[Here is a collection of files on what I found to be a fascinating question in plane geometry. The following question was posted to sci.math: The problem, posed by Zeljko.Vrba@oleh.srce.hr (Zeljko Vrba), assumed two circles given in the plane, and asked for a characterization of the triangles which have these two circles as their in- and circum-scribed circles. I followed up with the observation that for most pairs of circles there are no such triangles, while otherwise there will be infinitely many; the cases are distinguished by a certain relation holding on the radii and the separation of the centers. (If the inner and outer radii are r and R respectively and if the distance between the centers is a then the existence of such a triangle implies a^2=R^2-2rR.) It turns out that the condition ("Chapple's identity") for a triangle to exist is old, and while there are geometric/trigonometric proofs, I observe that the condition can be stated that certain line segments are congruent, yet I have no Euclidean proof of that fact. (Undoubtedly this is made harder by the non-uniqueness of the triangles, but this does suggest a question: if a true statement can be expressed in Euclidean terms, does it have a Euclidean proof?) The existence of infinitely many solutions is "Poncelet's theorem" ("Poncelet's Porism") and is connected to some statements about elliptic curves. Here you will find nine files related to this topic. First, a statement of the problem (the original post is lost, but was unclear anyway). I had played around with the problem a bit; for example, here is some Maple code seeking triangles. I made an initial post summarizing the condition for existence and the infinitude of solutions in the positive cases. A response is attached (which was more on-the-mark than I thought at the time). Meanwhile, I did some reading on the problem. Since I was at the time teaching a course on elliptic curves, when I found the relation of this problem to that topic I subjected my students to it. Here are some course notes clarifying details of the relationship. For the benefit of the USENET community I posted a summary of the whole matter in a lengthy second post. This prompted an email exchange with the original poster in which he outlines a geometric (albeit not "Euclidean") proof of Chapple's identity. ============================================================================== From: Zeljko.Vrba@oleh.srce.hr (Zeljko Vrba) Subject: Re: Triangle construction To: rusin@math.niu.edu (Dave Rusin) Date: Sat, 25 Mar 1995 09:45:43 +0100 (MET) > > you will have to tell us what properties you want this triangle to have, > otherwise we will just ignore the circles and draw any triangle we wish. > > Sorry for the delay... I was thinking about two circles so that one is INSIDE the other. Explanation: Draw a triangle. Inscribe and circumscribe a circle to it (it is always possible to do that). Now erase the triangle, but leave the circles. Given just these circles, (one is inscribed and the other is circumscribed to the triangle we want to construct) is it possible to construct ANY triangle? If yes, how? Will the constructed triangle be congruent to the erased one or are there infinitely many solutions? -- zvrba@oleh.srce.hr ============================================================================== #given two nested circles, find a triangle touching one, tangent to other. # #given point (x0,y0). Assume other point is symmetric wrt vertical axis x1:=-x0: y1:=y0: #draw two lines of any slope; find intersection solve({m=(y-y0)/(x-x0),n=-(y-y1)/(x-x1)},{x,y}): xi:=rhs("[1]): yi:=rhs(""[2]): #third point is on lines with double these angles A:=2*m/(1-m^2): B:=2*n/(1-n^2): solve({A=(ynew-y0)/(xnew-x0),B=-(ynew-y1)/(xnew-x1)},{xnew,ynew}): x2:=rhs("[1]): y2:=rhs(""[2]): #P0 and P2 lie on same circle. This pins down y0 solve(simplify((x2^2+y2^2)-(x0^2+y0^2)),y0): y0:=factor("): #Now we try to match the given data. The inner radius is supposed to be r: solve(simplify(yi-y0-r), x0): x0:=factor("): #Then the lengths of (x0,y0) and (xi,yi) pin down R, Q respectively cons1:=simplify((x0^2+y0^2)/r^2-Rbar): cons2:=simplify((x0^2+y0^2)/r^2 - (xi^2+yi^2)/r^2 -Qbar): #So we should get cons1=Rbar=(R/r)^2, cons2=Qbar=(R^2-Q^2)/r^2. But readlib(eliminate): eliminate({cons1,cons2},m): #shows that there is only a consistent solution if (n=0, n^2+1=0 or) condition:=simplify("[2][1]/(n*(n^2+1))): solve(",Rbar): Rbar:="; cons3:=factor(simplify(cons1)): cons4:=numer(factor(simplify(cons2))); #We