Orthocenter lies on circumcircle

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975 posts

#1 h Jun 26, 2017, 7:03 AM • 13 Y

Y by doxuanlong15052000, Tumon2001, Davi-8191, Centralorbit, thczarif, amar_04, Piano_Man123, nima.sa, Adventure10, Mango247, Rounak_iitr, NO_SQUARES, eggymath

Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$

Proposed by Michael Ren

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EulerMacaroni

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EulerMacaroni

#2 h Jun 26, 2017, 7:06 AM • 27 Y

Y by whatshisbucket, rkm0959, MLMC, hwl0304, Submathematics, JNEW, BobaFett101, AlastorMoody, Centralorbit, thczarif, AntaraDey, khina, Piano_Man123, Smkh, salengend, Infinityfun, ike.chen, sabkx, puntre, nargesrafi, Adventure10, Mango247, GrantStar, PRMOisTheHardestExam, Funcshun840, Aryan-23, MS_asdfgzxcvb

What a beautiful problem!

Let $T$ be the center of $\odot(AH)$ ; since $H$ is the antipode of $A$ in $\odot(AH)$ , notice that it is equivalent to showing that the midpoint of $PQ$ lies on the dilation of $\odot(ABC)$ at $H$ with ratio $\frac{1}{2}$ , which is the nine-point circle. But this locus is just the projection of $T$ onto $PQ$ , or the circle with diameter $\overline{TM}$ , as desired.

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rmtf1111

698 posts

rmtf1111

#3 h Jun 26, 2017, 7:08 AM • 7 Y

Y by yiwen, BobaFett101, Centralorbit, HolyMath, Adventure10, Mango247, PRMOisTheHardestExam

Let $E$ and $F$ be the projections of $B$ and $C$ on the opposite sides of $\triangle{ABC}$ .Because $BFEC$ is cyclic with diameter $BC$ , so respectively with circumcenter $M$ we have the following:
$\angle{FEM} = \angle{FEC}-\angle{MEC}=\angle{A}+\angle{C}-\angle{C}=\angle{A}$ Respectively we have that $\angle{FEM}=\angle{MFE}=\angle{A}$ so $MF$ and $ME$ are tangent to $(AFE) \implies$ $(QP;FE)=-1$ .

http://dl3.joxi.net/drive/2017/06/11/0019/0812/1270572/72/96e691ea7d.jpg

Inverting with pole $A$ and radius $\sqrt{AF\cdot AB}$ (that sends $F$ to $B$ , $E$ to $C$ , $(ABC)$ to $EF$ and $(AEF)$ to $BC$ ) and using the fact that this inversion will preserve $(QP;FE)=-1$ , the problem becomes the following:

Let $ABC$ be a triangle, let $F$ and $E$ be the projections of $B$ and $C$ on $ACB$ and $AB$ respectively. Let $Q$ and $P$ be on $BC$ such that $(QP;BC)=-1$ , Let the perpendicular to $AQ$ at $Q$ meet $AP$ at $Y$ and let the perpendicular to $AP$ at $P$ meet $AQ$ at $X$ . Prove that $EF$ passes through the Miquel point of $QXPY$ .*
Let $XY\cap PQ=S$ . Let $T=PX \cap QY$ and let $M$ be the Miquel point of quadrilateral $QXPY$ . Because $QXPY$ is cyclic with diameter $XY$ , and using the well-know fact that $M$ is the image of $S$ under inversion wrt $(XPYQ)$ and that $T-M-A$ (which is also well-known, because $PXQY$ is cyclic), we have that $M=AT \cap XY$ and $YM$ is perpendicular to $AT$ , so $\triangle{MQP}$ is the orthic triangle of $\triangle{TAY}$ , so $A$ is the $Q$ -excenter of $\triangle{MQP}$ .Let $\omega$ be the $Q$ -excircle of $\triangle{MQP}$ .Let $D$ , $K$ , $L$ be the tangency point of $\omega$ with $QP$ , $QM$ and $PM$ respectively.Let $H$ be the orthocenter of $ABC$ . Let $F'$ be the reflection of $F$ over $AB$ . Note that $BDFAF'$ is cyclic with diameter $AB$ .
$\angle{EDA}=\angle{EDH}=\angle{EBF}=\angle{EBF'}=\angle{F'DA} \implies$ $F'$ , $E$ and $D$ are colinear, so after reflecting over $AB$ we have that $EF$ passes through $W$ , where $W$ is a point of $\omega$ such that $W$ is different from $D$ and $BW$ is tangent to $\omega$ . If we let $V$ be a point on $\omega$ such that $CV$ is tangent to $\omega$ , repeating the same argument as above we have that $EF$ passes through $V$ . Now inverting wrt $\omega$ we have that the image of $Q$ will be the midpoint of $KD$ , the image of $B$ will be the midpoint of $WD$ , the image of $P$ will be of $LD$ and the image of $C$ will be the midpoint of $DV$ . Now let $Z`$ be the image of $Z$ under this very inversion. We have that $IQ`B`P`C`$ must be cyclic and that it also passes through $D`=D$ because $Q,B,P,D,C$ are colinear. But $-1=(QP,BC)=D(Q`P`,B`C`)=D(KL,WV) \implies WV$ passes through $M \implies EF$ passes through $M \blacksquare$

http://dl4.joxi.net/drive/2017/06/11/0019/0812/1270572/72/af19944ce4.jpg

* I did not preserve the name of points after inversion.

