Not a "reduced" fractional representation, but do you like this one? Using all digits from 0 to 9. Each once. 2340819/576 = 4063.921875 Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : mardi 19 septembre 2006 19:34 À : 'math-fun' Objet : RE: [math-fun] Decimal question 5/2 = 2.5 Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de David Wilson Envoyé : mardi 19 septembre 2006 18:29 À : math-fun Objet : [math-fun] Decimal question Is there a finite decimal that uses the same digits as its reduced fractional representation?

Take any triangle, and trisect the three edges. Connect each corner to the left trisection mark. What is the ratio of the full triangle to the interior triangle? Draw 3 circles that are tangent to each other, and which are also tangent to a line. Draw a unit circle tangent to the three other circles. What is the distance from the center of the unit circle to the line? What are some of the other amazing appearances of the number 7? Ed Pegg Jr

Draw 3 circles that are tangent to each other, and which are also tangent to a line. Draw a unit circle tangent to the three other circles. What is the distance from the center of the unit circle to the line?

1, of course: http://lunkwill.org/tmp/circle_silliness.png Although if you meant 3 /distinct/ circles, then it's 0: http://lunkwill.org/tmp/circle_silliness2.png -J If you are drawing three circles and a line, all externally tangent, then one could draw a circle in the space between the three circles that is also externally tangent to those three circles. If one assumes that this smallest circle has unit radius, one might be able to find the distance to the line, using, e.g., Descarte's Theorem. http://www.answers.com/topic/descartes-theorem --Bill

While on the subject of trisection and the number 7, here is a good problem from the 2000 World Puzzle Championship: http://wpc.puzzles.com/history/tests/2000wpc/Problem15.pdf Nick mcintosh@servidor.unam.mx wrote:

Quoting Jason Holt <jason@lunkwill.org>:

...

...

Take any triangle, and trisect the three edges. Connect each corner to the left trisection mark. What is the ratio of the full triangle to the interior triangle?

Everyone knows trisections are impossible: http://en.wikipedia.org/wiki/Compass_and_straightedge#Angle_trisection

Wasn't the trisection applied to an edge, ehich is easily possible, rather than to an angle? And isn't this somebody's famous construction, although I forget who?

- hvm

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On 9/24/06, Ed Pegg Jr <ed@mathpuzzle.com> wrote:

Take any triangle, and trisect the three edges. Connect each corner to the left trisection mark. What is the ratio of the full triangle to the interior triangle?

Draw 3 circles that are tangent to each other, and which are also tangent to a line. Draw a unit circle tangent to the three other circles. What is the distance from the center of the unit circle to the line?

What are some of the other amazing appearances of the number 7?

Ed Pegg Jr

I'm not sure these are 'amazing', but maybe worth a look. (1) there are 7 topologically distinct hexahedra (2) exotic spheres of dimension 7 exist, but not of smaller dimension, except maybe 4?? (3) 7 equal non-overlapping circles fit into a circle twice as large (4) 7 colors are needed to color maps on a torus (5) there is a torus consisting of 7 planar hexagons that requires 7 colors (the Szilassi polyhedron) (6) The minimal finite projective plane has 7 points and 7 lines (the Fano plane) (7) There are 7 deadly sins: pride, envy, gluttony, lust, anger, avarice and sloth. Jim Buddenhagen http://www.buddenbooks.com/jb

Sorry. What n maximizes n is as clear as infinity can be. I meant what n maximizes A(n)? Steve, stevebg wrote:

Being guilty of the eighth deadly sin, the compulsion to correct, I must point out that number (3) should be "a circle three times as large [in diameter]." Of course if there really are eight deadly sins, that makes (7) wrong too. But the newness (presumably to most of us) of the others more than makes up for it. If A(n) is the number of "amazing" (defined to taste, within limits) facts about the integer n, what n maximizes n? nA(n)? n+A(n)? Etc. Which choice would be best to make n "the most amazing number?"

Steve Gray

James Buddenhagen wrote:

...

I'm not sure these are 'amazing', but maybe worth a look.

(1) there are 7 topologically distinct hexahedra (2) exotic spheres of dimension 7 exist, but not of smaller dimension, except maybe 4?? (3) 7 equal non-overlapping circles fit into a circle twice as large (4) 7 colors are needed to color maps on a torus (5) there is a torus consisting of 7 planar hexagons that requires 7 colors (the Szilassi polyhedron) (6) The minimal finite projective plane has 7 points and 7 lines (the Fano plane) (7) There are 7 deadly sins: pride, envy, gluttony, lust, anger, avarice and sloth.

Jim Buddenhagen http://www.buddenbooks.com/jb

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On 9/26/06, Fred lunnon <fred.lunnon@gmail.com> wrote:

I'm afraid I have to withdraw the "Seven Circle Theorem" as a contender for this list. When I sat down to examine the proof, I realised it is actually the "Eight Circle Theorem", and ought to be rendered

Let A,C,D be circles tangent to one circle P, and B,E,F circles tangent to another circle Q [which may coincide with P or not]. If A,B,C,D,E,F,A are tangent in consecutive pairs, then the lines AD, BE, CF meet in a common point.

