Cyclic points

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MithsApprentice

2390 posts

MithsApprentice

#1 h Oct 27, 2005, 7:51 PM • 3 Y

Y by Adventure10, mathlearner2357, Rounak_iitr

An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\, AB \,$ intersects altitude $\, CC' \,$ and its extension at points $\, M \,$ and $\, N \,$ , and the circle with diameter $\, AC \,$ intersects altitude $\, BB' \,$ and its extensions at $\, P \,$ and $\, Q \,$ . Prove that the points $\, M, N, P, Q \,$ lie on a common circle.

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Virgil Nicula

7054 posts

Virgil Nicula

#2 h Oct 27, 2005, 8:43 PM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

$D\in BC,\ AD\perp BC;\ E\in CA,\ BE\perp CA;\ F\in AB,\ CF\perp AB\Longrightarrow$

$HM\cdot HN=HB\cdot HE=HA\cdot HD=HP\cdot HQ\Longrightarrow HM\cdot HN=HP\cdot HQ\Longrightarrow$

The points $M,N,P,Q$ lie on a common circle.

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darij grinberg

6555 posts

darij grinberg

#3 h Oct 27, 2005, 8:44 PM • 2 Y

Y by Adventure10, Mango247

See http://www.mathlinks.ro/Forum/viewtopic.php?t=42613 .

darij

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Virgil Nicula

7054 posts

Virgil Nicula

#4 h Oct 27, 2005, 9:01 PM • 1 Y

Y by Adventure10

Dear Darij, this problem is very easily and its author made a mistake because he posted it here. The offered solution (by me and you) - is classical and, I think, one solution.

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dgreenb801

1896 posts

dgreenb801

#5 h Oct 23, 2009, 12:41 PM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

By the converse of the radical axis theorem, we have to show MN,PQ, and the perpendicular from A to BC concur, but this is true as they are the altitudes of triangle ABC.

If you want a more direct proof, let the circumcircle of PMQ intersect the circle diameter AB at X. By the radical axis theorem, PQ, the altitude from A to BC, and MX concur. But they must concur at the orthocenter of ABC, so X=N.

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quantumbyte

547 posts

quantumbyte

#6 h Apr 27, 2011, 12:43 AM • 5 Y

Y by Math1331Math, TheHimMan, Adventure10, Mango247, Ar2023

Nice and Easy

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USAMOREAPER

70 posts

USAMOREAPER

#7 h Feb 23, 2013, 1:50 AM • 2 Y

Y by Adventure10, Mango247

Because $AB$ and $AC$ are diameters of the corresponding circles we know that the foot of the perpendicular from $A$ to $BC$ which we denote as $A'$ lies on both $c_1$ and $c_2$ where they are the circles with diameters $AB$ and $AC$ respectively. Because $c_1$ and $c_2$ intersect at points $A$ and $A'$ we know that $AA'$ is the radical axis of the two circles. So, to prove that quadrilateral $MPNQ$ is cyclic we want to show that $MH*HN=PH*HQ$ , where $H$ is the orthocenter of the triangle. However, $H$ lies on the radical axis so by the definition of being on the radical axis we have that power of point $H$ with respect to both circles is the same or that $MH*HN=PH*HQ$ as desired.

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jayme

9795 posts

jayme

#8 h Feb 23, 2013, 6:18 AM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

Dear Mathlinkers,
this problem is an application of the converse of the three chords theorem discovered by Monge.
Sincerely
Jean-Louis

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JSGandora

4216 posts

JSGandora

#9 h Apr 14, 2013, 6:49 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be closer to $C$ than $N$ and $P$ be closer to $B$ than $Q$ , now draw the circumcircle of $\triangle MNQ$ . The radical axis between $\odot MNQ$ and $\odot AMBN$ is $MN$ and the radical axis between $\odot AMBN$ and $\odot AQCP$ is $AA'$ where $A'$ is the foot of the altitude from $A$ . Since the radical axis of ever pair of circles in a set of three circles are concurrent, then we must have the radical axis of $\odot MNQ$ and $\odot AQCP$ to be concurrent with $AA'$ and $MN$ . However, we know that those two lines concur at the orthocenter since they are distinct altitudes of $\triangle ABC$ so the radical axis between $\odot MNQ$ and $\odot AQCP$ must contain the orthocenter, $H$ , and and the two points of intersection between the two circles. The two circles intersect at $Q$ so the radical axis between those two circles is $QH$ . $QH$ intersects $\odot AQCP$ at $P$ and so $P$ must also be on $\odot MNQ$ and so $M, N, Q, P$ are concyclic.

