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Re: [EMHL] Squares inscribed in triangle
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On Saturday, December 1, 2001, at 07:52 AM, Hyacinthos wrote:
>
What ??? I didn't send any nontext or has my mailer itself sent it ?
> [Nontext portions of this message have been removed]
>
>
APH 0 Attachment
Dear Atul and other Hyacinthists
> And also is there a point P for which the lengths of the
sides of
> inscribed squares in each of the triangles PBC,PCA and PAB equal?
Let P =(x,y,z) in barycentrics.
As the length of the sides of the square inscribed in ABC is
2ad/(aa+2d)where d=area(ABC), the length of the sides of a square
inscribed in PBC is 2dax/(aa(x+y+z)+2dx).
The three length are equal iff
u = v = w = 0
where u = bc(cybz)(x+y+z)+2d(bc)yz; v,w circular.
Now, as 1/(2d)(aau+bbv+ccw)=aa(bc)yz+bb(ca)zx+cc(ab)xy, P lies
necessarily on the circumhyperbola through K and I.
Putting x=aa/(at), y=..., we get
(E) : aa/(ta)+bb/(tb)+cc/(tc)+2d/t = 0.
If we suppose, for example, a<b<c, (E) has three real roots,
one in ]0,a[, one in ]a,b[, one in ]b,c[.
Only the first one gives a point inside ABC and t is the common value
of the lengths of the sides of the three squares.
Conclusion :
There exists an unique point P inside ABC for which the lengths of
the sides of the three squares are equal; the common length is the
smallest root t of (E) and P is barycentric(aa/(at),...).
P is not constructible and lies on the circumhyperbola through I and
K.
Note that we easily find a conicconstruction of P, using the three
common points of the mesh generated by u=0, v=0, w=0.
Another interesting problem could be :
If Pa,Pb,Pc are the centers of the three squares inscribed in PBC,
PCA, PAB, find the locus of P for which PaPbPc and ABC are
perspective.
This locus is, if I didn't mistake, the selfisogonal circumcubic
with pivot U, where
Vect(KU) = 4d/(aa+bb+cc) Vect(OH).
I think that this point U can be interesting for Clark because he
lies on the parallel through K to the Euler line and it is the first
time that I meet a triangle center with an interesting geometric
property on this line (except, of course, K and the infinite point).
Of course, putting d instead of d, we get similar results for the
the three other squares inscribed in the side lines of PBC,...
I hope that there are not too many mistakes.
Friendly. JeanPierre 0 Attachment
 In Hyacinthos@y..., "Antreas P. Hatzipolakis" <xpolakis@o...>
wrote:> Let ABC be a triangle and P = (x:y:z) a point in normals.
rectangles inscribed
> Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR
> in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp.,
whose
> base / height = t.
^^^^
T
>
of
> A
> /\
> / \ AbAc / AbA'b =T
> / \
> / \
> / \
> / P \
> / \
> / A'b M'a A'c \
> / Pa \
> BAbMaAcC
>
>
> Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c
> AbAcA'cA'b,
such that:
> and similarly MbM'b, McM'c.
>
> Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp.
>
^^^^
> M'aPa M'bPb M'cPc
>  =  =  = T
> M'aMa M'bMb M'cMc
t
>
normals:
>
> ABC, PaPbPc are perspective <==>
>
> (1+t)cosB + TsinB
> x() + z
> 2
>  * CYCLIC = 1
> (1+t)cosC + TsinC
> x() + y
> 2
>
> ==> The locus of P is the isogonal pivotal cubic with pivot in
>
Antreas
> (1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
> (()  [() * ()] ::)
> 2 2 2
>
> The case of the squares and their centers is for T = 1, t = 1/2.
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Dear JeanPierre,
[JPE]>Another interesting problem could be :

>If Pa,Pb,Pc are the centers of the three squares inscribed in PBC,
>PCA, PAB, find the locus of P for which PaPbPc and ABC are
>perspective.
>This locus is, if I didn't mistake, the selfisogonal circumcubic
>with pivot U, where
>Vect(KU) = 4d/(aa+bb+cc) Vect(OH).
>I think that this point U can be interesting for Clark because he
>lies on the parallel through K to the Euler line and it is the first
>time that I meet a triangle center with an interesting geometric
>property on this line (except, of course, K and the infinite point).
>
>Of course, putting d instead of d, we get similar results for the
>the three other squares inscribed in the side lines of PBC,...
I think that we have seen this problem before, and also its
parametrizations.
