ANOTHER ONE

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

ANOTHER ONE

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: MOP 1997

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

arandomperson123

430 posts

arandomperson123

#1 h Feb 16, 2017, 11:21 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle, and let $D,E,F$ be points where the incircle touches sides $BC, CA, AB,$ respectively. The parallel to $AB$ through $E$ meets $DF$ at $Q$ , and the paralleled to $AB$ through $D$ meets $EF$ at T. Prove that the lines $CF, DE, QT$ are concurrent!!!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

243 posts

#3 h Feb 17, 2017, 12:14 AM • 2 Y

Y by Andrew_P, Adventure10

Let $QT$ and $ED$ intersect at $X$ ,clearly $QDTE$ IS trapezium so $\frac{EX}{XD}=\frac{ET}{QD}$ .From sine law on $\triangle FET$ we have $ET=\cos(\frac{C}{2})\cdot \frac{EF}{\cos(\frac{B}{2})}$ ,from $\triangle QFD$ we have $QD=\cos(\frac{C}{2})\cdot \frac{FD}{\cos(\frac{A}{2})}$ so
$\frac{EX}{XD}=\frac{\cos(\frac{A}{2})\cdot EF}{FD\cdot \cos(\frac{B}{2})}$ .Now let $CF$ and $ED$ intersect at $Y$ ,clearly $FC$ is symedian of $\triangle EFD$ so $\frac{EY}{DY}=\frac{EF^2}{DF^2}=\frac{EX}{DX}$ because $\frac{EF}{DF}=\frac{\cos(\frac{A}{2})}{\cos(\frac{B}{2})}$ so $Y\equiv X$ .

This post has been edited 2 times. Last edited by Garfield, Feb 17, 2017, 8:44 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

763 posts

#4 h Feb 17, 2017, 3:33 AM • 2 Y

Y by Adventure10, Mango247

Garfield wrote:

Let $QT$ and $ED$ intersect at $X$ ,clearly $QDTE$ IS trapezium so $\frac{EX}{XD}=\frac{ET}{QD}$ .From sine law on $\triangle FET$ we have $ET=\cos(\frac{C}{2})\cdot \frac{EF}{\cos(\frac{B}{2})}$ ,from $\triangle QFD$ we have $QD=\cos(\frac{C}{2})\cdot \frac{FD}{\cos(\frac{A}{2})}$ so
$\frac{EX}{XD}=\frac{\cos(\frac{A}{2})\cdot EF}{FD\cdot \cos(\frac{B}{2})}$ .Now let $CF$ and $ED$ intersect at $Y$ ,clearly $FC$ is symedian of $\triangle EFD$ so $\frac{EY}{DY}=\frac{EF^2}{DF^2}=\frac{EX}{DX}$ because $\frac{EX}{DX}=\frac{\cos(\frac{A}{2})}{\cos(\frac{B}{2})}$ so $Y\equiv X$ .

could you descrpit a little about the sine law part? I don't get that well...

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9787 posts

jayme

#5 h Feb 17, 2017, 6:18 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

see also

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=359366

Sincerely
Jean-Louis

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

243 posts

#6 h Feb 17, 2017, 8:51 AM • 2 Y

Y by Andrew_P, Adventure10

Skravin wrote:

Garfield wrote:

Let $QT$ and $ED$ intersect at $X$ ,clearly $QDTE$ IS trapezium so $\frac{EX}{XD}=\frac{ET}{QD}$ .From sine law on $\triangle FET$ we have $ET=\cos(\frac{C}{2})\cdot \frac{EF}{\cos(\frac{B}{2})}$ ,from $\triangle QFD$ we have $QD=\cos(\frac{C}{2})\cdot \frac{FD}{\cos(\frac{A}{2})}$ so
$\frac{EX}{XD}=\frac{\cos(\frac{A}{2})\cdot EF}{FD\cdot \cos(\frac{B}{2})}$ .Now let $CF$ and $ED$ intersect at $Y$ ,clearly $FC$ is symedian of $\triangle EFD$ so $\frac{EY}{DY}=\frac{EF^2}{DF^2}=\frac{EX}{DX}$ because $\frac{EX}{DX}=\frac{\cos(\frac{A}{2})}{\cos(\frac{B}{2})}$ so $Y\equiv X$ .

could you descrpit a little about the sine law part? I don't get that well...

Sine low on $\triangle EFT$ $\angle EFD=180^{\circ}-\angle AFE-\angle BFD=180^{\circ}-90^{\circ}+\angle(\frac{A}{2})-90^{\circ}+\angle(\frac{B}{2})=\angle(\frac{A+B}{2})$ .
$\angle ETF=\angle TFB=90^{\circ}-\angle(\frac{B}{2})$ . Now $\frac{EF}{\sin(90^{\circ}-\angle(\frac{B}{2}))}=\frac{ET}{\sin(\frac{A+B}{2})}$ and we know that $\sin(\angle(\frac{A+B}{2}))=\cos(\angle(\frac{C}{2}))$ and $\sin(90^{\circ}-\angle(\frac{B}{2})=\cos(\frac{B}{2})$ .It's exactly same on $\triangle QFD$ .

