## Re: X(5390)

Expand Messages
• Dear Friends, I made a structured Mathematica-file with the calculation of the coordinates of X(5390). When you are interested please let me know so that I can
Message 1 of 26 , Jan 9, 2013
• 0 Attachment
Dear Friends,

I made a structured Mathematica-file with the calculation of the coordinates of X(5390).
When you are interested please let me know so that I can send it to you.

Best regards,

Chris van Tienhoven

www.chrisvantienhoven.nl

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
>
> Dear Friends,
>
> I finally managed to find the coordinates of X(5390).
> I used a new method that for as far as I know- never was used before.
> I already had found the coordinates and needed to scroll 16 times on my screen to reach the end of the coordinates.
> All Trig-commands in Mathematica could not help simplifying these coordinates.
> Still they were right, because when I substituted real values the outcome was perfect.
>
> The theory I used:
> Points that come forth from related subjects often are linearly related in the triangle field.
> I called this triangle field before a Perspective Field. It is a kind of coordinate system that can be used to describe all related points with real coordinates (often rational numbers).
> This relationship manifests itself in the property that each participating point in a Perspective Field can be described as the (weighted) sum of 3 other points in the field, whereas another 4th point is used to determine the standard counting weights, also called the compliance factors.
> So I reasoned that when I could find a Perspective Field and the location of X(5390) in this field, then it should be possible to calculate its coordinates from the 3+1 chosen reference points.
>
> I followed these steps:
> 1. I gathered all Morley related points in ETC (31 points).
> 2. I gathered some Euler line related points (13 points).
> 3. I made a program in Mathematica in which I determined all permutations of 4 of these 44 points, no three of which were collinear.
> 4. Per set of 4 points I calculated if X5390 could be the weighted sum of 3 of these points. The result would be 3 real / rational numbers (as coordinates), using the 4th point to determine the compliance factors.
> 5. Of all 990 possible sets of 4 points only one set gave a result. It said:
> X5390 = 2.cf1.X2 + 1.cf2.X1136 + 2.cf3.X1137,
> the compliance factors cf1, cf2, cf3 being determined by the 4th point X(3273):
> cf1 = Det[X3273, X1136, X1137]
> cf2 = Det[ X2, X3273, X1137]
> cf3 = Det[ X2, X1136, X3273]
>
> I know this will appear pretty technically and it is of course.
> I think I will write a third document on this subject.
> The first two documents on Perspective Fields can be downloaded at:
> http://www.chrisvantienhoven.nl/index.php/mathematics/perspective-fields.html
> Especially the 2nd document explains the algebraic implications.
>
> The final result for the first BAYCENTRIC coordinate of X(5390) is:
>
