Hong Kong 2017 TST1 P1

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YanYau

133 posts

YanYau

#1 h Aug 22, 2016, 4:45 AM • 2 Y

Y by mymathboy, Adventure10

In $\triangle ABC$ , let $AD$ be the angle bisector of $\angle BAC$ , with $D$ on $BC$ . The perpendicular from $B$ to $AD$ intersects the circumcircle of $\triangle ABD$ at $B$ and $E$ . Prove that $E$ , $A$ and the circumcenter $O$ of $\triangle ABC$ are collinear.

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ABCDE

1963 posts

ABCDE

#2 h Aug 22, 2016, 5:26 AM • 2 Y

Y by mymathboy, Adventure10

WLOG $AB<AC$ . Let $F$ be the reflection of $D$ over $BE$ . Note that $F$ is the orthocenter of $ABE$ , so it suffices to show that $\angle ABF=\angle C$ by circumcenter angles. But $\angle ADB=180^\circ-\frac{\angle A}{2}-\angle B$ , so $\angle EBD=\frac{\angle A}{2}+\angle B-90^\circ$ , so $\angle FBD=\angle A+2\angle B-180^\circ=\angle B-\angle C$ , as desired.

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Arab

612 posts

Arab

#3 h Aug 22, 2016, 6:11 AM • 1 Y

Y by Adventure10

WLOG, assume that $\triangle ABC$ is acute and $AB>AC$ . Let $P=DE\cap AC, Q=BE\cap AD$ .

Then we have that $\angle DAP=\angle BAD=\angle DEQ$ , so $A, P, Q, E$ are concyclic.

Hence $\angle AQE=90^\circ\implies \angle APD=90^\circ$ . Now $\angle BAE=\angle BDE=\angle CDP=90^\circ-\angle ACB\implies O\in AE$ .

This post has been edited 4 times. Last edited by Arab, Aug 22, 2016, 6:40 AM

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rmtf1111

698 posts

rmtf1111

#4 h Aug 22, 2016, 8:00 AM • 4 Y

Y by Georg.Cantor, Ryoma.Echizen, Adventure10, Mango247

Let angles $A,B,C$ be equal to $2\alpha, 2\beta, 2\gamma$ so $\angle OAB=\alpha+\beta-\gamma$ $\angle DAO=\beta-\gamma$ $\angle DAE=\angle DBE=2\beta-\angle ABE=2\beta-(180-90-\alpha)=2\beta-90+\alpha=\beta-\gamma$ . The rest is trivial.

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jayme

9772 posts

jayme

#5 h Aug 22, 2016, 10:14 AM • 3 Y

Y by Georg.Cantor, Adventure10, Mango247

Dear Mathlinkers,
have a look at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=605458

Sincerely
Jean-Louis

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lucasxia01

908 posts

lucasxia01

#6 h Aug 27, 2016, 7:44 PM • 1 Y

Y by Adventure10

Solution

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ethical_man

13 posts

ethical_man

#7 h Aug 29, 2016, 2:31 AM • 2 Y

Y by Adventure10, Mango247

Solution

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WizardMath

2487 posts

WizardMath

#8 h Aug 29, 2016, 1:14 PM • 1 Y

Y by Adventure10

Simple angle chasing using cyclic quads.

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CountofMC

838 posts

CountofMC

#9 h Sep 6, 2016, 11:12 PM • 1 Y

Y by Adventure10

This doesn't work.

This post has been edited 1 time. Last edited by CountofMC, Sep 7, 2016, 4:37 PM
Reason: Solution didn't work

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nicky-glass

239 posts

nicky-glass

#10 h Sep 7, 2016, 2:20 AM • 5 Y

Y by Georg.Cantor, Ryoma.Echizen, Med_Sqrt, Adventure10, Mango247

simple solution

This post has been edited 1 time. Last edited by nicky-glass, Sep 7, 2016, 2:34 AM
Reason: Add figure

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Generic_Username

1088 posts

Generic_Username

#11 h Sep 7, 2016, 2:44 AM • 2 Y

Y by Adventure10, Mango247

@2above

Doesn't

Quote:

$\angle BOE=180^\circ-\angle OBE-\angle OEB$ $180^\circ-\left(\angle C-\frac{\angle A}{2}\right)-\angle ADB$

assume what we want to prove?

