Balkan Mathematical Olympiad 2018 P1

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microsoft_office_word

66 posts

microsoft_office_word

#1 h May 9, 2018, 1:24 PM • 9 Y

Y by estoyanovvd, Amir Hossein, Mathuzb, Pluto04, HWenslawski, Adventure10, Mango247, KhaiMathAddict, ItsBesi

A quadrilateral $ABCD$ is inscribed in a circle $k$ where $AB$ $>$ $CD$ ,and $AB$ is not paralel to $CD$ .Point $M$ is the intersection of diagonals $AC$ and $BD$ , and the perpendicular from $M$ to $AB$ intersects the segment $AB$ at a point $E$ .If $EM$ bisects the angle $CED$ prove that $AB$ is diameter of $k$ .
Proposed by Emil Stoyanov,Bulgaria

This post has been edited 2 times. Last edited by microsoft_office_word, May 9, 2018, 7:02 PM

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GGPiku

402 posts

GGPiku

#2 h May 9, 2018, 1:28 PM • 7 Y

Y by estoyanovvd, argente_loup, AlastorMoody, Iora, Adventure10, Mango247, KhaiMathAddict

Let $AB\cap CD=G$ , and $AD\cap BC=H$ . Since $EM$ is the bisector of $DEM$ and $ME\perp GB$ , with $Q=EM\cap DC$ , then $(G,Q,C,D)=-1$ . But $(G,F,C,D)=-1$ , where $F=HM\cap CD$ , since those 2 have 3 common points, we have $E=F$ , so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$ Letting $BD'\perp AD$ and $AC'\perp CB$ , $C'D'\cap AB=G'$ , we have $C'D'\parallel CD$ , $(G,E,A,B)=(G',E,A,B)=-1$ , so $G=G'$ , so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter.

This post has been edited 3 times. Last edited by GGPiku, May 9, 2018, 3:55 PM

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Hamel

392 posts

Hamel

#3 h May 9, 2018, 1:35 PM • 2 Y

Y by Adventure10, KhaiMathAddict

Yep. That simple. Today's competition was horrible in terms of originality.

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reality95

12 posts

reality95

#4 h May 9, 2018, 1:36 PM • 14 Y

Y by XxProblemDestroyer1337xX, Snakes, miguell, aqwxderf, estoyanovvd, rmtf1111, Starlex, AlastorMoody, VedantTheGreat, puntre, Mathlover_1, ismayilzadei1387, Adventure10, ehuseyinyigit

Do trigonometric ceva in $DEC$ and obtain $\sin \angle DCA \cdot \sin (\angle DEA - \angle DCA) = sin \angle CDB \cdot \sin (\angle DEA - \angle CDB)$ $\iff$ $\frac{1}{2}(\cos \angle DEA - \cos (\angle DEA - 2 \angle DCA)) = \frac{1}{2}(\cos \angle DEA - \cos (\angle DEA - 2 \angle CDB))$
$AB$ is not parallel to $CD$ so $\angle ACD + \angle BDC = \angle DEA \iff$ $M$ is incenter in $DEC \iff BCME$ is cyclic $\iff$ $AB$ diameter.

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Math-Ninja

3749 posts

Math-Ninja

#5 h May 9, 2018, 3:02 PM • 2 Y

Y by Adventure10, Mango247

GGPiku wrote:

Let $AB\cap CD=G$ , and $AD\cap BC=H$ . Since $EM$ is the bisector of $DEM$ and $ME\perp GB$ , with $Q=EM\cap DC$ , then $(G,Q,C,D)=-1$ . But $(G,F,C,D)=-1$ , where $F=HM\cap CD$ , since those 2 have 3 common points, we have $E=F$ , so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$ . Letting $BD'\perp AD$ and $AC'\perp CB$ , $C'D'\cap AB=G'$ , we have $C'D'\parallel CD$ , $(G,E,A,B)=(G',E,A,B)=-1$ , so $G=G'$ , so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter.