write the constraints more symmetrically: simplify(",{m*n=p}): simplify(",{m^2+n^2=s^2-2*p}): simplify(",{m+n=s}): cons5:=factor(simplify(")); #Note:the same Rbar and Qbar result from using any pair of m,n,p=(1-mn)/(m+n). #The three symm functions of m,n,p are sum=(s^2-p+1)/s, two=1, prod=p*(1-p)/s. simplify(cons5,{s^2=s*newsum+p-1}): simplify(",{p^2=p-s*newprod}): cons6:=factor(")/s: #(Here s = m+n >0 is not zero). solve(",newsum): newsum:=factor("); #So! m,n,and p are three positive roots of the cubic cubic:=X^3-newsum*X^2+X-newprod; #Check for real roots with Sturm seq: a:=1: b:=-newsum/3: c:=1/3: d:=-newprod: G:=d-3*b*c+2*b^3: H:=c-b^2: oppdisc:=G^2+4*H^3: #condition for three real roots is that H and this opposite-of-discr. be neg. #Note: cubic(-X) has no pos. roots by Vincent's rule, so all real roots pos. #(Assuming newprod >0, which is necessary for 3 pos roots!) #Note that newprod only shows up as squares. isneg1:=factor(simplify(H,{newprod^2=pp})): isneg2:=factor(simplify(oppdisc,{newprod^2=pp})): #We wish to analyze the positive pp which make these both negative. Qbar:=2+(1+t^2)/t: #w=Qbar-2 is (R^2-Q^2)/r^2-2 = (2Rr)/r^2-2 = 2R/r -2 > 2, so = t+1/t,some t>1 w:=5+2*(1+t^2)/t: pp:=(3+(pr/2)*(w-9))/w^2: newneg1:=factor(simplify(isneg1))*9*t/(t-1)^2; #which shows that the solution is for pq to be positive newneg2:=factor(simplify(isneg2))*27*(t+2)^2*(2*t+1)^2*(2-4*t+2*t^2)*t; solve(",pr): pr1:="[1]; pr2:=""[2]; #As both roots are positive, then newneg2<0 iff pr is betweem these. Check: factor(simplify(subs(pr=(pr1+pr2)/2,newneg2))); #As both roots are positive, the other conditions newneg1<0 is irrelevant. #Thus we have limits on the values of newprod: positive square roots of lowp:=simplify(subs(pr=pr1,pp)); highp:=simplify(subs(pr=pr2,pp)); ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Triangle Construction Date: 1 Apr 1995 02:05:22 GMT OK, I'm stumped. A few weeks ago, zvrba@oleh.srce.hr (Zeljko Vrba) posted a terse query which he clarified for me in email as follows. Given any triangle in the plane, we know we can circumscribe and inscribe a pair of circles. Now, given conversely a nested pair of circles, must there be a triangle which leads to that pair? Is it unique (up to congruence, say) ? Unable to see an elegant solution I proceeded to coordinatize the problem. One can argue that there is the right number of degrees of freedom in either configuration to allow a one-to-one correspondence. (Triangles are represented by three numbers, such as the lengths of the sides; the circles are also decribed by three numbers - the inner and outer radii r and R and the distance Q between their centers). Imagine my surprise, then, when the algebraic approach revealed that instead there is a relation among Q, r, and R which must be satisfied if a corresponding triangle exists; conversely, when one such triangle exists, there must be infinitely many of them. Here's the challenge: provide a nice geometric proof of the required relation. That is, if a triangle is inscribed in a circle of radius R, and contains an inscribed circle of radius r, then the distance Q between their centers satisfies Q^2 + r^2 = (R-r)^2. Naturally, we ought to be able to draw a right triangle in the picture to prove this, but for the life of me I can't see how to use the existence of the original triangle in the proof. (Remember, that triangle is _not_ unique, although that's not clear at the outset; so the right triangle in question ought not be uniquely defined by the configuration of circles.) The relation may also be written (2r) / (R - Q) = (R + Q ) / R, but I can't see any way to introduce similar triangles either. Incidentally, the construction of the other triangles once one is obtained is easy: draw any line tangent to the inner circle; where it meets the larger circle we have two of the vertices. Then draw a line from one of these again tangent to the inner circle to obtain the third vertex as the other point of intersection with the outer circle. What is not clear from this description is that the third vertex so drawn will not depend on which of the first two vertices is used in the last part of the construction. At least this indicates, though, that the triangles in the solution set are not all congruent. Finally, I note that the process of selecting one of these triangles to satisfy some additional criterion is in general not possible with