From: Steve Yang Subject: A problem Date: Sat, 18 Aug 2001 12:57:32 -0300 (ADT) Newsgroups: [missing] To: Dave Rusin Summary: [missing] Keywords: [missing] Dear Dave, The Triangle ABC is isosceles with equal sides AC and BC. Two of its angles measure 40 degree. the interior point M is such that Subject: Re: A problem Date: Thu, 30 Aug 2001 18:06:08 -0500 (CDT) To: hyng@mta.ca You recently wrote: " The Triangle ABC is isosceles with equal sides AC and BC. Two of its angles measure 40 degree. the interior point M is such that It is a good problem from my point of view because it destroys my > self-confidence in geometry. You don't want to destroy self-confidence, only hubris :-) dave ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Adventitious angles Date: 3 Sep 2001 21:04:09 GMT Newsgroups: sci.math Inside a triangle ABC, pick a point X, and draw the line segments AX, BX, and CX. This partitions ABC into three smaller triangles, and in particular defines angles which can be labeled cyclically around the perimeter of the triangle as a1, a2, b1, b2, c1, c2. It is an elementary exercise to determine c1 and c2 from a1 a2 b1 b2 if, for example, AC=BC, a1=a2, and b1=b2. It is also a more-or-less straightforward application of the law of sines to determine c1 and c2 in general from the other four angles (assuming a1+a2+b1+b2 < \pi, of course). These two angles are determined by the equations a1 + a2 + b1 + b2 + c1 + c2 = \pi sin(a1) sin(b1) sin(c1) = sin(a2) sin(b2) sin(c2) I was asked to compute c1 and c2 in the case a1=30, a2=10, b1=b2=20 degrees, and found in this case that the other two angles are 20 and 80 degrees. Or at least, so I glean from numerical calculations -- I don't have a proof that the angles are _exactly_ 20 and 80 degrees. Indeed, a proof is not likely to be particularly simple, since it _must_ take advantage of the special nature of the numbers involved -- it is not generally true that c1 and c2 can be obtained as, for example, simple integer linear combinations of the other four angles. In fact, I just did an exhaustive search of all the sets of angles a1, a2, b1, b2 which are multiples of 10 degrees, and found that c1 and c2 are only close to a whole number of degrees if they are (numerically) multiples of 10 degrees, and _only_ for some combinations of a1, a2, b1, b2. (The winning combinations are attached at the end of this post. There are 70 up to reflection and rotation, if I did this right.) For example, I see some 80- and 100-degree angles in the list but not a single 90-degree angle. Of course there is nothing special about multiples of 10 degrees; one could similarly find examples with each of the six angles being other small rational multiples of \pi, e.g. with angles of 35, 30, 10, 5, 25, and 75 degrees. Still, only _some_ combinations of rational multiples of \pi for a1, a2, b1, b2 seem to make c1/\pi also rational. Now, this is a phenomenon which shows up from time to time in geometry puzzles, and in this newsgroup, under the name 'adventitious angles'. My questions are, (a) what is the algebraic relationship which must be satisfied in order for (for example) c1 and c2 to be integer linear combinations of a1, a2, b1, b2 ? I can express it as a statement that a certain polynomial has roots which are roots of unity, but I'm hoping for something prettier or quicker to check. (b) what is a nice elementary-geometry proof of the exact congruence of the angle c1 in these cases (take, e.g. the special case which was sent to me as an example). Do the proofs necessarily get increasingly complex in other cases (e.g. the 35, 30,... case)? dave Following are all congruence classes of partitioned triangles in which all angles are the indicated multiples of 10 degrees. They are grouped and ordered informally; it is clear that certain patterns occur frequently (e.g. something-or-other