FTFY

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sbealing

308 posts

sbealing

#6 h May 9, 2018, 4:59 PM • 5 Y

Y by Anar24, estoyanovvd, ElementalHERO, Iora, Adventure10

We can do this with areals. Let:
$E=(1,0,0) \, , \, C=(0,1,0) \, , \, D=(0,0,1) \quad \quad |CD|=a \, , \, |DE|=b \, , \, |CE|=c$ $M$ lies on the internal angle bisector of $\angle CED$ so as $ME \bot AB$ it follows $A,B$ lie on the internal angle bisector. Let:
$A=(\lambda,-b,c) \, , \, B=(\mu,-b,c) \quad \lambda,\mu \in \mathbb{R}$ From this we get:
$M=CA \cap BD=(\lambda \mu,-b \lambda,c \mu)$ And as $M$ is on the $E$ internal angle bisector:
$\frac{-b \lambda}{c \mu}=\frac{b}{c} \Leftrightarrow \mu=-\lambda$ .
Now use the standard circle equation for $\odot ABCD$ . As $C,D$ on the circle we get $v=w=0$ . Solving for $u$ using $A$ gives:
$u=\frac{bc \left ((b-c) \lambda-a^2 \right)}{\lambda(\lambda-b+c}$ And using $B$ and the fact $\mu=-\lambda$ we see:
$\frac{bc \left ((b-c) \lambda-a^2 \right)}{\lambda(\lambda-b+c}=\frac{bc \left((b-c) \lambda +a^2 \right)}{\lambda (c-b- \lambda)} \Leftrightarrow (b-c)(\lambda-a)(\lambda+a)=0$ But as $b-c \Leftrightarrow CE=ED$ which would mean $EM$ is the perpendicular bisector of $CD$ giving $AB \parallel CD$ contradiciting the problem statement we must have $\lambda=\pm a$ . WLOG $\lambda=a$ :
$A=(a,-b,c)\, , \, B=(-a,-b,c)$ But this gives $A,B$ are the $C,D$ excentres in $\triangle ECD$ and hence as $M$ is therefore the incentre of the triangle so:
$\angle BCA=\angle BDA=90^{\circ}$ And hence $\odot ABCD$ has diameter $AB$ as desired.

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atmargi

23 posts

atmargi

#7 h May 9, 2018, 6:01 PM • 6 Y

Y by estoyanovvd, nidzo, Mobashereh, Infinityfun, Adventure10, Mango247

Let $AB$ and $CD$ intersect at $X$ , $AD$ and $BC$ intersect at $Y$ , and $ME$ and $CD$ intersect at $F$ .
Since $EF$ is the bisector of $\angle DEC$ , and $\angle FEX=90^{\circ}$ , we have that $(D, C; F, X)=-1$ .
Now, if we let $F'$ be the intersection of $YM$ and $DC$ , we have that $(D, C; F', X)=-1$ . Thus, $F=F'$ , so $Y, F, M, E$ are collinear. Let $O$ be the center of $ABCD$ . By Brokard, $O$ is the orthocenter of $\triangle XYM$ , so $YM \perp OX$ . But $YM \perp AB$ , so $O$ lies on $AB$ .

This post has been edited 1 time. Last edited by atmargi, May 9, 2018, 8:00 PM
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sbealing