compass and straightedge, since the calculation of the coordinates of the points requires solving a cubic which is not generically reducible, and so is irreducible for some (most) combinations of R, r, and a third parameter used to distinguish the triangles in the solution set. Details available upon request. dave ============================================================================== Date: Fri, 31 Mar 95 19:15:33 PST From: [Anonymous] To: rusin@washington.math.niu.edu Subject: Re: Triangle Construction well, here's just a completely stupid wild guess: the construction can _always_ be carried out even if the two circles don't bear the prescribed relationship to each other; it's just that the polygon won't always end up being a triangle. sometimes it will be a square or a pentagon or a biangle, or, almost all of the time, a "star" with some limiting ratio of "sides per winding number". and in fact not only doesn't the inner circle have to bear the prescribed relationship to the outer circle, but also the inner circle doesn't even have to be a circle; any smooth curve enclosing a convex region would probably yield a definite real limiting ratio of sides per winding number. and then the stupid wild guess is, maybe this limiting ratio is some nice simple monotone analytic function f of the hyperbolic area of the convex region, or something like that. (f(0)=2, it seems.) and maybe putting the problem in this larger context would lead to a nicer "geometric interpretation" of what's going on. ============================================================================== (Actually that idea turned out to be really the right one, suitably interpreted! -- djr 3/1/96) ============================================================================== [These were notes I prepared for some lectures on this topic to a class studying Elliptic Curves. Unfortunately I have misplaced the first part of the notes establishing the notation. The initial curve E1 is, I think, the set of x-coordinates x, x' of pairs of points related by the conditions that: both points lie on one circle, and the line segment joining the points is tangent to another circle inside the first (radii r < R and centers offset by an amount a. Sorry for the incompleteness. -- djr, 1/26/96. Notes timestamped: -rw-r--r-- 1 rusin math 8734 Apr 13 1995 summary --djr] Let's first simplify the equation defining E1. Clearing terms we get E1={ (x, x') : (2a) x' x^2 + ((R^2-a^2) x^2 - 2 r^2 x x' + r^2 x'^2) -(2a)r^2 x' + (r^4 - r^2R^2+r^2a^2) = 0 } To simplify we set \alpha = 2a/r, \beta= (R^2-a^2-r^2)/r^2, both "dimensionless" constants; then E1 is E1={ (x, x') : \alpha (x' x^2) +r(\beta+1) x^2 +r(x-x')^2 - r^2\alpha x' - r^3\beta = 0 } Observe that this cubic equation has no cubed variables; we can adjust this with a linear change of variables. It is clear from the geometric picture that points with x=x' look different, so we choose a linear change of variables that introduces x2=x'-x at the expense of x'. The transformed curve is E2={ (x, x2): \alpha(x^3+x'x2) + r(\beta+1) x^2 +rx2^2 -r^2\alpha(x+x2) -r^3\beta = 0} Now we have an x^3 term. We'd like to have the other variable show up only as a perfect square. We can do this by completing the square. Use the invertible polynomial transformation x2=x3-(\alpha/(2r))(x^2-r^2) to get the transformed curve E3={ (x, x3): x3^2 + (1/4r^2)(x^2-r^2)( -\alpha x^2 + 4r\alpha x + r^2(\alpha^2+4\beta)) =0} Now we have x3^2 equal to a quartic in x. This can be reduced back to a cubic by noting that (x+r) (say) is a factor of the quartic. If we set x3=(1/2) x4 (x+r)^2 (the factor (1/2) was chosen in hindsight) then we have a birational map which is an isomorphism away from the point (-r,0) in E3. The transformed curve is equivalent, away from the line x=-r, to the curve E4={ (x, x4): x4^2 - (1/r^2)(r-x)( -\alpha x^2 + 4r\alpha x + r^2(\alpha^2+4\beta))/(x+r)^3 =0} This equation involves a cubic in x with a root at x=r and a triple pole at x=-r. We can use a fractional linear transformation to move the root and pole to 0 and \infty respectively. (Since x4 has "dimensions of inverse length" we choose the new variable to be scaled similarly. This is a matter of taste but leads to the following choice.) We let x = r(1-ur)/(1+ur). This gives E5={ (u, x4): x4^2 - (1/r) u ( (\beta-\alpha)u^2 + (\alpha^2+2\beta) ru + (\beta+\alpha) r^2) = 0} This now can be simply written in the form E5={ (u, x4): x4^2 = u ( A u^2 + B u + C ) }, which is a canonical form for elliptic curves with an element of order 2. Addition is calculated by the traditional chord-and-tangent process, negation is simply -(u, x4) = (u, -x4), and the given element of order 2 is (u, x4) = (0, 0). The identity element of the group is the point at infinity (there is just one; it has homogeneous coordinates [0:1:0].) We can trace back and compute these in other coordinate systems. In E4, the identity element has x=-r, the point of order 2 is x=r (x4=0), and inversion again inverts the x4 coordinate. In E3, the identity has (x, x3)=(-r, 0), the point of order 2 is (x,x3)=(r,0), and inversion is again the same. In E2, the two points have coordinates (-r, 0) and (r,0) again, but inversion sends (x, x2) to (x, -x2-(\alpha/r)(x^2-r^2) ). Finally, in E1, the identity is at (-r, -r), the element of order 2 at (r,r), and inversion sends (x,x') to (x, x'') where x''=-x'+2x-(\alpha/r)(x^2-r^2). It is interesting to observe that our natural calculations have selected as two points of interest the same two points we first noted on the elliptic curve E1, namely the points corresponding to the pairs (P,Q) on E which lie on vertical lines. Now, inversion is an automorphism of a curve of order 2. We now see that it is the automorphism which preserves the x coordinate of E1. It is clear that this uniquely specifies this map, since the equation defining E1 is a quadratic in x'. Moreover, it is clear what this automorphism involves geometrically: E itself has an automorphism \phi of order 2 which preserves the x coordinate, namely phi(P,Q) = (P,Q'), where Q' is the other point on the conic D lying on the (tangent) line PQ . Likewise, E has another automorphism psi of order 2: psi(P,Q) = (P',Q), where P' is the other point on the inner circle C where the line P'Q is tangent to C. Again it is clear that this map commutes with the automorphism s of E, that is, that it passes to an automorphism (also called psi ) on E2. Now we use some general results on rational maps defined on elliptic curves. If f is such a map, then f will, in general, move the identity element O of the curve. Let P=f(O). Then if g is the translation by -P (i.e. g(Q)=(-P)+Q for all points Q in the curve) then the composite g o f is a rational map which fixes the origin O. A general result on elliptic curves shows that any such map is in fact an _isogeny_, meaning that it is a homomorphism of groups. For curves defined over the real or complex field, there are usually few isogenies. Apart from the curves admitting complex multiplication, all isogenies on an elliptic curve are of the form Q --> mQ for all Q, where m is a fixed integer. None of these maps are invertible except in the cases m = 1 or -1. Combining these results, we see that in general an isomorphism of an elliptic curve must be of the form f = trans(P) or f = -trans(P) for some P in the group. We have already seen than phi is, indeed, inversion in the group. Now, psi fixes the pairs (P,Q) in which y=0 (that is, psi is not trivial on E, but for these points permutes the two elements in the fibre of the projection to E1, so that it fixes their image (x,x') in E1.) Hence we know that psi cannot be a translation in E1. From the previous paragraph, we expect that psi = -trans(P) for some point P in E1. Combining these results, we see that the isomorphism gamma = psi o phi ought to be a translation in E1. To test this, we compute P = gamma(O) and then see if gamma = trans(P). (This is the only possible translation gamma could be.) It is not difficult to compute psi in the coordinates of E1, since the equation E1={ (x, x') : \alpha (x' x^2) +r\beta x^2 +r(x-x')^2 - r^2\alpha x' - r^3\beta = 0 } that defines E1 is quadratic in x and since psi preserves the x' coordinate. Indeed, given x', the two possible x coordinates of points on E1 sum to (2rx')/(alpha x' + r beta). Thus, psi(x, x') = (-x + 2rx'/(alpha x' + r (beta+1)) , x'). Taking psi o phi (-r, -r), then , we see the identity element of the group is taken to P = r( (beta-alpha-1)/(beta-alpha+1), -1). We will show that psi o phi is precisely translation by this P. This is most easily done in the coordinates of E5, where phi(u, x4) = (u, -x4). In those coordinates we find P = ((1/r)(1/(beta-alpha), (-1/r)(1+beta)/(beta-alpha)) [Computation shows trans(P) agrees with psi o phi]. Now we address the connection with the triangle problem. The critical observation is that the isomorphism gamma is