EDIT(1/7/2019): this solution is the most horrible thing I've ever done

This post has been edited 1 time. Last edited by rmtf1111, Jan 7, 2019, 8:46 AM

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EulerMacaroni

851 posts

EulerMacaroni

#4 h Jun 26, 2017, 7:16 AM • 6 Y

Y by BobaFett101, lgkarras, ike.chen, sabkx, Adventure10, Mango247

^ what the actual overkill

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anantmudgal09

1980 posts

anantmudgal09

#5 h Jun 26, 2017, 7:26 AM • 11 Y

Y by kk108, rkm0959, Mathuzb, thczarif, Kamran011, ILOVEMYFAMILY, Adventure10, Mango247, PRMOisTheHardestExam, Funcshun840, TheHimMan

ELMO 2017/2 wrote:

Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$

Proposed by Michael Ren

Let $\overline{BE}$ and $\overline{CF}$ be the altitudes in $\triangle ABC$ . It is well-known that $\overline{ME}, \overline{MF}$ are tangent to $\odot(AH)$ . Let $N, L$ be midpoints of $\overline{PQ}$ and $\overline{AH}$ .

Claim: $N$ lies on the nine-point circle of $\triangle ABC$ .

(Proof) Note that $\measuredangle LEM=\measuredangle LFM=\measuredangle LNM=90^{\circ}$ hence $N$ lies on $\odot(MEF)$ as desired. $\blacksquare$

Construct parallelogram $PHQT$ ; since $N$ lies on the nine-point circle, $T$ lies on the circumcircle of $\triangle ABC$ . However, $T$ is the orthocenter of $\triangle APQ$ , hence we conclude that the result holds. $\blacksquare$

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rkm0959

1721 posts

rkm0959

#6 h Jun 26, 2017, 7:34 AM • 6 Y

Y by kk108, anantmudgal09, AntaraDey, Adventure10, Mango247, PRMOisTheHardestExam

Here's a complex bash solution that (I think) works together with other solutions above.
I did not actually turned in this solution but I enjoyed it.

So set up usual coordinates - let $H'$ be the orthocenter of $APQ$ and we will use lowercase for the complex numbers.
Denote $N$ as the midpoint of $AH$ .

Clearly, $h'=(a-n)+(p-n)+(q-n)+n = a+p+q-2n=p+q+a-(a+h) = p+q-h$ .
So it suffices to prove $|p+q-h|=1$ .

Now note $|p-n|=|q-n|=\frac{1}{2} |AH| = \frac{1}{2} |b+c|$ .
Also, note that $p-\frac{b+c}{2}$ and $q-\frac{b+c}{2}$ are parallel vectors, since $P, Q, M$ are colinear.

Set $p'=p-\frac{b+c}{2}$ and $q'=q-\frac{b+c}{2}$ . We now have $|p'-a|=|q'-a| = \frac{1}{2} |b+c|$ and $p', q'$ are parallel.
Now what we need to show is that $p+q-h=p'+q'-a$ has magnitude $1$ .

Now we will completely look at this problem with vectors.

Take $O$ as the origin. Take a point $A$ with coordinate $a$ , and draw a circle with radius $\frac{1}{2} |b+c|$ .
Draw a line from $O$ and let it hit the circle at two points $P', Q'$ .

We know that $p', q'$ are the coordinates for $P', Q'$ . This is because $|p'-a|=|q'-a| = \frac{1}{2} |b+c|$ and $p', q'$ are parallel.

Now $p'+q'-a$ is just the reflection of $A$ wrt $P'Q'$ , so $|p'+q'-a| =|a| = 1$ , as desired.

This post has been edited 1 time. Last edited by rkm0959, Jun 26, 2017, 7:34 AM

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SAUDITYA

250 posts

SAUDITYA

#7 h Jun 26, 2017, 7:48 AM • 4 Y

Y by vjdjmathaddict, DashTheSup, Adventure10, Mango247

Let $Y$ and $X$ be the circumcenter of $\odot ABC$ and $\odot AH$ .Let $H_1$ be the orthocenter of $\triangle APQ.$
Let $K$ be the midpoint of $PQ$ and $L$ be the reflection of $M$ across $K$

It's well known that $AYMX$ , $HPH_1Q$ and $YXHM$ are parallelogram.
(Basically the reflection of orthocenter across the midpoint of a side is the antipode of the other vertex)

Now,
See that $MH_1LH$ is a parallelogram !
$=>LH_1 \parallel HM => LH_1 \parallel XY$ .Also $LH_1 = XY => LH_1YX$ is a parallelogram!
$=> H_1Y = LX = MX = AY$ .Done !
Motivation
Click to reveal hidden text

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NHN

342 posts

NHN

#8 h Jun 26, 2017, 7:51 AM • 2 Y

Y by Adventure10, Mango247

antipode of $A$ in $\odot(ABC)$ is $O$ , line through $A$ and perpendicular $PQ$ cut $(ABC),(AH)$ at $X,Y$ so $XO//YH$ so perpendicular bisector of $XY$ cut $HO$ at midpoint of $OH$ is $M$ so $M$ in perpendicular bisector of $XY$ so $P,Q$ in perpendicular bisector of $XY$ so $X$ is orthocenter of $APQ$