is isoceles). Example: [8,2,2,2,1,3] represents the question originally posed to me. [3,3,3,3,3,3],[3,3,2,4,4,2],[3,3,4,2,2,4],[3,3,1,5,5,1],[3,3,5,1,1,5], [3,4,4,2,2,3],[4,3,2,4,3,2], [5,2,2,5,2,2],[2,5,5,2,2,2],[4,3,3,4,2,2],[3,4,4,3,2,2],[6,1,1,6,2,2], [1,6,6,1,2,2],[5,2,4,3,1,3],[4,3,5,2,1,3], [7,1,2,3,1,4],[7,1,1,4,2,3],[6,2,3,2,1,4],[6,2,1,4,3,2], [4,4,3,2,2,3],[4,4,2,3,3,2],[4,4,1,4,4,1],[4,4,4,1,1,4], [3,5,2,4,3,1],[3,5,5,1,1,3],[5,3,1,5,3,1],[5,3,4,2,1,3], [1,7,4,2,3,1],[7,1,2,4,1,3],[4,4,3,3,2,2], [2,6,4,3,2,1],[2,6,6,1,1,2],[6,2,1,6,2,1],[6,2,3,4,1,2], [3,5,3,4,2,1],[5,3,4,3,1,2], [7,1,1,7,1,1],[1,7,7,1,1,1],[6,2,2,6,1,1],[2,6,6,2,1,1], [5,3,3,5,1,1],[3,5,5,3,1,1],[4,4,4,4,1,1], [8,2,1,3,2,2],[8,2,2,2,1,3],[5,5,1,3,3,1],[5,5,3,1,1,3],[5,5,2,2,2,2], [8,2,2,3,1,2],[2,8,3,2,2,1],[7,3,1,4,2,1],[3,7,4,1,1,2], [6,4,3,2,1,2],[4,6,2,3,2,1], [7,3,2,4,1,1],[3,7,4,2,1,1],[5,5,3,3,1,1], [10,1,1,3,1,2],[8,3,2,2,1,2],[7,4,1,3,2,1],[4,7,3,1,1,2], [3,8,2,2,2,1],[1,10,3,1,2,1], [7,4,2,3,1,1],[4,7,3,2,1,1], [6,6,2,1,1,2],[6,6,1,2,2,1], [10,2,1,3,1,1],[2,10,3,1,1,1],[6,6,2,2,1,1], [10,3,1,2,1,1],[3,10,2,1,1,1], [7,7,1,1,1,1] ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Adventitious angles Date: 5 Sep 2001 01:54:57 GMT Newsgroups: sci.math I gave an elementary geometry problem: > Inside a triangle ABC, pick a point X, and draw the line segments > AX, BX, and CX. This partitions ABC into three smaller triangles, and > in particular defines angles which can be labeled cyclically around > the perimeter of the triangle as a1, a2, b1, b2, c1, c2. One can determine two of these angles from the other four. I asked about cases in which all six angles could be "nice" at once. I have some more details now about when this happens. In article <9n1258$4nv9a$1@ID-54909.news.dfncis.de>, Rainer Rosenthal wrote: >This is what Wolfgang Kirschenhofer wrote proving a conjecture: >Hallo Rainer ! >Behauptung: ctg(100)+ctg(30)=ctg(70)+ctg(40) (1) > >Beweis: [proof omitted] > >Well I am awfully sorry, I couldn't find these lovely postings regarding >the 9-gon (with an excellent proof of Klaus Nagel) and the 18-gon. >I am quite sure these would fit into this thread. I thought so too, but now I am not so sure: the phenomenon is not limited to those angles, each of which results from trisecting a 30-, 60-, or 90- degree angle in some way. I will give another example below which is quite different. Also, this proof is still algebraic, and not geometric as it "should" be. Here I count trigonometry as a part of algebra because sin(z) = (w- 1/w)/(2i) for some root of unity w. For example if w = exp(Pi i / 18) then ctg(30) = i (w^3 + w^(-3))/(w^3 - w^(-3)) and likewise for the others; a simple but lengthy calculation shows ctg(100)+ctg(30)-ctg(70)-ctg(40) to be a multiple of w^12 - w^6 + 1, which is the minimal polynomial of w. In exactly the same way I can prove that some sets of six angles "work", that is, I can find some sets of six positive rational numbers r_i with sum( r_i ) = 1 product( sin( r_i \pi )^{ (-1)^i } ) = 1 by writing all those sines in terms of powers of a primitive n-th root of unity. This is sufficient to prove that the angles r1 \pi, r2\pi, ... may be taken respectively as a1, a2, b1, b2, c1, c2 in my diagram. This is the sort of proof Gerry Myerson was suggesting. For example when the angles are (as in my example) 3, 1, 2, 2, 2, and 8 times a 10-degree angle, we need only check that (w^3-w^(-3))/(w-w^(-1))*(w^2-w^(-2))/(w^2-w^(-2))*(w^2-w^(-2))/(w^8-w^(-8)) equals 1 when w is a primitive 36th root of unity, which is actually a fairly easy bit of algebra. Very well, then, let us forgo the pleasure of a purely geometric proof that a given set of six angles solves the original diagram. Let us use the algebra to find patterns among the sets of six angles. As in the preceding, I will stick to angles which are rational