> Sin[A] (Cos[ C/3 -Pi]/6] (-Sin[A] + Cos[1/3 (B + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[B]) Sin[C] + Cos[B/3 -Pi]/6] Sin[B] (Sin[A] - Cos[1/3 (C + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[C]) + Cos[A/3 -Pi]/ 6] (Cos[1/3 (C + 4Pi])] (-Sec[1/3 (A + 4Pi])] Sin[A] + 2 Sec[1/3 (B + 4Pi])] Sin[B]) Sin[C] + Cos[1/3 (B + 4Pi])] Sin[ B] (Sec[1/3 (A + 4Pi])] Sin[A] - 2 Sec[1/3 (C + 4Pi])] Sin[C]) + 2 Cos[1/3 (A + 4Pi])] Sin[ A] (-Sec[1/3 (B + 4Pi])] Sin[B] + Sec[1/3 (C + 4Pi])] Sin[C])))
> There still could be a simpler expression for this.
>
> There is also another way to describe the result.
> Let Det1 = Determinant:
> | Cos[1/3(A+4Pi)] Cos[1/3(B+4Pi)] Cos[1/3(C+4Pi)] |
> | Sec[1/3(A+4Pi)] Sec[1/3(B+4Pi)] Sec[1/3(C+4Pi)] |
> | Sin[1/3(A+4Pi)] Sin[1/3(B+4Pi)] Sin[1/3(C+4Pi)] |
>
> Let Det2 = Determinant:
> | 1 1 1 |
> | a Sec[1/3(A+4Pi)] b Sec[1/3(B+4Pi)] c Sec[1/3(C+4Pi)] |
> | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
>
> Let Det3 = Determinant:
> | 1 1 1 |
> | a Cos[1/3(A+4Pi)] b Cos[1/3(B+4Pi)] c Cos[1/3(C+4Pi)] |
> | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
>
> then the 1st TRILINEAR coordinate of X5390 is:
> 2 b.c.Det1 - 2 Cos[1/3(A+4Pi)].Det2 + Sec[1/3(A+4Pi)].Det3
>
> A note from Peter Moses:
> X5390 is a point constructed from the first Morley Triangle.
> The equivalent points for the Second Morley Triangle can be derived by substituting 4Pi->2Pi.
> The equivalent points for the Third Morley Triangle can be derived by substituting 4Pi->0.
>
> Best regards,
>
> Chris van Tienhoven
>
> www.chris vantienhoven.nl
>
>
>
> --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
> >
> > Dear friends,
> >
> > I managed to calculate the coordinates of X(5390).
> > Indeed the outcome is horrendous.
> > You need several pages to print it.
> > Just like Bernard I found a SEARCH nr of: -0.3549673534054897101.
> > I calculated in Mathematica and used my observation that X(5390) was lying on the line X(357).X(1136).X(3273).
> > I also calculated that this line is the only connecting line of ETC-points where X(5390) is on.
> > If someone wants the outcome to simplify it, please let me know.
> > I have a strong feeling that this point should have simple coordinates.
> >
> > Best regards,
> >
> > Chris van Tienhoven
> >
> > www. chrisvantienhoven.nl
> >
> > --- In Hyacinthos@yahoogroups.com, Bernard Gibert wrote:
> > >
> > > Dear Antreas,
> > >
> > > > I am wondering if there is ANY known proof that they are concurrent.
> > > > Drawing evidence is not a proof...!
> > >
> > > Indeed !...
> > >
> > > After some very ugly and painful computations, I did prove that the 3 Euler lines are concurrent.
> > >
> > > The SEARCH number of X(5390) is : -0.35496735340549.
> > >
> > > I expect the coordinates of X(5390) are horrendous...
> > >
> > > I will try the APH conjecture but I'm pretty sure it's true also.
> > >
> > > Best regards
> > >
> > > Bernard
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
>
• Dear Friends, Seiichi Kirikami did send me another property of X(5390). Construction: • Let A1,B1,C1 be the vertices of the 1st Morley Triangle. • Let La,
Message 2 of 26 , Jan 16, 2013
• 0 Attachment
Dear Friends,

