Re: [EMHL] Quadrilateral problem.
- Dear Darij,
>The theorem of 3 circles states that their homothety centers
> This is very interesting: you state that the
> external homothetic center of the incircles of
> triangles PAB and PBC coincides with the external
> homothetic center of the incircles of triangles
> PCD and PDA. My sketches confirm this, but do you
> have a simple proof?
( external ones, and the circles have to be outside of
each other) are collinear. I do not know who's name this
theorem carries ( no doubt some French guy), but with all this
excess homothety the result of Menelaus theorem probably was used
to prove it.
The diagonals of the quadrilateral keep the entire groop
of 4 circles nicely together, being common tangents.
> >> When you complete the ABCD quadrilateralYou are right. I do not know what I had in mind.
> >> also, the 2 new vertices will be on this line
> >> too.
> Not on my figure!
>You probably realized that I'm not a professional mathematician,
> >> In fact, the homothety centers of the 4
> >> circles together with the point P make an
> >> auto-polar triangle. The orthocenter of this
> >> triangle is outside of it, and is the center
> >> of the circle in question. It is on the line
> >> connecting the P and the incircle O.
> Unfortunately, I have lost the thread in these
> investigations, but my dynamic sketches seem to
> contradict this. In fact, in "very obtuse"
> circumscribable quadrilaterals ABCD, the center
> of the circle doesn't lie on PO.
> It seems that I am misunderstanding something.
> Can you please elaborate more on the
> explanation of the results?
just an amature, who likes math. I'm sorry, I do not always have
rigorous proofs to the statements I make.
I should have taken the sketch to the 'extreme', you are right,
some lines only looked parallel.
But still, the perpendiculars from the homothety centers on the
continuations of the diagonals onto the other diagonal do cross
in the center of the resulting circle , being the orthocenter
of the auto-polar triangle.
Another point: if you take a Gergonne line and rotate it a little
around it's originating vertex, the distance its foot travels along
the opposite side of the triangle is congruent to the distance
that separates now the tangent points of the pair of incircles
on the new position of the 'Gergonne' line. The proof should
be similar to the one that helped M. Pitot to prove his theorem
a+c=b+d for the quadrilateral with a circle inside.
Are you sure there is no very simple solution using angle
property of inscribed quadrilaterals A+C=B+D=2d? There must exist
some pair of pivotal angles that affects all others when the
nice square being twisted into irregular quadrilateral squeezing
without crushing the very round egg inside?
Sorry for the nonsense, old, present, and future!
- Dear Charles,
In Hyacinthos message #8929, you wrote:
>> This is intriguing. I'd be very interested toUnfortunately, I am not the expert on general
>> know if there is a general principle here about
>> whether an angle can possibly be 1/3 of another
>> or not, simply because there is an
>> impossibility of trisecting an angle with
>> Euclidean tools--I suppose it must hinge on
>> whether or not the diagram can be constructed
>> with Euclidean tools, right? Regardless, I'm
>> sure in this case that a 60 degree angle can
>> be found in ways other than
>> trisecting--wouldn't that fact preclude the
>> application of such a principle anyway? (In
>> fact, Euclid's very first theorem is a
>> construction of an equilateral triangle,
>> unless I'm mistaken!)
constructability questions and angle
trisections you can consult. What I wrote was
an empiric fact with a plausible explanation
rather than a proof. My assertion was that
starting with a general triangle ABC, and
(a) rational constructions
(b) angle bisections
only, we will never find a triangle XYZ which
is equilateral for all triangles ABC. Hereby,
a "rational construction" is an operation
assigning to some given points another point
whose Cartesian coordinates are given
rational functions of the coordinates of the
You can see that, for instance, the Napoleon
theorem does not fall under my assertion: In
fact, in order to construct the Napoleon
triangle, it is necessary to draw three
equilateral triangles, hence there is no
contradiction with my principle. But, as I
have said, I await a specialist in these
>> But I bow to your judment, Darij, that thereUnfortunately, any kind of averaging other than
>> is no equilateral triangle in the situation
>> described. What I noticed that made me
>> think there might be one was that, in my
>> constructions, connecting one vertex from
>> each "local" pair to one from the next pair
>> always gave me a triangle with very nearly
>> (but not quite) 60 degree angles (even in
>> the limiting cases). So it seemed like the
>> angles might "average" to 60 degrees in the
>> two triangles, if only I could find the
>> right way to "average" them. Arithmetic,
>> geometric, and various constructive
>> "averages" got close, but nothing got me
>> equilaterals. Alas!
arithmetic means is out of place in elementary
geometry - since angles are usually not
multiplied or divided by each other. But I
don't know, maybe there is an application.
