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Greg Markowsky
guest

May1207, 07:45 AM (EST) 

3. "RE: Challenge problems"
In response to message #2

alexb, Thanks for making that applet on the parabola, and noticing that M and K also lie on CD. This whole thing is just a projection of the same situation in a circle, which is how the idea came to me in the first place(while looking at Archimedes Book of Lemmas, Proposition 12). I have since realized, though, that one should be very wary of parabolas in composing problems, since generally it's so easy to deal with the equations y=x^2 and y'=2x, and that usually one can just mess around with equations to prove almost anything true. This problem is not too hard if you just write down equations and crank. Oh well. As for the second one, probably it's no surprise that I noticed that while reading about Sangaku. It's roughly equivalent to an old Sangaku problem. Greg 

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alexb
Charter Member
2031 posts 
May1207, 07:53 AM (EST) 

4. "RE: Challenge problems"
In response to message #3

>alexb, > >equations and crank. Oh well. Thank you for posting here. They both are truly nice problems. The second one combines quantities that I never before even saw used. But it is pretty simple. Configuration of the first one admits a generalization that I placed at https://www.cuttheknot.org/Curriculum/Geometry/Markowsky2.shtml I expect there to be some generalization anglewise, perhaps for a different direction. I am away right now. Will be looking into that when back. Alex 

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Greg Markowsky
guest

May1207, 10:24 PM (EST) 

5. "RE: Challenge problems"
In response to message #4

That generalization is really pretty, I hadn't thought of that. It can be proved by letting points come together in Pascals hexagon theorem, is that how you did it? Also, if you wouldn't mind I would be interested to see how you proved(a sketch would be enough) the second problem I posted. My proof gets kinda heavy into trig, but I've always preferred proofs that don't use trig, so I'd be curious to see how you do it. Greg 

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Greg Markowsky
guest

May1307, 07:57 PM (EST) 

7. "RE: Challenge problems"
In response to message #6

Thanks. My proof was longer because I wasn't aware of that identity that you used, so I had to derive something that probably is basically equivalent, and the whole thing was more involved. Yours is a bit nicer tho, I think. 

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alexb
Charter Member
2031 posts 
Jun1107, 02:29 PM (EST) 

15. "RE: Challenge problems"
In response to message #4

>The second one combines quantities that I never before even >saw used. It appears the quantity is well known and is called versine (as a piece of since rotated 90°). If AB is a segment in a (unit) circle with center O and midpoint M. If angle AOM = α and the radius through M is OS, then MS is the versine of α:

MS  = vers(α)  MS  = 1  cos(α)  MS  = 2·sin^{2}(α/2). 

I ran into this term in an article in Math Horizons (Sept., 2006). Here's a hole in my education. 

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mpdlc
Member since Mar1207

May1907, 12:03 PM (EST) 

8. "RE: Challenge problems"
In response to message #0

An Analytic Geometry approach The answer given below to the parabola problem , is based on Analytic Geometry and Calculus even it is simple, does not have the charm of pure Geometry where in my opinion Math and Art melt altogether. So using the same graph posted by G. Markowsky the equality between angles ADO and BDE implies also the equality between angles ADC and BDC. Taking under consideration that points B and A belong to a parabola y =X2 and point C is in line connecting the origin with point B m=Yb/Xb we can easily write the coordinates for the above points as follows : For point B Xb = x ; Yb = x2 For point A Xa = (x – q) ; Yb =(x  q)2 For point C Xc = (x – q) ; Yc = (x – q) Yb/Xb or Yc = x(x – q) Now tangent of angle BCD = (Yb Yc)/Xb = (x 2 – x(x – q))/x = q and tangent of angle CDA = (Yc Ya)/Xc = (x(x – q) – (x – q)2)/ (x – q) =q so both angle are equals. Now let us prove that if angle BCD and CDA are the same point B and A must both lay in a parabola vertex in origin of equation y = Kx2
Since point A and B can be as close as we want we take B a point having the coordinates B(x, y) and for A(xdx , ydy) being on the straight line connecting the origin with point B we obtain the coordinates C(xdx , yy dx/x ) So tangent CDB = (ydx/x)x and tangent ADC = (ydx/x+dy)/(xdx) equalizing both tangents and eliminating denominators we get y(x – dx) dx = x2 dy – yxdx if we discard (dx)2 we will get the differential equation 2y dx = x dy which render Ln(y)= 2L(x) + Ln(K) or more simple y = Kx2 So both point A and B belong to the parabola mpdlc 