precisely the one which determines each vertex of the triangle from the previous one. If the given conics are to be inscribed and circumscribed about the polygon, then consecutive directed edges of the polygon form points (P,Q) and (P',Q') of E such that the latter is the image of the former under gamma. Thus the condition that P be a vertec of such a triangle is the condition that (P,Q) = gamma^3(P,Q). Since gamma = trans(T), gamma^3 = trans(3T); since E is a group, the condition (P,Q) = trans(3T) (P,Q) is the condition that 3T=O. In particular, we have Poncelet's theorem: if there is one triangle between the circles, there is one through each point of D. (This follows since the condition 3T=O is independent of (P,Q).) Furthermore, we have a way to check that 3T=O. This is equivalent to the condition that 2T = -T. We need only compute 2T and -T in the coordinate system E5. Recall the formula for doubling in the curve y^2=x(Ax^2+Bx+C): the tangent at (x,y) has slope m = -(3Ax^2+2Bx+C)/(2y), so that the tangent line meets the curve where 0=A(x+t)^3 + B(x+t)^2+C(x+t)-(y+mt)^2 =At^3 + (3Ax+B-m^2)t^2 + ... But we know t=0 is already a double root as this is a tangent line, so the remaining root is t = (1/A)(m^2 - 3Ax - B), and then the x coordinate of the doubled point is x* = x+t = -2x + (m^2-B)/A. On the other hand, the inverse of (x,y) is (x, -y), so the condition that 2T = (-T or T) is the condition that x*=x. This requires (3Ax + B) = m^2, i.e. (3Ax+B)(4sx)(Ax^2+Bx+C) = (3Ax^2+2Bx+C)^2 If we substitute in our given values for A, B, C, and x (the u-coordinate of T) we get the necessary and sufficient condition for the cycle to have length 3: \beta+1 = +- \alpha or \beta^2-2\beta-\alpha^2-3=0. Substituting in turn the definitions of alpha and beta we get R^2 - a^2 = +- (2 a R or 2 R r) The condition R^2 - a^2 = +- 2 a R results in R/r = +-1 +- sqrt(2). ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Triangle construction (summary: not so trivial!) Date: 12 Apr 1995 20:58:49 GMT Summary: Euclidean problem relates to fancy math Keywords: Poncelet, geometry, dynamical systems, elliptic curves I'd like to summarize a plane-geometry question which arose in this newsgroup recently. Quite unexpectedly, I found the topic related to a graduate course I'm teaching on elliptic curves! The problem, posed by Zeljko.Vrba@oleh.srce.hr (Zeljko Vrba), assumed two circles given in the plane, and asked for a characterization of the triangles which have these two circles as their in- and circum-scribed circles. I followed up with the observation that for most pairs of circles there are no such triangles, while otherwise there will be infinitely many; the cases are distinguished by a certain relation holding on the radii and the separation of the centers. I asked for an 'elementary' proof of this relation. I found a thorough and delightful reference in Expositiones Mathematicae 5 (1987) 289-364, by Bos, Kers, Oort, and Raven. In summary: (a) The necessary condition on the parameters is "Chapple's identity". (b) Still no geometric proof is available to me. (c) The existence of infinitely many solutions is "Poncelet's theorem". (d) There is a connection to elliptic curves by Griffiths and Harris. I would like to clarify these points a bit, although I can't hope to duplicate what is found in that survey article. (a) If the inner and outer radii are r and R respectively and if the distance between the centers is a then as I noted in a previous post, the existence of such a triangle implies a^2=R^2-2rR. This identity was first published in 1746 by Chapple, but apparently was not widely circulated. I particularly liked the characterization of Chapple's work: "It is not easy to summarize Chapple's article because almost all the logic in it is wrong. Apparently Chapple was an enthusiastic amateur of math, but he was, even by the standards of his time, very weak in formal mathematical argument." Thus, Chapple anticipated USENET by two and a half centuries! Correct proofs may be deduced in elementary fashion (a la Euler). The above-mentioned summary does better justice to these ideas than I can. This is Chapple's idea: if there is a triangle between the circles, then (and this is really Poncelet's "Closure Theorem") there is another one which has a vertex on the line joining the centers of the points. From _this_ triangle we can make elementary arguments: it's easy to see that it's isoceles, which makes the calculations much