Geometry

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IstekOlympiadTeam

542 posts

IstekOlympiadTeam

#1 h Dec 12, 2015, 4:33 PM • 3 Y

Y by jhu08, Adventure10, Mango247

An acute-angled $ABC \ (AB<AC)$ is inscribed into a circle $\omega$ . Let $M$ be the centroid of $ABC$ , and let $AH$ be an altitude of this triangle. A ray $MH$ meets $\omega$ at $A'$ . Prove that the circumcircle of the triangle $A'HB$ is tangent to $AB$ . (A.I. Golovanov , A.Yakubov)

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sunken rock

4384 posts

sunken rock

#2 h Dec 12, 2015, 7:38 PM • 3 Y

Y by Arjun167, jhu08, Adventure10

Hint: Let $P\in\omega|AP\parallel BC$ and $Q$ the projection of $P$ onto $BC$ ( $HAPQ$ is a rectangle) and $N$ midpoint of $BC$ . $N$ is also midpoint of $HQ$ , thus $H-M-P$ are collinear. Now it is easy to see $\angle BA'P=\angle BCP=\angle ABC$ , done.

Best regards,
sunken rock

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Ankoganit

3070 posts

Ankoganit

#3 h Jul 22, 2016, 9:31 AM • 4 Y

Y by jhu08, Adventure10, Mango247, COCBSGGCTG3

This is also India IMOTC 2016, Practice Test 1, Problem 1.

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WizardMath

2487 posts

WizardMath

#4 h Jul 23, 2016, 1:40 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Follows from a homothety at the centroid with ratio -2 and angle chasing.

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jayme

9782 posts

jayme

#5 h Jul 23, 2016, 8:22 AM • 4 Y

Y by govind7701, jhu08, Adventure10, Mango247

Dear Mathlinkers,

1. A" the second point of intersection of MH with (O)
2. AA'' is parallel to BC
3. according to theReim's theorem, we are done...

Sincerely
Jean-Louis

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SidVicious

584 posts

SidVicious

#6 h Jul 23, 2016, 8:41 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Let $A''$ be point on $BC$ such that $BH=CA''$ . Let $A_1$ be midpoint of $BC$ , then $A_1$ is also midpoint of $HA''$ . As $AM : MA_1=2$ we have that $\Delta ABC$ and $\Delta AHA''$ share centroid $M$ . Let $L$ be midpoint of $AA''$ . Then obviously $L,H,A''$ are collinear. Let $O$ be circumcenter of $\odot (\Delta ABC)$ . We have that $O,L,A_1$ are collinear so $\angle HLO=\angle A''LO$ and $OA'=OY$ and $OL=OL$ thus $\Delta A'LO=\Delta LYO$ thus $LA'=LY$ so $A'Y || BC$ (where $Y=AA'' \cap \omega$ ). Let $X=\omega \cap A'L$ . Then because of symmetry we have $AX||A'Y||BC$ thus $AX||BC$ so $\angle ABC=\pi - \angle BAX=\angle BA'X$ and we are done

This post has been edited 1 time. Last edited by SidVicious, Jul 23, 2016, 8:45 AM

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jayme

9782 posts

jayme

#7 h Aug 6, 2016, 2:22 PM • 2 Y

Y by jhu08, Adventure10

Dear Mathlinkers,
this problem is also based on

http://www.artofproblemsolving.com/community/c6h1161858_two_parallels

Sincerely
Jean-Louis

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MonsterS

148 posts

MonsterS

#8 h Mar 9, 2017, 1:04 PM • 2 Y

Y by jhu08, Adventure10

Let Ray $HM$ intersects circumcircle at $A''$
Applying euler circle and a homothety at point M send euler circle to circumcircle .We get $AA'||BC$ .
So $\angle AA'A''=\angle ACB=\angle ABC$

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phymaths

264 posts

phymaths

#9 h Sep 15, 2017, 11:31 AM • 2 Y

Y by jhu08, Adventure10

Just consider a Homothety taking medial triangle to Triangle $ABC$ with $M$ as centre of homothety and Ratio $1/2$