Let ABC be a triangle and P = (x:y:z) a point in normals.
Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR rectangles inscribed
in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp., whose
base / height = t.
A
/\
/ \ AbAc / AbA'b = t
/ \
/ \
/ \
/ P \
/ \
/ A'b M'a A'c \
/ Pa \
BAbMaAcC
Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c of
AbAcA'cA'b,
and similarly MbM'b, McM'c.
Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp. such that:
M'aPa M'bPb M'cPc
 =  =  = T
M'aMa M'bMb M'cMc
ABC, PaPbPc are perspective <==>
(1+t)cosB + TsinB
x() + z
2
 * CYCLIC = 1
(1+t)cosC + TsinC
x() + y
2
==> The locus of P is the isogonal pivotal cubic with pivot in normals:
(1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
(()  [() * ()] ::)
2 2 2
The case of the squares and their centers is for T = 1, t = 1/2.
Quoting you but applying it for me:
>I hope that there are not too many mistakes.
Greetings from Athens
Antreas 0 Attachment
Let ABC be a triangle, and AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a
the three similar rectangles inscribed in ABC, based on BC, CA,
AB, resp., whose Height / Base = t.
Let AA', BB', CC' be the three altitudes of ABC.
A
/\
/ \ A'bAb / AbAc = t
/ \
/ \
/ \
A'b A'c
/ \
/ \
/ \
BAbA'AcC
If (1/AbAc) + (1/BcBa) + (1/CaCb) = (1/AA') + (1/BB') + (1/CC')
then find t.
Antreas
PS: The t that I found equals to:
sinBsinC + sinCsinA + sinAsinB
1  (  )
sinA + sinB + sinC
APH 0 Attachment
Dear Antreas
> Let ABC be a triangle and P = (x:y:z) a point in normals.
inscribed
> Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR rectangles
> in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp., whose
that:
> base / height = t.
>
> A
> /\
> / \ AbAc / AbA'b = t
> / \
> / \
> / \
> / P \
> / \
> / A'b M'a A'c \
> / Pa \
> BAbMaAcC
>
>
> Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c of
> AbAcA'cA'b,
> and similarly MbM'b, McM'c.
>
> Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp. such
>
normals:
> M'aPa M'bPb M'cPc
>  =  =  = T
> M'aMa M'bMb M'cMc
>
>
> ABC, PaPbPc are perspective <==>
>
> (1+t)cosB + TsinB
> x() + z
> 2
>  * CYCLIC = 1
> (1+t)cosC + TsinC
> x() + y
> 2
>
> ==> The locus of P is the isogonal pivotal cubic with pivot in
>
)] ::)
> (1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
> (()  [() * (
> 2 2 2
I'm sorry but I didn't remember this discussion.
>
> The case of the squares and their centers is for T = 1, t = 1/2.
Note that the locus of the pivot is the hyperbola through G, the
Steiner point, the Fermat points and the reflection of G wrt O
(surely, this has been discussed too).
If we take P inside ABC and lying on the circumhyperbola through I
and K, we can inscribe three congruent rectangles in PBC, PCA, PAB.
How can we construct these three rectangles?
I have another question related to the following problem :
Find P inside ABC such as the incircles of PBC, PCA, PAB are
congruent.
I presume that this problem has been discussed and solved (it doesn't
seem to be not an easy one)
Have you any reference?
Friendly. JeanPierre 0 Attachment
Dear sir
>I have another question related to the following problem :
It's a special case of Elkies point and has been named as "Congruent
>Find P inside ABC such as the incircles of PBC, PCA, PAB are
>congruent.
>I presume that this problem has been discussed and solved (it doesn't
>seem to be not an easy one)
>Have you any reference?
>Friendly. JeanPierre
incircles point"
References:
1)http://mathworld.wolfram.com/CongruentIncirclesPoint.html
2)Kimberling, C. "Central Points and Central Lines in the Plane of a
Triangle." Math. Mag. 67, 163187, 1994.
Yours faithfully
Atul.A.Dixit
_________________________________________________________________
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Dear Atul> >I have another question related to the following problem :
doesn't
> >Find P inside ABC such as the incircles of PBC, PCA, PAB are
> >congruent.
> >I presume that this problem has been discussed and solved (it
> >seem to be not an easy one)
as "Congruent
> >Have you any reference?
> >Friendly. JeanPierre
>
> It's a special case of Elkies point and has been named
> incircles point"
a
>
> References:
> 1)http://mathworld.wolfram.com/CongruentIncirclesPoint.html
> 2)Kimberling, C. "Central Points and Central Lines in the Plane of
> Triangle." Math. Mag. 67, 163187, 1994.