In my opinion, we would rather likely find
triangles not equilateral but similar to ABC.
In Hyacinthos message #8930, you wrote:
>> The theorem of 3 circles states that their^^^^^^^^^^
>> homothety centers ( external ones, and the
>> circles have to be outside of each other)
Not necessarily, by the way.
>> are collinear. I do not know who's name thisIt is called Monge theorem.
>> theorem carries ( no doubt some French guy),
>> but with all this excess homothety theWe actually have to prove now that the lines
>> result of Menelaus theorem probably was used
>> to prove it.
YZ, WX and BD concur. I fear this will be not
too easy since the condition that the
quadrilateral ABCD has an incircle is necessary
- but how do we involve it?
>> The diagonals of the quadrilateral keep theI am an amateur too. Triangle geometry is not
>> entire groop of 4 circles nicely together,
>> being common tangents.
>> You probably realized that I'm not a
>> professional mathematician, just an amature,
for professionals only, it is a topic where
amateurs can discover lots of new things. I
am glad to participiate in a newsgroup like
Hyacinthos which gets together professionals
as well as hobby geometers.
>> But still, the perpendiculars from theAuto-polar with respect to the circle passing
>> homothety centers on the continuations of
>> the diagonals onto the other diagonal do
>> cross in the center of the resulting
>> circle , being the orthocenter of the
>> auto-polar triangle.
through the four incenters. This is true and
a very nice fact!
Moreover, you and Charles use the same
perpendiculars defining them in different
ways: You state that the circle through the
4 incenters is centered at the meet of the
perpendiculars from Q to AC and from R to BD,
where Q and R are the external homothetic
centers on BD and AC, respectively. Charles
states that the center is the meet of the
perpendiculars from U to AC and from T to
BD, where U is the common point of tangency
of the incircles of triangles CDA and ABC
with AC, and T is the similar point on BD.
These perpendiculars are the same two
Another note: The internal common tangent
of the incircles of triangles PBC and PCD
different from the line AC passes through T.
>> Are you sure there is no very simpleNot sure, but there is obviously no
>> solution using angle property of inscribed
>> quadrilaterals A+C=B+D=2d?
straightforward way to solve the problem. I
will see if I will "get" it with your and
- Dear Darij,
>I think we have to use the fact proved by Charles that
> It is called Monge theorem.
> We actually have to prove now that the lines
> YZ, WX and BD concur. I fear this will be not
> too easy since the condition that the
> quadrilateral ABCD has an incircle is necessary
> - but how do we involve it?
a quadrilateral with incircle is equivalent to
two triangular tangent circles. These 2 pairs of circles
have homothety centers in the same spots as the 4 incircles.
M. Monge will help us. We have to look at 3 circles out of 4
at a time and dance around replacing one of the 3 circle by
the 4th one. There will be lines of collinear 3 points, 2 of which
belong to another line of collinear 3 points. At the end of the
music we will have all the points on one line.
This homothety line seems to be the radical line of the
original incircle and the solution incircle ?
Is the perpendicular from the homethety center to the diagonal
also an angle bisector ?
>I thought that when you degenerate the quadrilateral, the point D
> Another note: The internal common tangent
> of the incircles of triangles PBC and PCD
> different from the line AC passes through T.
ends up in the point P (AC x BD). The two small circles tangent
to BD do not touch each other, the incircle of the ABC touches
AC at M, MP=the distance on BP between the tangent points of 2 small
circles. What you do is degenerating quadrilateral with point D
sliding into foot of Gergonne line.
