JBMO 2013 Problem 2

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Source: Proposed by Macedonia

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137 posts

Igor

#1 h Jun 23, 2013, 10:50 AM • 7 Y

Y by Miku_, ImSh95, Adventure10, Mango247, ItsBesi, DEKT, and 1 other user

Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$ . Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$ . Let $E$ be the second point of intersection of $\omega$ and the line $AD$ . If $M$ , $N$ and $P$ are the midpoints of the line segments $BE$ , $OD$ and $AC$ , respectively, show that the points $M$ , $N$ and $P$ are collinear.

This post has been edited 1 time. Last edited by Igor, Jun 24, 2013, 12:17 AM

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#2 h Jun 23, 2013, 11:29 AM • 7 Y

Y by MrOreoJuice, Miku_, ImSh95, Math_legendno12, Adventure10, ehuseyinyigit, and 1 other user

Let $H$ be the orthocenter of $ABC$ .Then it is well known that $AO$ and $AH$ are isogonal WRT $\angle BAC$ ,so $A$ , $H$ , $D$ are collinear.
It is also well known that $BH=BE$ ,so $MD=MB=1/2BH=OP$ .
Let $MD$ intersect $AC$ at $K$ .
$\angle DKA=180-\angle KAD-\angle KDA=180-90+\gamma -(90-\angle MDB)=90+\gamma -(90-\angle MDB)=90+\gamma -(90-\angle CAD)=90+\gamma -\gamma =90$ .
So $OP$ and $MD$ are parallel.So triangles $MDN$ and $NOP$ are congruent and conclusion follows.

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563 posts

BBAI

#3 h Jun 23, 2013, 12:32 PM • 5 Y

Y by ImSh95, Adventure10, Mango247, Orazmuhammet, and 1 other user

Let $N =MP \cap OD$ .We will prove $N$ is the midpoint of $OD$ .As $AD$ is isogonal to $AO$ ,so $H$ lies on $AD$ .By the well known property , $D$ is the midpoint of $HE$ and $M$ is the midpoint of $BE$ So $MD \mid\mid BH$ Also $BH$ and $OP$ are parallel So $MD \mid\mid OP$ .Now $MD=BM =\frac{1}{2}BH=\frac{1}{2} .2OP=OP$ So $\triangle MND$ and $\triangle NOP$ are congruent $\Longrightarrow NO=OD$ .Hence done.

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252 posts

MMEEvN

#4 h Jun 23, 2013, 1:14 PM • 4 Y

Y by ImSh95, Adventure10, Rotten_, Orazmuhammet

Let $H$ be the orthocenter.A little angle chasing proves that $MD||OP$ Also since $\Delta EDB \sim \Delta CDA \Rightarrow \frac{MD}{DH}=\frac{MD}{DE}=\frac{DP}{DC}$ But $\Delta HCD \sim \Delta OAP \Rightarrow \frac{DP}{DC}=\frac{AP}{DC}=\frac{PO}{DH}\Rightarrow MD=OP \Rightarrow MDPO$ a parallelogram.Done!

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344 posts

seby97

#5 h Jun 23, 2013, 2:23 PM • 4 Y

Y by AwesomeYRY, ImSh95, Adventure10, Mango247

Analitic solution:
Let D(0,0),A(0,a),B(b,0),C(c,0).We find $O(\dfrac{b+c}{2},\dfrac{a^2+bc}{2a})$ ,and $E(0,\dfrac{bc}{a})$ .So, $M(\dfrac{b}{2},\dfrac{bc}{2a}),N(\dfrac{b+c}{4},\dfrac{a^2+bc}{4a}),P(\dfrac{c}{2},\dfrac{a}{2})$ .Because MN,MP has the same slope(equal with $\dfrac{a^2-bc}{a(c-b)}$ ),the conclusion follows

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44 posts

#6 h Jun 24, 2013, 4:08 AM • 2 Y

Y by ImSh95, Adventure10

Hello,
Can someone explain why $\frac{1}{2}BH=OP$ ?

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304 posts

#7 h Jun 24, 2013, 6:06 AM • 2 Y

Y by ImSh95, Adventure10

It can be easily proved using trigonometry that $\frac{1}{2}BH=OP= \frac{1}{2}\cot \beta$

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15142 posts

#8 h Jun 24, 2013, 7:30 AM • 4 Y

Y by ohmcfifth, ImSh95, Adventure10, and 1 other user

The following analytical solution has some unexpected consequences, detailed in the sequel.