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sunken rock

4394 posts

sunken rock

#10 h Jul 14, 2013, 8:39 PM • 2 Y

Y by Adventure10, ehuseyinyigit

Strange! Nobody has proven $AM=AN=AP=AQ$ ! (I was not able to open all the links, but I surely proved it this way:)

Let's $BD, CE$ be altitudes of the triangle $\triangle ABC$ . Then $AN^2=AM^2=AE\cdot AB$ , while $AP^2=AQ^2=AD\cdot AC$ .
By pop $A$ w.r.t. circle $\odot(BCDE$ we have $AE\cdot AB=AD\cdot AC$ and our claim has been proven.

Best regards,
sunken rock

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sayantanchakraborty

505 posts

sayantanchakraborty

#11 h Jan 19, 2014, 6:19 AM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

Let H be the orthocentre.D,E,F be the foot of altitudes to BC,CA,AB.
By power of point theorem in circle APC
$AH*HD=PH*HQ$
Similarly for circle ABM we get
$MH*HN=AH*HD$
Combining these we get
$MH*HN=PH*HQ$
Hence M,N,P,Q are concyclic.

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john10

134 posts

john10

#12 h Aug 23, 2015, 10:39 PM • 1 Y

Y by Adventure10

hello , if we show that H the orthocentre lies on the radical axii of the two circles it's true ?

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MSTang

6012 posts

MSTang

#13 h Mar 28, 2016, 3:27 AM • 2 Y

Y by Adventure10, Mango247

Note that the circle with diameter $AB$ passes through $A'$ and $B'$ , and the circle with diameter $AC$ passes through $A'$ and $C'$ . Let the circumcircle of $QMP$ intersect the circle with diameter $AB$ at points $M$ and $N'$ . Then lines $PQ$ , $MN'$ , and $AA'$ intersect at the radical center of the three circles, which must be $H$ since $AA'$ and $PQ$ meet at $H$ . Thus, $N'$ lies on line $HM$ and the circle with diameter $AB$ . But $N$ also lies on line $HM$ and the circle with diameter $AB$ , so $N' = N$ , and $Q, M, P, N$ are concyclic.

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Calculus123

313 posts

Calculus123

#14 h Jun 30, 2016, 9:01 PM • 1 Y

Y by Adventure10

USAMOREAPER wrote:

Because $AB$ and $AC$ are diameters of the corresponding circles we know that the foot of the perpendicular from $A$ to $BC$ which we denote as $A'$ lies on both $c_1$ and $c_2$ where they are the circles with diameters $AB$ and $AC$ respectively. Because $c_1$ and $c_2$ intersect at points $A$ and $A'$ we know that $AA'$ is the radical axis of the two circles. So, to prove that quadrilateral $MPNQ$ is cyclic we want to show that $MH*HN=PH*HQ$ , where $H$ is the orthocenter of the triangle. However, $H$ lies on the radical axis so by the definition of being on the radical axis we have that power of point $H$ with respect to both circles is the same or that $MH*HN=PH*HQ$ as desired.

Wait why do we need to show $H$ is the orthocenter? We know that $AA'$ is the radical axis, and that $PQ$ and $MN$ intersect there, meaning it's power is equal. So, by the converse of the power of point theorem, they are cyclic. Where does the orthocenter come into play?

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Calculus123

313 posts

Calculus123

#15 h Jul 1, 2016, 6:27 AM • 2 Y

Y by Adventure10, Mango247

$Bump$

Sorry, this is really bugging me.