Once we know that the incenters are cyclic we can see which of
the inscribed angles are equal. Why is it so hard to see before
Is it possible to show that each of the incenters is symmetrical
to orthocenter of the other 3 wrt diagonal?
The sides of the incircles quadrilateral are anti-parallel.
It would be nice to have some Lemoine connection here.
Could it be that the original incircle tangent points with
the quadrilateral are symedians ?
- Dear Rafi,
In Hyacinthos message #8936, you wrote:
>> I think we have to use the fact proved by CharlesWOW! This is something really nice! For instance,
>> that a quadrilateral with incircle is equivalent
>> to two triangular tangent circles. These 2 pairs
>> of circles have homothety centers in the same
>> spots as the 4 incircles.
the external homothetic center of the incircles of
triangles ABC and CDA is Q, the external homothetic
center of the incircles of triangles PDA and PAB
and the external homothetic center of the incircles
of triangles PBC and PCD. This means, three
homothetic centers coincide at Q !
Moreover, once we have proven that Q is the
external homothetic center of the incircles of
triangles ABC and CDA, it follows that the line
joining their centers - i. e., the perpendicular
to AC through U - passes through Q: another step
towards the proof of your theorem.
>> M. Monge will help us. We have to look at 3Good idea. But could you make a complete proof
>> circles out of 4 at a time and dance around
>> replacing one of the 3 circle by the 4th one.
>> There will be lines of collinear 3 points, 2
>> of which belong to another line of collinear
>> 3 points. At the end of the music we will
>> have all the points on one line.
out of this? Note that there are eight circles
to consider: the incircles of triangles PAB,
PBC, PCD, PDA, ABC, BCD, CDA, DAB.
>> This homothety line seems to be the radicalDo you mean the line QR ? No, unfortunately,
>> line of the original incircle and the
>> solution incircle ?
it is not even perpendicular to OO', where O'
is the center of the circle through the four
incenters. It is perpendicular to O'P.
>> Is the perpendicular from the homethetyDo you mean: Is the perpendicular from Q to AC
>> center to the diagonal also an angle bisector ?
the angle bisector between the lines WX and YZ ?
No. But it is the reflection of AC in this angle
bisector. Just another proposition to prove.
>> Is it possible to show that each of theWhich diagonal?
>> incenters is symmetrical to orthocenter of
>> the other 3 wrt diagonal?
Let me finally systemize the results and
conjectures you, Charles and me have obtained:
(1) Given a quadrilateral ABCD with an incircle
centered at O. The diagonals AC and BD meet
at P. Let X, Y, Z, W be the incenters of
triangles PAB, PBC, PCD, PDA; then, these
incenters X, Y, Z, W lie on one circle.
The center of this circle will be called O'.
(2) The lines XZ and YW are the angle bisectors
of the angles APB = CPD and DPA = BPC and
pass through P.
[This is clear.]
(3) The homothetic center of the incircles of
triangles PBC and PCD coincides with the
homothetic center of the incircles of
triangles PDA and PAB.
This homothetic center is denoted by Q.
(4) This Q is also the homothetic center of
the incircles of triangles ABC and CDA.
(5) Q lies on the lines BD, WX and YZ.
[This is clear, since the line BD is an
external common tangent of the incircles of
triangles PBC and PCD, and the lines WX and
YZ are the joins of the centers of these and
the two other circles.]
(6) The incircles of triangles ABC and CDA
touch AC at one point U.
[This was proven by Charles.]
(7) The incenters of triangles ABC and CDA
lie on a line perpendicular to AC and
passing through Q and U.
[This follows from (4) and (6).]
(8) This line is the reflection of BD in
the angle bisector between the lines
WX and YZ.
[I guess this can be proven using the
properties of cyclic quadrilaterals.]
(9) This line passes through O'.
[This follows from (10) below.]
(10) If R is defined in a similar fashion
to Q, (hence R lies on AC, XY and ZW,)
then the triangle PQR is autopolar with
respect to the circle centered at O'
and passing through X, Y, Z, W.