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sa2001

281 posts

sa2001

#10 h Feb 10, 2018, 11:39 AM • 3 Y

Y by jhu08, Adventure10, Mango247

A solution using a little of complex numbers and a little of synthetic geometry-

This post has been edited 4 times. Last edited by sa2001, Feb 10, 2018, 11:42 AM

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Vrangr

1600 posts

Vrangr

#12 h Jun 28, 2018, 6:49 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Let $K \in \omega$ such that $AK \parallel BC$ .
Consider the homothety with ratio $-2$ at $M$ . Under this homothety $H \to K$ . Thus, $H, M, K$ are collinear. Note that $AKBC$ is an isosceles trapezium.
$\measuredangle BA'H = \measuredangle BA'K = \measuredangle BCK = \measuredangle ABC = \measuredangle ABH\quad\blacksquare$

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math_pi_rate

1218 posts

math_pi_rate

#13 h Dec 20, 2018, 4:55 PM • 2 Y

Y by jhu08, Adventure10

ARO Restated wrote:

An acute-angled $\triangle ABC \ (AB<AC)$ is inscribed into a circle $\omega$ . Let $G$ be the centroid of $\triangle ABC$ , and let $\overline{AD}$ be an altitude of this triangle. Ray $\overline{GD}$ meets $\omega$ at $X$ . Prove that the circumcircle of the triangle $XDB$ is tangent to $\overline{AB}$ .

Here's my solution: Let $P$ be the point such that $APCB$ is an isosceles trapezoid. Then, by an easy homothety argument, we can see that $P$ lies on line $\overline{GDX}$ . Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the angle bisector of $\angle BAC$ . Then $D$ gets sent to the antipode of $A$ in $\omega$ (say $A'$ ), while $B$ gets swapped with $C$ . Also, $P$ goes to a point $T \in \overline{BC}$ such that $\overline{AT}$ is tangent to $\omega$ , and so $X$ gets sent to the point $X'=\odot (AA'T) \cap \overline{BC}$ . As $\angle A'AT=90^{\circ}$ , so we get that $\overline{A'X'} \perp \overline{BC}$ . Then the center of $\odot (A'CX')$ lies on $\overline{A'C}$ , and so $\odot (A'CX')$ is tangent to $\overline{AC}$ . Inverting back, we get the desired result. Hence, done. $\blacksquare$

This post has been edited 1 time. Last edited by math_pi_rate, Dec 20, 2018, 5:18 PM

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AlastorMoody

2125 posts

AlastorMoody

#15 h Jul 26, 2019, 9:46 AM • 3 Y

Y by jhu08, Adventure10, Mango247

We'll restate the whole problem and solve: Let $H$ be the foot from $A$ to $\overline{BC}$ in $\Delta ABC$ with $AB<AC$ . Let $G$ be centroid of $\Delta ABC$ . Let $M$ be midpoint of $\overline{BC}$ . Let $AM \cap \odot (ABC)$ $=$ $S$ . Let $A'$ $\in$ $\odot (ABC)$ , such, $AA'$ $||$ $BC$ and Let $A'H$ $\cap$ $\odot (ABC)$ $=$ $X$ $\implies$ $XHMS$ is cyclic. Let $H'$ be reflection of $H$ over $M$ $\implies$ $G$ is centroid WRT $\Delta AHH'$ $\implies$ $G$ $\in$ $\overline{XA'}$ $\implies$ $\angle BXH$ $=$ $\angle BCA'$ $=$ $\angle ABH$

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amar_04

1915 posts

amar_04

#16 h Sep 6, 2019, 8:23 PM • 2 Y

Y by jhu08, Adventure10

Beautiful problem.

