external bisectors in a quadrilateral

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Remike

126 posts

Remike

#1 h Mar 23, 2005, 5:54 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be an arbitrary quadrilateral. The bisectors of external angles $A$ and $C$ of the quadrilateral intersect at $P$ ; the bisectors of external angles $B$ and $D$ intersect at $Q$ . The lines $AB$ and $CD$ intersect at $E$ , and the lines $BC$ and $DA$ intersect at $F$ . Now we have two new angles: $E$ (this is the angle $\angle{AED}$ ) and $F$ (this is the angle $\angle{BFA}$ ). We also consider a point $R$ of intersection of the external bisectors of these angles. Prove that the points $P$ , $Q$ and $R$ are collinear.

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mecrazywong

606 posts

mecrazywong

#2 h Mar 23, 2005, 6:32 AM • 2 Y

Y by Adventure10, Mango247

Well....what comes to my mind immediately is I prefer solving it by co-geom/complex number rather than drawing the diagram accurately, no matter by hand or geometry software

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prowler

312 posts

prowler

#3 h Mar 24, 2005, 7:24 PM • 2 Y

Y by Adventure10, Mango247

We denote centers of excircles of different triangles as $I_1, I_2, I_3, I_4, J_1, J_2, J_3, J_4$
(from sketch)
It is not hard to prove that $I_1I_2I_3I_4$ is concyclic by angle chasing.

$< I_4I_1I_3 = < J_4 J_2J_3$ from concyclity of $J_2EDI_1$

$< I_4I_1I_3 = < I_4I_2I_3 = < FI_2B = < BJ_1F = < J_4J_1J_3$ (from concyclity $J_1FBI_2$ )

Hence $< J_4 J_2J_3 = < J_4J_1J_3$ and $J_1J_2J_3J_4$ is concyclic.

Let $U$ be the intersection of $I_1I_4$ and $I_2I_3$ , similarly $V$ - $J_1J_4$ and $J_2J_3$ .

We will demonstrate that $P,U,V$ are collinear.

$\frac {\sin UPI_2}{\sin UPI_3}= \frac {UI_2\sin I_1I_2I_3}{UI_3\sin I_4I_3I_2}= \frac{\sin I_2I_3I_1}{\sin I_4I_1I_3}$

$\frac {\sin VPJ_2}{\sin VPJ_3}= \frac{\sin J_2J_3J_1}{\sin J_4J_1J_3}$

As we showed $< I_4I_1I_3 = < J_4J_1J_3$

And we can see that $< I_2I_3I_1 = 180- < DI_3I_2 = < DJ_3F = 180 - J_2J_3J_1$

So we can say fractions are equal hence $P,U,V$ are collinear.

Last step is from projective geometry:
$PU$ is polar to $Q$ through angle $I_3PI_1$ so it divides it in harmonic ratio.
So lines $PI_1, PU, PI_3, PQ$ are in fascicol.
Analogously for - $PJ_1, PV, PJ_3, PR$ are in fascicol.
Three lines are identical so $P,Q,R$ are collinear.

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prowler

312 posts

prowler

#4 h Mar 26, 2005, 8:31 PM • 2 Y

Y by Adventure10, Mango247

I think I solved a little bit different problem about intersection
of internal bisectors of $A,C$ and $B,D$ .

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prowler

312 posts

prowler

#5 h Apr 5, 2005, 11:50 AM • 2 Y

Y by Adventure10, Mango247

Continuing solution:
Note $T_1,T_2,T_3,T_4$ intersections of external bisectors of angles $A,B,C,D$

Let $W$ be the intersection of external bisector of $A$ and $C$ .
From problem we need to prove that $W,R,V$ are collinear.
(sorry, but in sketch they are not)

Again same idea we easily can prove that $T_1,T_2,T_3,T_4$ is concyclic.
It’s $V,T_1,T_2,T_3,T_4,W$ is complete quadrilateral and it’s diagonal intersection is $U$

It means $VW,VT_3,VU,VT_1$ is fascilcol armonic.
From first post we had $VR,VJ_3,VU,VJ_1$ is fascilcol armonic.

Three lines coinside - $V,R,W$ are collinear.

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prowler

312 posts

prowler

#6 h Apr 5, 2005, 11:51 AM • 2 Y

Y by Adventure10, Mango247

I hate monologs. Can anybody accept my solution ?

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Valiowk

374 posts

Valiowk

#7 h Apr 6, 2005, 1:53 PM • 2 Y

Y by Adventure10, Mango247

prowler, do you mind editing your diagram slightly to use different colours for the sides of the quadrilateral, the external angle bisectors at $A, B, C, D$ and the external angle bisectors at $E$ and $F$ (something like what Darij does for his diagrams)? It would be much easier to compare your solution with the diagram in that case. Right now, all I see is a whole lot of lines...

Would you please state exactly which triangles $I_1, I_2, I_3, I_4, J_1, J_2, J_3, J_4$ are excentres of? It's really hard to tell from your diagram at present. Perhaps if the diagram was coloured, it might be easier. Thanks!

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prowler

312 posts

prowler

#8 h Apr 6, 2005, 5:40 PM • 2 Y

Y by Adventure10, Mango247

$I_1$ - excenter $AED$
$I_2$ - excenter $AFB$
$I_3$ - incenter $CFD$
$I_4$ - incenter $BEC$

$J_1$ - excenter $ABF$
$J_2$ - excenter $ADE$
$J_3$ - excenter $DCF$
$J_4$ - incenter $BCE$

$T_1$ - excenter $BEC$
$T_2$ - incenter $ABF$
$T_3$ - excenter $DCF$
$T_4$ - incenter $AED$

My advice to print sketch.
P.S Which program Darij uses to draw sketches with colors, or dinamic one ?