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Ankoganit

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Ankoganit

#9 h Jun 26, 2017, 8:08 AM • 6 Y

Y by SAUDITYA, richrow12, like123, coldheart361, Adventure10, PRMOisTheHardestExam

$[asy]size(7cm); pair B=(0,0),A=(7,11),C=(10,0),H,M,P,Q,X,Y,Z; H=orthocenter(A,B,C);M=(B+C)/2;X=(A+H)/2; Q=X+(A-X)*dir(45);Y=foot(X,Q,M);P=2*Y-Q; Z=orthocenter(A,P,Q); draw(circumcircle(A,P,Q)^^circumcircle(Y,X,M)^^circumcircle(A,B,C),green); D(MP("A",A,N)--MP("B",B,S)--MP("M",M,S)--MP("C",C,SE)--cycle); D(MP("H",H,S)--MP("X",X,NE)--A--MP("Q",Q,W)--MP("Y",Y,SSW)--MP("P",P,W)--M,magenta); D(X--Y,magenta);D(P--A--MP("Z",Z,S),magenta); dot(A^^B^^C^^H^^M^^X^^Q^^Y^^P^^Z); [/asy]$

Let $X$ be the midpoint of $AH$ (and hence the circumcenter of $\odot(AH)$ ), $Y$ be the foot of perpendicular from $X$ to $PQ$ , and $Z$ be the orthocenter of $APQ$ . Also let $k$ be the circle with diameter $XM$ , which is therefore the nine-point circle of $ABC$ .

Let $f$ be the homothety centered at $H$ and ratio $2$ ; it sends $k\to\odot(ABC)$ , and $X\to A$ . Clearly $XY||AZ$ (both are $\perp$ to $PQ$ ), and $AZ=2\cdot XY$ by a well-known property of orthocenters. So $f$ sends $Y\to Z$ . But $Y\in k$ since $\angle XYM=90^\circ$ , so $Z=f(Y)\in f(k)=\odot(ABC)$ , as desired. $\blacksquare$

This post has been edited 1 time. Last edited by Ankoganit, Jun 26, 2017, 8:08 AM

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SAUDITYA

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SAUDITYA

#10 h Jun 26, 2017, 8:47 AM • 2 Y

Y by Adventure10, Mango247

Here's another way to complex bash !
Let $Y$ and $X$ be the circumcenter of $\odot ABC$ and $\odot AH$ .Let $H_1$ be the orthocenter of $\triangle APQ$
We will work in the complex plane and use the general notations.
Now,
Set $m = 0$ .
It's well known that $AYMX$ and $HPH_1Q$ are parallelogram. (this is the main observation the rest is trivial !)
(Basically the reflection of orthocenter across the midpoint of a side is the antipode of the other vertex)

Now,
$a = x+y => y = a-x$
Also $H$ is the reflection of $A$ across $X$ so $h = 2x - a$
Now,
$h_1 = (p+q) - h => h_1 - y = (p+q) - (2x-a) -(a-x) = (p+q) -x$
It suffices to show that
$|h_1-y| = |a-y| => |(p+q) -x| = |x|$
As $M-P-Q$ are collinear $=> \frac{p}{\overline{p}} = \frac{q}{\overline{q}}$
So ,
Eq. of line $PQ$ is $\overline{z} = kz$ for some constant $k(\neq 0)$ . Also see that $|k| = 1$ .
Eq. of circle $\odot AH$ is $|z -x| = |a-x| => |z|^2 - z\overline{x} - \overline{z}x + {|x|^2 - |a-x|^2} = 0$
Now,
See that the quadratic equation $kz^2 - z(\overline{x} + kx)+ {|x|^2 - |a-x|^2} = 0$ has roots $p,q$ .
So,
$p+q = \frac{\overline{x}+ kx}{k} => (p+q)-x = \frac{\overline{x}}{k} =>|p+q-x| = \frac{|\overline{x}|}{|k|} = |x|$
Hence, proved

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TheDarkPrince

3042 posts

TheDarkPrince

#11 h Jun 26, 2017, 8:53 AM • 8 Y

Y by RMO17geek, Maths_Guy, AlgebraFC, lifeisgood03, RudraRockstar, sabkx, Adventure10, Mango247

Probably the worst solution:

Let $H'$ be the orthocenter of $\triangle APQ$ . As $P, Q$ lie on the circle with diameter $AH$ , $\angle APH = \angle AQH = 90^{\circ}$ . So, $AP\perp PH$ and $AQ\perp QH$ . As $H'$ is the orthocenter of $\triangle APQ$ , $H'P \perp AQ$ and $H'Q \perp AP$ .
$\Rightarrow PH' || HQ, QH'||HP$ . So, $PH'QH$ is a parallelogram.

Now we apply coordinate bash.
Let $B=(-1,0), C = (1,0), A(a_1,a_2)$ .
So $M=(0,0)$ .

Coordinates of $H$ : Slope of $AC = \frac{a_2}{a_1 - 1}$ . As $AC\perp BH$ , slope of $BH = \frac{1-a_1}{a_2}$ .
Equation of $BH = y - 0 = (x-(-1))\left(\frac{1-a_1}{a_2}\right)$
or $y = (x+1)\left(\frac{1-a_1}{a_2}\right)$ .