There is nothing in the Weinsstein CRC except the definition and I
don't have Clark's article.
What I wanted to know is :
Can we find a trigonometric expression of the coordinates of the
point?
What is the algebraic equation (wrt a,b,c) of lowest degree with root
the common radius ? Is there a known curve going through this point?
Friendly. JeanPierre 0 Attachment
Dear Antreas:
[JPE]:> > If we take P inside ABC and lying on the circumhyperbola through I
PAB.
> > and K, we can inscribe three congruent rectangles in PBC, PCA,
> > How can we construct these three rectangles?
[APH]
> >
>
> Do you mean congruent = with both equal bases and equal heights or
I mean congruent = with both equal bases and equal heights
> do you mean equal [in area] ?
> That is, having their bases inversely proportional to their heights.
Friendly. JeanPierre 0 Attachment
Dear Antreas and other Hyacinthists
> [JPE]:
through I
> > > If we take P inside ABC and lying on the circumhyperbola
> > > and K, we can inscribe three congruent rectangles in PBC, PCA,
heights.
> PAB.
> > > How can we construct these three rectangles?
> > >
> >
>
> [APH]
> > Do you mean congruent = with both equal bases and equal heights or
> > do you mean equal [in area] ?
> > That is, having their bases inversely proportional to their
[JPE]> I mean congruent = with both equal bases and equal heights
May be, this needs a little explanation.
I've discovered (but I guess that I'm not the first one) that when
I've looked at congruent inscribed squares (Recent problem from Atul
Dixit)
Let m = height/basis.
The length of the basis of such a rectangle inscribed in ABC (with
basis on BC ) is 2ad/(maa+2d) where d = area(ABC).
Thus the length of the basis of the rectangle inscribed in PBC (for a
given ratio m) is
La = 2adx/(maa(x+y+z)+2dx) if p is barycentric (x,y,z).
We have La = Lb = Lc iff
u = v = w = 0 where
u = bcm(x+y+z)(cybz)+2(bc)dyz; v and w : circular.
As 1/(2d)(aau+bbv+ccw) = aa(bc)yz+bb(ca)zx+cc(ab)xy,
necessarily P lies on the circumhyperbola through I and K.
Putting P = (aa/(aL),bb/(bL),cc/(cL)), we have u=v=w=0 iff
(E) : m(aa/(aL)+bb/(bL)+cc/(cL))=2d/L
This way, we get three congruent rectangles inscribed in PBC, PCA,
PAB with common basis L and common height mL, where m is given by (E).
Now, the problem is : find an easy construction of the three
rectangles for given P.
May be, it can be useful to note that
L = 1/3(a+b+c)(IP*/GP*) where P*=isogonal(P)  on the line IG 
Friendly. JeanPierre 0 Attachment
Dear Antreas,
> [JPE]:
PAB.
> > If we take P inside ABC and lying on the circumhyperbola through I
> > and K, we can inscribe three congruent rectangles in PBC, PCA,
> > How can we construct these three rectangles?
[APH]
> >
>
If P = I, the rectangles degenerate in three segments (basis = 0,
> A
> /\
> / \
> / \
> / \
> / \
> / I \
> / A'b A'c\
> / \
> BAbAcC
>
> Assume that P = I.
>
> Let AbAcA'cA'b, BcBaB'aB'c, CaCbC'bC'a be the three
> congruent rectangles inscribed in IBC, ICA, IAB, resp.
> (based on BC, CA, AB, resp.)
>
> Denote AbAc = BcBa = CaCb = f
>
> AbA'b = AcA'c = BcB'c = BaB'a = CaC'a = CbC'b = g
>
> We have:
>
> BC = BAb + AbAc + AcC = gcot(B/2) + f + gcot(C/2) or
>
> a = f + g(cot(B/2) + cot(C/2))
>
> Similarly:
>
> b = f + g(cot(C/2) + cot(A/2))
>
> c = f + g(cot(A/2) + cot(B/2))
>
> Hmmm.... What do we get from this?
height = r)
Friendly. JeanPierre 0 Attachment
Dear JeanPierre,
[APH]:
[parameters corrected]>> Let ABC be a triangle and P = (x:y:z) a point in normals.
[JPE]:
>> Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR rectangles
>> inscribed
>> in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp., whose
>> base / height = T.