[This can be proven as follows: The triangle
PQR is the diagonal triangle of the cyclic
quadrilateral XYZW; hence, it is autopolar
with respect to the circumcircle of this
cyclic quadrilateral. (In fact, it is not
hard to prove that the diagonal triangle of
a cyclic quadrilateral is autopolar with
respect to the circumcircle.)
Of course, this proof makes use of the
earlier facts (1), (3) and (5), but, at
least, I see some success!]
(11) The internal common tangent of the
incircles of triangles PCD and PDA
different from the line BD passes
And finally, a fact not related to this
configuration I have mentioned earlier:
(X) The perpendicular bisectors of AB, BC, CD,
DA form another inscriptable quadrilateral.
This was a problem in a booklet published
by the Moscow University in 1959 containing
training problems for the participants of the
23th Moscow Mathematics Olympiad. It was one
of the problems proposed for the 7th form.
And I am in the 11th form and don't succeed
to prove it!!
- Dear Darij,
>If you take X,Y,Z,W points, then W is a reflection
> >> Is it possible to show that each of the
> >> incenters is symmetrical to orthocenter of
> >> the other 3 wrt diagonal?
> Which diagonal?
of the XYZ orthocenter over diagonal XZ.
>Where do the angle bisectors of the internal tangent lines
> (11) The internal common tangent of the
> incircles of triangles PCD and PDA
> different from the line BD passes
> through U.
to the 3 smaller incircles (out of 4 ) intersect?
It is upsetting that the O' is not on the OP line. Well it is
due to the fact that the quadrilateral made by the tangent points
of the original quadrilateral has it's opposite side cross
on the same diagonals in points M and N further away from points
Q and R, and the line QR is very CLOSE to be parallel to MN,
Some kind of reflective transformation can correct this
'flaw'. There is an angle at point P between the bisectors
connecting the 4 incenters and the diagonals of the
quadrilateral inscribed in the original circle O at the
But all the prior and above is way too much fancy geometry.
There should be something simple, related to the
angles the sides of the initial square or trapecia are twisted
> And finally, a fact not related to thisI saw some similar problems: the common tangent lines to circles
> configuration I have mentioned earlier:
> (X) The perpendicular bisectors of AB, BC, CD,
> DA form another inscriptable quadrilateral.
> This was a problem in a booklet published
> by the Moscow University in 1959 containing
> training problems for the participants of the
> 23th Moscow Mathematics Olympiad. It was one
> of the problems proposed for the 7th form.
> And I am in the 11th form and don't succeed
> to prove it!!
S1-S2, S2-S3,S3-S4, and S4-S1 form an inscriptable quadrilateral.
It is proved simply by comparing the lengths of common tangents
and getting the a+c=b+d for the quadrilateral.
Your perpendiculars are to the midpoints and it is hard
(close to impossible ?) to make them common tangents of S1-S4.
The picture has to be inverted somehow. That is why I suspected
that the tangent points of the inscriptable quadrilateral
are the symedians points.
In this case an isogonal transformation will make it possible to
have tangent circles, and the quadrilateral in the intersection
will not ( hopefully ?!) be out of the original shape.
There is also another one where you have 2 pairs of vertical lines
and 2 pairs of horizontal lines. If you can inscribe a circle
in the external quadrilateral and into the 4 coner quadrilaterals,
then the internal quadrilateral is also inscriptable.
- Dear Rafi,
In Hyacinthos message #8941, you wrote:
>> If you take X,Y,Z,W points, then W is a reflectionYes. This follows from the fact that the point where
>> of the XYZ orthocenter over diagonal XZ.
an altitude of a triangle meets the circumcircle is
the reflection of the orthocenter in the
corresponding sideline of the triangle. (This is,
actually, a very simple fact.) Now, apply this to
triangle XYZ; the line YP is an altitude of this
triangle; hence, W is the reflection of the
orthocenter of triangle XYZ in the sideline XZ.
>> Where do the angle bisectors of the internalWhy do you take the three smaller incircles?
>> tangent lines to the 3 smaller incircles
>> (out of 4 ) intersect?
They are all equal in rights! But - you are
completely right - taking any three of the
four internal tangent lines, the angle
bisectors meet at O'. In fact,
(12) The internal common tangent of the
incircles of triangles PAB and PBC
different from BD, and the three other
analogous tangents touch a circle
centered at O'.