Seiichi Kirikami did send me another property of X(5390).
Construction:
 Let A1,B1,C1 be the vertices of the 1st Morley Triangle.
 Let La, Lb, Lc be the Euler Lines of triangles A.B1.C1, B.C1.A1, C.A1.B1.
 Then La, Lb, Lc concur in X(5390).
 The lines through A,B,C parallel to La, Lb, Lc also concur in a point.
 This point happens to be X(1136) !
When the parallel lines are drawn through the vertices of the Medial Triangle or the AntiComplementary Triangle, these lines also are concurrent, however not in ETC-points.
This parallel-construction is similar to the construction method of points X(3647) to X(3652) in ETC.

I suppose there will be more Morley related points that can be constructed this way.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
>
> X(5390) = EULER-MORLEY-ZHAO POINT
>
> Barycentrics (unknown)
> Let DEF be the classical Morley triangle. The Euler lines of the three
> triangles AEF, BFD, CDE
> appear to concur in a point for which barycentric coordinates remain
> to be discovered.
> Construction by Zhao Yong of Anhui, China, October 2, 2012.
>
> http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
>
• Dear Chris Please check this whether it is true: The reflections of the Euler Lines La,Lb,Lc (concurrent in X5390) in the respective angle bisectors of ABC,
Message 3 of 26 , Jan 16, 2013
• 0 Attachment
Dear Chris

Please check this whether it is true:

The reflections of the Euler Lines La,Lb,Lc (concurrent in X5390)
in the respective angle bisectors of ABC, are concurrent.

APH

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
>
> Dear Friends,
>
> Seiichi Kirikami did send me another property of X(5390).
> Construction:
>  Let A1,B1,C1 be the vertices of the 1st Morley Triangle.
>  Let La, Lb, Lc be the Euler Lines of triangles A.B1.C1, B.C1.A1, C.A1.B1.
>  Then La, Lb, Lc concur in X(5390).
>  The lines through A,B,C parallel to La, Lb, Lc also concur in a point.
>  This point happens to be X(1136) !
> When the parallel lines are drawn through the vertices of the Medial Triangle or the AntiComplementary Triangle, these lines also are concurrent, however not in ETC-points.
> This parallel-construction is similar to the construction method of points X(3647) to X(3652) in ETC.
>
> I suppose there will be more Morley related points that can be constructed this way.
>
> Best regards,
>
> Chris van Tienhoven
>
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
> >
> > X(5390) = EULER-MORLEY-ZHAO POINT
> >
> > Barycentrics (unknown)
> > Let DEF be the classical Morley triangle. The Euler lines of the three
> > triangles AEF, BFD, CDE
> > appear to concur in a point for which barycentric coordinates remain
> > to be discovered.
> > Construction by Zhao Yong of Anhui, China, October 2, 2012.
> >
> > http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
> >
>
• Dear Antreas, They are not concurrent. Chris
Message 4 of 26 , Jan 16, 2013
• 0 Attachment
Dear Antreas,

They are not concurrent.

Chris

--- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
>
> Dear Chris
>
> Please check this whether it is true:
>
> The reflections of the Euler Lines La,Lb,Lc (concurrent in X5390)
> in the respective angle bisectors of ABC, are concurrent.
>
> APH
>
>
> --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
> >
> > Dear Friends,
> >
> > Seiichi Kirikami did send me another property of X(5390).
> > Construction:
> >  Let A1,B1,C1 be the vertices of the 1st Morley Triangle.
> >  Let La, Lb, Lc be the Euler Lines of triangles A.B1.C1, B.C1.A1, C.A1.B1.
> >  Then La, Lb, Lc concur in X(5390).
> >  The lines through A,B,C parallel to La, Lb, Lc also concur in a point.
> >  This point happens to be X(1136) !
> > When the parallel lines are drawn through the vertices of the Medial Triangle or the AntiComplementary Triangle, these lines also are concurrent, however not in ETC-points.
> > This parallel-construction is similar to the construction method of points X(3647) to X(3652) in ETC.
> >
> > I suppose there will be more Morley related points that can be constructed this way.
> >
> > Best regards,
> >
> > Chris van Tienhoven
> >
> >
> > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
> > >
> > > X(5390) = EULER-MORLEY-ZHAO POINT
> > >
> > > Barycentrics (unknown)
> > > Let DEF be the classical Morley triangle. The Euler lines of the three
> > > triangles AEF, BFD, CDE
> > > appear to concur in a point for which barycentric coordinates remain
> > > to be discovered.
> > > Construction by Zhao Yong of Anhui, China, October 2, 2012.
> > >
> > > http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
> > >
> >
>
• If A*B*C* is the Morley triangle, the Euler Lines of AB*C*, BC*A*, CA*B* are concurrent [X(5390] A similar construction: I have constructed a central triangle
Message 5 of 26 , Mar 6, 2013
• 0 Attachment
If A*B*C* is the Morley triangle, the Euler Lines
of AB*C*, BC*A*, CA*B* are concurrent [X(5390]

A similar construction:

I have constructed a central triangle A*B*C* (using the bisectors), with the same property ie the Euler Lines of AB*C*, BC*A*, CA*B*
are concurrent.

