MEMO 2010, Problem T-5: Parallel lines

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Martin N.

434 posts

Martin N.

#1 h Sep 12, 2010, 1:33 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

The incircle of the triangle $ABC$ touches the sides $BC$ , $CA$ , and $AB$ in the points $D$ , $E$ and $F$ , respectively. Let $K$ be the point symmetric to $D$ with respect to the incenter. The lines $DE$ and $FK$ intersect at $S$ . Prove that $AS$ is parallel to $BC$ .

(4th Middle European Mathematical Olympiad, Team Competition, Problem 5)

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jgnr

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jgnr

#2 h Sep 13, 2010, 7:16 AM • 1 Y

Y by Adventure10

Let $T$ be the point on $DE$ such that $AT\parallel BC$ . So $\frac{DT}{DE}=\frac{AC}{CE}$ , $DT=b\cdot\frac{DE}{CE}=2b\sin\frac{C}2$ . We also have $DF=2BD\sin\frac{B}2=(a-b+c)\sin\frac{B}2$ . Thus $\begin{align*}\frac{DF}{DT}&=\frac{(a-b+c)\sin\frac{B}2}{2b\sin\frac{C}2}\\&=\frac{(\sin A-\sin B+\sin C)\sin\frac{B}2}{2\sin B\sin\frac{C}2}\\&=\frac{\sin(B+C)-2\cos\frac{B+C}2\sin\frac{B-C}2}{4\cos\frac{B}2\sin\frac{C}2}\\&=\frac{2\sin\frac{B+C}2\cos\frac{B+C}2-2\cos\frac{B+C}2\sin\frac{B-C}2}{2\sin\frac{B+C}2-2\sin\frac{B-C}2}\\&=\cos\frac{B+C}2\\&=\cos\angle FDT.\end{align*}$ Hence $\angle DFT=90^{\circ}$ , which gives $T,K,F$ are collinear and $T=S$ . So $AS\parallel BC$ .

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jayme

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jayme

#3 h Sep 13, 2010, 10:43 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
the result can be obtains by using three times the Pascal's in a degenerated form...
Sincerely
Jean-Louis

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limes123

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limes123

#4 h Sep 13, 2010, 1:45 PM • 1 Y

Y by Adventure10

It's known that $DK$ , $EF$ and $A$ median intersect in one point. Using this we can easily prove that $AS$ is polar of this point with respect to inscribed circle, which means that $AS\perp DK$ which we wanted to prove.

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armpist

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armpist

#5 h Sep 13, 2010, 11:36 PM • 2 Y

Y by Adventure10, Mango247

Dear J-L,

What about not using Pascal or Stanley Rabinowitz?

A new challenge ...

M.T.

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vladimir92

212 posts

vladimir92

#6 h Sep 14, 2010, 12:58 AM • 2 Y

Y by Adventure10 and 1 other user

Martin N. wrote:

The incircle of the triangle $ABC$ touches the sides $BC$ , $CA$ , and $AB$ in the points $D$ , $E$ and $F$ , respectively. Let $K$ be the point symmetric to $D$ with respect to the incenter. The lines $DE$ and $FK$ intersect at $S$ . Prove that $AS$ is parallel to $BC$ .

(4th Middle European Mathematical Olympiad, Team Competition, Problem 5)

My solution

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jayme

9782 posts

jayme

#7 h Sep 14, 2010, 5:01 AM • 1 Y

Y by Adventure10

Dear Armpist (M.) and Mathlinkers,
your idea seems to be very nice, but I don't see the way you have in mind.
Sincerely
Jean-Louis

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StefanS

149 posts

StefanS

#8 h Apr 26, 2011, 9:38 AM • 2 Y

Y by reveryu, Adventure10

armpist wrote:

Dear J-L,

What about not using Pascal or Stanley Rabinowitz?

A new challenge ...

M.T.