>>
>> A
>> /\
>> / \ AbAc / AbA'b = T
>> / \
>> / \
>> / \
>> / P \
>> / \
>> / A'b M'a A'c \
>> / Pa \
>> BAbMaAcC
>>
>>
>> Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c of
>> AbAcA'cA'b,
>> and similarly MbM'b, McM'c.
>>
>> Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp. such that:
>>
>> M'aPa M'bPb M'cPc
>>  =  =  = t
>> M'aMa M'bMb M'cMc
>>
>>
>> ABC, PaPbPc are perspective <==>
>>
>> (1+t)cosB + TsinB
>> x() + z
>> 2
>>  * CYCLIC = 1
>> (1+t)cosC + TsinC
>> x() + y
>> 2
>>
>> ==> The locus of P is the isogonal pivotal cubic with pivot in normals:
>>
>> (1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
>> (()  [() * ()]
>> ::)
>> 2 2 2
>>
>> The case of the squares and their centers is for T = 1, t = 1/2.
>
> I'm sorry but I didn't remember this discussion.
Probably yes, probably no! I don't remember.
> Note that the locus of the pivot is the hyperbola through G, the
> Steiner point, the Fermat points and the reflection of G wrt O
> (surely, this has been discussed too).
[JPE]:>
Do you mean congruent = with both equal bases and equal heights or
> If we take P inside ABC and lying on the circumhyperbola through I
> and K, we can inscribe three congruent rectangles in PBC, PCA, PAB.
> How can we construct these three rectangles?
>
do you mean equal [in area] ?
That is, having their bases inversely proportional to their heights.
[JPE]:> I have another question related to the following problem :
Let P = (x:y:z) in actual normals.
> Find P inside ABC such as the incircles of PBC, PCA, PAB are
> congruent.
> I presume that this problem has been discussed and solved (it doesn't
> seem to be not an easy one)
> Have you any reference?
>
Let R_a, R_b, R_c be the inradii of the triangles PBC, PCA, PAB.
R_a = R_b = R_c <==>
area(PBC)/semiper.(PBC) = area(PCA)/semiper.(PCA) =
area(PAB)/semiper.(PAB)
<==> ax/(a + PB + PC) = by/(b + PC + PA) = cz/(c + PA + PB) <==>
xsinA/(a + PB + PC) = ysinB/(b + PC + PA) = zsinC/(c + PA + PB) [1]
Now, we have a = (xsinA + ysinB + zsinC)/sinBsinC, b = ..., c = ...
sqrt(y^2 + z^2 + 2yzcosA)
PA = , PB = ..., PC =....
sinA
The system [1] now can be homogenized but, as Paul would say, neither I
nor my computer could solve it!
Antreas
[Nontext portions of this message have been removed] 0 Attachment
[Atul Dixit]:> It's a special case of Elkies point ...
For a plane triangle A1A2A3, call two circles within the triangle
<i>companion incircles</i> if they are the incircles of two triangles
formed by dividing A1A2A3 into two smaller triangles by passing a line
through one of the vertices and some point on the opposite side.
(a) Show that any chain of circles C1,C2,... such that Ci, Ci+1 are
companion incircles for every i consists of at most six distinct circles.
(b) Give a ruler and compass construction for the unique chain which
has only three distinct circles.
(c) For such a chain of three circles show that the three subdividing
lines are concurrent.
(The American Mathematical Monthly 94(1987) 877 by Noam Elkies)
Antreas 0 Attachment
Dear JeanPierre,
[JPE]:> If we take P inside ABC and lying on the circumhyperbola through I
A
> and K, we can inscribe three congruent rectangles in PBC, PCA, PAB.
> How can we construct these three rectangles?
>
/\
/ \
/ \
/ \
/ \
/ I \
/ A'b A'c\
/ \
BAbAcC
Assume that P = I.
Let AbAcA'cA'b, BcBaB'aB'c, CaCbC'bC'a be the three
congruent rectangles inscribed in IBC, ICA, IAB, resp.
(based on BC, CA, AB, resp.)
Denote AbAc = BcBa = CaCb = f
AbA'b = AcA'c = BcB'c = BaB'a = CaC'a = CbC'b = g
We have:
BC = BAb + AbAc + AcC = gcot(B/2) + f + gcot(C/2) or
a = f + g(cot(B/2) + cot(C/2))
Similarly:
b = f + g(cot(C/2) + cot(A/2))
c = f + g(cot(A/2) + cot(B/2))
Hmmm.... What do we get from this?
Antreas
[Nontext portions of this message have been removed]
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