Geometry problem

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JANMATH111

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JANMATH111

#1 h Mar 10, 2018, 3:01 PM • 1 Y

Y by Adventure10

Let $O$ be the circumcenter of triangle $ABC$ and $H$ be it's orthocenter where is $D$ the foot from $C$ . Perpendicular line to $OD$ in the point $D$ intersects side $BC$ at the point $E$ . Circumcircle of $BCH$ intersects the line $AB$ at points $B$ and $F$ . Prove, that $H$ , $E$ , $F$ are collinear.

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JANMATH111

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JANMATH111

#2 h Mar 10, 2018, 5:27 PM • 2 Y

Y by Adventure10, Mango247

bumppppppp

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SomeCanadian

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SomeCanadian

#3 h Mar 10, 2018, 5:28 PM • 2 Y

Y by Adventure10, Mango247

Can you use Menelaus?

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JANMATH111

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JANMATH111

#4 h Mar 10, 2018, 5:45 PM • 1 Y

Y by Adventure10

Yessssss

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Wizard_32

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Wizard_32

#5 h Mar 10, 2018, 6:07 PM • 2 Y

Y by Adventure10, Mango247

The complex solution goes something like this:
Let $(ABC)$ be the unit circle. Then $d=\frac{1}{2}(h+a+b-ab\bar{h})=\frac{1}{2} \left(a+b+c-\frac{ab}{c} \right)$ .
Now note that $F$ is the reflection of $A$ over $D$ and so $f=2d-a=b+c-\frac{ab}{c}$ .
Also, $\frac{e-d}{\bar{e}-\bar{d}}=-\frac{d}{\bar{d}} \implies e\bar{d}+\bar{e}d=2d\bar{d}$ and $\frac{b-e}{\bar{b}-\bar{e}}=\frac{b-c}{\bar{b}-\bar{c}}=-bc \implies b-e=-c-bc\bar{e} \implies bc\bar{e}-e=-b-c$ .
Solve these 2 equations to get the coordinates of $E$ and then apply the simple collinearity test.

Edit: Fixed the silly mistake.

This post has been edited 3 times. Last edited by Wizard_32, Mar 11, 2018, 3:25 AM

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WizardMath

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WizardMath

#6 h Mar 10, 2018, 6:17 PM • 3 Y

Y by GoldGirl, Adventure10, Mango247

@above, your coordinate of $F$ is wrong.
$D$ isn't opposite $A$ as it usually is.

This post has been edited 1 time. Last edited by WizardMath, Mar 10, 2018, 6:42 PM

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SomeCanadian

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SomeCanadian

#7 h Mar 10, 2018, 7:49 PM • 4 Y

Y by lsi, JANMATH111, Adventure10, Mango247

Sorry I wrote this really quickly I'm not sure if I made a mistake somewhere. Can someone verify my solution?

By Menelaus Theorem, $F$ , $E$ , and $H$ are collinear iff $\frac{DH}{HC} \times \frac{CE}{BE} \times \frac{BF}{FD} = 1$ .
$\frac{S_{\triangle DEH}}{S_{\triangle CEH}} \times \frac{S_{\triangle CEF}}{S_{\triangle BEF}} \times \frac{S_{\triangle BEF}}{S_{\triangle DEF}} = 1$ $\frac{S_{\triangle DEH}}{S_{\triangle CEH}} \times \frac{S_{\triangle CEF}}{S_{\triangle DEF}} = 1$ $\frac{S_{\triangle DEH}}{S_{\triangle DEF}} = \frac{S_{\triangle CEH}}{S_{\triangle CEF}}$ Which is trivial to show.

This post has been edited 1 time. Last edited by SomeCanadian, Mar 10, 2018, 8:10 PM

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JANMATH111

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JANMATH111

#8 h Mar 11, 2018, 1:36 AM • 2 Y

Y by Adventure10, Mango247

SomeCanadian wrote:

Sorry I wrote this really quickly I'm not sure if I made a mistake somewhere. Can someone verify my solution?