This post has been edited 1 time. Last edited by Garfield, Feb 17, 2017, 8:54 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

763 posts

#7 h Feb 17, 2017, 9:26 AM • 2 Y

Y by Adventure10, Mango247

Garfield wrote:

Skravin wrote:

Garfield wrote:

Let $QT$ and $ED$ intersect at $X$ ,clearly $QDTE$ IS trapezium so $\frac{EX}{XD}=\frac{ET}{QD}$ .From sine law on $\triangle FET$ we have $ET=\cos(\frac{C}{2})\cdot \frac{EF}{\cos(\frac{B}{2})}$ ,from $\triangle QFD$ we have $QD=\cos(\frac{C}{2})\cdot \frac{FD}{\cos(\frac{A}{2})}$ so
$\frac{EX}{XD}=\frac{\cos(\frac{A}{2})\cdot EF}{FD\cdot \cos(\frac{B}{2})}$ .Now let $CF$ and $ED$ intersect at $Y$ ,clearly $FC$ is symedian of $\triangle EFD$ so $\frac{EY}{DY}=\frac{EF^2}{DF^2}=\frac{EX}{DX}$ because $\frac{EX}{DX}=\frac{\cos(\frac{A}{2})}{\cos(\frac{B}{2})}$ so $Y\equiv X$ .

could you descrpit a little about the sine law part? I don't get that well...

Sine low on $\triangle EFT$ $\angle EFD=180^{\circ}-\angle AFE-\angle BFD=180^{\circ}-90^{\circ}+\angle(\frac{A}{2})-90^{\circ}+\angle(\frac{B}{2})=\angle(\frac{A+B}{2})$ .
$\angle ETF=\angle TFB=90^{\circ}-\angle(\frac{B}{2})$ . Now $\frac{EF}{\sin(90^{\circ}-\angle(\frac{B}{2}))}=\frac{ET}{\sin(\frac{A+B}{2})}$ and we know that $\sin(\angle(\frac{A+B}{2}))=\cos(\angle(\frac{C}{2}))$ and $\sin(90^{\circ}-\angle(\frac{B}{2})=\cos(\frac{B}{2})$ .It's exactly same on $\triangle QFD$ .

thanks

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

48 posts

Gao-JX

#8 h Feb 17, 2017, 2:05 PM • 2 Y

Y by Adventure10, Mango247

easy by harmonic pencil of lines

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

763 posts

#9 h Feb 17, 2017, 11:43 PM • 1 Y

Y by Adventure10

Garfield wrote:

Skravin wrote:

Garfield wrote:

Let $QT$ and $ED$ intersect at $X$ ,clearly $QDTE$ IS trapezium so $\frac{EX}{XD}=\frac{ET}{QD}$ .From sine law on $\triangle FET$ we have $ET=\cos(\frac{C}{2})\cdot \frac{EF}{\cos(\frac{B}{2})}$ ,from $\triangle QFD$ we have $QD=\cos(\frac{C}{2})\cdot \frac{FD}{\cos(\frac{A}{2})}$ so
$\frac{EX}{XD}=\frac{\cos(\frac{A}{2})\cdot EF}{FD\cdot \cos(\frac{B}{2})}$ .Now let $CF$ and $ED$ intersect at $Y$ ,clearly $FC$ is symedian of $\triangle EFD$ so $\frac{EY}{DY}=\frac{EF^2}{DF^2}=\frac{EX}{DX}$ because $\frac{EX}{DX}=\frac{\cos(\frac{A}{2})}{\cos(\frac{B}{2})}$ so $Y\equiv X$ .

could you descrpit a little about the sine law part? I don't get that well...

Sine low on $\triangle EFT$ $\angle EFD=180^{\circ}-\angle AFE-\angle BFD=180^{\circ}-90^{\circ}+\angle(\frac{A}{2})-90^{\circ}+\angle(\frac{B}{2})=\angle(\frac{A+B}{2})$ .
$\angle ETF=\angle TFB=90^{\circ}-\angle(\frac{B}{2})$ . Now $\frac{EF}{\sin(90^{\circ}-\angle(\frac{B}{2}))}=\frac{ET}{\sin(\frac{A+B}{2})}$ and we know that $\sin(\angle(\frac{A+B}{2}))=\cos(\angle(\frac{C}{2}))$ and $\sin(90^{\circ}-\angle(\frac{B}{2})=\cos(\frac{B}{2})$ .It's exactly same on $\triangle QFD$ .

If following the condition of the problem I wonder that ETF really makes triangle...

This post has been edited 1 time. Last edited by Skravin, Feb 17, 2017, 11:43 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

arandomperson123

430 posts

arandomperson123

#10 h Feb 18, 2017, 12:39 AM • 1 Y

Y by Adventure10

No it is not a triangle... I do not understand his solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

243 posts

#11 h Feb 18, 2017, 8:44 AM • 2 Y

Y by Adventure10, Mango247

arandomperson123 wrote:

No it is not a triangle... I do not understand his solution.

I'm sorry I didn't used notations as in problema and forgot to mention that..

Attachments:

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a