB = \angle DNC$ .

Finally, $\angle ANC + \angle DNC = \angle DMB + \angle AMB$ , so $\angle AMD = \angle DNA$ and $AMND$ is cyclic.

Then the pairwise radical axis of $(AMND)$ , the circle with diameter $AC$ , and the circle with diameter $BD$ must be concurrect, which implies our result.

Can someone check my proof as I am new to radical axis?

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