As $BC$ is parallel to the $x-axis$ , $H$ has the same x-coordinate as that of $A$ .
So, $H = \left(a_1,\frac{1-a_1^2}{a_2}\right)$ .

Equation of circle with diameter $AH$ : Let $O$ be the center of circle with diameter $AH$ . As $O$ is the midpoint of $AH$ , $O = \left(a_1, \frac{1-a_1^2+a_2^2}{2a_2}\right)$ . Let $R = \frac{AH}{2}$ .
So, equation of circle with diameter $AH$ is: $(x-a_1)^2 + \left(y-\frac{1-a_1^2+a_2^2}{2a_2}\right)^2 = R^2$ .

Coordinates of $P, Q$ : As $M$ is the origin, let the equation of line $PM$ be $y = mx$ . So, let $P = (p,mp), Q = (q,mq), \text{where} p\neq q$ Simplification of coordinates of $P, Q$ As $P, Q$ lie on the circle with diameter $AH$ ,

$(p-a_1)^2 + \left(mp-\frac{1-a_1^2+a_2^2}{2a_2}\right)^2 = R^2$

$(q-a_1)^2 + \left(mq - \frac{1-a_1^2 + a_2^2}{2a_2}\right)^2 = R^2$ .

Subtracting these equations and using the identity $a^2 - b^2 = (a-b)(a+b)$ ,
$(p-q)(p+q-2a_1) + (p-q)\left(p+q - 2\left(\frac{1-a_1^2 + a_2^2}{2a_2}\right)\right) = 0$

As $p \neq q$ we get after simplifying, $p+q = \frac{m(1-a_1^2+a_2^2)+2a_1a_2}{a_2(m^2+1)}$ Coordinates of $H'$ : As $PH'QH$ is a parallelogram, $H' = \left(p+q - a_1,mp+mq - \frac{1-a_1^2}{a_2}\right)$ .
Substituting the value for $p+q$ we get, $H' = \left(\frac{m(1-a_1^2+a_2^2)+a_1a_2(1-m^2)}{a_2(m^2+1)}, \frac{m^2a_2^2+2ma_1a_2-1+a_1^2}{a_2(m^2+1)}\right)$ Equation of circumcircle of $ABC$ : Let $O'$ be the center of circle $(ABC)$ . As $M$ is on the y-axis and $BC||x-axis$ , the x-coordinate of $O'$ is $0$ . Let the y-coordinate of $O'$ be $y_1$ . Let $B'$ be the midpoint of $AC$ . So, $B' = \left(\frac{a_1+1}{2},\frac{a_2}{2}\right)$ .
As $O'B'\perp AC$ , the product of the slopes of the two lines is $-1$ .
So, $\frac{\frac{a_2}{2}-y_1}{\frac{a_1+1}{2}} = -1$ .
Solving for $y_1$ we get $y_1 = \frac{a_1^2 + a_2^2 - 1}{2a_2}$ . So, $O'=\left(0,\frac{a_1^2+a_2^2-1}{2a_2}\right)$ .

$(\text{Radius of} (ABC))^2 = OM^2 + MB^2 = \left(\frac{a_1^2+a_2^2-1}{2a_2}\right)^2 + 1 = \frac{(1+m^2)^2\cdot((a_1^2+a_2^2-1)^2+4a_2^2)}{(2a_2)^2(1+m^2)^2}$ .

So equation of $(ABC)$ is $x^2 + \left(y-\frac{a_2^2+a_1^2-1}{2a_2}\right)^2 = \frac{(1+m^2)^2\cdot((a_1^2+a_2^2-1)^2+4a_2^2)}{(2a_2)^2(1+m^2)^2}$
Putting the coordinates of $H'$ in the equation of the circle:

We need to show:
$x^2 + \left(y-\frac{a_2^2+a_1^2-1}{2a_2}\right)^2 = \left(\frac{2m(1-a_1^2+a_2^2)+2a_1a_2(1-m^2)}{2a_2(m^2+1)}\right)^2 + \left(\frac{(m^2-1)(a_2^2+1-a_1^2+4ma_1a_2}{2a_2(m^2+1)}\right)^2 = \frac{(1+m^2)^2\cdot((a_1^2+a_2^2-1)^2+4a_2^2)}{(2a_2)^2(1+m^2)^2}$ We cancel the denominators.

$LHS = (2m(1-a_1^2+a_2^2)+2a_1a_2(1-m^2))^2 + ((m^2-1)(a_2^2+1-a_1^2)+4ma_1a_2)^2 = 4m^2(1-a_1^2 + a_2^2)^2 + 4a_1^2a_2^2(1-m^2)^2 + 8ma_1a_2(1-a_1^2+a_2^2)(1-m^2) + (m^2-1)^2(a_2^+1-a_1^2)^2 + 16m^2a_1^2a_2^2+8ma_1a_2(a_2^2+1-a_1^2)(m^2-1)$ Simplifying and taking similar terms together we get,
$LHS = (m^2+1)^2((1-a_1^2+a_2^2)^2+4a_1^2a_2^2) = (m^2+1)^2((a_1^2+a_2^2-1)^2+4a_2^2) = (\text{Radius of (ABC)})^2$ So, $H'$ lies on $(ABC)$ .