See:
http://anthrakitis.blogspot.gr/2013/03/concurrent-euler-lines.html

Which is the point of concurrence?

APH

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> X(5390) = EULER-MORLEY-ZHAO POINT
>
> Barycentrics (unknown)
> Let DEF be the classical Morley triangle. The Euler lines of the three
> triangles AEF, BFD, CDE
> appear to concur in a point for which barycentric coordinates remain
> to be discovered.
> Construction by Zhao Yong of Anhui, China, October 2, 2012.
>
> http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
>
• From ETC ... three triangles AEF, BFD, CDE to concur in X5390), ... of Anhui, China, October 2, 2012. ... barycentric coordinates ... Rewards. ... The
Message 6 of 26 , Apr 6, 2013
• 0 Attachment
From ETC
>  X(5390) = EULER-MORLEY-ZHAO POINT
>   Barycentrics   (unknown)
>   Let DEF be the classical Morley triangle. The Euler lines
> of the
three triangles AEF, BFD, CDE to concur in X5390),
> as discovered by Zhao Yong
of Anhui, China, October 2, 2012.
> For a construction and derivation of
barycentric coordinates
> by Shi Yong, see Problem 20 at Unsolved Problems and
Rewards.
> For further developments type X(5390) into Search at Hyacinthos
The coordinates of X(5390) are listed as
unknown, but
some Hyacinthos messages a while ago shows that they
aren't really
unknown, just too long. See message 21367.
I used complex numbers to get fairly long
real-valued expressions
of complex parameters for these coordinates, and then
some ideas
to symmetrize and simplify them. I'll use the abbreviations
<m,n> =  sin(x+2my+2nz) +
sin(x+2ny+2mz)
and
<m> = <m,m>/2.
Define
f(x,y,z)=<-3,-1>+5*<-2,-1>+<-2,0>-5*<-1,1>-3*<-1,2>
-<-1,3>+<0,1>+2*<0,2>-<0,3>-2*<1,3>-2*<2,3>
+<-3>-2*<-2>+3*<-1>+3*<0>+3*<1>+3*<2>-2*<3>-<4>

Then the barycentrics of X(5390) are
(
a*f(A/3,B/3,C/3) : b*f(B/3,C/3,A/3) : c*f(C/3,A/3,B/3) )
Antreas, can you forward this to Clark Kimberling? Everything
I've found from Clark hides his address.
--
Barry Wolk

[Non-text portions of this message have been removed]
• OK, Barry, done. The email addresses are hiden for protection reasons. Clark s is: ck6(AT)evansville.edu where (AT) stands for @ Antreas
Message 7 of 26 , Apr 6, 2013
• 0 Attachment
OK, Barry, done.

The email addresses are hiden for protection reasons.

Clark's is:

