Back about 30-40 years ago, I posed a problem in Mathematics Magazine (or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity. Can anyone (a) figure out what X was, or (b) locate my problem and the solution? Jim Propp PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar.
If the middle S is a chord of a circle, then the end S can be a chord going either way on that circle and the angle A will be equal either way. So I'm guessing X is the circumcircle. --ms On 06-Apr-15 15:51, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine (or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Thanks, but I'm not seeing this. Can you tell me how to construct your diagram step by step, since the one I'm drawing is apparently different in important respects and doesn't show me what you're seeing? Jim Propp On Mon, Apr 6, 2015 at 4:40 PM, Mike Speciner <ms@alum.mit.edu> wrote:
If the middle S is a chord of a circle, then the end S can be a chord going either way on that circle and the angle A will be equal either way. So I'm guessing X is the circumcircle.
--ms
On 06-Apr-15 15:51, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine (or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Let the three vertices of the triangle be A,B,C, so AB is the middle S and BC is the end S. Constructing the triangle given the three parameters involves drawing AB, drawing a line from A at the specified angle (where there are two choices--on the left or on the right), and drawing a circle at B of length BC (which intersects the line from A at two points, resulting in two ABC triangles). The circumcircle of either ABC is the same, because the angle A is the same so BC is the same length. Is that any clearer? --ms On 06-Apr-15 17:51, James Propp wrote:
Thanks, but I'm not seeing this. Can you tell me how to construct your diagram step by step, since the one I'm drawing is apparently different in important respects and doesn't show me what you're seeing?
Jim Propp
On Mon, Apr 6, 2015 at 4:40 PM, Mike Speciner <ms@alum.mit.edu> wrote:
If the middle S is a chord of a circle, then the end S can be a chord going either way on that circle and the angle A will be equal either way. So I'm guessing X is the circumcircle.
--ms
On 06-Apr-15 15:51, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine (or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I still don't see it. In my picture of a triangle in which "AB is the middle S and BC is the end S", there are two triangles, ABC and ABC', with A, C , and C' collinear, and where B is the apex of isosceles triangle BCC'. The circle centered at B with radius BC does indeed go through C', but I don't see why it goes through A, so I don't see why it's the common circumcircle of the two triangles. Jim Propp On Monday, April 6, 2015, Mike Speciner <ms@alum.mit.edu> wrote:
Let the three vertices of the triangle be A,B,C, so AB is the middle S and BC is the end S. Constructing the triangle given the three parameters involves drawing AB, drawing a line from A at the specified angle (where there are two choices--on the left or on the right), and drawing a circle at B of length BC (which intersects the line from A at two points, resulting in two ABC triangles). The circumcircle of either ABC is the same, because the angle A is the same so BC is the same length.
Is that any clearer?
--ms
On 06-Apr-15 17:51, James Propp wrote:
Thanks, but I'm not seeing this. Can you tell me how to construct your diagram step by step, since the one I'm drawing is apparently different in important respects and doesn't show me what you're seeing?
Jim Propp
On Mon, Apr 6, 2015 at 4:40 PM, Mike Speciner <ms@alum.mit.edu> wrote:
If the middle S is a chord of a circle, then the end S can be a chord
going either way on that circle and the angle A will be equal either way. So I'm guessing X is the circumcircle.
--ms
On 06-Apr-15 15:51, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine
(or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Ah. The circle at B is not the circumcircle of ABC; it's just part of the construction. How about a different way.... Start with a triangle ABC. Draw it's circumcircle. Now, intersect the circumcircle with the circle centered at B through C. This gives you a C'. BC' = BC, so A = A' because the angle from any point on the circle, that is subtended by a chord of a given length is always the same (and equal to half the angle of the arc delimited by the chord). --ms On 06-Apr-15 18:52, James Propp wrote:
I still don't see it.