Proved

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tarzanjunior

849 posts

tarzanjunior

#13 h Jun 26, 2017, 12:25 PM • 3 Y

Y by tiwarianurag021999, Adventure10, Mango247

Problem 2:
Let $H_1$ be the orthocenter of triangle $APQ$ , $N$ be the midpoint of $PQ$ and $R$ be the midpoint of $AH$
$PQ$ is the simson's line of point $H$ . So, $MH$ = $MH_1$ .
Since the reflection orthocenter in midpoint of side lies on the circumcircle of the triangle, $H_1NH$ is a straight line.
$\angle RNP = \angle RNM = 90$ . So, $M$ lies on the nine-point circle of triangle $ABC$ . Now, consider homothety taking nine-point circle to the circumcircle of $\triangle ABC$ , $N$ is taken to $H_1$ as desired.

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FabrizioFelen

241 posts

FabrizioFelen

#14 h Jun 26, 2017, 12:52 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Let $O$ the circumcenter of $\triangle ABC$ , let $N$ the midpoint of $AH$ , let $H'$ the orthocenter of $\triangle APQ$ , let $M'$ the midpoint of $PQ$ and let $N'$ the midpoint of $AH'$ , so from $H'M'=M'H$ and $H'N'=N'A$ we get $M'N'$ is the midline of $\triangle AH'H$ . $\Longrightarrow 2M'N'=HA=2OM$ So from $OM\parallel AH\parallel M'N'$ we get $OMN'M'$ is a paralellogram $\Longrightarrow$ $ON'\parallel MM'$ , but $MM'\perp AH'$ $\Longrightarrow$ $ON'\perp AH'$ in $N'$ , hence $OH'=OA$ $\Longrightarrow$ $H'\in \odot (ABC)$ .

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v_Enhance

6876 posts

v_Enhance

#15 h Jun 26, 2017, 8:29 PM • 14 Y

Y by anantmudgal09, liekkas, AlastorMoody, MeowX2, Smkh, v4913, ike.chen, Jalil_Huseynov, sabkx, Adventure10, Mango247, Rounak_iitr, Ritwin, Funcshun840

Here are the various official solutions provided by the organizers:

First solution (Michael Ren): Let $R$ be the intersection of $(AH)$ and $(ABC)$ , and let $D$ , $E$ , and $F$ respectively be the orthocenter of $APQ$ , the foot of the altitude from $A$ to $PQ$ , and the reflection of $D$ across $E$ . Note that $F$ lies on $(AH)$ and $E$ lies on $(AM)$ . Let $S$ and $H'$ be the intersection of $AH$ with $BC$ and $(ABC)$ respectively. Note that $R$ is the center of spiral similarity taking $DEF$ to $H'SH$ , so $D$ lies on $(ABC)$ , as desired.

Second solution (Vincent Huang, Evan Chen): Let $DEF$ be the orthic triangle of $ABC$ . Let $N$ and $S$ be the midpoints of $PQ$ and $AH$ . Then $\overline{MS}$ is the diameter of the nine-point circle, so since $\overline{SN}$ is the perpendicular bisector of $\overline{PQ}$ the point $N$ lies on the nine-point circle too. Now the orthocenter of $\triangle APQ$ is the reflection of $H$ across $N$ , hence lies on the circumcircle (homothety of ratio $2$ takes the nine-point circle to $(ABC)$ ).

Third solution (Zack Chroman): Let $R$ be the midpoint of $PQ$ , and $X$ the point such that $(M,X;P,Q)=-1$ . Take $E$ and $F$ to be the feet of the $B,C$ altitudes. Recall that $ME,MF$ are tangents to the circle $(AH)$ , so $EF$ is the polar of $M$ .

Then note that $MP \cdot MQ=MX \cdot MR=ME^2$ . Then, since $X$ is on the polar of $M$ , $R$ lies on the nine-point circle --- the inverse of that polar at $M$ with power $ME^2$ . Then by dilation the orthocenter $2 \vec R - \vec H$ lies on the circumcircle of $ABC$ .

Fourth solution (Zack Chroman): We will prove the following more general claim which implies the problem:

Claim: For a circle $\gamma$ with a given point $A$ and variable point $B$ , consider a fixed point $X$ not on $\gamma$ . Let $C$ be the second intersection of $XB$ and $\gamma$ , then the locus of the orthocenter of $ABC$ is a circle

Proof. Complex numbers is straightforward, but suppose we want a more synthetic solution. Let $D$ be the midpoint of $BC$ . If $O$ is the center of the circle, $\angle OMX=90$ , so $M$ lies on the circle $(OX)$ . Then $H=4O-A-B-C=4O-A-2D.$ So $H$ lies on another circle. (Here we can use complex numbers, vectors, coordinates, whatever; alternatively we can use the same trick as above and say that $H$ is the reflection of a fixed point over $D$ ). $\blacksquare$

Fifth solution (Kevin Ren): Let $O$ be the midpoint of $AH$ and $N$ be the midpoint of $PQ$ . Let $K$ be the orthocenter of $APQ$ .

Because $AP \perp KQ$ and $KP \perp HP$ , we have $KQ \parallel PH$ . Similarly, $KP \parallel QH$ . Thus, $KPHQ$ is a parallelogram, which means $KH$ and $PQ$ share the same midpoint $N$ .