308 posts

sbealing

#8 h May 9, 2018, 8:07 PM • 4 Y

Y by estoyanovvd, Zoom, Adventure10, Mango247

We can also do this with complex. Let $\odot ABCD$ be the unit circle with $A=a, B=b, \cdots$ . Then it's easy to get:
$m=\frac{bd(a+c)-ac(b+d)}{bd-ac} \quad \quad \overline{m}=\frac{(b+d)-(a+c)}{bd-ac}$ Then dropping the perpendicular from $M$ to $AB$ gives:
$e=\frac{1}{2} \left (a+b+m-ab \overline{m} \right) \quad \quad \overline{e}=\frac{1}{2ab} \left (a+b+ab \overline{m}-m \right)$ The angle bisector condition is equivalent to:
$\frac{(e-m)^2}{(e-c)(d-e)} \in \mathbb{R} \Leftrightarrow \left (\frac{e-m}{\overline{(e-m)}} \right)^2 \left ( \frac{\overline{(e-c)}}{e-c} \right) \left ( \frac{\overline{(e-d)}}{e-d} \right)=1$ Now focussing on $(e-m)$ we see as $EM \bot AB$ :
$\frac{e-m}{a-b}+\overline{\left (\frac{e-m}{a-b} \right)}=0 \Rightarrow \left (\frac{e-m}{\overline{(e-m)}} \right)^2=\left (\frac{a-b}{\overline{(a-b)}} \right)^2=\left(\frac{a-b}{\frac{b-a}{ab}} \right)^2=a^2 b^2$
Now we look at $(e-c)$ :
$2(e-c)=a+b+m-ab \overline{m}-2c=\frac{(b-c)(a^2+bd+ad-ab-2ac)}{bd-ac}$ $2 \overline{(e-c)}=\frac{(b-c)(bcd+a^2c+abc-acd-2abd)}{(bd-ac)(abc)}$ Combining these we get:
$\left ( \frac{\overline{(e-c)}}{e-c} \right)=\frac{(bcd+a^2c+abc-acd-2abd)}{abc(a^2+bd+ad-ab-2ac)}$ And similarly for $d$ by interchanging $c \leftrightarrow d \, , \, a \leftrightarrow b$ :
$\left ( \frac{\overline{(e-d)}}{e-d} \right)=\frac{(acd-bcd+b^2 d+abd-2abc)}{abd(b^2+ac+bc-ab-2bd)}$
Hence our angle condition is equivalent to:
$1=a^2 b^2 \cdot \frac{(bcd+a^2c+abc-acd-2abd)}{abc(a^2+bd+ad-ab-2ac)} \cdot \frac{(acd-bcd+b^2 d+abd-2abc)}{abd(b^2+ac+bc-ab-2bd)}$ $cd(a^2+bd+ad-ab-2ac)(b^2+ac+bc-ab-2bd)=(bcd+a^2c+abc-acd-2abd)(acd-bcd+b^2 d+abd-2abc)$ Which we can expand out cancel a lot of terms then factorise to give:
$2(a+b)(c-d)(ac-bd)(ab-cd)=0$ Clearly $c \neq d$ .
$ac=bd \Leftrightarrow \frac{a}{b}=\frac{d}{c} \Leftrightarrow \angle BOA=\angle COD \Leftrightarrow AB=CD \Leftrightarrow AC \parallel BD$ Similarly:
$ab=cd \Leftrightarrow AB \parallel CD$ .
The former is not true as the diagonals intersect and the latter is forbidden by the problem statement. Hence we must have:
$\boxed{a+b=0 \Leftrightarrow a=-b \Leftrightarrow AB \text{ diamter of } \odot ABCD}$

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InCtrl

871 posts

InCtrl

#9 h May 9, 2018, 11:32 PM • 3 Y

Y by cip999, Adventure10, Mango247

Let $EM$ intersect the perpendicular bisector of $CD$ at $P$ . This is clearly the midpoint of arc $CD$ in $(EDC)$ . However, notice that $MP$ is isogonal to the $M-$ altitude in $\triangle MDC$ , so $P$ is also the circumcenter of $(MDC)$ . It follows that $M$ is the incenter of $\triangle EDC$ .
From this we easily conclude: $\angle CDM=\angle MDC=\angle MDC=\angle BAC=\angle BAM$ so $AEMD$ is cyclic $\implies \angle MDA=\frac{\pi}{2}\implies AB \text{is a diameter.} \Box$

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krenkovr

183 posts

krenkovr

#10 h May 10, 2018, 1:46 PM • 4 Y

Y by Lukaluce, Tellocan, Adventure10, Mango247

microsoft_office_word wrote:

A quadrilateral $ABCD$ is inscribed in a circle $k$ where $AB$ $>$ $CD$ ,and $AB$ is not paralel to $CD$ .Point $M$ is the intersection of diagonals $AC$ and $BD$ , and the perpendicular from $M$ to $AB$ intersects the segment $AB$ at a point $E$ .If $EM$ bisects the angle $CED$ prove that $AB$ is diameter of $k$ .
Proposed by Emil Stoyanov,Bulgaria