In my picture of a triangle in which "AB is the middle S and BC is the end S", there are two triangles, ABC and ABC', with A, C , and C' collinear, and where B is the apex of isosceles triangle BCC'. The circle centered at B with radius BC does indeed go through C', but I don't see why it goes through A, so I don't see why it's the common circumcircle of the two triangles.
Jim Propp
On Monday, April 6, 2015, Mike Speciner <ms@alum.mit.edu> wrote:
Let the three vertices of the triangle be A,B,C, so AB is the middle S and BC is the end S. Constructing the triangle given the three parameters involves drawing AB, drawing a line from A at the specified angle (where there are two choices--on the left or on the right), and drawing a circle at B of length BC (which intersects the line from A at two points, resulting in two ABC triangles). The circumcircle of either ABC is the same, because the angle A is the same so BC is the same length.
Is that any clearer?
--ms
On 06-Apr-15 17:51, James Propp wrote:
Thanks, but I'm not seeing this. Can you tell me how to construct your diagram step by step, since the one I'm drawing is apparently different in important respects and doesn't show me what you're seeing?
Jim Propp
On Mon, Apr 6, 2015 at 4:40 PM, Mike Speciner <ms@alum.mit.edu> wrote:
If the middle S is a chord of a circle, then the end S can be a chord
going either way on that circle and the angle A will be equal either way. So I'm guessing X is the circumcircle.
--ms
On 06-Apr-15 15:51, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine
(or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On 06/04/2015 23:52, James Propp wrote:
I still don't see it.
In my picture of a triangle in which "AB is the middle S and BC is the end S", there are two triangles, ABC and ABC', with A, C , and C' collinear, and where B is the apex of isosceles triangle BCC'. The circle centered at B with radius BC does indeed go through C', but I don't see why it goes through A, so I don't see why it's the common circumcircle of the two triangles.
What's common is not the *circumcircle* but the *circumradius*. Two ass-congruent triangles have equal circumradius, so no sequence of ass-congruences can produce similar but incongruent triangles. (No, I don't know why so many other respondents have been saying "same circumcircle" when they mean "same circumradius".) -- g
Well, it actually is the same circumcircle if you have AB be common to both triangles (ABC and ABC') if you have C and C' on opposite sides of chord AB. --ms On 06-Apr-15 19:05, Gareth McCaughan wrote:
On 06/04/2015 23:52, James Propp wrote:
I still don't see it.
In my picture of a triangle in which "AB is the middle S and BC is the end S", there are two triangles, ABC and ABC', with A, C , and C' collinear, and where B is the apex of isosceles triangle BCC'. The circle centered at B with radius BC does indeed go through C', but I don't see why it goes through A, so I don't see why it's the common circumcircle of the two triangles.
What's common is not the *circumcircle* but the *circumradius*. Two ass-congruent triangles have equal circumradius, so no sequence of ass-congruences can produce similar but incongruent triangles.
(No, I don't know why so many other respondents have been saying "same circumcircle" when they mean "same circumradius".)
On 07/04/2015 00:11, Mike Speciner wrote:
Well, it actually is the same circumcircle if you have AB be common to both triangles (ABC and ABC') if you have C and C' on opposite sides of chord AB.
Oh, neat. I hadn't thought of drawing it that way. (This may reveal that I didn't read every message in this thread in detail and draw diagrams as I did so. Sorry.) -- g
Could someone draw a picture for me and post it somewhere on the web (or just email it to me)? I'm having trouble converting people's verbal descriptions into pictures. Thanks, Jim Propp On Mon, Apr 6, 2015 at 7:45 PM, Gareth McCaughan <gareth.mccaughan@pobox.com
wrote:
On 07/04/2015 00:11, Mike Speciner wrote:
Well, it actually is the same circumcircle if you have AB be common to
both triangles (ABC and ABC') if you have C and C' on opposite sides of chord AB.
Oh, neat. I hadn't thought of drawing it that way.
(This may reveal that I didn't read every message in this thread in detail and draw diagrams as I did so. Sorry.)