Since $N$ is the midpoint of chord $PQ$ , we have $\angle ONM = 90^\circ$ . Hence $N$ lies on the 9-point circle. Take a homothety from $H$ mapping $N$ to $K$ . This homothety maps the 9-point circle to the circumcircle, so $K$ lies on the circumcircle.

This post has been edited 1 time. Last edited by v_Enhance, Jun 26, 2017, 8:30 PM
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v_Enhance

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v_Enhance

#16 h Jun 26, 2017, 8:37 PM • 9 Y

Y by tiwarianurag021999, Mathuzb, yunseo, omriya200, itslumi, v4913, sabkx, Adventure10, Mango247

Oops, one more I remembered just now:

Sixth solution (Evan Chen, complex numbers): We use complex numbers with $(AHEF)$ the unit circle, centered at $N$ . Let $a$ , $e$ , $f$ denote the coordinates of $A$ , $E$ , $F$ , and hence $h = -a$ . Since $M$ is the pole of $\overline{EF}$ , we have $m = \frac{2ef}{e+f}$ . Now, the circumcenter $O$ of $\triangle ABC$ is given by $o = \frac{2ef}{e+f} + a$ , due to the fact that $ANMO$ is a parallelogram.

The unit complex numbers $p$ and $q$ are now known to satisfy $p + q = \frac{2ef}{e+f} + \frac{2pq}{e+f}$ so $(a + p + q) - o = \frac{2pq}{e+f} \qquad \text{and}\qquad a - o = \frac{2ef}{e+f}$ which clearly have the same magnitude. Hence the orthocenter of $\triangle APQ$ and $A$ are equidistant from $O$ .

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shendrew7

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shendrew7

#79 h Feb 11, 2024, 11:14 PM

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Define the intersections of the altitude from $A$ to $PQ$ with $(AH)$ and $PQ$ as $J$ and $K$ , respectively. Also suppose $L$ is the orthocenter of $\triangle APQ$ and $A'$ as the antipode of $A$ on $(ABC)$ . Then

$HJ$ and $MK$ are both perpendicular to $AJ$ .
$K$ is the midpoint of $JL$ and $M$ is the midpoint of $HA'$ .

Hence we also have $A'L \perp AJ \implies \angle ALA' = 90$ , so $L$ lies on $(ABC)$ . $\blacksquare$

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ihatemath123

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ihatemath123

#80 h Feb 11, 2024, 11:40 PM • 1 Y

Y by OronSH

Let $N$ be the midpoint of $\overline{AH}$ . It's well known that $N$ and $M$ are antipodes on the nine-point circle of $\triangle ABC$ , so if $R$ be the foot from $N$ to line $PQ$ , then $R$ lies on the nine-point circle. A homothety of factor $2$ sends $R$ to the orthocenter of $\triangle AQP$ (since it's the midpoint of $\overline{PQ}$ ), as well as sending the nine-point circle to $(ABC)$ , hence the orthocenter of $\triangle AQP$ lies on $(ABC)$ .

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AlanLG

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AlanLG

#81 h Feb 27, 2024, 9:12 PM

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Let $O$ be the midpoint of $AH$ and $K$ the midpoint of $PQ$ , as $H$ is the $A-$ antipode of $(AH)$ then the orthocenter of $APQ$ is the reflection of $H$ over $K$ , so it suffices to prove that this point is on $(ABC)$ , but as the homothety with factor $\frac{1}{2}$ with center $H$ sends $(ABC)$ to the nine-point circe, it suffices to prove that $K$ is on the nine-point circle, but this is true as $\angle OKM=90^\circ$

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Shreyasharma

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Shreyasharma

#82 h Feb 27, 2024, 11:08 PM

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Hee hee hee haw, complex.