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Bui Quang Tuan
guest

Jun0807, 09:02 PM (EST) 

9. "RE: Challenge problems"
In response to message #0

Dear Greg and Alex, We can study some other interesting facts related "A Relation in Triangle". For convenience I use some detail notations: A triangle ABC cuts off from the circumcircle three circular segments. Let A', B', C' are midpoints of these circular segments with respect to BC, CA, AB respectively. k, l, m are distances from A', B', C' to sidelines BC, CA, AB respectively. Greg's result: 2*k*l*m = R*r^2 where R is the circumradius, r is the inradius. Now let p, q, r are distances from A, B, C to sidelines B'C', C'A', A'B' respectively then we can have similar result: 2*p*q*r = R*r^2 This result can be derivered from one more general following fact (may be new): Let A, B, C, A', B', C' are any six concyclic points k, l, m are distances from A', B', C' to lines BC, CA, AB respectively. p, q, r are distances from A, B, C to lines B'C', C'A', A'B' respectively. The result: k*l*m = p*q*r I hope that we can find nice synthetic proof for this fact. Best regards, Bui Quang Tuan


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Bui Quang Tuan
guest

Jun0907, 10:25 PM (EST) 

12. "RE: Challenge problems"
In response to message #9

Dear Greg and Alex, I have found one following proof, please check for me if I make some mistakes. Let A, B, C, A', B', C' are concyclic on one circle (O), radius R Let A'', B'', C'' are orthogonal projection of A'B'C' on BC, CA, AB respectively. Let A*, B*, C* are orthogonal projection of ABC on B'C', C'A', A'B' respectively. We can easy show some following concyclic point sets: A, A*, B', B'' are concyclic on circle (Ab) (center Ab = midpoint of AB'), radius = Rab A, A*, C', C'' are concyclic on circle (Ac) (center Ac = midpoint of AC'), radius = Rac B, B*, C', C'' are concyclic on circle (Bc) (center Bc = midpoint of BC'), radius = Rbc B, B*, A', A'' are concyclic on circle (Ba) (center Ba = midpoint of BA'), radius = Rba C, C*, A', A'' are concyclic on circle (Ca) (center Ca = midpoint of CA'), radius = Rca C, C*, B', B'' are concyclic on circle (Cb) (center Cb = midpoint of CB'), radius = Rcb Moreover: A, Ac, Ab, O are concyclic on circle (Oa), (center Oa = midpoint of OA), radius R/2 B, Bc, Ba, O are concyclic on circle (Ob), (center Ob = midpoint of OB), radius R/2 C, Ca, Cb, O are concyclic on circle (Oc), (center Oc = midpoint of OC), radius R/2 By some similar triangles we can also easy show following fact: The common chord d of two given circles with radius R1, R2 can be calculated as: d = R1*R2/R3 where R3 = radius of the circle passing through two centers of given circle and one their intersection point. We now consider three circles at vertex A: (Ac), (Ab), (Oa), we can show: AA* = 2*Rab*Rac/R (1) Similarly we can have BB* = 2*Rbc*Rba/R (2) CC* = 2*Rca*Rcb/R (3) Do the same with triangle A'B'C' we can have: A'A'' = 2*Rca*Rba/R (1') B'B'' = 2*Rab*Rcb/R (2') C'C'' = 2*Rbc*Rac/R (3') From (1), (2), (3), (1'), (2'), (3') we have the result: AA* x BB* x CC* = A'A'' x B'B'' x C'C'' or k*l*m = p*q*r Best regards, Bui Quang Tuan