-- g
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/centershow {gsave dup stringwidth pop -.5 mul 0 rmoveto show grestore} bind def /Helvetica .5 selectfont 1 setlinejoin 1 setlinecap 0 setlinewidth 8.5 36 mul 11 36 mul translate 25 25 scale 0 0 5 0 360 arc closepath stroke -3 -4 moveto 0 5 lineto -3 4 lineto closepath stroke -3 -4 moveto 0 5 lineto 3 4 lineto closepath stroke -3 -4.5 moveto (A) centershow 0 5.1 moveto (B) centershow -3 4.1 moveto (C') centershow 3.1 4.1 moveto (C) centershow showpage On 06-Apr-15 23:18, James Propp wrote:
Could someone draw a picture for me and post it somewhere on the web (or just email it to me)? I'm having trouble converting people's verbal descriptions into pictures.
Thanks,
Jim Propp
On Mon, Apr 6, 2015 at 7:45 PM, Gareth McCaughan <gareth.mccaughan@pobox.com
wrote: On 07/04/2015 00:11, Mike Speciner wrote:
Well, it actually is the same circumcircle if you have AB be common to
both triangles (ABC and ABC') if you have C and C' on opposite sides of chord AB.
Oh, neat. I hadn't thought of drawing it that way.
(This may reveal that I didn't read every message in this thread in detail and draw diagrams as I did so. Sorry.)
-- g
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For those unable to easily view Mike Speciner's postscript, here is his pic as a png file: http://1drv.ms/1ag7ic3 Let me know if the link does not work. On Tue, Apr 7, 2015 at 6:19 AM, Mike Speciner <ms@alum.mit.edu> wrote:
/centershow {gsave dup stringwidth pop -.5 mul 0 rmoveto show grestore} bind def
/Helvetica .5 selectfont
1 setlinejoin 1 setlinecap 0 setlinewidth 8.5 36 mul 11 36 mul translate 25 25 scale 0 0 5 0 360 arc closepath stroke -3 -4 moveto 0 5 lineto -3 4 lineto closepath stroke -3 -4 moveto 0 5 lineto 3 4 lineto closepath stroke -3 -4.5 moveto (A) centershow 0 5.1 moveto (B) centershow -3 4.1 moveto (C') centershow 3.1 4.1 moveto (C) centershow
showpage
On 06-Apr-15 23:18, James Propp wrote:
Could someone draw a picture for me and post it somewhere on the web (or just email it to me)? I'm having trouble converting people's verbal descriptions into pictures.
Thanks,
Jim Propp
On Mon, Apr 6, 2015 at 7:45 PM, Gareth McCaughan < gareth.mccaughan@pobox.com
wrote: On 07/04/2015 00:11, Mike Speciner wrote:
Well, it actually is the same circumcircle if you have AB be common to
both triangles (ABC and ABC') if you have C and C' on opposite sides of chord AB.
Oh, neat. I hadn't thought of drawing it that way.
(This may reveal that I didn't read every message in this thread in detail and draw diagrams as I did so. Sorry.)
-- g
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* James Buddenhagen <jbuddenh@gmail.com> [Apr 07. 2015 16:38]:
For those unable to easily view Mike Speciner's postscript, here is his pic as a png file:
Alternatively open a file, say foo.ps, in emacs, copy everything from /centershow to showpage into it, then hit ctrl-c twice (same key combo gets you back and forth between rendering and source). This also works for svg, pdf and others. Very useful for tweaking.
Let me know if the link does not work.
On Tue, Apr 7, 2015 at 6:19 AM, Mike Speciner <ms@alum.mit.edu> wrote:
/centershow {gsave dup stringwidth pop -.5 mul 0 rmoveto show grestore} bind def
/Helvetica .5 selectfont
1 setlinejoin 1 setlinecap 0 setlinewidth 8.5 36 mul 11 36 mul translate 25 25 scale 0 0 5 0 360 arc closepath stroke -3 -4 moveto 0 5 lineto -3 4 lineto closepath stroke -3 -4 moveto 0 5 lineto 3 4 lineto closepath stroke -3 -4.5 moveto (A) centershow 0 5.1 moveto (B) centershow -3 4.1 moveto (C') centershow 3.1 4.1 moveto (C) centershow
showpage
[...]