ck6(AT)evansville.edu

where (AT) stands for @

Antreas

--- In Hyacinthos@yahoogroups.com, Barry Wolk <wolkbarry@...> wrote:
>
> From ETC
> >  X(5390) = EULER-MORLEY-ZHAO POINT
> >   Barycentrics   (unknown)
> >   Let DEF be the classical Morley triangle. The Euler lines
> > of the
> three triangles AEF, BFD, CDE to concur in X5390),
> > as discovered by Zhao Yong
> of Anhui, China, October 2, 2012.
> > For a construction and derivation of
> barycentric coordinates
> > by Shi Yong, see Problem 20 at Unsolved Problems and
> Rewards.
> > For further developments type X(5390) into Search at Hyacinthos
> The coordinates of X(5390) are listed as
> unknown, but
> some Hyacinthos messages a while ago shows that they
> aren't really
> unknown, just too long. See message 21367.
> I used complex numbers to get fairly long
> real-valued expressions
> of complex parameters for these coordinates, and then
> some ideas
> to symmetrize and simplify them. I'll use the abbreviations
>    <m,n> =  sin(x+2my+2nz) +
> sin(x+2ny+2mz)
> and
>    <m> = <m,m>/2.
> Define
> f(x,y,z)=<-3,-1>+5*<-2,-1>+<-2,0>-5*<-1,1>-3*<-1,2>
>    -<-1,3>+<0,1>+2*<0,2>-<0,3>-2*<1,3>-2*<2,3>
>    +<-3>-2*<-2>+3*<-1>+3*<0>+3*<1>+3*<2>-2*<3>-<4>
>
> Then the barycentrics of X(5390) are
>    (
> a*f(A/3,B/3,C/3) : b*f(B/3,C/3,A/3) : c*f(C/3,A/3,B/3) )
> Antreas, can you forward this to Clark Kimberling? Everything
> I've found from Clark hides his address.
> --
> Barry Wolk
• Dear Friends, I refined the coordinates of X(5390) that I mentioned earlier in message # 21379. The trilinear coordinates are: Cos[B - C] - Cos[B + C] -
Message 8 of 26 , Apr 7, 2013
• 0 Attachment
Dear Friends,

I refined the coordinates of X(5390) that I mentioned earlier in message # 21379.
The trilinear coordinates are:

Cos[B - C] - Cos[B + C]
- Cos[B/3 + C/3] + Cos[5B/3 + 5C/3]
- Sin[C - B/3 - Pi/6] - Sin[B - C/3 - Pi/6]
+ Sin[B + 5C/3 - Pi/6] + Sin[C + 5B/3 - Pi/6]