Choose a coordinatesystem $xOy$ where $O(0,0), A(a,u), B(-b,v), C(b,v), A'(-a,-u)$ .
From $\angle BAD = \angle CAO$ , i.e. $\angle BAE = \angle CAA'$ , follows $E(a,-u)$ (thus $AD\perp BC$ ), hence $D(a,v)$ .
But then $M\left (\dfrac{a-b}{2},\dfrac{v-u}{2}\right )$ , $N\left (\dfrac {a} {2},\dfrac{v}{2}\right )$ , $P\left (\dfrac{a+b}{2},\dfrac{u+v}{2}\right )$ .
The slope of the line $MN$ is $\displaystyle \dfrac {\dfrac {v} {2} - \dfrac {v-u} {2}} {\dfrac {a} {2} - \dfrac {a-b} {2}} = \dfrac {u} {b}$ , while the slope of the line $NP$ is $\displaystyle \dfrac {\dfrac {u+v} {2} - \dfrac {v} {2}} {\dfrac {a+b} {2} - \dfrac {a} {2}} = \dfrac {u} {b}$ ; as seen, they coincide, proving the colinearity.

We can certify now a suspicion lingering from the very beginning, namely that $AB<AC$ is a useless restriction (not even the isosceles case produces any discomfort). Neither the requirement $\triangle ABC$ be acute-angled is instrumental (true, some degenerate cases may appear, but they only increase the wealth of the configurations; also, point $D$ may now be outside the side $BC$ ). It is strange the choice of the jury to abide by these irrelevant conditions !?!

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563 posts

BBAI

#9 h Jun 24, 2013, 9:05 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

trigisfun wrote:

Hello,
Can someone explain why $\frac{1}{2}BH=OP$ ?

See this an important part of proving the existence of the euler line i.e $O,G,H$ are collinear.
if $BO \cap \odot ABC =X$ First prove that $HAXC$ is a parallelogram and then you can see that $H ,P,X$ are collinear .Then by using similarity of $\triangle BHX$ and $OMX$ ,we get the required result.

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2572 posts

#10 h Jun 24, 2013, 3:55 PM • 2 Y

Y by ImSh95, Adventure10

This post has been edited 1 time. Last edited by exmath89, Jun 24, 2013, 3:55 PM

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Igor

#11 h Jun 24, 2013, 3:55 PM • 4 Y

Y by fleurdelis, ImSh95, Adventure10, Mango247

mavropnevma wrote:

We can certify now a suspicion lingering from the very beginning, namely that $AB<AC$ is a useless restriction (not even the isosceles case produces any discomfort). Neither the requirement $\triangle ABC$ be acute-angled is instrumental (true, some degenerate cases may appear, but they only increase the wealth of the configurations; also, point $D$ may now be outside the side $BC$ ). It is strange the choice of the jury to abide by these irrelevant conditions !?!

In my opinion, it was done on purpose so that students won't lose time checking other configurations.

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1412 posts

#12 h Jun 25, 2013, 4:17 AM • 3 Y

Y by ImSh95, Adventure10, antimonio

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Lemma: It is well known that the reflection of the orthocentre $H$ through the midpoint of $BC$ and $AC$ is the antipode of $A, B$ respectively.

If we let primes denote the antipodes of points, we have $BA'CH$ (Bach hehe) and $AB'CH$ are parallelograms.

Part 1: Through dilations with ratio $2$ about $E$ and $A$ , we take $MD \to BH$ and $OP \to A'C$ . But, $BA'CH$ is a parallogram, so $BH \parallel A'C \implies MD \parallel OP$ .

Part 2: Through dilations with ratio $2$ about $E$ and $H$ , we take $OM \to E'B$ and $DP \to EB'$ . But $EB'BE'$ is a rectangle (points and their antipodes), so $E'B \parallel EB' \implies OM \parallel DP$ .

Therefore, $OMDP$ is a parallelogram, so the $MP$ goes through the midpoint of $OD$ , or $M, N, P$ are collinear, as desired $\blacksquare$ .

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1525 posts

mathuz

#13 h Jun 25, 2013, 8:03 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, and 1 other user

We have $ACEB$ is cyclic quadriteral and dioganals perpendicular - $AE\bot BC.$
So, $N$ - center of the eight points circle.