On 2015-04-07 07:52, Joerg Arndt wrote:
* James Buddenhagen <jbuddenh@gmail.com> [Apr 07. 2015 16:38]:
For those unable to easily view Mike Speciner's postscript, here is his pic as a png file:
Alternatively open a file, say foo.ps, in emacs, copy everything from /centershow to showpage into it, then hit ctrl-c twice (same key combo gets you back and forth between rendering and source). This also works for svg, pdf and others. Very useful for tweaking.
My emacs teasingly says Type C-c C-c to toggle between editing or viewing the document. The previous line says (PostScript DocView). I wound up saving the file and running it through Preview. --rwg
Let me know if the link does not work.
On Tue, Apr 7, 2015 at 6:19 AM, Mike Speciner <ms@alum.mit.edu> wrote:
/centershow {gsave dup stringwidth pop -.5 mul 0 rmoveto show grestore} bind def
/Helvetica .5 selectfont
1 setlinejoin 1 setlinecap 0 setlinewidth 8.5 36 mul 11 36 mul translate 25 25 scale 0 0 5 0 360 arc closepath stroke -3 -4 moveto 0 5 lineto -3 4 lineto closepath stroke -3 -4 moveto 0 5 lineto 3 4 lineto closepath stroke -3 -4.5 moveto (A) centershow 0 5.1 moveto (B) centershow -3 4.1 moveto (C') centershow 3.1 4.1 moveto (C) centershow
showpage
[...]
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Ah! Now I see it! Thanks to all. Jim On Tuesday, April 7, 2015, James Buddenhagen <jbuddenh@gmail.com> wrote:
For those unable to easily view Mike Speciner's postscript, here is his pic as a png file:
Let me know if the link does not work.
On Tue, Apr 7, 2015 at 6:19 AM, Mike Speciner <ms@alum.mit.edu <javascript:;>> wrote:
/centershow {gsave dup stringwidth pop -.5 mul 0 rmoveto show grestore} bind def
/Helvetica .5 selectfont
1 setlinejoin 1 setlinecap 0 setlinewidth 8.5 36 mul 11 36 mul translate 25 25 scale 0 0 5 0 360 arc closepath stroke -3 -4 moveto 0 5 lineto -3 4 lineto closepath stroke -3 -4 moveto 0 5 lineto 3 4 lineto closepath stroke -3 -4.5 moveto (A) centershow 0 5.1 moveto (B) centershow -3 4.1 moveto (C') centershow 3.1 4.1 moveto (C) centershow
showpage
On 06-Apr-15 23:18, James Propp wrote:
Could someone draw a picture for me and post it somewhere on the web (or just email it to me)? I'm having trouble converting people's verbal descriptions into pictures.
Thanks,
Jim Propp
On Mon, Apr 6, 2015 at 7:45 PM, Gareth McCaughan < gareth.mccaughan@pobox.com <javascript:;>
wrote: On 07/04/2015 00:11, Mike Speciner wrote:
Well, it actually is the same circumcircle if you have AB be common to
both triangles (ABC and ABC') if you have C and C' on opposite sides of chord AB.
Oh, neat. I hadn't thought of drawing it that way.
(This may reveal that I didn't read every message in this thread in detail and draw diagrams as I did so. Sorry.)