Best regards,

Chris van Tienhoven
www.chrisvantienhoven.nl

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear Friends,
>
> I finally managed to find the coordinates of X(5390).
> I used a new method that for as far as I know- never was used before.
> I already had found the coordinates and needed to scroll 16 times on my screen to reach the end of the coordinates.
> All Trig-commands in Mathematica could not help simplifying these coordinates.
> Still they were right, because when I substituted real values the outcome was perfect.
>
> The theory I used:
> Points that come forth from related subjects often are linearly related in the triangle field.
> I called this triangle field before a Perspective Field. It is a kind of coordinate system that can be used to describe all related points with real coordinates (often rational numbers).
> This relationship manifests itself in the property that each participating point in a Perspective Field can be described as the (weighted) sum of 3 other points in the field, whereas another 4th point is used to determine the standard counting weights, also called the compliance factors.
> So I reasoned that when I could find a Perspective Field and the location of X(5390) in this field, then it should be possible to calculate its coordinates from the 3+1 chosen reference points.
>
> I followed these steps:
> 1. I gathered all Morley related points in ETC (31 points).
> 2. I gathered some Euler line related points (13 points).
> 3. I made a program in Mathematica in which I determined all permutations of 4 of these 44 points, no three of which were collinear.
> 4. Per set of 4 points I calculated if X5390 could be the weighted sum of 3 of these points. The result would be 3 real / rational numbers (as coordinates), using the 4th point to determine the compliance factors.
> 5. Of all 990 possible sets of 4 points only one set gave a result. It said:
> X5390 = 2.cf1.X2 + 1.cf2.X1136 + 2.cf3.X1137,
> the compliance factors cf1, cf2, cf3 being determined by the 4th point X(3273):
> cf1 = Det[X3273, X1136, X1137]
> cf2 = Det[ X2, X3273, X1137]
> cf3 = Det[ X2, X1136, X3273]
>
> I know this will appear pretty technically and it is of course.
> I think I will write a third document on this subject.
> The first two documents on Perspective Fields can be downloaded at:
> http://www.chrisvantienhoven.nl/index.php/mathematics/perspective-fields.html
> Especially the 2nd document explains the algebraic implications.
>
> The final result for the first BAYCENTRIC coordinate of X(5390) is:
>
> Sin[A] (Cos[ C/3 -Pi]/6] (-Sin[A] + Cos[1/3 (B + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[B]) Sin[C] + Cos[B/3 -Pi]/6] Sin[B] (Sin[A] - Cos[1/3 (C + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[C]) + Cos[A/3 -Pi]/ 6] (Cos[1/3 (C + 4Pi])] (-Sec[1/3 (A + 4Pi])] Sin[A] + 2 Sec[1/3 (B + 4Pi])] Sin[B]) Sin[C] + Cos[1/3 (B + 4Pi])] Sin[ B] (Sec[1/3 (A + 4Pi])] Sin[A] - 2 Sec[1/3 (C + 4Pi])] Sin[C]) + 2 Cos[1/3 (A + 4Pi])] Sin[ A] (-Sec[1/3 (B + 4Pi])] Sin[B] + Sec[1/3 (C + 4Pi])] Sin[C])))
> There still could be a simpler expression for this.
>
> There is also another way to describe the result.
> Let Det1 = Determinant:
> | Cos[1/3(A+4Pi)] Cos[1/3(B+4Pi)] Cos[1/3(C+4Pi)] |
> | Sec[1/3(A+4Pi)] Sec[1/3(B+4Pi)] Sec[1/3(C+4Pi)] |
> | Sin[1/3(A+4Pi)] Sin[1/3(B+4Pi)] Sin[1/3(C+4Pi)] |
>
> Let Det2 = Determinant:
> | 1 1 1 |
> | a Sec[1/3(A+4Pi)] b Sec[1/3(B+4Pi)] c Sec[1/3(C+4Pi)] |
> | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
>
> Let Det3 = Determinant:
> | 1 1 1 |
> | a Cos[1/3(A+4Pi)] b Cos[1/3(B+4Pi)] c Cos[1/3(C+4Pi)] |
> | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
>
> then the 1st TRILINEAR coordinate of X5390 is:
> 2 b.c.Det1 - 2 Cos[1/3(A+4Pi)].Det2 + Sec[1/3(A+4Pi)].Det3
>
> A note from Peter Moses:
> X5390 is a point constructed from the first Morley Triangle.
> The equivalent points for the Second Morley Triangle can be derived by substituting 4Pi->2Pi.
> The equivalent points for the Third Morley Triangle can be derived by substituting 4Pi->0.
>
> Best regards,
>
> Chris van Tienhoven
>
> www.chris vantienhoven.nl
>
>
>
> --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
> >
> > Dear friends,
> >
> > I managed to calculate the coordinates of X(5390).
> > Indeed the outcome is horrendous.
> > You need several pages to print it.
> > Just like Bernard I found a SEARCH nr of: -0.3549673534054897101.
> > I calculated in Mathematica and used my observation that X(5390) was lying on the line X(357).X(1136).X(3273).
> > I also calculated that this line is the only connecting line of ETC-points where X(5390) is on.
> > If someone wants the outcome to simplify it, please let me know.
> > I have a strong feeling that this point should have simple coordinates.
> >
> > Best regards,
> >
> > Chris van Tienhoven
> >
> > www. chrisvantienhoven.nl
> >
> > --- In Hyacinthos@yahoogroups.com, Bernard Gibert wrote:
> > >
> > > Dear Antreas,
> > >
> > > > I am wondering if there is ANY known proof that they are concurrent.
> > > > Drawing evidence is not a proof...!
> > >
> > > Indeed !...
> > >
> > > After some very ugly and painful computations, I did prove that the 3 Euler lines are concurrent.
> > >
> > > The SEARCH number of X(5390) is : -0.35496735340549.
> > >
> > > I expect the coordinates of X(5390) are horrendous...
> > >
> > > I will try the APH conjecture but I'm pretty sure it's true also.
> > >
> > > Best regards
> > >
> > > Bernard
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
>
Your message has been successfully submitted and would be delivered to recipients shortly.