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Bui Quang Tuan
guest

Jun1007, 02:02 AM (EST) 

13. "RE: Challenge problems"
In response to message #12

>AA* = 2*Rab*Rac/R (1) >Similarly we can have >BB* = 2*Rbc*Rba/R (2) >CC* = 2*Rca*Rcb/R (3) >Do the same with triangle A'B'C' we can have: >A'A'' = 2*Rca*Rba/R (1') >B'B'' = 2*Rab*Rcb/R (2') >C'C'' = 2*Rbc*Rac/R (3') Dear Greg and Alex, From my proof we can show one interesting fact for any inscribed hexagon (not necessary convex) AB'CA'BC'. From (1), (2), (3), (1'), (2'), (3') k*l*m = p*q*r = 8*Rab*Rac*Rbc*Rba*Rca*Rca = 8*(AB'/2)*(AC'/2)*(BC'/2)*(BA'/2)*(CA'/2)*(CB'/2)/R^3 If we denote six sides of inscribed hexagon AB'CA'BC' as a, b, c, d, e, f then k*l*m = p*q*r = a*b*c*d*e*f/(2*R)^3 = a*b*c*d*e*f / W^3 or a*b*c*d*e*f = k*l*m*W^3 = p*q*r*W^3 (4) where W = diameter of circumcircle of the hexagon p, q, r = distances from A, B, C to sidelines of A'B'C' k, l, m = distances from A', B', C' to sidelines of ABC If can we somehow generalize (4) for other inscribed polygon? Best regards, Bui Quang Tuan


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Bui Quang Tuan
guest

Jun1007, 09:18 AM (EST) 

14. "RE: Challenge problems"
In response to message #13

Dear Greg and Alex, I just have generalized (and it is easy) the result as following: n >=3 is one integer. X1, Y1, X2, Y2, X3, Y3, ... Xn, Yn are any 2n points concyclic on one circle O with diameter W y1, y2, y3, ... yn are distances from Y1, Y2, Y3, ... Yn to X1X2, X2X3, X3X4, ... XnX1 respectively. x1, x2, x3, ... xn are distances from X1, X2, X3, ... Xn to YnY1, Y1Y2, Y2Y3, ... Yn1Yn respectively. a1, a2, a3, ... a2n are sides of polygon X1Y1X2Y2X3Y3 ... XnYn (by this order). It means: a1 = X1Y1 a2 = Y1X2 a3 = X2Y2 a4 = Y2X3 ... a(2n1) = XnYn a2n = YnX1 We can use the same my posted proof to prove the following result: a1*a2*a3* ... *a2n = x1*x2*x3* ... *xn*W^n = y1*y2*y3* ... *yn*W^n Best regrads, Bui Quang Tuan


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Bui Quang Tuan
guest

Jun1907, 02:39 PM (EST) 