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34 posts

liimr

#14 h Jun 26, 2013, 3:06 PM • 2 Y

Y by ImSh95, Adventure10

useful lemma:
Given cyclic quadrilateral $ABCD$ with perpendicular diagonals. Let $AC\cap BD=E$ and foot of perpendicular from $E$ to the side $AB$ be $X_{1}$ and that perpendicular line meets $CD$ at $Y_{1}$ . Define $X_{2}, Y_{2}, X_{3}, Y_{3}, X_{4}, Y_{4}$ . Then these 8 points are concyclic.

This post has been edited 1 time. Last edited by liimr, Jun 26, 2013, 4:49 PM

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563 posts

BBAI

#15 h Jun 26, 2013, 4:21 PM • 2 Y

Y by ImSh95, Adventure10

liimr wrote:

useful lemma:
Given cyclic quadrilateral $ABCD$ with perpendicular diagonals. Let $AC\cap BD=E$ and foot of perpendicular to the side $AB$ be $X_{1}$ and meets $CD$ at $Y_{1}$ . Define $X_{2}, Y_{2}, X_{3}, Y_{3}, X_{4}, Y_{4}$ . Then these 8 points are concyclic.

From Where the foot of the perpendicular is drawn to $AB$ ?

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34 posts

liimr

#16 h Jun 26, 2013, 4:51 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

BBAI wrote:

From Where the foot of the perpendicular is drawn to $AB$ ?

I edited my post I think now it is understandable

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304 posts

#17 h Jun 26, 2013, 6:04 PM • 2 Y

Y by ImSh95, Adventure10

liimr can you prove the lemma?

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670 posts

#18 h Jun 26, 2013, 6:09 PM • 3 Y

Y by liimr, ImSh95, Adventure10

It's just angle chasing , and maybe Brahmagupta's lemma can be use here.

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liimr

#19 h Jun 26, 2013, 7:20 PM • 4 Y

Y by ImSh95, sayemsub15, Adventure10, Mango247

Math-lover123 wrote:

liimr can you prove the lemma?

proof is just angle chasing. Firstly prove that $X_{1}, X_{2}, X_{3}, X_{4}$ are concyclic then try to prove $X_{1}, X_{2}, X_{3}, Y_{3}$ are concyclic then others will follow directly.

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1963 posts

ABCDE

#20 h Jun 27, 2013, 6:29 PM • 5 Y

Y by PatrikP, raven_, ImSh95, Adventure10, Mango247

Not too hard.

Click to reveal hidden text

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803 posts

#21 h Jul 6, 2013, 5:01 PM • 4 Y

Y by Mehmet_Koca, ImSh95, Adventure10, Mango247

it's obvious to see that , $AD$ is perpendicular to $BC$ and $E$ is the reflection of $H$ w.r.t. $BC$ . now , WLOG , assume that , the circumcircle of $ABC$ is the unit circle centred at origin. then $h=a+b+c$ . also , we get , $d=(a+b+c-bc/a)/2$ . so , $e=-bc/a$ . hence , $p=\frac{a+c}{2}$ . $n=[a+b+c-bc/a]/4$
$m=b(a-c)/2$ . then it is easy to see that $\mu_{NP}=(b-a-c-bc/a)/(1/b-a/bc-1/a-1/c)$ and ,
$\mu_{PM}=\frac{a+c-b+bc/a}{a/bc+1/a+1/c-1/b}$ . hence $M,P,N$ are collinear.

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207 posts

Fitim

#22 h Jul 16, 2013, 3:31 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Author of this problem is Stefan Lozanovski (Стефан Лозановски).

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4402 posts

#23 h Jul 16, 2013, 11:03 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

$\triangle BDE\sim\triangle ADC$ and being right-angled means the median $DP$ of $\triangle ADC$ is altitude in $\triangle BDE$ , so $DP\parallel OM$ ( $OM$ is perpendicular bisector of $BE$ ), and similarly the median $DM$ is altitude of $\triangle ADC$ , hence $DM\parallel OP$ ( the latter being perpendicular bisector of $AC$ ), hence $OMDP$ is parallelogram, done.

Best regards,
sunken rock

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759 posts

#24 h Jan 30, 2014, 2:54 AM • 2 Y

Y by ImSh95, Adventure10

It's well known that the orthocenter is the isogonal conjugate of the circumcenter; therefore, we know that $AD \perp BC$ . Then we know that $BM=MD = ME$ , and $\angle BCA = \angle BEA = \angle MDE$ . Extend $ME$ to intersect $AC$ at $P'$ . Therefore, $\angle ADP' = \angle BCA = 90^{\circ} - \angle EAC$ , so we see that $\angle DP'A = 90^{\circ}$ , so $DM || OP$ . Let $AO$ intersect $\omega$ at $Q$ . Then, we know that $CQ = BE$ , so $OP = MD$ . Therefore, $MDPO$ is a parallelogram, so the diagonals bisect each other, and we are done.