This is a conjecture, too.
>> There should be something simple, relatedI don't think this could lead to a simple
>> to the angles the sides of the initial
>> square or trapecia are twisted about.
proof. Of course, you can imagine any
quadrilateral as an image of a square in a
continuous transformation - but how can we
use something like this in a proof? We
don't know anything about this transformation
other than the fact that it is continuous.
>> The picture has to be inverted somehow.Of what? Do you mean that the point where
>> That is why I suspected that the tangent
>> points of the inscriptable quadrilateral
>> are the symedians points.
the incircle of ABCD touches AB is the foot
of the P-symmedian in triangle PAB ? This is,
unfortunately, incorrect - I cannot even
imagine what is the role of the cevian
through this point in triangle PAB.
>> In this case an isogonal transformationApplied to the whole figure??
>> will make it possible to have tangent[...]
>> There is also another one where you haveYes, I knew this one. The fact can be
>> 2 pairs of vertical lines and 2 pairs of
>> horizontal lines. If you can inscribe a
>> circle in the external quadrilateral and
>> into the 4 coner quadrilaterals, then the
>> internal quadrilateral is also inscriptable.
restated as follows: If we have eight lines
a, b, c, d, a', b', c', d' such that the lines
b, c, b', c' touch one circle, the lines
c, d, c', d' touch one circle, the lines
a, b, a', b' touch one circle, the lines
a, b, c', d' touch one circle, and the lines
c, d, a', b' touch one circle, then the lines
d, a, d', a' touch one circle.
Note that this fact is true for directed lines
and circles only; else there may be cases the
theorem doesn't apply. I. e., there are some
restrictions to the arrangement of the lines.
- Dear Rafi, Darij, Jean-Pierre and other colleagues!
>This fact is known. It was proposed by Vajnstejn as a problem for "Kvant" in
> Inscriptable quadrilateral is divided into 4
> non-overlapping triangles by it's diagonals.
> Prove that the 4 incenters are on a circle.
> A simple solution is preferred.
1995. I proved an additional fact. If ABCD is a circumscribed quadrilateral,
I1, I2, I3, I4 - the incenters of ABM, BCM, CDM, DAM (M - the common point
of AC and BD), then the tangents to circle I1I2I3I4 in I1,I2,I3,I4 form the
quadrilateral witn vertex in AC and BD.
Happy new year!
- Dear Alexey,
Thanks for the reply.
I do not know about others, but as for me, the only interest I have
in this problem is the solution, as simple as possible.
Unfortunately, I do not have access to the old 'Kvant' issues to
satisfy my curiousity.
Is it possible for you to describe the proofs of the problem and
your addition to it in general terms?
- Dear Darij,
I was using the 'Mediterranean' method of proof: assuming it is
true what facts surface that can be used for the proof; that
explains the reflexion of the orthocenter and the antiparallel
There is no explanation for symmedians, other than pure
adventuristic speculation and a wish to bring H. Lemoine into
Another idea, and this one has some thought behind it:
the 3 incenters are projected into points
on a straight (the homothety centers) line using the 4th incenter
as the projection center.
Do we have to verify that the double ratious do not change? There
are only 3 points, with the 4th at infinity .
The fact that it is a line should place all 4 of the incenters on
- Dear friends,
I didn't follow your messages so excuse me for repeating the following:
it is known that AI*AJ = AB*AC where I, J are the incenter and A_excenter
If the quadrilateral ABCD is circumscribed to a circle P is the intersection
of the diagonals
and I1, I2, I3, I4 are the incenters and J1, J2, J3, J4 are the
P_excenters of the triangles
PAB, PBC, PCD, PDA then since
PI1*PJ1 = PA*PB
PI3*PJ3 = PC*PD
PI2*PJ2 = PB*PC
PI4*PJ4 = PD*PA
PI1*PI3*PJ1*PJ3 = PI2*PI4*PJ2*PJ4
PI1*PI3 = PI2*PI4
PJ1*PJ3 = PJ2*PJ4
or that the excenters are also cyclic points.