A proof via Laguerre transformations is essentially trivial. Sorry about that, Ed! --- WFL

Well, that statement was fairly garbled --- but it hardly matters, since I now have to withdraw the withdrawal --- sadly, it ain't so! It eventually dawned on me that the configuration of eight circles admits a more symmetric formulation: Take four arbitrary (oriented) circles P1,P2,P3,P4 [ex A,C,E,Q]; construct (choosing one of each pair possible) the circles Q1,Q2,Q3,Q4 [ex B,D,F,P] tangent to (P2,P3,P4), (P3,P4,P1), (P4,P1,P2), (P1,P2,P3) resp; join by lines the points of tangency of one opposing pair --- e.g. (Q4,P4) --- with (P1,Q1), (P2,Q2), (P3,Q3) resp. These lines are claimed to be concurrent, in a point R4; furthermore by symmetry, there should be in total four such points R1,R2,R3,R4. Consideration of a simple bilaterally symmetric example having just two such points establishes that this unrestricted claim is fallacious. A deceptively convincing diagram seems to have been an instance of its author's inadvertant capacity to generate apparently random examples that support unjustifiable hypotheses. I trust nobody spent too much time attempting to reconstruct the allegedly trivial proof. However, a special case for which the claim does hold true is a triangle together with its circum- and in-circle: the joins from vertices to in-circle tangencies meet the sides at intercepts s-a, s-b, s-c twice over, hence are are concurrent by Ceva's theorem. Six of the circles have zero or infinite radii; and the other three concurrences at the vertices are trivial. This instance, along with the Seven Circle Theorem, suggests that there may yet be lurking an elegant common generalisation, subject to some further constraint: what might this look like? Fred Lunnon (Dept. of Red Faces)

----------------David wrote: So, are there any of these fractions on (0, 1)? Starting with 5/2 = 2.5, you can find an infinite number of solutions on (0, 1) -> 0: 5/2 = 2.5 5/20 = 0.25 5/200 = 0.025 5/2000 = 0.0025 ... Starting with 5/2 = 2.5, you can find an infinite number of solutions -> oo 5/2 = 2.5 59/2 = 29.5 599/2 = 299.5 5999/2 = 2999.5 ... Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de David Wilson Envoyé : vendredi 22 septembre 2006 19:29 À : math-fun Objet : Re: [math-fun] Decimal question Oh, that is pretty. That deserves to be enshrined somewhere on the net. Anyway, the original question occured to me just before I had to go to work, as have many of my earth-shattering ideas, and I popped it off to math-fun in a hurry. Otherwise I would have forgotten all about it. I suppose if I had taken the time, I would have found 5/2 = 2.5. So, are there any of these fractions on (0, 1)? ----- Original Message ----- From: "Christian Boyer" <cboyer@club-internet.fr> To: "'math-fun'" <math-fun@mailman.xmission.com> Sent: Thursday, September 21, 2006 6:00 AM Subject: RE: [math-fun] Decimal question Not a "reduced" fractional representation, but do you like this one? Using all digits from 0 to 9. Each once. 2340819/576 = 4063.921875 Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : mardi 19 septembre 2006 19:34 À : 'math-fun' Objet : RE: [math-fun] Decimal question 5/2 = 2.5 Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de David Wilson Envoyé : mardi 19 septembre 2006 18:29 À : math-fun Objet : [math-fun] Decimal question Is there a finite decimal that uses the same digits as its reduced fractional representation? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.1.405 / Virus Database: 268.12.8/455 - Release Date: 9/22/2006 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

---- Franklin wrote Yes, but those aren't in lowest terms any more. ---- Using only irreducible fractions is a difficult problem for "small" values. Not only in [0, 1] asked by David. You have seen the family of solutions using irreducible fractions: 2.5 = 5/2 29.5 = 59/2 299.5 = 599/2 ... Here is a new puzzle using the 2 first terms of this family: Who will find a solution < 29.5 other than 2.5 and using an irreducible fraction? I have only one solution to this puzzle (and not in the lowest part of the range...), other solutions should exist... but seems difficult to find. Christian.

I'm now fairly convinced that there is no reduced fraction a/b on (0, 1) with the same digits in the fraction and its decimal expansion. I argue as follows. If reduced a/b has a finite decimal expansion, then b = 2^k or b = 5^k. If b = 2^k, the decimal expansion has k digits (or k+1 digits if we count the zero before the decimal point, it makes little difference to the argument). b = 2^k has about log10(2) k ~= 0.7 k. Since a and b together have k digits, a has about 0.7 k digits. This means that for sufficient k, specifically k >= 3, a will have strictly more digits than b, so that a/b > 1 hence not on (0, 1). We can exhaustively rule out solutions with k < 3, so there are no solutions with b = 2^k. If b = 5^k, b has about log(5) k = 0.7 k digits. This means a has about 0.3 k digits, and a/b < 1, so a/b is on (0, 1) as required. This means that a is about 10^(0.3 k) while b is about 10^(0.7 k), so that a/b is about 10^(-0.4 k), and for large k, about 0.4k digits after the decimal point will be zeroes. But in the fraction a/b, 0.7 k of the digits are part of b = 5^k, which, for large k, appears to have equidistributed digits, so that we can expect about 9/10 of them to be nonzero, that is, about 0.63 k nonzero digits. For large k, the law of averages would dictate that b = 2^k has > 0.6 k nonzero digits, that is, more nonzero digits than the decimal expansion. I'm guessing that for some moderately large k, we can be all but certain that the fraction includes more nonzero digits than the decimal, which rules out solutions. We need only rule out the small k to get whatever degree of certainty we require that no solution exists. ----- Original Message ----- From: "Christian Boyer" <cboyer@club-internet.fr> To: "'math-fun'" <math-fun@mailman.xmission.com> Sent: Saturday, September 23, 2006 9:52 AM Subject: RE: [math-fun] Decimal question

Cancel my previous message.

Here is a family of solutions -> 0 using irreducible fractions:

17537/12800 = 1.370078125 17537/128000 = 0.1370078125 17537/1280000 = 0.01370078125 17537/12800000 = 0.001370078125 ... = ...

Christian.

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Sorry: 17537 = 13 * 19 * 71 ... not a prime number... But it remains of course a reduced fraction. Christian.