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#25 h Jan 12, 2017, 10:22 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Triangles POC and DEC are similar, using PC:CD=PO:DE and AC:BE=CD:DE. Since AC=2PC and BE=2ME=2MD we get MD=PO.
N is midpoint of DO so DN=NO. We will prove angle MND=PNO, respectively triangles MDN and NOP are congruent.
Easy to prove that angle MDN=PON. And we have MD=PO , DN=NO and angle MDN=PON, so triangles MDN and NOP are congruent.

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AlexLewandowski

157 posts

AlexLewandowski

#26 h Jan 18, 2017, 3:58 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Using some simple trig bash, we find that OPDM is a ||gm. If PM meets OD in N', then N'D = N'O.
This implies that N and N' coincides and hence the result follows.

This post has been edited 1 time. Last edited by AlexLewandowski, Jan 18, 2017, 3:59 PM

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MathStudent2002

934 posts

MathStudent2002

#27 h Feb 21, 2018, 9:20 AM • 3 Y

Y by v_Enhance, ImSh95, Adventure10

Very challenging and beautiful problem! Solution with v_Enhance, msinghal, linpaws, and mathcool2009.

By $\sqrt{bc}$ inversion, $AD$ , and $AO$ map to each other, so $AD$ passes through the reflection of $A$ over $BC$ , hence $D$ is the foot of the $A$ -altitude. Consider a homothety centered at $D$ with factor two, so that $M$ maps to $M'$ , $P$ maps to $P'$ , and $N$ to $O$ . Also, let $Q,R$ be the reflections of $D$ over the midpoints of $AB$ , $CE$ , so that $QP'RM'$ is a rectangle. Since its center is the intersection of the perpendicular bisectors of $P'R$ and $RM'$ , which are precisely the perpendicular bisectors of $AE$ and $BC$ , which meet at $O$ , $O$ is on the diagonal $P'M'$ . $\blacksquare$

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2487 posts

#28 h Feb 21, 2018, 11:41 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Let $H$ be the orthocenter of $ABC$ . From Brahmagupta's theorem, $MD$ is perpendicular to $AC$ , which is perpendicular to $OP$ .
Also we have $MD = 0.5 BE = 0.5 BH = OP$ , so $MDOP$ is a parallelogram, and thus we are done.

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328 posts

#29 h Feb 21, 2018, 12:03 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Straight-forward with complex numbers.
Set $(ABC)$ as unit circle.

$m=\frac{b+(\frac{-bc}a)}2$
$p=\frac{a+c}2$
$d=\frac{1}2 \left(a+b+c-bc\overline{a} \right)$
Midpoint of $MP$ is $\frac{1}2 \left (\frac{b+(\frac{-bc}a)}2+\frac{a+c}2 \right)=\frac{1}2 \left(\frac{1}2(a+b+c-bc\overline{a}) \right)=\frac{o+d}2=n$

Hence, $N$ is midpoint of $MP \implies M,N,P$ are collinear, as desired. $\blacksquare$

This post has been edited 4 times. Last edited by Drunken_Master, Feb 21, 2018, 12:15 PM

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93 posts

#30 h Feb 25, 2018, 6:55 PM • 3 Y

Y by Circumcircle, ImSh95, Adventure10

(A non-guessing solution using Complex numbers,isogonal conjugates)
The orthocenter $H$ is the isogonal conjugate of the circumcenter $O$ , combining this with the given angle condition we have $AD \perp BC$ .Working in the complex plane we set $|A|=|B|=|C|=|E|=1$ .
Since $AE \perp BC$ we have $ae+bc=0$ so $e=\frac{-bc}{a}$ .
$d=\frac{1}{2} (a+b+c-\frac{bc}{a})$ ,
$n=\frac{1}{4}(a+b+c-\frac{bc}{a})$ ,
$m=\frac{1}{2}(b-\frac{bc}{a})$
$p=\frac{1}{2}(a+c)$
$P,M,N$ are collinear if $\frac{p-n}{p-m}$ is a real number. $\frac{p-n}{p-m}=\frac{\frac{1}{2}(a+c)-\frac{1}{4}(a+b+c-\frac{bc}{a})}{\frac{1}{2}(a+c)-\frac{1}{2}(b-\frac{bc}{a})}=\frac{\frac{1}{4}(a+c-b-e)}{\frac{1}{2}(a+c-b-e)}=\frac{1}{2}$ So $M,N$ and $P$ are collinear.