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mahaler

3084 posts

mahaler

#64 h Jul 18, 2023, 10:21 PM

Y by

Solution: Notice that our two circles intersect at the foot of the altitude $A'$ (by inscribed angle theorem). Since both circles also contain $A$ itself, $AA'$ forms the radical axis of the two circles. Noticing that the intersection of $PQ$ and $NM$ also intersect $AA'$ at the orthocenter $H$ , we are done since the powers of both circles are equal and hence $PH \cdot HQ = NH \cdot HM$ , implying that $P, Q, M, N$ are concyclic.

Comments: From EGMO Chapter 2.

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AshAuktober

1007 posts

AshAuktober

#65 h Sep 21, 2023, 1:56 PM

Y by

My solution using trig:
Click to reveal hidden text

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ismayilzadei1387

219 posts

ismayilzadei1387

#66 h Nov 26, 2023, 5:11 PM • 1 Y

Y by FriIzi

We know that from power $QH.HP=C'H.CH$
$B'H.HB=NH.HM$
obviously $C'H.HC=B'H.HB$ so that $M,N,P,Q$ is cyclic

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blueberryfaygo_55

340 posts

blueberryfaygo_55

#67 h Mar 24, 2024, 3:52 PM • 1 Y

Y by megarnie

Let $A'$ be the foot of the altitude of $A$ on $BC$ ( $A' \in BC$ ). Then, $A' \in (AB)$ and $A' \in (AC)$ , so it follows that $AA'$ is the radical axis of $(AB)$ and $(AC)$ . Let $H$ be the orthocenter of $\Delta ABC$ . Since $H \in AA'$ , we have $\text{Pow}_{(AB)} (H) = \text{Pow}_{(AC)} (H)$ which is equivalent to $HM \cdot HN = HP \cdot HQ$ and the result follows by the Converse of Power of a Point. $\blacksquare$

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G0d_0f_D34th_h3r3

22 posts

G0d_0f_D34th_h3r3

#68 h Jun 15, 2024, 1:28 PM

Y by

We can think of the given problem in some other way which eventually makes it trivial.

We can think of $M$ , $N$ to be two points on $CC'$ also on $(ABB')$ . Similarly we define $P$ , $Q$ .

Lemma: The point of intersection of $MN$ and $PQ$ which is $H$ the orthocentre, lies on the radical axis of $(ABB')$ and $(ACC')$ .
Proof We know that if we extend $AH$ to $A'$ which is $\perp BC$ .
So, $A'$ lies on both $(ABB')$ and $(ACC')$ . Therefore, $AA'$ is the radical axis of the two circles.
Hence, we prove our claim.

Since, lines $MN$ and $PQ$ intersect on the radical axis of the two circles, hence, points $M$ , $N$ , $P$ and $Q$ lie on a circle.

Attachments:

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LuisFonseca123

71 posts

LuisFonseca123

#70 h Jun 17, 2024, 2:03 PM

Y by

Is this correct?
Let $X$ be the second intersection point of the circles with $A \neq X$ . Note by radical axis theorem that the problem is equivalent to showing that $MN \cap PQ$ lies on $AX$ . $\angle BC'C=\angle CB'B = 90^{\circ}$ so points $B,C',B',C$ are concyclic. Thus by radical axis theorem, $BB' \cap CC'$ lies on $AX$ and due to $PQ$ and $BB'$ being the same line, and $MN$ and $CC'$ being the same line, $MN \cap PQ$ lies on $AX$ , as desired.

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dudade

139 posts

dudade

#71 h Jun 24, 2024, 4:13 AM

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Let $X$ and $Y$ be points on $AB$ and $AC$ , respectively, such that $BY \perp AC$ and $CX \perp AB$ . Let $H$ be the orthocenter of $\triangle ABC$ . As $\angle BXC = \angle BYC = 90^{\circ}$ , then $BXYC$ is cyclic.