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Wictro

119 posts

Wictro

#11 h May 10, 2018, 4:43 PM • 2 Y

Y by Adventure10, ehuseyinyigit

By $Blanchet's$ $Theorem$ we get that $AD,BC$ and $EM$ concurr. So the problem is equivalent to this: [Let $ABC$ be a triangle and $AD$ an altitude. $BE$ and $CF$ are cevians concurrent with this altitude at $H$ . Then $BCFE$ is cyclic if and only if $H$ is the orthocenter of $ABC$ .]
Let $ABC$ be the reference triangle and use barycentric coordinates. Let $H$ be a point on $AD$ and $BH$ intersect $AC$ at $E$ thus: $H= ( t : S_{AC} : S_{AB})$ and $E=( t : 0 : S_{AB})$ . Let $(BCE)$ intersect $AB$ at $F$ . We easily get $F=( c^{2}(S_{AB}+t)-b^{2}S_{AB} : b^{2}S_{AB} : 0 )$ . Now $C,H,F$ are collinear if and only if $\frac {c^{2}(S_{AB}+t)-b^{2}S_{AB}}{b^{2}S_{AB}} = \frac {t}{S_{AC}}$ from which we find $t=S_{BC}$ which means that $H$ is the orthocenter of $\triangle{ABC}$ as needed $\blacksquare$
(Of course we get the result for t if amd only if b is not equal to c which is true from the problem statement)

This post has been edited 2 times. Last edited by Wictro, May 12, 2018, 9:22 PM
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DimPlak

44 posts

DimPlak

#12 h May 11, 2018, 6:00 AM • 2 Y

Y by Adventure10, Mango247

InCtrl wrote:

Let $EM$ intersect the perpendicular bisector of $CD$ at $P$ . This is clearly the midpoint of arc $CD$ in $(EDC)$ . However, notice that $MP$ is isogonal to the $M-$ altitude in $\triangle MDC$ , so $P$ is also the circumcenter of $(MDC)$ . It follows that $M$ is the incenter of $\triangle EDC$ .
From this we easily conclude: $\angle CDM=\angle MDC=\angle MDC=\angle BAC=\angle BAM$ so $AEMD$ is cyclic $\implies \angle MDA=\frac{\pi}{2}\implies AB \text{is a diameter.} \Box$

MP is isogonal to the M - altitude? Please explain a little more! Thanks!

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BronzaniCrnogorac

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BronzaniCrnogorac

#13 h May 11, 2018, 4:25 PM • 65 Y

Y by steppewolf, Wolowizard, Zoom, VB1999, pumkin_3.14, mne-bu2, petar2018, mathMNE, nikolapavlovic, takoma, GGPiku, sakiscaki, Pajaraja, HECAM-CA-CEBEPA, nidzo, teletabis, GocMathics, Kirilbangachev, unknown9, Lamp909, rmtf1111, lifeisgood03, cip999, MilosMilicev, like123, anantmudgal09, m2017m, Lukaluce, dooo203, FISHMJ25, atmchallenge, Snakes, miguell, dmitar_15, harapan57, IljaUB, bosanac007, MK4J, Angelaangie, wu2481632, aats411, ramirahma, mihnea10, Wizard_32, ShR, e_plus_pi, Polynom_Efendi, WindTheorist, Mr.Mister, Night_Witch123, parola, Supermathlet_04, Jordan21, RamtinVaziri, KST2003, hakN, BVKRB-, Jalil_Huseynov, jocker, Iora, Adventure10, StCs, ATGY, ehuseyinyigit, L13832

Let's assume that such a cyclic quadrilateral exists so that $AB$ is not a diameter. Let $A'$ and $B'$ be points on the diagonals $AC$ and $BD$ respectively such that $\angle A'DB = \angle B'CA = 90^{\circ}$ . Since $A'B'CD$ is cyclic we get that $\angle B'A'C = \angle BDC = \angle BAC$ which further implies $AB \parallel A'B'$ . That means $ME$ must be perpendicular to $A'B'$ . Let $E'$ be the point of intersection of $ME$ and $A'B'$ . From the fact that $A'E'MD$ and $B'E'MC$ are cyclic by angle chasing we get that $\angle DE'M = \angle CE'M$ . Since we assumed that $\angle DEM = \angle CEM$ triangles $\triangle CEE'$ and $\triangle DEE'$ must be equal, or in other words $CE = DE$ . Because $\triangle CED$ is isosceles and since line $EM$ is its angle bisector it is also perpendicular to $CD$ which is not possible because $AB$ and $CD$ are not parallel. We got a contradiction!
PS: If this post gets more than 40 likes im going to IMO