This post has been edited 1 time. Last edited by Leartia, Aug 1, 2019, 11:07 AM
Reason: ...

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#32 h Nov 27, 2019, 7:46 AM • 2 Y

Y by ImSh95, Adventure10

this problem becom very easy in the moment when you realized a known confiraguration. That MDPO is a parallelogram.

This post has been edited 1 time. Last edited by itslumi, Jul 26, 2020, 1:46 PM

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#34 h Apr 22, 2021, 3:27 PM • 1 Y

Y by ImSh95

It is well known that the isogonal of $AC$ is the $A$ -Altitude of $\triangle ABC \implies AD$ is the $A$ -Altitude.
Let $(ABC)$ be the unit circle.
$h = a + b + c$ $d = \frac{a + b + c - \frac{bc}{a}}{2}$ Also since $E$ is the reflection of $H$ across $BC \in (ABC)$
$\implies d = \frac{h + e}{2} \implies e = 2d - h$ $p = \frac{a+c}{2}$ $n = \frac{d + o}{2} = \boxed{\frac{d}{2}}$ $m = \frac{b + e}{2} = \frac{b - h + 2d}{2} = d - \frac{a+c}{2} = d - p$ $\implies m + p = d$ $\implies \frac{m+p}{2} = \frac d2 = n$ Hence $N$ is the midpoint of $PM$ and we are done.

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#35 h Apr 29, 2021, 9:02 AM • 1 Y

Y by ImSh95

@mathur and @liimr in posts 13 and 14 respectively, I read the 8 point theorem and I found it really impressive! But when it comes to the accual problem, I couldnt combine these two.... It is clear that the ABEC has perpendicular diameters and N is the midpoint of BE and P the midpoint of AC, but I cant keep applying this theorem. Can anybody helps me with that???

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JustKeepRunning

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JustKeepRunning

#36 h Jun 3, 2021, 8:38 PM • 1 Y

Y by ImSh95

It is well known that $N$ is the center of the eight point circle of quadrilateral $ABCD$ with diameter $MP,$ and the conclusion follows.

Note: This problem is also killed by the so-called parallelogram trick(which is in author's book).

This post has been edited 1 time. Last edited by JustKeepRunning, Jun 3, 2021, 8:46 PM

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#37 h Jan 25, 2022, 2:15 AM • 1 Y

Y by ImSh95

Note that: $AD$ is altitude from $A$ to $BC$
So:
$MO \perp BE$ and $DP \perp BE$ $\implies MO \parallel DP$
$OP \perp AC$ and $MD \perp AC$ $\implies OP \parallel MD$
$\implies MDOP$ is paralelogram.
Then: $M,N,P$ are collinear. $\blacksquare$

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#38 h Apr 28, 2022, 9:49 AM • 1 Y

Y by ImSh95

Extend $AO$ so that it meets $\omega$ at $F$ . As $\angle B A E=\angle C A F$ so, $BE=CF$ $\Rightarrow$ $BCEF$ is an isoceles trapezoid with $BC \parallel EF$ . Now $\angle A E F=90^{\circ} \Rightarrow \angle A D C=90^{\circ}$ .
Let $\angle B A E=\angle C A O=x$ and $\angle M E D=\angle M D E=\theta$ , then by some angle chase, we get $\angle B M D=2 \theta$ $\Rightarrow$ $\angle D M O=90-2 \theta$ . Similarly, we get $\angle D P O=\angle D P C-\angle O P C=180-2 \theta-90=90-2 \theta$ . Also $\angle M D P=360-(\angle A D P+\angle A D B+\angle B D M)=90+2 \theta$ and $\angle M O P=360-(\angle M O E+\angle C O E+\angle P O C)=90+2 \theta$ . As the opposite angles are equal, $MDOP$ is a parallelogram and $N$ is the midpoint of the diagonal $OD$ and so we get $M,N,P$ are collinear.

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tony88

#39 h May 1, 2022, 7:56 AM • 2 Y

Y by ImSh95, Mango247

trigisfun wrote:

Hello,
Can someone explain why 1/2BH=OP?

see Property 10.3.2.