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CountofMC

838 posts

CountofMC

#12 h Sep 7, 2016, 1:29 PM • 2 Y

Y by Adventure10, Mango247

No, because it only uses the fact that $BOE$ is a triangle but doesn't assume that $O$ and $E$ are on the same line as $A$ . Triangles $AOE$ and $BOE$ don't have to have their bases on the same line.

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Generic_Username

1088 posts

Generic_Username

#13 h Sep 7, 2016, 2:48 PM • 1 Y

Y by Adventure10

Then how do you get $\angle{OEB} = \angle{ADB}$ ?

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Virgil Nicula

7054 posts

Virgil Nicula

#14 h Sep 7, 2016, 3:59 PM • 1 Y

Y by Adventure10

See the proposed problem PP19 from here

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CountofMC

838 posts

CountofMC

#15 h Sep 7, 2016, 4:36 PM • 2 Y

Y by Adventure10, Mango247

@Generic_Username: Wait, never mind. You're right. It's $\angle AEB=\angle ADB$ but then my solution wouldn't work.

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zacchro

179 posts

zacchro

#16 h Sep 10, 2016, 8:32 PM • 1 Y

Y by Adventure10

Let $H$ be the orthocenter of $\triangle ABC$ , and $H_b$ the orthocenter of $ABD$ . Then $A,H,H_b$ collinear on the perpendicular from $A$ to $BC$ . Then $AO$ is the reflection of $AH$ in $AD$ , and $AE$ is the reflection of $AH_b$ in $AD$ , giving the desired result.

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Kezer

986 posts

Kezer

#17 h Sep 23, 2016, 9:33 PM • 2 Y

Y by Adventure10, Mango247

Have never done Hong Kong TSTs before, so it's hard for me to judge, whether this is an easy or average #1. Let's see, how rusty I am with directed angles, it's been ages since I've used them. So here is a simple angle chasing solution.

Let $S = AD \cap BE$ and $L=DE \cap CA$ . Use directed angles modulo $180^{\circ}$ . It is easy to show that $\measuredangle BAO = 90^{\circ}-\measuredangle ACB$ . Also, $\measuredangle BAE = \measuredangle BDE = \measuredangle CDE$ in cyclic quadrilateral $ABDE$ . But $\measuredangle BAD = \measuredangle DAL$ as $AD$ is an angle bisector and $\measuredangle SBA = \measuredangle EBA = \measuredangle EDA = \measuredangle LDA$ in cyclic quadrilateral $ABDE$ . So $\triangle ADL \sim \triangle ABS$ and therefore $\measuredangle ALD = \measuredangle ASB = 90^{\circ}$ . Thus $\measuredangle BAE = \measuredangle CDE = 90^{\circ} - \measuredangle LCD = 90^{\circ}-\measuredangle ACB.$ So $\measuredangle BAO = \measuredangle BAE$ which implies that $A,O,E$ are collinear. Done. $\hfill \square$ .

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MathStudent2002

934 posts

MathStudent2002

#18 h Nov 25, 2016, 5:11 PM • 4 Y

Y by va2010, Phoenix_The_Cat, Adventure10, Mango247

Solution

Better Solution

This post has been edited 1 time. Last edited by MathStudent2002, Nov 28, 2016, 3:26 AM

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math1181

41 posts

math1181

#27 h Jan 28, 2017, 5:40 AM • 2 Y

Y by Adventure10, Mango247

We show that $\measuredangle BAE = \measuredangle BAO$
$\measuredangle BAE = \measuredangle BAD +\measuredangle DAE$ $= \measuredangle BAD + \measuredangle DBE$ $= \measuredangle BAD + \measuredangle ABC - \measuredangle EBA$ $= \measuredangle BAD +\measuredangle ABC - (90^{\circ} - \measuredangle BAD)$ $= 2 \measuredangle BAD + \measuredangle ABC -90^{\circ}$ $=\measuredangle BAC + \measuredangle ABC -90^{\circ}$ $= \measuredangle BCA -90^{\circ} = \measuredangle BAO$ $\implies$ $A E O$ collinear.