We complex bash with respect to $(AEF)$ . Set $a = 1$ and $h = -1$ and note $m = \frac{2ef}{e+f}$ . Arbitrarily choose $p$ and $q$ and note that the collinearity condition is the same as,
$\begin{align*} \begin{vmatrix} p & 1/p & 1\\ q & 1/q & 1\\ \frac{2ef}{e+f} & \frac{2}{e + f} & 1 \end{vmatrix} &= 0\\ \iff p/q + \frac{2ef}{p(e + f)} + \frac{2q}{e+f} - q/p - \frac{2ef}{q(e + f)} - \frac{2p}{e + f} &= 0\\ \iff \frac{(p - q)(p + q)}{pq} + \frac{2ef}{e+f}\left(\frac{q - p}{pq} \right) + \frac{2}{e + f}(q - p) &= 0\\ \iff \frac{p + q}{pq} - \frac{2ef}{pq(e+f)} - \frac{2}{e+f} &= 0\\ \iff (p + q)(e + f) - 2ef - 2pq &= 0\\ \iff (p + q)(e + f) &= 2(ef + pq) \end{align*}$ Now note that the orthocenter of $\triangle APQ$ has coordinates $1 + p + q$ , $B$ has coordinates $\frac{2ef + e - f}{e + f}$ and $C$ has coordinates $\frac{2ef - e + f}{e + f}$ . Thus we need to show,
$\begin{align*} \frac{1 - \frac{2ef + e - f}{e + f}}{1 - \frac{2ef - e + f}{e + f}} \div \frac{1 + p + q - \frac{2ef + e - f}{e + f}}{1 + p + q - \frac{2ef - e + f}{e + f}} \in i\mathbb{R} \end{align*}$ Upon expansion we have,
$\begin{align*} \frac{1 - \frac{2ef + e - f}{e + f}}{1 - \frac{2ef - e + f}{e + f}} \div \frac{1 + p + q - \frac{2ef + e - f}{e + f}}{1 + p + q - \frac{2ef - e + f}{e + f}} &= \frac{2f - 2ef}{2e - 2ef} \div \frac{(1 + p + q)(e + f) - 2ef - e + f}{(1 + p + q)(e + f) - 2ef + e - f}\\ &= \frac{2f(1 - e)}{2e(1 - f)} \div \frac{(e + f) + 2(ef + pq) - 2ef - e + f}{(e + f) + 2(ef + pq) - 2ef + e - f}\\ &= \frac{f(1 - e)}{e(1 - f)} \div \frac{pq + f}{pq + e} \end{align*}$ Now taking the conjugate of this expression we find,
$\begin{align*} \overline{\left( \frac{f(1 - e)}{e(1 - f)} \cdot \frac{pq + e}{pq + f} \right)} &= \frac{1/f(1 - 1/e)}{1/e(1 - 1/f)} \cdot \frac{1/pq + 1/e}{1/pq + 1/f}\\ &= \frac{(e - 1)}{(f - 1)} \cdot \frac{ef + pqf}{ef + pqe}\\ &= \frac{f(1 - e)}{e(1 - f)} \cdot \frac{pq + e}{pq + f} \end{align*}$ and we're done.

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naonaoaz

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naonaoaz

#83 h Feb 29, 2024, 2:52 AM

Y by

Let $APQ$ be the unit circle. Since $M \in \overline {PQ}$ , we have $m+pq \overline{m} = p+q$ .

Letting $H'$ be the reflection of $H$ over M, we want the orthocenter of $\Delta APQ$ to lie on the circle with diameter $\overline {AD}$ . This is equivlent to $LA \perp LD$ where $L$ is the orthocenter of $\Delta APQ$ .

Note $H' = 2m+a$ , so it suffices to show
$\frac{p+q}{p+q-2m} = \overline{\left(\frac{p+q}{p+q-2m}\right)} = \frac{p+q}{p+q-2pq \overline{m}}$ and we're done since $pq\overline{m} = p+q-m$ .

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Pyramix

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Pyramix

#84 h Mar 2, 2024, 2:58 PM

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Let $J$ be the orthocenter of $\triangle APQ$ and $K$ be the mid-point of $\overline{PQ}$ .

Here are a few well-known facts: (some of the outline proofs here)
Fact 1: The $A-$ antipode, mid-point $\overline{BC}$ and orthocentre are collinear. Hence, $K,J,H$ are collinear.
Fact 2: $\overline{ME}$ and $\overline{MF}$ are tangents to the circle $(AEF)$ , and $\angle MFE=\angle BAC$ .
Fact 3: If $Q_A$ is the $A-$ Dumpty Point and $R$ is the reflection of $A$ in $Q_A$ , then $R\in ABC$ , $AR$ is symmedian and $\angle BQ_AR=\angle RQ_AC=\angle BAC$ .

Note that $M,P,Q$ are collinear, so $(PFQE)$ is a harmonic bundle. Hence, $PQ$ is the $P-$ symmedian of $\triangle PEF$ . So, $K$ is the $P-$ Dumpty Point. Hence, $\angle FKM=\angle FKQ=\angle FPE=\angle FAE=\angle BAC=\angle FEM$ and it follows that $K,F,M,E$ are concyclic. Hence, $K$ lies on the nine-point circle of $ABC$ . Take homothety at $H$ with factor 2 to get that $J$ lies on circumcircle, as required. $\blacksquare$

Attachments:

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HamstPan38825

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HamstPan38825

#85 h Mar 2, 2024, 4:42 PM

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Surprisingly clean. Let $K$ be the midpoint of $\overline{AH}$ , $N$ be the midpoint of $\overline{PQ}$ , and $H_1$ be the desired orthocenter. Then $\angle KNM = 90^\circ$ , thus $N$ lies on the nine-point circle. However, note that $N = \overline{HH_1} \cap \overline{PQ}$ is the midpoint of $\overline{HH_1}$ ; thus, the homothety sending the nine-point circle to $(ABC)$ sends $N$ to $H_1$ , and $H_1$ lies on $(ABC)$ .

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shendrew7

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shendrew7

#86 h Jun 14, 2024, 9:30 PM • 1 Y

Y by teomihai

Define $J$ as the orthocenter of $\triangle APQ$ and $K = AJ \cap PQ$ . Notice the circles $(AH)$ , $(AM)$ , $(ABC)$ are coaxial, so if we aim to prove $J \in (ABC)$ , our Coaxiality Lemma tells us this is equivalent to
$\frac{\operatorname{pow}(J,(AH))}{\operatorname{pow}(J,(AM))} = \frac{\operatorname{pow}((B,(AH))}{\operatorname{pow}(B,(AM))} \iff \frac{-AJ \cdot 2JK}{-AJ \cdot JK} = \frac{BF \cdot BA}{BD \cdot BM} \iff 2 = 2. \quad \blacksquare$

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OronSH

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OronSH

#87 h Jun 21, 2024, 3:08 PM • 1 Y

Y by ihatemath123

ermmmm what the hyperbola

Claim: $P,Q$ are antigonal conjugates.
Proof. Let $P'$ be the reflection of $P$ over $M.$ Then $MP'\cdot MQ=MP\cdot MQ=MB\cdot MC$ by three tangents lemma, so $-\measuredangle BPC=\measuredangle BP'C=\measuredangle BQC.$ Also we have $\measuredangle APH=90^\circ=-\measuredangle AQH,$ so the antigonal conjugate of $P$ is given by one of the intersection points of two circles, one of which is $Q.$ We can show it must be $Q$ by considering the special cases $P=Q$ and using continuity.