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darij grinberg

6555 posts

darij grinberg

#9 h May 3, 2005, 6:56 AM • 2 Y

Y by Adventure10, Mango247

Prowler, your solution is correct and nice, but it can be simplified. In fact, you proved the following three assertions (where I am using your notations, not the notations of the original problem):

- The point of intersection P of the internal angle bisectors of the angles A and C, the point of intersection V of the external angle bisectors of the angles B and D, and the point of intersection U of the internal angle bisectors of the angles E and F are collinear. (This is the collinearity of the points P, U, V, proven by you in post #3.)

- The point of intersection P of the internal angle bisectors of the angles A and C, the point of intersection Q of the internal angle bisectors of the angles B and D, and the point of intersection R of the external angle bisectors of the angles E and F are collinear. (This is the collinearity of the points P, Q, R, proven by you in post #3. This is also problem 6 of the 239MO 1997, classes 10-11.)

- The point of intersection W of the external angle bisectors of the angles A and C, the point of intersection V of the external angle bisectors of the angles B and D, and the point of intersection R of the external angle bisectors of the angles E and F are collinear. (This is the collinearity of the points V, R, W proven by you in post #5, and this is also the original problem of this thread.)

But actually, from the more convenient viewpoint of directed lines, these three assertions are all equivalent to each other. They all can be formulated in one theorem:

Theorem 1. Let x, y, z, w be four directed lines. Then, the points $\left[x;\;y\right]\cap\left[z;\;w\right]$ , $\left[y;\;z\right]\cap\left[w;\;x\right]$ and $\left[x;\;z\right]\cap\left[y;\;w\right]$ are collinear.

Here, for any two directed lines g and h, the abbreviation [g; h] means the exterior angle bisector between the lines g and h. This angle bisector is defined as the locus of all points P such that d(P; g) = - d(P; h) (this locus is one line, not two lines, since we are using directed distances!). Here, the notation d(U; t) stands for the directed distance from a point U to a directed line t. Note that, if two directed lines g and h are parallel and have the same direction, then their exterior angle bisector [g; h] is the line parallel to them and lying in the middle between them; if two directed lines g and h are parallel and have opposite directions, then their exterior angle bisector [g; h] doesn't exist, strictly speaking; we can say that this exterior angle bisector is the line at infinity.

I have explained the necessary fundamentals about directed lines and directed distances in http://www.mathlinks.ro/Forum/viewtopic.php?t=6559 post #5. You can now take x = DA, y = AB, z = BC and w = CD, and you will get all three assertions you showed as consequences of Theorem 1 for different directions of the lines x, y, z, w. So it is enough to prove Theorem 1.

And here goes a particularly beautiful proof of Theorem 1, which, if I remember correctly, was not my idea, but I don't remember anymore where I learnt it:

Proof of Theorem 1. Let $\lambda$ be the locus of all points P such that d(P; x) + d(P; y) + d(P; z) + d(P; w) = 0. This locus $\lambda$ is a linear locus, i. e. it is either the empty set, or a straight line, or the whole plane. This is because, by the formula for the directed distance from a point to a line in a cartesian coordinate system, the directed distances d(P; x), d(P; y), d(P; z) and d(P; w) can be written as linear functions of the cartesian coordinates of the point P (in some fixed cartesian coordinate system), and thus, the equation d(P; x) + d(P; y) + d(P; z) + d(P; w) = 0 is a linear equation in the coordinates of P and therefore represents a linear locus. [You can also show this without the use of Cartesian coordinates.]

So, as we said, the locus $\lambda$ is either the empty set, or a straight line, or the whole plane. Now, we can easily see that the case when the locus $\lambda$ is the whole plane can be left out of consideration - for an explanation why, open the following hide tag:

Why we can assume that the locus is not the whole plane

Hence, it remains to consider the case when the locus $\lambda$ is either the empty set, or a straight line.

In this case, consider the point $T=\left[x;\;y\right]\cap\left[z;\;w\right]$ . Since this point T lies on the exterior angle bisector [x; y], we have d(T; x) = - d(T; y). Since this point T lies on the exterior angle bisector [z; w], we have d(T; z) = - d(T; w). Thus, d(T; x) + d(T; y) + d(T; z) + d(T; w) = (- d(T; y)) + d(T; y) + (- d(T; w)) + d(T; w) = 0. Hence, the point T lies on the locus $\lambda$ . Hence, the locus $\lambda$ is a line (in fact, we know that the locus $\lambda$ is either the empty set, or a line; but it cannot be an empty set, since the point T lies on this locus, so it must be a line), and the point $T=\left[x;\;y\right]\cap\left[z;\;w\right]$ lies on this line. Similarly, the points $\left[y;\;z\right]\cap\left[w;\;x\right]$ and $\left[x;\;z\right]\cap\left[y;\;w\right]$ lie on the same line $\lambda$ . Thus, the points $\left[x;\;y\right]\cap\left[z;\;w\right]$ , $\left[y;\;z\right]\cap\left[w;\;x\right]$ and $\left[x;\;z\right]\cap\left[y;\;w\right]$ lie on one line, and Theorem 1 is proven.

prowler wrote:

P.S Which program Darij uses to draw sketches with colors, or dinamic one ?

I use Euklid DynaGeo, a very nice shareware dynamic geometry program. Alas, it is avaliable in German only.

Darij

This post has been edited 1 time. Last edited by darij grinberg, Jul 3, 2021, 6:10 PM
Reason: Hereby -> Here

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enescu

741 posts

enescu

#10 h May 3, 2005, 8:32 PM • 2 Y

Y by Adventure10, Mango247

prowler wrote:

P.S Which program Darij uses to draw sketches with colors, or dinamic one ?

Prowler, try this program it's colorful, dynamic, in English and freeware

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