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jayme

9772 posts

jayme

#28 h Jan 28, 2017, 9:14 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

1. the tangent to (ABD) at B
2. A' the second point of intersection of AE with (O)
3. by applying the Reim's theorem two times we are done...

Sincerely
Jean-Louis

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Tumon2001

449 posts

Tumon2001

#29 h Apr 2, 2017, 2:16 PM • 2 Y

Y by reun, Adventure10

Let $F$ be the intersection of $AE$ with the circumcircle of $\Delta ABC$ . Observe that $DE\perp AC$ . We have $\angle EDC = \angle EAB = \angle BCF$ . So, $DE || FC$ . This gives $\angle ACF = 90^{\circ}$ and hence the result follows.

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WolfusA

1900 posts

WolfusA

#30 h Mar 30, 2018, 12:04 PM • 2 Y

Y by Adventure10, Mango247

Argentina Cono Sur TST 2014

Leicich wrote:

In an acute triangle $ABC$ , let $D$ be a point in $BC$ such that $AD$ is the angle bisector of $\angle{BAC}$ . Let $E \neq B$ be the point of intersection of the circumcircle of triangle $ABD$ with the line perpendicular to $AD$ drawn through $B$ . Let $O$ be the circumcenter of triangle $ABC$ . Prove that $E$ , $O$ , and $A$ are collinear.

I'd like to give some warning. Maybe Mister YanYau didn't translate the problem carefully, but we need assumption weaker than in Cono Sur very much: $\angle ACB\le 90^\circ$ . It's easy to observe that this equality holds in reverse we have $O$ on the other side of line $AB$ than point $E$ , which contradicts collinearity $A,E,O$ , because foot of $O$ on segment $AB$ is a center of this segment.

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Arkmmq

214 posts

Arkmmq

#31 h Mar 30, 2018, 6:33 PM • 2 Y

Y by Adventure10, Mango247

Extend $AD$ to met $(ABC)$ at $M$ By angle chasing we get that $DE$ is perpendicular to $AC$ and by Reim's theorem we have $FC$ is parallel to $DE$ then $\angle AFC$ =90.

This post has been edited 1 time. Last edited by Arkmmq, Mar 30, 2018, 6:33 PM
Reason: ...

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Mr.Chem-Mathy

49 posts

Mr.Chem-Mathy

#32 h Mar 30, 2018, 7:41 PM • 1 Y

Y by Adventure10

We can mark the intersection of CA with Circumcircle of ABD as Q and then proceed to prove BDA'=BQC which isn't too hard to do with angle chasing.

Explanation: AA' is the diameter of ABC; The problem is equivalent to proving BEO+BEA'=180*.

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sbealing

307 posts

sbealing

#33 h Mar 30, 2018, 8:19 PM • 2 Y

Y by Adventure10, Mango247

Slight overkill but we can use areals. Let $P$ be the foot of the perpendicular from $B$ to $AD$ then $P=(k,b,c)$ for some $k$ . Using EFFT we see:
$P=(b-c,b,c) \, , \, D=(0,b,c)$ Then solving for circle $\odot ABD$ we get $u=v=0 \, , \, w=\frac{a^2 b}{b+c}$ . Letting $E=(b-c,k,c)$ we can solve for $k$ and see:
$k=\frac{b^2 S_B}{c S_C} \implies E=(c S_C (b-c),b^2 S_B,c^2 S_C)$ And as $O=(a^2 S_A,b^2 S_B,c^2 S_C)$ it is then obvious $A,O,E$ are colinear.

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WolfusA

1900 posts

WolfusA

#34 h May 16, 2018, 6:47 PM • 2 Y

Y by Adventure10, Mango247

What's EFFT?

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