-- g
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We'll continue to talk. Happy to visit areas with you. This is one case where the old real estate adage is paramount--location, location, location... --R -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of James Propp Sent: Monday, April 06, 2015 11:19 PM To: math-fun Subject: Re: [math-fun] SSA congruence Could someone draw a picture for me and post it somewhere on the web (or just email it to me)? I'm having trouble converting people's verbal descriptions into pictures. Thanks, Jim Propp On Mon, Apr 6, 2015 at 7:45 PM, Gareth McCaughan <gareth.mccaughan@pobox.com
wrote:
On 07/04/2015 00:11, Mike Speciner wrote:
Well, it actually is the same circumcircle if you have AB be common to
both triangles (ABC and ABC') if you have C and C' on opposite sides of chord AB.
Oh, neat. I hadn't thought of drawing it that way.
(This may reveal that I didn't read every message in this thread in detail and draw diagrams as I did so. Sorry.)
-- g
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Given an increasing sequence of positive integers a_1 < a_2 < a_3 < ... we can look at their residues mod p for p prime. In order to compare the results, we might want to think of these p-residues as divided by p and viewed in the unit interval [0,1). Someone did just that for the sequence f_n := Sum_{1<=k<=n} k! and the resulting sequence of fractions x_n := {(Sum_{1<=k<=n} k! mod p) / p as he displayed it in [0,1) varied surprisingly little across primes. * * * ALSO: This same increasing sequence f_n of integers seems to have an interestingly limited set of prime factors 3, 11, 17, 73, 97, 141, 347, 467,... (not necessarily excluding smaller ones). Also, Wolfram ( http://mathworld.wolfram.com/FactorialSums.html ) states that the sequence fsq_n := f_n := Sum_{1<=k<=n} (k!)^2 contains only a finite set of primes among its terms. Amazing! Is there some easy way to see this? --Dan
The Wolfram statement: OEIS A104344 ... is prime for indices 2, 3, 4, 5, 7, 8, 10, 18, 21, 42, 51, 91, 133, 177, 182, 310, 3175, 9566, ... (OEIS A100289). It is known that there can be only a finite number of such primes, but it is not known what the last term is. Going to https://oeis.org/A100289 we find this comment: Write the sum as S(2,n)-1, where S(k,n) = sum_{i=0..n} (i!)^k. Let p=1248829. Because p divides S(2,p-1)-1, p divides S(2,n)-1 for all n >= p-1. Hence there are no primes for n >= p-1.
On Apr 9, 2015, at 3:02 PM, Dan Asimov <asimov@msri.org> wrote:
Also, Wolfram ( http://mathworld.wolfram.com/FactorialSums.html ) states that the sequence
fsq_n := f_n := Sum_{1<=k<=n} (k!)^2
contains only a finite set of primes among its terms. Amazing!
Is there some easy way to see this?
Even more from OEIS A100289: ----- Numbers n such that (1!)^2 + (2!)^2 + (3!)^2 +...+ (n!)^2 is prime. 2, 3, 4, 5, 7, 8, 10, 18, 21, 42, 51, 91, 133, 177, 182, 310, 3175, 9566 OFFSET 1,1 COMMENTS All n <= 310 yield provable primes. No other n < 4000. Write the sum as S(2,n)-1, where S(k,n) = sum_{i=0..n} (i!)^k. Let p=1248829. Because p divides S(2,p-1)-1, p divides S(2,n)-1 for all n >= p-1. Hence there are no primes for n >= p-1. ----- ((( Let's see: (1!)^2 + (2!)^2 + (3!)^2 +...+ (6!)^2 = 31 * 17207 (prime factorization). ))) But it's quite interesting that there are only finitely many primes on this list. --Dan
On Apr 10, 2015, at 8:02 AM, Hans Havermann <gladhobo@teksavvy.com> wrote:
The Wolfram statement:
OEIS A104344 ... is prime for indices 2, 3, 4, 5, 7, 8, 10, 18, 21, 42, 51, 91, 133, 177, 182, 310, 3175, 9566, ... (OEIS A100289). It is known that there can be only a finite number of such primes, but it is not known what the last term is.