multiples of pi (but not necessarily whole numbers of degrees). In that case, all the angles are integer multiples of a small angle pi/n, and these integers must sum to n. In fact this is precisely the issue: given any six positive integers s_i we can let n = sum( s_i ) and then observe that the six angles (s1/n) pi, (s2/n) pi, ... solve the geometric problem iff the polynomial (w^(2s1) - 1)(w^(2s3) - 1)(w^(2s5) - 1)w^(s2+s4+s6) - (w^(2s2) - 1)(w^(2s4) - 1)(w^(2s6) - 1)w^(s1+s3+s5) is a multiple of the minimal polynomial for a primitive n-th root of unity. This will give _all_ examples of adventitious angles which are rational multiples of pi. Observation #1: for any integers s1, s3, s5 we have a solution [a1,a2,b1,b2,c1,c2]=[s1,s1,s3,s3,s5,s5]. Proof: immediate from the preceding algebra, but also a consequence of the fact that the angle bisectors in a triangle always meet in a point. (Here the geometric proof is superior since it does not require the three angles to be commensurable.) Observation #2: if [a1,a2,b1,b2,c1,c2] is any solution, there are (3!)^2 related solutions obtained by permuting {a1,b1,c1} and {a2,b2,c2} among themselves. Proof: again immediate from the algebra. Is there a geometric proof? For example the third tells us any assignment of {a1,b1,c1}= {pi/14, 2pi/14, 4pi/14} and {a2,b2,c2}={pi/14, 2pi/14, 4pi/14} will work. Some of these give pretty pictures which look difficult to prove geometrically! With these two observations we can begin to itemize all solutions consisting of six angles which are multiples of pi / n. For each way to write n as a sum of two positive integers n = n1 + n2 we consider all (unordered) partitions of n1 and n2 into 3 parts each, say n1 = s1+s3+s5 and n2=s2+s4+s6, and then check to see whether the polynomial above is a multiple of the minimal polynomial for a 2n-th root of unity. Here are some Maple procedures to do this: par3:=proc(n) local a,A,B: #Gives all partitions of n = x + y + z A:=partition(n):B:={}:for a in A do if nops(a)=3 then B:=[op(B),a]:fi:od:B:end: splits:=proc(n) local A,X,Y,i,x,y: #Gives all solutions n=(x+y+z)+(p+q+r) A:=[]: for i from 1 to floor(n/2) do X:=par3(i):Y:=par3(n-i): for x in X do for y in Y do A:=[op(A),[x,y]]:od:od: od: A:end: att:=proc(p) local s: #Computes the polynomial shown above s:= [p[1][1],p[2][1],p[1][2],p[2][2],p[1][3],p[2][3]]: (w^(2*s[1]) - 1)*(w^(2*s[3]) - 1)*(w^(2*s[5]) - 1)*w^(s[2] + s[4] + s[6]) - (w^(2*s[2]) - 1)*(w^(2*s[4]) - 1)*(w^(2*s[6]) - 1)*w^(s[1] + s[3] + s[5]); expand("):end: #So we can simply check all possible sets of 6 integers summing to n: e.g. n:=18: S:=splits(n): for s in S do if simplify(att(s),{cyclotomic(2*n,w)})=0 then lprint(s) fi:od: Now here are the empirical findings. Conjecture #1: There are no solutions when n is odd. Example: there is no triangle in which the six angles are all multiples of pi / 17. I verified the conjecture for n < 30. Conjecture #2: When n is even but not a multiple of 6, then in every solution, n1 = n2. Example: When using multiples of pi/16, both sets of angles a1+b1+c1 and a2+b2+c2 _must_ sum to 90 degrees. This I checked for n < 30. This conjecture is especially relevant because I had for a while another: Conjecture #3: When n is even and n1 = n2, for every partition n1 = x + y + z the only solution uses the _same_ partition n2 = x + y + z. That is, when a1+b1+c1 is a right angle, then the other three angles a2,b2,c2 would be some permutation of these _same_ a1, b1, c1. This conjecture checks for n < 30 but when n=30 there is one solution {{2, 5, 8}, {3, 3, 9}} , i.e. {a1,b1,c1} are 2, 5, and 8 times pi/30, in some order, and similarly for a2, b2, c2. Apart from the cases discussed in these conjectures, we need consider only n = a multiple of 6 with the sets {a1,b1,c1}, {a2,b2,c2} splitting the 180 degrees unequally. I looked for examples of these