Now the midpoint of $PQ$ lies on the nine point circle. Since $H$ is the antipode of $A$ on $(APQ)$ we get the orthocenter of $APQ$ is the reflection of $H$ over the midpoint of $PQ,$ which lies on the circumcircle by homothety.

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Eka01

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Eka01

#88 h Aug 30, 2024, 2:38 PM

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Since $H$ is the $A$ antipode in $(APQ)$ , it lies on the reflection of orthocenter $N$ of $\Delta APQ$ through midpoint of $PQ$ . Taking a homothety through $H$ with factor of $\frac{1}{2}$ , it suffices to prove that midpoint $K$ of $PQ$ lies on the nine point circle which is the circle with diameter $(OM)$ where $O$ is center of $(AH)$ /midpoint of $AH$ .
This is obvious because $OK \perp PQ \equiv MK$ due to $PQ$ being a chord on a circle centered at $O$ .

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ErTeeEs06

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ErTeeEs06

#89 h Nov 20, 2024, 10:00 PM

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Let $D$ be the orthocenter of $\triangle APQ$ and $H'$ the reflection of $H$ in $M$ . Obviously $D$ is reflection of $H$ in midpoint of $PQ$ , since $A$ , $H$ are antipodes on $(APQ)$ and also $H'$ is antipode of $A$ on $(ABC)$ . So the midpoint of $HD$ is on $PQ$ . But $M$ is on $PQ$ so by midparallel we have $PQ\parallel DH'$ . Also $AD\perp PQ$ . Combining gives $AD\perp DH'$ and therefore $\angle ADH'=90^\circ$ which implies $H'$ on $(ABC)$ .

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cj13609517288

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cj13609517288

#90 h Dec 6, 2024, 3:33 PM

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Let $N$ be the midpoint of $AH$ , let $X$ be the $A$ -queue point, and define $E,F$ as normal.

Note that there exists a fixed linear transformation that takes the midpoint of $PQ$ to the orthocenter of $\triangle APQ$ by first taking a $\frac23$ -homothety from $A$ to get to the centroid, then taking a $3$ -homothety from $N$ . Since the midpoint of $PQ$ lies on the circle with diameter $NM$ , it suffices to check three distinct such midpoints to show that the circle with diameter $NM$ maps to $(ABC)$ under the linear transformation.

If $P=Q=F$ , then the midpoint is $F$ , so the image after the $\frac23$ -homothety is $\frac23 F+\frac13 A$ , so the image after the $3$ -homothety is $2F+A-2N$ . Indeed, this point is the reflection of $H$ over $AB$ , say $H'$ , because $H'=2F-H$ and $A=2N-H$ .

Similarly, $P=Q=E$ also works.

Finally, $P=X$ , $Q=H$ works since then the orthocenter is just $H$ itself. $\blacksquare$

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ihategeo_1969

208 posts

ihategeo_1969

#91 h Jan 22, 2025, 9:42 PM

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As usual we define some new points.

$\bullet$ Let $W$ be the orthocenter of $\triangle APQ$ .
$\bullet$ Let $T$ be foot of $A$ onto $\overline{PQ}$ .
$\bullet$ Let $W'=\overline{AT} \cap (ABC)$ .
$\bullet$ Let $Q_A$ be the $A$ -Queue point of $\triangle ABC$ .

See that $T \in (AMQ_A)$ and since $(AM)$ , $(AH)$ , $(ABC)$ are coaxial; we have by the coaxiality lemma that $-1=\frac{\text{Pow}(M,(AH))}{\text{Pow}(M,(ABC))}=\frac{\text{Pow}(T,(AH))}{\text{Pow}(T,(ABC))} \iff TA \cdot TW'=-TP \cdot TQ$ And see that obviously $TA \cdot TW=-TP \cdot TQ$ and so $W \equiv W'$ (here lengths are directed).

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Saucepan_man02

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Saucepan_man02

#92 h Feb 8, 2025, 8:04 AM

Y by

Nice problem..

Lets denote some points:

$H'$ : Orthocenter of $\triangle APQ$
$T$ : Midpoint of $AH$
$X$ : Midpoint of $PQ$

Then, notice that: $PHQH'$ is a parallelogram. In-order to have $H' \in (ABC)$ , it suffice to show $X$ lies in the nine-point circle of $\triangle ABC$ (due to homothety of factor $1/2$ at $H$ ).
Notice that: $TX \perp PQ$ and thus: $TX \perp XM$ which implies $X$ lies on the circle with diameter $TM$ (which is the nine-point circle).