This post has been edited 5 times. Last edited by BronzaniCrnogorac, May 12, 2018, 8:49 PM

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Pinko

437 posts

Pinko

#14 h May 11, 2018, 7:12 PM • 3 Y

Y by Mr.Mister, Adventure10, Mango247

GGPiku wrote:

Let $AB\cap CD=G$ , and $AD\cap BC=H$ . Since $EM$ is the bisector of $DEM$ and $ME\perp GB$ , with $Q=EM\cap DC$ , then $(G,Q,C,D)=-1$ . But $(G,F,C,D)=-1$ , where $F=HM\cap CD$ , since those 2 have 3 common points, we have $E=F$ , so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$ Letting $BD'\perp AD$ and $AC'\perp CB$ , $C'D'\cap AB=G'$ , we have $C'D'\parallel CD$ , $(G,E,A,B)=(G',E,A,B)=-1$ , so $G=G'$ , so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter.

$"(G,Q,C,D)=-1"$ What's the term for this in English? I think I know what it means but I want to be sure.

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Hunter01234

90 posts

Hunter01234

#15 h May 11, 2018, 7:26 PM • 3 Y

Y by Pinko, Adventure10, Mango247

Pinko wrote:

GGPiku wrote:

Let $AB\cap CD=G$ , and $AD\cap BC=H$ . Since $EM$ is the bisector of $DEM$ and $ME\perp GB$ , with $Q=EM\cap DC$ , then $(G,Q,C,D)=-1$ . But $(G,F,C,D)=-1$ , where $F=HM\cap CD$ , since those 2 have 3 common points, we have $E=F$ , so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$ Letting $BD'\perp AD$ and $AC'\perp CB$ , $C'D'\cap AB=G'$ , we have $C'D'\parallel CD$ , $(G,E,A,B)=(G',E,A,B)=-1$ , so $G=G'$ , so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter.

$"(G,Q,C,D)=-1"$ What's the term for this in English? I think I know what it means but I want to be sure.

It means that the four points are harmonic. It is called "cross-ratio".

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math31415926535

5617 posts

math31415926535

#38 h Dec 9, 2021, 7:51 PM

Y by

Let $F=EM \cap CD,$ by Lemma 9.18, we have $(X, F; D, C)=-1.$ Now let $F'=YM \cap DC, E'=AB \cap YF'.$ From Lemma 9.11, $(X, E'; A, B)=-1,$ projecting through $Y,$ we have $(X, F'; D, C)=-1.$ This is enough to show that $F=F', E=E'.$ By Brocard, we have $XO \perp XM,$ but since $YM \perp AB,$ $O$ lies on $AB.$ $\blacksquare$

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Albert123

204 posts

Albert123

#39 h Dec 26, 2021, 8:58 PM

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Let $AB \cap CD=X; AD \cap BC=Y$
$E,M,Y$ are collinear.
By Brokard: $E$ is center of $k$ .
$\implies AB$ is diameter of $k$ . $\blacksquare$

This post has been edited 1 time. Last edited by Albert123, Jan 28, 2022, 5:02 AM

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aleksijt

44 posts

aleksijt

#40 h Jan 20, 2022, 12:10 PM • 1 Y

Y by riadok

Dual w.r.t. a circle centered at $M$ , with an arbitrary radius.
We get the following equivalent problem:

Quote:

Let $ABC$ be a non-isosceles triangle. Then let $M$ be a point on the $BC$ bisector such that $M$ defines the same angles with the midpoints of $AB$ and $AC$ with $A$ . Prove that $M$ is the circumcenter.