types too, and for n <= 30 I find only these: n=12: {{a1,b1,c1},{a2,b2,c2}} are these multiples of 15 degrees: {1, 2, 2} and {1, 1, 5} only. n=18: {{a1,b1,c1},{a2,b2,c2}} are these multiples of 10 degrees: {1, 2, 3}, {1, 1, 10} {2, 2, 3}, {1, 2, 8} <- This is the example I started with! {2, 3, 3}, {1, 4, 5} {2, 2, 4}, {1, 3, 6} {1, 3, 4}, {1, 2, 7} n=24: {{a1,b1,c1},{a2,b2,c2}} are these multiples of 7.5 degrees: {1, 4, 6}, {1, 3, 9} only. n=30: {{a1,b1,c1},{a2,b2,c2}} are these multiples of 6 degrees: {{1, 2, 4}, {1, 1, 21}} {{2, 3, 3}, {1, 2, 19}} {{2, 3, 5}, {1, 3, 16}} {{1, 4, 5}, {1, 2, 17}} {{3, 4, 4}, {1, 5, 13}} {{2, 4, 5}, {1, 4, 14}} {{1, 3, 7}, {1, 2, 16}} {{3, 4, 5}, {1, 8, 9}} * {{3, 4, 5}, {2, 3, 13}} {{3, 3, 6}, {1, 6, 11}} {{2, 4, 6}, {1, 5, 12}} * {{2, 3, 7}, {2, 2, 14}} {{2, 2, 8}, {1, 3, 14}} {{3, 5, 5}, {2, 4, 11}} {{2, 5, 6}, {2, 3, 12}} {{2, 4, 7}, {1, 7, 9}} {{2, 3, 8}, {1, 5, 11}} * {{1, 4, 8}, {1, 3, 13}} {{4, 5, 5}, {3, 4, 9}} {{3, 5, 6}, {2, 6, 8}} {{3, 4, 7}, {2, 5, 9}} * {{1, 6, 7}, {2, 2, 12}} {{1, 5, 8}, {1, 4, 11}} {{2, 3, 9}, {1, 7, 8}} * ([*] Note these examples in which all six angles are distinct.) So here are the new challenges: (1) Prove conjectures #1 and #2 (2) Characterize the 'interesting' examples when n is a multiple of 6 (3) I still await purely synthetic proofs that say c1 is what is claimed in these examples when only a1,a2,b1,b2 are given (e.g. to show geometrically that angle c1 is congruent to a1 or a2+b1 or whatever.) Any interesting case will do! dave ============================================================================== From: ayatollah potassium Subject: Re: Adventitious angles Date: Wed, 05 Sep 2001 02:21:26 -0400 Newsgroups: sci.math Dave Rusin wrote: > > Inside a triangle ABC, pick a point X, and draw the line segments > > AX, BX, and CX. This partitions ABC into three smaller triangles, and > > in particular defines angles which can be labeled cyclically around > > the perimeter of the triangle as a1, a2, b1, b2, c1, c2. (where the angles are commensurable with pi). This is expressible as an algebraic equation in roots of unity. The general theory of these is developed by Conway in a paper with "trigonometric diophantine equations" in the title. There is one advantage to the geometric approach: if you verify that a given solution works, but the proof doesn't use all of the information in the diagram, then there is a family of solutions including the one given. I remember the surprise of noticing this when I worked out the famous diagram with (a1,a2,b1,b2) = (30,50,60,20). [remainder of quoted original article deleted --djr] ============================================================================== From: Klaus Nagel Subject: Re: Adventitious angles Date: Wed, 05 Sep 2001 18:03:28 +0200 Newsgroups: sci.math Dave Rusin schrieb: > > I gave an elementary geometry problem: > > > Inside a triangle ABC, pick a point X, and draw the line segments > > AX, BX, and CX. > Apart from the cases discussed in these conjectures, we need consider > > > n=18: {{a1,b1,c1},{a2,b2,c2}} are these multiples of 10 degrees: > {1, 2, 3}, {1, 1, 10} > {2, 2, 3}, {1, 2, 8} <- This is the example I started with! > {2, 3, 3}, {1, 4, 5} > {2, 2, 4}, {1, 3, 6} > {1, 3, 4}, {1, 2, 7} > > (3) I still await purely synthetic proofs that say c1 is what is > claimed in these examples when only a1,a2,b1,b2 are given > (e.g. to show geometrically that angle c1 is congruent to a1 > or a2+b1 or whatever.) Any interesting case will do! Hi Dave, as Rainer Rosenthal wrote, a couple months ago there was a thread about triangles related to the regular 18-gon in de.sci.mathematik. The example you started with is a special case and I just found a geometrical proof. Draw a regular 18-gon with center C and denote the corners by P0, P1, P2 ... P17. (You may print a sketch from