Synthetic, Simple and Detailed Solution

Attachments:

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hatchguy

555 posts

hatchguy

#9 h Sep 1, 2011, 3:27 AM • 3 Y

Y by reveryu, Adventure10, Mango247

Let $R \equiv DF \cap KE$ . Apply Pascal theorem to degenerate hexagon $KFFDEE$ to obtain that $A,S$ and $R$ are collinear. Since $KD$ is a diameter of the incircle we have $\angle KFD = \angle KED = 90$ and therefore $K$ is the orthocenter of triangle $DRS$ which implies $DK \perp AS => AS \parallel BC$

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reveryu

218 posts

reveryu

#10 h Feb 5, 2017, 5:29 AM • 2 Y

Y by Adventure10, Mango247

by angle chasing can be seen easily that FAE=2(FSE) and A lies on perpendicular bisector of FE imply A is circumcenter of FES so AS=AE
, angle AES=DEC and DEC is isosceles imply AES similar to CED so the results.

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Tumon2001

449 posts

Tumon2001

#11 h Feb 5, 2017, 7:50 AM • 2 Y

Y by Adventure10, Mango247

Solution:

Join $AI$ and $IF$ .
Clearly, $K$ lies on the incircle. We have,
$\angle AFI$ = $90$ and $\angle AIF$ = $\angle FDS$ .
So, $\Delta AFI$ ~ $\Delta SFD$ .
Thus, $\angle FAI$ = $\angle FSD$ ,
or, $\angle FSD$ = $\frac{\angle A}{2}$ .
Also, $AF$ = $AE$ .
So, $A$ is the circumcenter of $\Delta FES$ .
This implies that $AE$ = $AS$ .
Therefore, $\angle ASE$ = $\angle AES$ = $\angle CED$ = $\angle CDE$ .
Hence, $AS$ is parallel to $BC$ .

This post has been edited 1 time. Last edited by Tumon2001, Feb 5, 2017, 7:53 AM

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llplp

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llplp

#12 h Jul 19, 2021, 1:44 AM

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How has noone beat me to this?
Complex bash

This post has been edited 1 time. Last edited by llplp, Jul 19, 2021, 1:44 AM

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lazizbek42

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lazizbek42

#13 h Jan 18, 2022, 10:22 AM

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Pascal theorem: $DEEKFF$

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EulersTurban

386 posts

EulersTurban

#14 h Jan 19, 2022, 11:29 PM

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Let $T$ be the point of intersection of the lines $EK$ and $DF$ .

By Pascal on $DEEKFF$ , we have that $T,A,S$ are colinear points.
Since we know that $\angle KFD = \angle KED = 90$ , we have that $I \in KD$ .

Let $G$ be the intersection points of $EF$ and $KD$ .
By Brokard's we have that $IG \perp TS$ , and since $IG \perp BC$ , we have that $AS \parallel BC$ .

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hqxaev

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hqxaev

#15 h Apr 28, 2022, 8:59 AM

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denote: $KE \cap DF=T$ , second intersection of AD with incircle $ABC$ as $Q$

let's prove that the points $T,A,S$ lie on the same line. Let it not be so. By the Brocard's theorem to the inscribed quadrilateral $DEKF$ , we know that $DK \bot TS$ , and by virtue of the fact that $DK$ is a diameter of the circumcircle of the triangle DEF, $DK \bot BC$ , so $TS \parallel BC$ .

and we know that $SF \bot TD$ , $TE \bot SD$ , so $K$ is the orthocenter of the triangle $TDS$

denote $TS \cap BC=P_{\infty}$ , $TS \cap AD=R$

note that the $DEQF$ is a harmonic quadrilateral. By taking perspectivity in point $D$ , $(F,E;Q,D)=(T,S;R,P_{\infty})=-1$ , so $TR=SR$ , so $TR=SR=FR=ER$

note that $\angle STE=\angle SFE$ , because of $AC$ is a tangent to circumcircle of $DEF$ , $\angle SFE=\angle TEA$ , and because of $RT=RE$ , $\angle RTE=\angle RET$ . so $\angle REA=\angle RET- \angle TEA=0$ , contradiction, so $T,A,S$ lie on the same line. And so because of $TS \parallel BC$ , $AS \parallel BC$

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