Now, to prove this, let there be such triangle $ABC$ and $M$ not the circumcenter $O$ . $P$ and $Q$ are the midpoints of $AB$ and $AC$ . We have that $\angle MPA = \angle MQA$ . By sine law in triangles $OMP$ , $OMQ$ and $APQ$ (we use that $OPBR$ and $OQCR$ are cyclic, where $R$ is the midpoint of $BC$ ), we get that $MP \cdot AP = MQ \cdot AQ$ . So the areas $AMP$ and $AMQ$ are equal and since their circumcircles are congruent, we have that $AMP \cong AMQ$ . So $AP = AQ$ , or the triangle $ABC$ is isosceles.

This post has been edited 1 time. Last edited by aleksijt, Jan 20, 2022, 12:17 PM

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Commander_Anta78

58 posts

Commander_Anta78

#41 h Jan 28, 2022, 4:35 AM

Y by

Sol

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Mahdi_Mashayekhi

695 posts

Mahdi_Mashayekhi

#42 h Feb 24, 2022, 12:06 PM

Y by

First Note that $E$ is not midpoint of $AB$ . It's well-known that if $\angle CEM = \angle DEM$ then $ME$ , $AD$ and $BC$ are concurrent at a point $P$ .
Claim1 : $PDMC$ is cyclic.
Proof : Let $B'$ be reflection of $B$ across $E$ we have $\angle PB'M = \angle PBM = \angle CBD = \angle MAP$ so $PAB'M$ is cyclic and $\angle DCM = \angle DCA= \angle MBB' = \angle MB'B = \angle MPA = \angle MPD$ so $PDMC$ is cyclic.

Now we have $\angle PCM = \angle 180 - \angle BCA = \angle 180 - \angle BDA = \angle PDM$ and $\angle PCM + \angle PDM = 180$ so $\angle PCM = \angle PDM = \angle 90$ so $\angle BCA = \angle BDA = \angle 90$ so $AB$ is the diameter as wanted.
we're Done.

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eibc

600 posts

eibc

#44 h Sep 5, 2023, 7:27 PM

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what in the world is blanchet

$[asy] size(12cm); pair B = dir(0); pair C = dir(60); pair D = dir(110); pair A = dir(180); pair M = extension(A, C, B, D); pair E = foot(M, A, B); pair X = extension(A, C, D, E); pair Y = IP(Line(D, E, 10), unitcircle, 1); draw(unitcircle, blue); draw(A--B--C--D--cycle, blue); markscalefactor=0.03; draw(anglemark(C,E,M), cyan); draw(anglemark(M,E,D), cyan); draw(C--E--M, cyan+blue); draw(D--Y, cyan+blue); draw(A--C, cyan+blue); draw(A-(0, 0.5)--A+(0, 1.25), cyan+blue+dashed); draw(B-(0, 0.5)--B+(0, 1.25), cyan+blue+dashed); draw(Y--C+(0, 0.4), cyan+blue+dashed); draw(E--M+(0, 0.6), cyan+blue+dashed); dot("$A$", A, W); dot("$B$", B, E); dot("$C$", C, NE); dot("$D$", D, N); dot("$E$", E, SW); dot("$M$", M, NE); dot("$X$", X, NE); dot("$Y$", Y, SE); [/asy]$

Let $X = \overline{AC} \cap \overline{DE}$ and $Y = \overline{DE} \cap (ABC)$ . By the right angle and bisectors lemma, we have
$-1 = (AM; XC) \overset{D}{=} (AB; YC).$ Thus, $ACBY$ is harmonic so $AA$ , $BB$ , and $YC$ concur at some point $T$ (possibly at infinity). By Pascal's on $ABBDYC$ we find that $T$ lies on line $EM$ . This is enough to imply that $T$ must be the point at infinity along line $EM$ , as otherwise $E$ must be the midpoint of $\overline{AB}$ , which would mean that $\overline{AB} \parallel \overline{CD}$ by symmetry, contradiction. Therefore $\overline{AA} \parallel \overline{BB}$ so $\overline{AB}$ is indeed a diameter of $k$ .

This post has been edited 3 times. Last edited by eibc, Nov 19, 2023, 2:08 AM

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shendrew7

796 posts

shendrew7

#45 h Nov 16, 2023, 4:13 AM

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Denote $X = AD \cap BC$ and $Y - AB \cap BC$ .