http://home.t-online.de/home/nagel.klaus/achtzehn.jpg ). The triangle {2, 2, 3}, {1, 2, 8} corresponds to P0, P5, C. The interior point corresponds to the intersection of P0-P6, P6-P16 and P4-C. It must be shown, that these three lines indeed intersect in one common point A. The angles (A,P5,C) and (A,C,P5) are both 20°, hence the triangle (P5,C,A) is isosceles and A is located on the perpendicular line bisecting P5-C. This perpendicular line is part of P2-P8, which is the reflection of P0-P6 with respect to P4-C. That completes the proof. Regards, Klaus Nagel ============================================================================== From: Nagel.Klaus@t-online.de (Klaus Nagel) Subject: Re: Adventitious angles Date: Wed, 05 Sep 2001 12:26:20 +0200 To: Dave Rusin Dave Rusin schrieb: > >Well I am awfully sorry, I couldn't find these lovely postings regarding > >the 9-gon (with an excellent proof of Klaus Nagel) and the 18-gon. > >I am quite sure these would fit into this thread. Hi Dave, Rainer Rosenthal wrote me about your posting. I found the proof you mentioned, it appeared in de.sci.mathematik on April 07 2001, 11:48 in the thread "Geometrischer Beweis zum 40-40-70 Dreieck". Das Dreieck ABC mit dem eingezeichneten Dreieck MTN spiegele ich an der Seite AC und erhalte die neuen Punkte B',M' und T'. Ich zeige, daß das Dreieck M'MT gleichschenklig ist mit der Basisseite M'T. Daraus folgt, daß die Verbindung TM' die Seite AC unter einem Winkel von 40° schneidet. Um die Gleichheit der Strecken MT und MM' zu zeigen, fälle ich das Lot von M auf BT, den Fußpunkt nenne ich U. Wegen des 30°-Winkels MTU ist MT doppelt so lang wie UM. Andrerseits hat UM die halbe Länge von MM'. Let me know, if you need a translation. Regards, Klaus ============================================================================== From: Dave Rusin Subject: Re: Adventitious angles Date: Wed, 5 Sep 2001 12:27:18 -0500 (CDT) To: Nagel.Klaus@t-online.de Thank you for your mail and your post to the newsgroup. This is excellent! The embedding of the triangle into a regular n-gon in general is probably the right way to give geometric proofs. For your next assignment, try one of the cases in which the six angles are distinct :-) I suppose I will assign this to my bright students some day: suppose the four bottom angles are [a1,a2,b1,b2]=[18,54,30,48] (degrees). The correct angles at the top are then 24 degrees and 6 degrees, but I guess I would phrase this geometrically as: "Prove that c1 equals the angle bisector of b2" or something like that. In terms of your regular polygons, the question is to show that the lines C-P9, P0-P20, and P12-P44 are concurrent in a regular 60-gon. (I hope I got that right!) Again, thank you for responding (auf Deutsch und in English, both OK for me). dave ============================================================================== From: Gerry Myerson Subject: Re: Adventitious angles Date: Thu, 06 Sep 2001 14:25:33 +1000 Newsgroups: sci.math In article <9n0r89$lkg$1@news.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) wrote: > These two angles are determined by the equations > a1 + a2 + b1 + b2 + c1 + c2 = \pi > sin(a1) sin(b1) sin(c1) = sin(a2) sin(b2) sin(c2) In addition to the Conway & Jones paper on trigonometric diophantine equations (aka vanishing sums of roots of unity), and my paper Rational products of sines of rational angles, Aeq. Math. 45 (1993) 70-82, (where I find all solutions of sin x sin y = sin z sin w with x, y, z, w rational multiples of pi) there's the paper of Poonen & Rubinstein, The number of intersection points made by the diagonals of a regular polygon, SIAM J. Discrete Math. 11 (1998), no. 1, 135--156, and also Gardner and Gritzmann, Discrete tomography: determination of finite sets by X-rays, Trans. Amer. Math. Soc. 349 (1997), no. 6, 2271--2295. Gerry Myerson (gerry@mpce.mq.edu.au)