By Menelaus Theorem, $F$ , $E$ , and $H$ are collinear iff $\frac{DH}{HC} \times \frac{CE}{BE} \times \frac{BF}{FD} = 1$ .
$\frac{S_{\triangle DEH}}{S_{\triangle CEH}} \times \frac{S_{\triangle CEF}}{S_{\triangle BEF}} \times \frac{S_{\triangle BEF}}{S_{\triangle DEF}} = 1$ $\frac{S_{\triangle DEH}}{S_{\triangle CEH}} \times \frac{S_{\triangle CEF}}{S_{\triangle DEF}} = 1$ $\frac{S_{\triangle DEH}}{S_{\triangle DEF}} = \frac{S_{\triangle CEH}}{S_{\triangle CEF}}$ Which is trivial to show.

I did the same thing but I don't see why the last equality would work? Is it just me?

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SomeCanadian

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SomeCanadian

#9 h Mar 11, 2018, 4:06 AM • 2 Y

Y by Adventure10, Mango247

JANMATH111 wrote:

SomeCanadian wrote:

Sorry I wrote this really quickly I'm not sure if I made a mistake somewhere. Can someone verify my solution?

By Menelaus Theorem, $F$ , $E$ , and $H$ are collinear iff $\frac{DH}{HC} \times \frac{CE}{BE} \times \frac{BF}{FD} = 1$ .
$\frac{S_{\triangle DEH}}{S_{\triangle CEH}} \times \frac{S_{\triangle CEF}}{S_{\triangle BEF}} \times \frac{S_{\triangle BEF}}{S_{\triangle DEF}} = 1$ $\frac{S_{\triangle DEH}}{S_{\triangle CEH}} \times \frac{S_{\triangle CEF}}{S_{\triangle DEF}} = 1$ $\frac{S_{\triangle DEH}}{S_{\triangle DEF}} = \frac{S_{\triangle CEH}}{S_{\triangle CEF}}$ Which is trivial to show.

I did the same thing but I don't see why the last equality would work? Is it just me?

$\frac{S_{\triangle DEH}}{S_{\triangle DEF}} = \frac{EH}{EF} = \frac{S_{\triangle CEH}}{S_{\triangle CEF}}$ using area method.

This post has been edited 1 time. Last edited by SomeCanadian, Mar 11, 2018, 4:06 AM

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pro_4_ever

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pro_4_ever

#10 h Mar 11, 2018, 5:10 AM • 2 Y

Y by SomeCanadian, Adventure10

SomeCanadian wrote:

$\frac{S_{\triangle DEH}}{S_{\triangle DEF}} = \frac{EH}{EF} = \frac{S_{\triangle CEH}}{S_{\triangle CEF}}$ using area method.

Aren't you assuming $E,F,H$ to be collinear (and proving the same)?

**Also, you did not use the fact that $\angle ODE = 90^\circ$ . Even if your proof was right, then it would work for any point $E$ on $BC$ , which is COMPLETELY ABSURD**

This post has been edited 4 times. Last edited by pro_4_ever, Mar 11, 2018, 2:38 PM

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ltf0501

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ltf0501

#11 h Mar 11, 2018, 8:18 AM • 2 Y

Y by Adventure10, Mango247

Butterfly thm simply solves it.

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georgeado17

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georgeado17

#12 h Mar 11, 2018, 9:13 AM • 3 Y

Y by SomeCanadian, Adventure10, Mango247

Let $\angle BCH=\angle BFH=x$ and let $\angle DBH=m=\angle DCA$ $\implies$ $\angle BHF=\angle BCF=m-x$ $\implies$ $\angle FCD=\angle DCA=m$ $\implies$ $\Delta ACF$ is isosceles and $AD=DF$ .Let $CD$ intersects circumcircle $(ABC)$ at $K$ and let $DE$ intersects $AK$ and $BC$ at $M$ and $E$ respectively,then by Butterfly theorem have that $MD=DE$ and note that cause of $MD=DE$ and $AD=DF$ $\implies$ quadrilateral $AMFE$ is parallelogram and $\angle KAB=\angle KCB=\angle AFE=\angle AFH=x$ hence $H,E,F$ are collinear.
Done.

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