Let $HM\cap\odot(ABC)=K$ , and let $KO\cap\odot(ABC)=L$ and let $H'$ be the orthocenter of $\triangle ABC$ . Draw the Euler line of $\triangle ABC$ .
Now it's easy to see that $M$ is the centroid of $\triangle H'LK$ . So, $H'H=HL\implies L\in\odot(ABC)$ . Hence $L$ is the $K-$ antipode of $\odot(ABC)$ , so $\angle HAK=90^\circ\implies AK\|BC$ .

Now the rest part is easy, $\angle ABC=\angle BCK=\angle BA'K$ . Hence, proved. $\blacksquare$ .

P.S

This post has been edited 11 times. Last edited by amar_04, Sep 7, 2019, 12:35 PM

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Steve12345

618 posts

Steve12345

#17 h Sep 6, 2019, 8:54 PM • 2 Y

Y by jhu08, Adventure10

This is also Romania JBMO TST 2018

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aops29

452 posts

aops29

#18 h Jan 9, 2020, 9:01 PM • 2 Y

Y by jhu08, Adventure10

Solution

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Feridimo

563 posts

Feridimo

#19 h Jan 10, 2020, 2:57 AM • 3 Y

Y by jhu08, Adventure10, Mango247

http://www.artofproblemsolving.com/community/c6h1159504

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EulersTurban

386 posts

EulersTurban

#20 h Nov 18, 2020, 8:28 PM • 1 Y

Y by jhu08

Let $X$ be the second intersection of $HM$ with $(ABC)$ .

We want to show that $BX=AC$ , or by angle chase we want to show that $AX \parallel BC$ .

Throw the configuration onto the complex plane and let the circumcircle be the unit circle, and denote with lowercase letters of the points their complex values.
Place then such that $c = \frac{1}{b}$ .
Then we have that:
$h=\frac{1}{2}\left( a+b+\frac{1}{b}-\frac{1}{a} \right)$ $m = \frac{1}{3}\left(a+b+\frac{1}{b}\right)$
We want to show that $x= \frac{1}{a}$ , but we see that this value satisfies:
$\frac{h-m}{\overline{h-m}}=\frac{h-\frac{1}{a}}{\overline{h-\frac{1}{a}}}$
Thus we have that $AX \parallel BC$ .

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MatBoy-123

396 posts

MatBoy-123

#21 h Apr 27, 2021, 4:28 AM • 4 Y

Y by jhu08, Mango247, Mango247, Mango247

EulersTurban wrote:

Let $X$ be the second intersection of $HM$ with $(ABC)$ .

We want to show that $BX=AC$ , or by angle chase we want to show that $AX \parallel BC$ .

Throw the configuration onto the complex plane and let the circumcircle be the unit circle, and denote with lowercase letters of the points their complex values.
Place then such that $c = \frac{1}{b}$ .
Then we have that:
$h=\frac{1}{2}\left( a+b+\frac{1}{b}-\frac{1}{a} \right)$ $m = \frac{1}{3}\left(a+b+\frac{1}{b}\right)$
We want to show that $x= \frac{1}{a}$ , but we see that this value satisfies:
$\frac{h-m}{\overline{h-m}}=\frac{h-\frac{1}{a}}{\overline{h-\frac{1}{a}}}$
Thus we have that $AX \parallel BC$ .

I just want to know that why putting $c= \frac{1}{b}$ doesn't alter the question condition ??

