$D\in BC,\ AD\perp BC;\ E\in CA,\ BE\perp CA;\ F\in AB,\ CF\perp AB\Longrightarrow$ $HM\cdot HN=HB\cdot HE=HA\cdot HD=HP\cdot HQ\Longrightarrow HM\cdot HN=HP\cdot HQ\Longrightarrow$ The points $M,N,P,Q$ lie on a common circle.

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

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.

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.

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.

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

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.

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

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.

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

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.

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?

$Bump$ Sorry, this is really bugging me.

Well, since $PQ$ and $MN$ are on the altitudes $BB'$ and $CC'$, the intersect at the orthocenter $H$. Also, $AA'$ is the radical axis because it is perpendicular to the line connecting the midpoints of $AB$ and $BC$ (which is parallel to $BC$), the centers of the two circles. Thus, $H$ is on the radical axis of the two circles, so $HM*HN=HP*HQ$, implying that $M$, $N$, $P$, and $Q$ are concyclic.

[asy][asy] import geometry; unitsize(1cm); pair A, B, C, a,b,c, H; A = (2,5); B = (6,0); C = (0,0); H=orthocenter(A,B,C);b=extension(A,C,B,H);c=extension(C,H,A,B);dot("$A$",A);dot("$B$",B);dot("$C$",C);dot("$C_1$",c);dot("$B_1$",b);dot("$H$",H); draw(A--B--C--A);draw(B--b,blue);draw(C--c,blue);draw(circumcircle(B,b,A));draw(circumcircle(C,c,A));a=extension(A,H,B,C);dot("$A_1$",a);draw(A--a,blue); [/asy][/asy] Let $A_1,B_1,C_1$ be the feet of the perpendiculars from $A,B,C$ respectively and $H$ be the orthocenter of $ABC$. Let $\omega_1$ be the circle with diameter $AB$ and $\omega_2$ be the circle with diameter $AC$. It suffices to prove by the Radical Lemma that $H$ lies on the radical axis of the two circles with diameters $AB$ and $AC$. Notice that $ABA_1B_1$ is concyclic since $\angle AA_1B = \angle AB_1B = \pi/2$. Similarly, $ACA_1C_1$ is concyclic. We conclude that $AA_1$ lies on the radical axis of $\omega_1$ and $\omega_2$. However, since it is the altitude from $A$, it passes through $H$, and we are done. $\blacksquare$

The conditions on $M$ imply \begin{align*} AM^2 + BM^2 & = AB^2\\ AM^2 - BM^2 & = AC^2 - BC^2 \end{align*}so $AM^2 = \tfrac{1}{2}(AB^2 + AC^2 - BC^2)$. Evidently $M$, $N$, $P$, $Q$ lie on a circle with center $A$.

Let $\omega_1$, $\omega_2$ be the circles with diameters $AB$, $AC$, respectively and define $D$ as the foot of the perpendicular from $A$ to $BC$. Additionally, let $H$ be the orthocenter of $\triangle ABC$. Claim: $AD$ is the radical axis of $\omega_1$ and $\omega_2$. Proof. Since $\angle ADB=\angle ADC=90^{\circ}$, $D$ lies on both $\omega_1$ and $\omega_2$, and the desired result follows. $\square$ Now, notice that since $MN \cap PQ=H$ and $H \in AD$, it follows that $M,N,P,Q$ are concyclic, as desired. $\blacksquare$

Because $BB'\in\omega_{AB},\omega_{BC}$ and $CC'\in\omega_{AC},\omega_{BC}$, \[HM\cdot HN\stackrel{\omega_{AB}}{=}HB\cdot HB'\stackrel{\omega_{BC}}{=}HC\cdot HC'\stackrel{\omega_{AC}}{=}HP\cdot HQ;\]done by converse of PoP. $\square$

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.

My solution using trig:

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

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. $\blacksquare$

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.

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.

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.

Can someone explain the motivation for considering power of point?

SomeonecoolLovesMaths wrote: Can someone explain the motivation for considering power of point? bump....

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.

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.

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