easier. (b) I must confess that a completely geometric proof still seems, in my opinion, to be wanting. Even in the isoceles case, I don't see how the identity may be obtained. For those who enjoy ruler-and-compass problems, here is the issue to be resolved: Begin with three points A, B, C in the plane. (1) Join these points with 3 line segments. [Now you have the triangle.] (2) Bisect AB and BC and find the intersection D of those bisectors. [This is the center of the circumscribed circle.] (3) Bisect angles BCA and BAC and find the intersection E of those bisectors. [This is the center of the inscribed circle.] (4) Drop a perpendicular from E to AC. Draw the circle S around E with this radius. [This is a the inscribed circle.] (5) Draw the perpendicular to the segment DE through E and let F be either of the points where this meets the circle S. (6) Draw the circle around D passing through F and locate the point G where it intersects the segment DB. [Note that this circle has radius sqrt(a^2+r^2) by the Pythagorean theorem.] Now show that the segments GB and EF are congruent. [Then their lengths will be equal, showing that sqrt(r^2+a^2)+r = R, as needed.] The problem may also be recast so as to use similar triangles instead of the Pythagorean theorem, or so that the goal is to prove various triangle isoceles, and so on. The point is to find a proof acceptable to Euclid (not just Descartes). The difficulty is that many triangles will (by Poncelet's theorem) lead to the same E, S, and F, so that the segment GB is "free" even though the segment EF is "fixed". I concede that this is not a critical issue -- it just sort of annoys me. (c) The 0-or-infinite characterization was proven by Poncelet in 1822; his proof is valid for general conics. A generalization to curves of higher order is due to Darboux, and once the algebraic geometers got hold of it, the multiplication and reinterpretations of the problem mushroomed. The theorem states that if there is any triangle which fits the circles, then given any point P on the outside circle there is a triangle which has P as a vertex, and which fits the two circles in the desired sense. One can think of this as a dynamical systems problem: one leaves the initial point P in a direction (there are two) which will be tangent to the inner circle. When intersecting the outer circle again, one repeats the process. The question is to show that the orbit closes up after precisely three iterations if and only if the relation a^2=R^2-2rR is met. This perspective makes it equally easy to address the question of when there is a quadrilateral (say) which fits the circle in the same sense the triangle does in the original problem. (d) The striking fact for me is the connection with elliptic curves. Given the two conics, define the set E to be the set of pairs of points, one from each curve, such that the tangent to the first curve through the first point meets the second curve at the second point (and other points, in general). One can show that E is an elliptic curve, meaning that it has a group law defined on it. E also has an isomorphism defined as follows: given a pair of points (P,Q) on E, let Q' be the other point on the second conic meeting that tangent line thru P; then let P' be the other line tangent to the first conic which passes thru Q'. (One may justify the use of the phrases "the other..." here). Then (P',Q') is another point of E. Well, guess what: the map which sends (P,Q) --> (P'Q') is just a translation in the elliptic curve! That is, there is a point (P0,Q0) in E such that (P',Q') = (P,Q) + (P0,Q0) (this is not component-wise addition here; rather, it's addition in the elliptic curve). Thus, this process terminates after n steps iff (P0,Q0) is an element of order n. This observation, as I understand it, dates to Griffiths and Harris in a set of papers around 1977. In particular, this construction makes it clear that the condition for closure ( 3(P0,Q0) = identity ) is independent of the starting point. It is possible, if not entirely trivial, to begin with a characterization of E as a subvariety of { (x,y,x',y') : x^2+y^2=r^2, (x-a)^2+y^2=R^2 } and compute the element (P0, Q0). Then, using the group operations in the elliptic curve, one may compare 2(P0, Q0) to -(P0, Q0) to deduce the condition that this automorphism have order 3. Of course, we already know the answer is that a^2 must equal R^2-2rR. Details available on request (except for my unfortunate