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Bratin_Dasgupta

598 posts

Bratin_Dasgupta

#22 h Nov 15, 2021, 3:30 AM • 1 Y

Y by Bgmi

$\measuredangle$ represents angles taken modulo $180^{\circ}$
$[asy] import olympiad; size(230); pair A = (-1 , 2) , B = (-3 , -2) , C = (4,-2); draw(A--B--C--cycle , green); draw(circumcircle(A,B,C) , cyan); pair D = (0.5 , -2); pair E = (1.5 , 0); pair F = (-2 , 0); pair G = centroid(A,B,C); pair H = foot(A,B,C); pair A_1 = (-1 , -4.5); draw(circumcircle(A_1, H , B) , orange); draw(A--H); dot("$A$" , A , N); dot("$B$" , B , W); dot("$C$" , C , NE); dot("$D$" , D , NE); dot("$E$" , E , NE); dot("$F$" , F , W); dot("$G$" , G , SW); dot("$H$" , H , SW); dot("$A_1$" , A_1 , S); [/asy]$
Let $X \in \omega$ such that $AX \parallel BC$ If we consider a homothety at $M$ with a factor of $-2$ . As a result we have $H \mapsto X$ and so we have $H-M-X$ . Also we have that $AXBC$ is an isosceles trapezium. Thus we get the following $\measuredangle BA^{\prime}H = \measuredangle BA^{\prime}X = \measuredangle BCX = \measuredangle ABC = \measuredangle ABH$ $\blacksquare$

This post has been edited 4 times. Last edited by Bratin_Dasgupta, Nov 22, 2021, 3:15 AM
Reason: Made a new diagram :D

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JAnatolGT_00

559 posts

JAnatolGT_00

#23 h Nov 15, 2021, 11:45 AM • 1 Y

Y by jhu08

Let homothety with coefficient $-2$ wrt $M$ maps $H\mapsto D;$ easy to conclude, that $ABCD$ is an isosceles trepezoid.
But then $\measuredangle BA'H=\measuredangle BA'D=\measuredangle BCD=\measuredangle CBA=\measuredangle HBA$ , so the conclusion follows.

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BVKRB-

322 posts

BVKRB-

#24 h Nov 15, 2021, 12:23 PM • 1 Y

Y by jhu08

One thing I have realised is that Russians love this config a lot
Well know that $A'-H-M-A''$ where $A''$ is such that $ABCA''$ is a cyclic isosceles trapezoid

$\measuredangle BA'D = \measuredangle BA'A'' = \measuredangle BCA'' = \measuredangle CBA = \measuredangle DBA \ \blacksquare$

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JAnatolGT_00

559 posts

JAnatolGT_00

#25 h Nov 15, 2021, 12:40 PM • 2 Y

Y by BVKRB-, jhu08

BVKRB- wrote:

One thing I have realised is that Russians love this config a lot

Same feeling

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REYNA_MAIN

41 posts

REYNA_MAIN

#26 h Dec 23, 2021, 7:15 AM • 3 Y

Y by jhu08, SPHS1234, amar_04

Shortage

Attachments:

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IAmTheHazard

5001 posts

IAmTheHazard

#27 h Aug 22, 2023, 7:58 PM • 1 Y

Y by centslordm

If $A''$ is the other intersection of $\overline{MH}$ with $\omega$ it's not hard to see that $AA''CB$ is an isosceles trapezoid (for instance, use coordinates to prove the converse), so $\angle BA'H=\angle BA'A''=\angle ABC=\angle ABH$ as desired.

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mcmp

53 posts

mcmp

#28 h Nov 8, 2024, 9:57 AM

Y by

What.

You’ve got to be serious.

By the configuration in ISL 2011/G4, if $A”=\overline{A’HM}\cap(ABC)\neq A’$ then it’s well known that $AA’’BC$ is an isosceles trapezium, so from there $\measuredangle BA’H=\measuredangle BA’A’’=\measuredangle BAA’’=\measuredangle ABC=\measuredangle ABH$ , thus done from alt seg.

:huh:

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ihategeo_1969

205 posts

ihategeo_1969

#29 h Apr 7, 2025, 5:40 AM

Y by

Why another Why point config. Rename $H$ to $D$ , $A'$ to $Y_A$ and let $A'$ be $A$ -antipode. If we $\sqrt{bc}$ invert at $A$ in $\triangle ABC$ then $Y_A$ goes to foot of altitude from $A'$ to $\overline{BC}$ and we need to prove $Y_A^*A'C$ is tangent to $\overline{AC}$ which is just because $\angle ACA'=\angle CY_A^*A'=90 ^{\circ}$ , done.