Claim 1: $XE$ is an altitude of $\triangle XAB$ , or $XME$ collinear.

Suppose $N = XM \cap CY$ and $N' = ME \cap CY$ . From the well known Ceva-Menalaus and Apollonian Circle Lemmas, we have
$(CD; NY) = (CD; N'Y) = -1,$
so $N = N'$ . ${\color{blue} \Box}$

Claim 2: $M$ is the orthocenter of $\triangle XAB$ .

We want to show the orthocenter is the unique point on altitude $XE$ such that $\angle XAM = \angle XBM$ . We can reflect $A$ over $E$ to $A'$ to form cyclic quadrilateral $A'BXM$ . Then
$\angle MBA = \angle A'XM = \angle MXA = 90 - \angle XAB,$
which implies $BD$ is an altitude, so $M$ is the orthocenter. We finish by noting $\angle BDA = 90$ , so $AB$ is a diameter. $\blacksquare$

$[asy] size(300); defaultpen(fontsize(10)); pair A, B, C, D, E, M, N, X, Y; A = dir(0); B = dir(180); C = dir(100); D = dir(50); M = extension(A, C, B, D); E = foot(M, A, B); N = extension(E, X, C, D); X = extension(A, D, B, C); Y = extension(A, B, C, D); draw(circumcircle(A, B, C)); draw(B--X--A--C--Y--B--D--E--C--X--E); dot("$A$", A, dir(315)); dot("$B$", B, dir(225)); dot("$C$", C, dir(120)); dot("$D$", D, dir(45)); dot("$E$", E, dir(270)); dot("$M$", M, dir(170)); dot(extension(E, X, C, D)); label("$N$", extension(E, X, C, D), dir(225)); dot("$X$", X, dir(90)); dot("$Y$", Y, dir(270)); [/asy]$

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dolphinday

1325 posts

dolphinday

#46 h Jun 6, 2024, 4:04 PM

Y by

Let $AD \cap BC = X$ and $CD \cap AB = Y$ . By right angles and bisectors we get that $(A, M; G, C) = -1$ and $(A, M; G, C) \overset{D}= (A, B; E, Y) = -1$ which implies that $EM$ , $AD$ and $BC$ concur by Ceva-Menelaus. Then by Brokard's we get that the center of $k$ is the orthocenter $\triangle XMY$ so $YE$ passes through $O \implies AB$ is a diameter as desired.

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cursed_tangent1434

628 posts

cursed_tangent1434

#47 h Jun 25, 2024, 2:55 PM

Y by

Really too easy for words. I first saw this problem as a G4 and nearly got a heart attack, Problem 1 makes a lot more sense. We let $P = \overline{AD} \cap \overline{BC}$ and $X= \overline{AB}\cap \overline{CD}$ . Now, we show the following key result.

Claim : Points $E$ , $M$ and $P$ are collinear.
Proof : Let $R_1$ and $R_2$ be the intersections of $\overline{CD}$ with $\overline{EM}$ and $\overline{PM}$ respectively. Then, due to the Right Angles/Bisectors picture, we know that $(DC;R_1X)=-1$ Further, from the Ceva/Menelaus picture we also have that
$(DC;R_2X)=-1$ Thus, $R_1=R_2$ and indeed points $E$ , $M$ and $P$ are collinear as claimed.

Now, let $O$ be the center of $\Gamma$ . Then, by Brokard's Theorem we know that $PM \perp XO$ . But, due to the above observation we also have that $\overline{PM} \perp \overline{AB}$ . Thus, the center $O$ must lie on line $\overline{AB}$ which implies that $AB$ is a diameter of $\Gamma$ as desired.

