JBMO 2013 Problem 2

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Igor

137 posts

Igor

#1 h Jun 23, 2013, 10:50 AM • 8 Y

Y by Miku_, ImSh95, Adventure10, Mango247, ItsBesi, DEKT, Rounak_iitr, 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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Math-lover123

304 posts

Math-lover123

#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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BBAI

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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MMEEvN

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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seby97

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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trigisfun

44 posts

trigisfun

#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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Math-lover123

304 posts

Math-lover123

#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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mavropnevma

15142 posts

mavropnevma

#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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BBAI

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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exmath89

2572 posts

exmath89

#10 h Jun 24, 2013, 3:55 PM • 2 Y

Y by ImSh95, Adventure10

Solution

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Igor

137 posts

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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IDMasterz

1412 posts

IDMasterz

#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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mathuz

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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liimr

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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BBAI

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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MrOreoJuice

595 posts

MrOreoJuice

#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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sttsmet

146 posts

sttsmet

#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

2958 posts

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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Albert123

204 posts

Albert123

#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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aansc1729

111 posts

aansc1729

#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

5 posts

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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john0512

4192 posts

john0512

#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

1757 posts

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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Spectator

657 posts

Spectator

#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

219 posts

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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Mathandski

775 posts

Mathandski

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

Y by

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

Attachments:

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cherry265

16 posts

cherry265

#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

21 posts

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

150 posts

Markas

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

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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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Ilikeminecraft

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Ilikeminecraft

#48 h Jul 4, 2025, 6:27 AM

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Solved with a_smart_alecks buh buh buh skibidi skuh

Angle chase to prove the midpoint of $BE,$ midpoint of $AC,$ $D$ , $O$ form a parallelogram. This implies the claim.

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