MOP 1997 [concurrent lines from intouch triangle]

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paul_mathematics

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paul_mathematics

#1 h Dec 31, 2004, 5:56 PM • 2 Y

Y by Adventure10, Mango247

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

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darij grinberg

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darij grinberg

#2 h Dec 31, 2004, 6:41 PM • 2 Y

Y by Adventure10, Mango247

Actually, the problem holds more generally for an arbitrary cevian triangle DEF. In other words, we have the rather general result:

Let ABC be an arbitrary triangle, and let D, E, F be any three points on the lines BC, CA, AB such that the lines AD, BE, CF concur. Let the parallel to the line AB through the point E meet the line DF at a point Q, and let the parallel to the line AB through the point D meet the line EF at a point T. Then, the lines CF, DE and QT are concurrent.

Proof. Let the parallel to the line AB through the point C meet the lines EF and DF at the points U and V, respectively. Since the lines AD, BE, CF concur, the Ceva theorem yields

$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$ .

On the other hand, since the line CUV is parallel to the line AB, Thales yields $\frac{BD}{DC}=\frac{FB}{CV}$ and $\frac{CE}{EA}=\frac{UC}{AF}$ , so that we have

$\frac{FB}{CV}\cdot\frac{UC}{AF}\cdot\frac{AF}{FB}=1$ .

This simplifies into $\frac{UC}{CV}=1$ . In other words, UC = CV, so that the point C is the midpoint of the segment UV.

Since DT || AB and UV || AB, we have DT || UV. Thus, if the line CF meets the line DT at a point R, Thales yields $\frac{TR}{RD}=\frac{UC}{CV}$ . Since $\frac{UC}{CV}=1$ , we therefore have $\frac{TR}{RD}=1$ , too.

We have DT || AB and EQ || AB. Hence, DT || EQ. Thus, by Thales, $\frac{DQ}{QF}=\frac{TE}{EF}$ . Together with $\frac{TR}{RD}=1$ , this yields

$\frac{TR}{RD}\cdot\frac{DQ}{QF}\cdot\frac{FE}{ET} = 1\cdot\frac{TE}{EF}\cdot\frac{FE}{ET} = \frac{TE}{EF}\cdot\frac{FE}{ET} = \frac{ET}{FE}\cdot\frac{FE}{ET} = 1$ .

Hence, by the Ceva theorem, applied to the triangle FTD, it follows that the lines FR, TQ and DE are concurrent. In other words, the lines CF, DE and QT are concurrent. $\blacksquare$

Darij

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RaMlaF

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RaMlaF

#3 h Dec 31, 2004, 6:49 PM • 2 Y

Y by Adventure10, Mango247

i'll show that $CF$ bisects segment $QE$ (and analogously bisects segment $DT$ ). Since $DT//QE$ , the midpoints of $DT$ , $QE$ , the point $F$ , and the intersection of $QT$ and $DE$ are all collinear (Ceva). And the problem follows..

To prove that $CF$ bisects $QE$ , let $S=QE\cap CF, P=BE\cap AD, R=DF\cap BE$ . Since $(E,R;P,B)=-1$ , then $-1=(FE,FR;FP,FB)=(FE,FQ;FS,FB)$ . But then, using that $EQ//AB$ , we have that the points $S$ and a point in the line at infinity are harmonic conjugates with respect to $\text{[math]}$ . Therefor $S$ is the midpoint of $\text{[math]}$ , and we are done.

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MysticTerminator

3697 posts

MysticTerminator

#4 h Jan 2, 2005, 5:54 AM • 2 Y

Y by Adventure10, Mango247

darij grinberg wrote:

Actually, the problem holds more generally for an arbitrary cevian triangle DEF. In other words, we have the rather general result:

Let ABC be an arbitrary triangle, and let D, E, F be any three points on the lines BC, CA, AB such that the lines AD, BE, CF concur. Let the parallel to the line AB through the point E meet the line DF at a point Q, and let the parallel to the line AB through the point D meet the line EF at a point T. Then, the lines CF, DE and QT are concurrent.

I think this can be generalized a bit more still: instead of definining Q and T such, I believe we can pick any point G on line AB and then define Q = DF ^ EG, T = EF ^ DG, and the result should still hold. In your case, of course, G is simply the point at infinity corresponding to AB. So here is my attempt at proving this statement:

Take line ABFG to infinity. If P = AD ^ BE, we know CDPE is a parallelogram, but we are given that AD, BE, CF concur, so P lies on line CF, so the midpoint of DE is the midpoint of CF and lies on line CF. Now, we see that DQET is in fact another parallelogram, and DE ^ QT will be the center of this parallelogram, i.e. the midpoint of DE. As this lies on CF, we have the result that DE, CF, and QT concur. Does this work?

Edit: By the way, the problem statement is equivalent to "C lies on the Steiner line of trapezoid DTEQ". 2nd Edit: Oh wow now I see actually both the above solutions use Steiner line (without actually mentioning it)

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darij grinberg

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darij grinberg

#5 h Jan 2, 2005, 12:38 PM • 2 Y

Y by Adventure10, Mango247

MysticTerminator wrote:

I think this can be generalized a bit more still: instead of definining Q and T such, I believe we can pick any point G on line AB and then define Q = DF ^ EG, T = EF ^ DG, and the result should still hold. In your case, of course, G is simply the point at infinity corresponding to AB. So here is my attempt at proving this statement:

Take line ABFG to infinity. If P = AD ^ BE, we know CDPE is a parallelogram, but we are given that AD, BE, CF concur, so P lies on line CF, so the midpoint of DE is the midpoint of CF and lies on line CF. Now, we see that DQET is in fact another parallelogram, and DE ^ QT will be the center of this parallelogram, i.e. the midpoint of DE. As this lies on CF, we have the result that DE, CF, and QT concur. Does this work?

Indeed, this works fine! So we have another solution now...

MysticTerminator wrote:

Edit: By the way, the problem statement is equivalent to "C lies on the Steiner line of trapezoid DTEQ". 2nd Edit: Oh wow now I see actually both the above solutions use Steiner line (without actually mentioning it)

Hmm, I didn't even know that the thing was called "Steiner line"

anyway, of course I did use it implicitely.

Darij

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grobber

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grobber

#6 h Jan 2, 2005, 2:13 PM • 2 Y

Y by Adventure10, Mango247

The only Steiner line I know (which also shows up with Google

) is the parallel through $H$ to the SDimson line of a point $P$ (called the Steiner line of $P$ wrt $ABC$ ).

Sort of off-topic though

. Sorry about that

.

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MysticTerminator

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MysticTerminator

#7 h Jan 2, 2005, 2:18 PM • 2 Y

Y by Adventure10, Mango247

Ah, yes. My apologies. It is not called "Steiner line" as such - the theorem saying that the midpoints of the bases of the trapezoid are collinear with the intersection of diagonals and non-parallel sides is called "Steiner's theorem"

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abdurashidjon

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abdurashidjon

#8 h May 6, 2006, 6:27 AM • 2 Y

Y by Adventure10, Mango247

Hi Darij You have used Gauss theorem, I couldn not find . PLease post ity to me , u used but I coud not understend at the usage part of theorem.
Abdurashid

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