danube junior angle chasing 2018 P2

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parmenides51

30652 posts

parmenides51

#1 h Dec 11, 2018, 7:20 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle such that in its interior there exists a point $D$ with $\angle DAC = \angle DCA = 30^o$ and $\angle DBA = 60^o$ . Denote $E$ the midpoint of the segment $BC$ , and take $F$ on the segment $AC$ so that $AF = 2FC$ . Prove that $DE \perp EF$ .

This post has been edited 1 time. Last edited by parmenides51, Dec 11, 2018, 7:28 PM
Reason: added junior in source

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omriya200

317 posts

omriya200

#2 h Dec 12, 2018, 4:10 AM • 1 Y

Y by Adventure10

It is $CGMO$ 2007/5
Assume that $AF=2$ and $FC=1$ so los on $\triangle ADC$ gives $AD=DC=\sqrt 3$ and let $O$ be the circumcentre of $\triangle ADC$ then $\angle AOD=120$ and los on $\triangle AOD$ gives $OD=AO=1$ and $OD \parallel AC$ so $ODFC$ is a parallelogram. Let OC intersects $AD$ at $X$ then $DX=XF$ . Now we need to prof that $DX=EX$ computation gives $XD=XF=1/2$ and in $\triangle OCB$ , $XE=1/2BO=1/2$ so $XD=DF=EF$ so proved

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zuss77

520 posts

zuss77

#3 h Jul 27, 2019, 12:23 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Take homothety at $C$ with $k=2$ , mapping $E \mapsto B, D \mapsto K, F \mapsto G$ . So now we need to prove that $\angle KBG = 90^\circ$ .
Very easy observations gives: $\triangle ADK$ - equilateral, $\angle CAK = 90^\circ$ , $DBKA$ - cyclic.
If $A$ midpoint of segment $KK'$ , you can see that $G$ - centroid of equilateral $\triangle CKK'$ ( $CG=2AG$ ) and so also its incenter, $\implies \angle AGK = 60^\circ$ , $G \in (DBKA)$ , $\angle KBG = 90^\circ$ .

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JustKeepRunning

2958 posts

JustKeepRunning

#4 h Jul 27, 2019, 12:26 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

zuss77 wrote:

Take homothety at $C$ with $k=2$ , mapping $E \mapsto B, D \mapsto K, F \mapsto G$ . So now we need to prove that $\angle KBG = 90^\circ$ .
Very easy observations gives: $\triangle ADK$ - equilateral, $\angle CAK = 90^\circ$ , $DBKA$ - cyclic.
If $A$ midpoint of segment $KK'$ , you can see that $G$ - centroid of equilateral $\triangle CKK'$ ( $CG=2AG$ ) and so also its incenter, $\implies \angle AGK = 60^\circ$ , $G \in (DBKA)$ , $\angle KBG = 90^\circ$ .

What is $\text{homothety}$ ?

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zuss77

520 posts

zuss77

#5 h Jul 27, 2019, 12:28 PM • 1 Y

Y by Adventure10

JustKeepRunning wrote:

What is $\text{homothety}$ ?

You are kidding, right?

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JustKeepRunning

2958 posts

JustKeepRunning

#6 h Jul 27, 2019, 12:46 PM • 1 Y

Y by Adventure10

zuss77 wrote:

JustKeepRunning wrote:

What is $\text{homothety}$ ?

You are kidding, right?

No...

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minageus

281 posts

minageus

#7 h Jul 27, 2019, 1:03 PM • 2 Y

Y by Adventure10, Mango247

Homothety is a brilliant geometric transformation that is very useful to solving difficult problems.
Definition:
Let $O$ a stable point in plane and a number $k>0$ . Also let $A$ a random point of the plane. In $OA$ we take $A'$ such that $OA'=kOA$ .
The transformation (image) such that in every point $A$ there is a point $A'$ such that $OA'=kOA$ is called homothety with centre $O$ and ratio $k$ .
https://en.wikipedia.org/wiki/Homothetic_transformation
https://brilliant.org/wiki/euclidean-geometry-homothety/
These will help you in the beginning.

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JustKeepRunning

2958 posts

JustKeepRunning

#8 h Jul 27, 2019, 1:04 PM • 2 Y

Y by Adventure10, Mango247

minageus wrote:

Homothety is a brilliant geometric transformation that is very useful to solving difficult problems.
Definition:
Let $O$ a stable point in plane and a number $k>0$ . Also let $A$ a random point of the plane. In $OA$ we take $A'$ such that $OA'=kOA$ .
The transformation (image) such that in every point $A$ there is a point $A'$ such that $OA'=kOA$ is called homothety with centre $O$ and ratio $k$ .
https://en.wikipedia.org/wiki/Homothetic_transformation
https://brilliant.org/wiki/euclidean-geometry-homothety/
These will help you in the beginning.

Oh ok thanks

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gnoka

245 posts

gnoka

#9 h Dec 16, 2019, 10:10 AM • 2 Y

Y by Adventure10, Mango247

https://artofproblemsolving.com/community/c6h247620p1358815

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