Power of Point on $MBNY$ , $QCPX$ , and $BXYC$ implies $MH \cdot HN = BH \cdot HY$ . $\begin{cases} MBNY &: BH \cdot HY = MH \cdot HN \\ QCPX &: QH \cdot HP = XH \cdot HC \\ BXYC &: BH \cdot HY = XH \cdot HC. \end{cases}$ Combining these equalities yields $MH \cdot HN = QH \cdot HP$ which imply $M$ , $P$ , $N$ , and $Y$ are conyclic points, as desired.

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SomeonecoolLovesMaths

3256 posts

SomeonecoolLovesMaths

#72 h Sep 12, 2024, 5:49 PM

Y by

Can someone explain the motivation for considering power of point?

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SomeonecoolLovesMaths

3256 posts

SomeonecoolLovesMaths

#73 h Sep 16, 2024, 3:09 PM

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SomeonecoolLovesMaths wrote:

Can someone explain the motivation for considering power of point?

bump....

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ehuseyinyigit

837 posts

ehuseyinyigit

#74 h Sep 27, 2024, 12:27 PM

Y by

Since $NAMA'B$ and $QAPA'C$ are cyclic pentagons, PoP in these gives
$PH\cdot HQ=AH\cdot HA'=BH\cdot HB'=NH\cdot HM$ implies $NPMQ$ being cyclic quadrilateral by Reverse PoP as desired.

Remark.
Actually we have $AN=AQ$ , $PB'=QB'$ and $MC'=NC'$ . Because
$AN^2=AC'\cdot AB=AB'\cdot AC=AQ^2$ Thus, $AN=AQ$ . On the other hand
$C'N^2=AC'\cdot BC'=NC'\cdot MC'$ $B'Q^2=AB'\cdot CB'=QB'\cdot PB'$ which implies points $N$ and $Q$ being reflection of $M$ and $P$ wrt $AB$ and $AC$ , respectively.

This post has been edited 2 times. Last edited by ehuseyinyigit, Sep 27, 2024, 12:32 PM

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alexanderhamilton124

397 posts

alexanderhamilton124

#75 h Oct 17, 2024, 12:48 PM

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By converse of radical axis theorem, it's equivalent to proving $MN \cap PQ$ , which is the orthocenter, lies on the radical axis of the two circles. Let $X$ the second intersection of the two circles, so $AX$ is the radical axis. Note that $\angle{AXB} = 90, \angle{AXC} = 90$ , so $X, B, C$ are collinear and $AX$ passes through the orthocenter, so we are done.

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Siddharthmaybe

115 posts

Siddharthmaybe

#76 h Oct 23, 2024, 9:00 AM

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Call the left circle W (AB diamand the right circle T(AC diameter)
Call the intersections of W and T :- A and K

Claim: AK intersects BC at a right angle

Proof: Let M1 be the midpt of AB and M2 be the midpoint AC , then by midpoint theorem M1M2 || BC
Also let the intersection of M1M2 and AK be X then AK perpendicular to M1M2 (M1, M2 are the centres of the circles) , also since M1-X-M2
M1X is also parallel to BC so after making AK meet BC at some point (say L) then BLA = 90 and hence the claim is true

Now H is the intersection of the three altitudes, hence it also lies on AK and H is the intersection of MN and PQ so the powers are the same, from which we trivially deduce that MNPQ is cyclic

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Vedoral

89 posts

Vedoral

#77 h Dec 18, 2024, 9:48 PM

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ali123456

52 posts

ali123456

#78 h Feb 25, 2025, 8:00 PM

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My Solution

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QueenArwen

109 posts

QueenArwen

#79 h Mar 20, 2025, 12:40 PM

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Let $A'$ be the foot of the altitude from $A$ to $BC$ . $A'$ lies on the circles with diameters $AB$ and $AC$ since $\angle AA'B = \angle AA'C = 90$ , so $AA'$ is the radical axis of these two circles. Since $CC'$ , $BB'$ and $AA'$ are concurrent as all three are altitudes, the intersection of $MN$ and $PQ$ lies on the radical axis, so by converse of power of a point $QMPN$ is cyclic.

This post has been edited 1 time. Last edited by QueenArwen, Mar 20, 2025, 12:41 PM

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