Going to https://oeis.org/A100289 we find this comment:
Write the sum as S(2,n)-1, where S(k,n) = sum_{i=0..n} (i!)^k. Let p=1248829. Because p divides S(2,p-1)-1, p divides S(2,n)-1 for all n >= p-1. Hence there are no primes for n >= p-1.
On Apr 9, 2015, at 3:02 PM, Dan Asimov <asimov@msri.org> wrote:
Also, Wolfram ( http://mathworld.wolfram.com/FactorialSums.html ) states that the sequence
fsq_n := f_n := Sum_{1<=k<=n} (k!)^2
contains only a finite set of primes among its terms. Amazing!
Is there some easy way to see this?
Oh--there are really 4 choices for the intersection of the circle at B with the two choices of lines. Choose the nearer one on one side and the farther one on the other side. (If there isn't a nearer and a farther, you've got a right triangle, and the chord AB of the circumcircle is a diameter.) On 06-Apr-15 18:38, Mike Speciner wrote:
Let the three vertices of the triangle be A,B,C, so AB is the middle S and BC is the end S. Constructing the triangle given the three parameters involves drawing AB, drawing a line from A at the specified angle (where there are two choices--on the left or on the right), and drawing a circle at B of length BC (which intersects the line from A at two points, resulting in two ABC triangles). The circumcircle of either ABC is the same, because the angle A is the same so BC is the same length.
Is that any clearer?
--ms
On 06-Apr-15 17:51, James Propp wrote:
Thanks, but I'm not seeing this. Can you tell me how to construct your diagram step by step, since the one I'm drawing is apparently different in important respects and doesn't show me what you're seeing?
Jim Propp
On Mon, Apr 6, 2015 at 4:40 PM, Mike Speciner <ms@alum.mit.edu> wrote:
If the middle S is a chord of a circle, then the end S can be a chord going either way on that circle and the angle A will be equal either way. So I'm guessing X is the circumcircle.
--ms
On 06-Apr-15 15:51, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine (or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On 4/6/2015 3:51 PM, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine (or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar.
Mathematics Magazine, Vol. 57, No. 5, November 1984, page 299. The problem appeared as a "quickie" (Q695); the triangles were called "skew-congruent". The one-line solution on page 305 is credited to John Horton Conway (who observes that circumradius = BC/(2 sin A)). -- Fred W. Helenius fredh@ix.netcom.com
Thanks, Fred! (I didn't see your post until after I replied to Mike's.) Jim Propp On Mon, Apr 6, 2015 at 5:26 PM, Fred W. Helenius <fredh@ix.netcom.com> wrote:
On 4/6/2015 3:51 PM, James Propp wrote:
Back about 30-40 years ago, I posed a problem in Mathematics Magazine (or maybe the Monthly) asking whether any triangle could be linked to a non-congruent similar triangle via a sequence of triangles, each "SSA-congruent" to the one before and the one after. There was a cute solution that pointed out that this is impossible because two SSA-congruent triangles have the same X, where X was some triangle statistic (like perimeter, inradius, or circumradius, but slightly less well-known) that scales linearly under similarity.
Can anyone (a) figure out what X was, or (b) locate my problem and the solution?
Jim Propp
PS: In my original submission I proposed the term "ASS-congruent", which struck me as both more pronounceable and more apt, but the stodgy problems editor who reigned at the time deemed this too vulgar.
Mathematics Magazine, Vol. 57, No. 5, November 1984, page 299. The problem appeared as a "quickie" (Q695); the triangles were called "skew-congruent". The one-line solution on page 305 is credited to John Horton Conway (who observes that circumradius = BC/(2 sin A)).
-- Fred W. Helenius fredh@ix.netcom.com
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participants (11)
-
Dan Asimov -
Dan Asimov -
Fred W. Helenius -
Gareth McCaughan -
Hans Havermann -
James Buddenhagen -
James Propp -
Joerg Arndt -
Mike Speciner -
Richard E. Howard -
rwg