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bjump

1027 posts

bjump

#48 h Jul 31, 2024, 2:26 AM

Y by

fakesolve $\implies$ hinted on pascal :oops:

Assume otherwise, then let $L = \overline{ED} \cap k$ , and $G= \overline{AC} \cap \overline{ED}$ . By Right Angles and Bisectors $-1=(AM;GC) \stackrel{D} = (AB ; LC)$ . let $\overline{A_{\perp}}$ , and $\overline{B_{\perp}}$ be the tangent lines to $k$ at $A$ , and $B$ respectively. Note that $\overline{A_{\perp}}$ , $\overline{B_{\perp}}$ , and $\overline{LC}$ must concur at a point call it $X$ , as $ABLC$ is harmonic. Pascal on $ABBDLC$ gives us $E$ , $M$ , $X$ collinear. However $X$ lies on the perpendicular bisector of $\overline{AB}$ , so $E$ must be the midpoint of $\overline{AB}$ , which implies $\overline{CD} \parallel \overline{AB}$ contradiction.

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Eka01

204 posts

Eka01

#49 h Aug 14, 2024, 9:00 AM • 1 Y

Y by Sammy27

Let $AD \cap BC=N$ , $AB \cap CD=P$ and $EM \cap CD=K$ .
It is easy to see that $(D,C;K,P)=-1$ due to the famous right angle-angle bisector-harmonic lemma.
This implies $K$ lies on the polar of P with respect to $(ABCD)$ hence $K$ and consequently $E$ lie on $NM$ due to brokard's.
$(D,C;K,P) \stackrel{N}{=} (A,B;E,P)=-1$ which implies $P$ is the $N$ expoint in $\Delta NAB$ which coupled with the given conditions implies that $C$ and $D$ are feet of altitudes which is what we needed to show.

This post has been edited 2 times. Last edited by Eka01, Aug 14, 2024, 9:05 AM

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mahmudlusenan

24 posts

mahmudlusenan

#50 h Sep 6, 2024, 5:58 PM

Y by

Harmonic ratioes kill this problem.

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optimusprime154

21 posts

optimusprime154

#51 h Jan 5, 2025, 2:17 PM • 1 Y

Y by SorPEEK

just projective geometry finishes this.
let $DB \cap EC$ = $K$ and $AC \cap ED$ = $L$ from the fact that $EM$ bisects $\angle CED$ and $\angle EDB = 90^\circ$ we obtain $(D, K, M , B) = -1)$ then project from $C$ the line $DB$ onto $ME$ to get $(BC \cap ME, M, DC \cap ME, E) = -1$ then do the same for the line $CA$ to get $AD, ME, CB$ concurrent the rest just finishes by brocard on $ADCB$

This post has been edited 1 time. Last edited by optimusprime154, Jan 5, 2025, 2:17 PM

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Nari_Tom

117 posts

Nari_Tom

#52 h Jan 19, 2025, 10:43 AM

Y by

Nice solution using Pappus and Desargues theorem.
Let $EM$ intersects $CD$ at $G$ . $GA$ and $DE$ intersects at $J$ and $GB$ and $EC$ intersects at $I$ . Then by Pappus theorem $J-M-I$ collinear, since $(D,C;G,F)=-1$ we know that $F$ lies on the line $J-M-I$ . If we use Desargues theorem on triangles $JAD$ and $CBI$ , we get that $AD, BC$ and $GE$ concurrent, let that concurrency point be $N$ . By the Brocards theorem $NM$ is the polar of $F$ , since $AB$ is perpendicular to $EN$ , we get that $AB$ is a diameter.

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EmersonSoriano

45 posts

EmersonSoriano

#54 h Apr 3, 2025, 11:07 PM

Y by

Let $N$ be the intersection point of $CE$ and $BD$ , let $T$ be the intersection point of $AB$ and $CD$ , and let $R$ be the intersection point of $AD$ and $BC$ . Since $\angle MEB = 90^\circ$ and $\angle DEM = \angle MEC$ , we have $(D, N; M, B) = -1$ . Then, since $(D, N; M, B) \overset{C}{=} (T, E; A, B)$ , it follows that $(T, E; A, B) = -1$ . Therefore, $R$ , $M$ , and $E$ are collinear. Finally, since the quadrilateral $ABCD$ is cyclic and $RM$ is perpendicular to $TA$ , by Brocard's theorem, we deduce that line $TA$ passes through the center of the circumcircle of $ABCD$ , concluding that $AB$ is a diameter.

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