students, who will be forced to sit through a discussion of this problem this week). I always love it when a problem in this newsgroup turns out to be more interesting than at first blush. dave ============================================================================== From: Zeljko.Vrba@oleh.srce.hr (Zeljko Vrba) Subject: Triangle construction To: rusin@math.niu.edu Date: Fri, 7 Apr 1995 12:02:46 +0200 (MET) Thanks for your reply. I'd like to see the details of your calculations. As for the proof of the 'distance of the centers' formula, i think that I have one at home (and it uses original triangle). I'll post it to you if you are interested. -- zvrba@oleh.srce.hr ============================================================================== Date: Fri, 7 Apr 95 09:10:35 CDT From: rusin (Dave Rusin) To: Zeljko.Vrba@oleh.srce.hr, rusin@math.niu.edu Subject: Re: Triangle construction >Thanks for your reply. I'd like to see the details of your calculations. Here's the maple input: #given two nested circles, find a triangle touching one, tangent to other. #We try to find a triangle P0=(x0,y0), P1=(x1,y1), P2=(x2,y2). # #Given point (x0,y0), assume other point is symmetric wrt vertical axis x1:=-x0: y1:=y0: #draw two lines of any slope; find intersection solve({m=(y-y0)/(x-x0),n=-(y-y1)/(x-x1)},{x,y}): xi:=rhs("[1]): yi:=rhs(""[2]): #third point is on lines with double these angles A:=2*m/(1-m^2): B:=2*n/(1-n^2): solve({A=(ynew-y0)/(xnew-x0),B=-(ynew-y1)/(xnew-x1)},{xnew,ynew}): x2:=rhs("[1]): y2:=rhs(""[2]): #P0 and P2 lie on same circle. This pins down y0 solve(simplify((x2^2+y2^2)-(x0^2+y0^2)),y0): y0:=factor("): # #Now we try to match the given data. The inner radius is supposed to be r: solve(simplify(yi-y0-r), x0): x0:=factor("): #Then the lengths of (x0,y0) and (xi,yi) pin down R, Q respectively cons1:=simplify((x0^2+y0^2)/r^2-Rbar): cons2:=simplify((x0^2+y0^2)/r^2 - (xi^2+yi^2)/r^2 -Qbar): #So we should get cons1=Rbar=(R/r)^2, cons2=Qbar=(R^2-Q^2)/r^2. But readlib(eliminate): eliminate({cons1,cons2},m): #shows that there is only a consistent solution if (n=0, n^2+1=0 or) condition:=simplify("[2][1]/(n*(n^2+1))): solve(",Rbar): Rbar:="; cons3:=factor(simplify(cons1)): cons4:=numer(factor(simplify(cons2))); #This is the condition that must be satisfied if the points exist. >As for the proof of the 'distance of the centers' formula, i think >that I have one at home (and it uses original triangle). >I'll post it to you if you are interested. I'd like to see it. Meanwhile, I have learned that this is classical geometry. I think the result is first due to Euler (of course), but the result is known as Poncelet's theorem: given two conics the number of triangles inscribed in one and circumscribed around the other is either 0 or infinite (parameterized by the points of the outer conic). I will be checking into this today, and I plan to post a summary of what I find. In modern times the result has been linked to elliptic curves. dave ============================================================================== From: Zeljko.Vrba@oleh.srce.hr (Zeljko Vrba) Subject: Re: Triangle construction To: rusin@math.niu.edu (Dave Rusin) Date: Thu, 13 Apr 1995 10:59:45 +0200 (MET) > >As for the proof of the 'distance of the centers' formula, i think > >that I have one at home (and it uses original triangle). > >I'll post it to you if you are interested. > > I'd like to see it. Meanwhile, I have learned that this is classical > geometry. I think the result is first due to Euler (of course), but > the result is known as Poncelet's theorem: given two conics the number of > triangles inscribed in one and circumscribed around the other is > either 0 or infinite (parameterized by the points of the outer conic). > I will be checking into this today, and I plan to post a summary of > what I find. In modern times the result has been linked to elliptic curves. > > dave > > This proof uses these theorems: 1. Steiner's law for moment of inertia: If P is any point in space, T center of mass of system S, Ip and It moments of inertia with respect to points P and T, then following holds: Ip=It + (sum {from i=1 to n} of {m(i)}) * |PT|^2 (sum is really the mass of the system S). 2. Jacobi's theorem: Let pair (X,m) denote a material point X with mass m. Given a system S={(A1,m1), (A2,m2),...,(An,mn)}, the moment of inertia with respect to the center of mass of the system (It) is given with: 1 It=- * SIGMA mi*mj*|AiAj|^2 m 1<=i