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#40 h Feb 10, 2023, 12:35 AM

Y by

Note that by isogonality $AD\perp BC$ .

Claim: $MDPO$ is a parallelogram. Extend $DM$ to meet $AC$ at $Q$ . Note that since $DM$ is a median in $\triangle DBE$ , $DQ$ is a symmedian in $\triangle ADC$ , and is therefore also an altitude (since it is a right triangle). Therefore, $DM\perp AC$ , so $DM\parallel OP$ .

Let $H$ be the orthocenter. Furthermore, $BH=2OP$ , so $BE=2OP$ since $E$ is the reflection of $H$ over $D$ . Then $MD=ME=\frac{1}{2}BE$ , so $MD=OP$ which shows the claim.

Since $MDPO$ is a parallelogram, $MP$ passes through the midpoint of $OD$ , so we are done.

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Taco12

#41 h Mar 12, 2023, 1:58 AM

Y by

Clearly, we have $AD \perp BC$ . Now, apply complex numbers with $(ABC)$ as the unit circle. We then have $d=\frac{a+b+c-\frac{bc}{a}}{2}, e=\frac{bc}{a}$ . The three midpoints are just

$\begin{align*} \frac{a+b+c-\frac{bc}{a}}{4} \\ \frac{ab-bc}{2a} \\ \frac{a+c}{2} \\ \end{align*}$
These are collinear by the Collinearity Criterion. $\blacksquare$

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#42 h Jul 20, 2023, 5:42 PM

Y by

Wow so nice

Consider the rectangle passing through $A, B, C$ and $E$ such that the sides are parallel to $BC$ and $AE$ . Note that a homothety of factor $2$ centered at $D$ maps $M$ and $P$ to opposite vertices of the rectangle and $N$ to $O$ . It suffices to prove that $O$ lies on the diagonal from the opposite vertices, which is true because $O$ lies on the perpendicular bisector of $BC$ and $AE$ .

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ismayilzadei1387

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ismayilzadei1387

#43 h Dec 30, 2023, 10:45 AM

Y by

from $\angle BAH$ = $\angle CAO$ lemma $AD$ is perpendicular to $BC$
from a little angle chasing you will observe that if $MDPO$ is parallelogram then we are done
we can see that $OM$ and $DP$ are perpendicular to $BE$
similarly $MD$ and $OP$ are perpendicular to $AC$
Thus $OM//DP$ and $MD//OP$ since that
$MDPO$ is parallelogram.

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#44 h Aug 28, 2024, 6:51 PM

Y by

Subjective Rating (MOHs) $\text{[math]}$

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#45 h Jan 25, 2025, 9:37 AM

Y by

It is well known that $D$ is the foot of the perpendicular from $A$ to $BC$ . Then $(ABC)$ unit circle and braindead complex bash easily kills.

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EVS383

#46 h Apr 10, 2025, 6:49 PM

Y by

Coordinates have never been easier.

First, $AD \perp BC$ as $O$ and $H$ are isogonal conjugates. Now, put everything on the unit circle using coordinates.
$\begin{align*} A &= (\cos A, \sin A) \\ B &=(-\cos T, \sin T) \\ C &= (\cos T, \sin T) \\ D &= (\cos A, \sin T) \\ E &= (\cos A, -\sin A) \\ \end{align*}$ It's clear that the midpoint of $M$ and $P$ is $N$ .

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Markas

#47 h May 11, 2025, 6:20 PM

Y by

We have that $\angle CAO = \angle BAD$ and since O and H are isogonal $\Rightarrow$ AD is the height from A to BC. Now BH = 2a = 2PO, where PO = a from properties of H. We have that BH = BE = 2a and since $\triangle BDE$ is a right triangle, we have that DM = BM = ME = a $\Rightarrow$ PO = DM. From P being midpoint of EH and M being a midpoint of BE $\Rightarrow$ $DM \parallel BH$ . Since $BH \perp AC$ and $OP \perp AC$ $\Rightarrow$ $BH \parallel PO$ $\Rightarrow$ $DM \parallel PO$ . We have that PO = DM and $DM \parallel PO$ $\Rightarrow$ POMD is parallelogram $\Rightarrow$ the diagonals of the parallelogram are dividing each other in halves and since N is the midpoint of DO, it follows that N is a midpoint of PM $\Rightarrow$ $N \in PM$ . We are ready.

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