18. "RE: Challenge problems"
In response to message #16

Dear Alex, Using this simple lemma and the same proof method we can show also following interesting facts: Given point P and n points P1, P2,â€¦ Pn (n>1) on the circle (O) with diameter d. s1, s2, â€¦ sn are lengths of segments PP1, PP2, â€¦ PPn respectively d1, d2, â€¦ dn are distances from P to segments P1P2, P2P3, â€¦ PnP1 respectively The result: (s1*s2*â€¦ *sn)^2 = d1*d2*â€¦ *dn*d^n (When n=2 so we have our lemma.) Please note that P1P2, P2P3, â€¦ PnP1 can be see as one close path passing all n point and each point one time (if I am not wrong it is Hamilton / Euler path in graph theory). For short we name it as "HE path". We can name each point P as a "vertex" and each segment PiPj as a "side" of this path. Now our result can be formulated as: Theorem 1: If P, P1, P2, ... Pn (n>1) are concyclic on one circle with diameter d then (s1*s2*â€¦ *sn)^2 = d1*d2*â€¦ *dn*d^n where si = segments from P to all vertices of one HE path from Pi (i=1 to n) di = distances from P to all n sides of this HE path. or by words The square of product of all segments from one point to all vertices of one HE path is equal product of all distances from this point to all sides of this HE path multiplied by nth power of diameter of the cycle on which the point and the HE path are concyclic. Of course, there are a lot of HE paths from n vertices and they have different distances from P to their sides. But they have the same one set of segments PP1, PP2, ... PPn or they have the same product of segments from P to n vertices: s1*s2*s3*...*sn. So we can have: Theorem 2: The product of all distances from one point to any HE path from given points are constant if all points are concyclic. d1*d2*â€¦ *dn = (s1*s2*â€¦ *sn)^2 / (d^n) = constant If we choose two HE paths with some common sides then they have also some common distances, therefore we can get some other equal products of distances. One special example when n = 4 s1 = PP1, s2 = PP2, s3 = PP3, s4 = PP4 From four points P1, P2, P3, P4 we can make three HE paths 1. P1P2P3P4  sides: P1P2, P2P3, P3P4, P4P1  distances from P to these sides: d12, d23, d34, d41 2. P1P2P4P3  sides: P1P2, P2P4, P4P3, P3P1  distances from P to these sides: d12, d24, d43, d31 3. P2P3P1P4  sides: P2P3, P3P1, P1P4, P4P2  distances from P to these sides: d23, d31, d14, d42 (by these dij notations, of course dij = dji) By our theorems: d12*d23*d34*d41 = d12*d24*d43*d31 = d23*d31*d14*d42 = (s1*s2*s3*s4)^2/d^4 From these we can have following other equal products of distances: d12*d34 = d14*d23 = d13*d24 This result can be formulated: Theorem 3: The products of two distances from one point on circumcircle of one inscribed quadrilateral to two opposite sides or two diagonals are equal. Best regards, Bui Quang Tuan


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Bui Quang Tuan
guest

Jun2107, 01:53 PM (EST) 

20. "RE: Challenge problems"
In response to message #19

Dear Alex, I continue study these distances from one points to segments of other points on one circle. Using the same lemma and Ptolemy's theorem I can prove following fact: Given one inscribed convex polygon X1X2X3...Xn1Xn (n>2) and one point P on its circumcircle We use following notations (i, j can be interchanged) aij = length of side XiXj dij = distance from P to sidelines XiXj Rij = ratio aij/dij then the set of ratios R12, R23, R34,...R(n1)n, Rn1 hold a property: one is sum of others First, we prove for n = 3, the triangle case Given triangle ABC and one point P on circumcircle of ABC a, b, c = sides of ABC da, db, dc = distances from P to sides BC, CA, AB The result: Three ratios a/da, b/db, c/dc hold a property: one is sum of other two. Proof: We use following lemma: in the triangle ABC h = b*c/(2*R) where h = altitude from A, b = AC, c = AB, R = circumradius of ABC With triangle PBC da = PB*PC/(2*R) Similarly we have: a/da = a/(PB*PC)*(2*R) = a*PA*(2*R)/(PA*PB*PC) b/db = b/(PC*PA)*(2*R) = b*PB*(2*R)/(PA*PB*PC) c/dc = c/(PA*PB)*(2*R) = c*PC*(2*R)/(PA*PB*PC) Triangle ABC divides its circumcircle into three arc, say (BC), (CA), (AB) If P on arc (BC) then by Ptolemy's theorem: a*PA = b*PB + c*PC therefore a/da = b/db + c/dc. It is what should be proved. For n>3, we use mathematical induction method. Suppose now our theorem is true for n1 and convex polygon X1X2...Xn divides circumcircle into n arcs and P is on arc X1X2. Consider triangle X1X2X3. This triangle devides circumcircle into three arcs and P is on arc X1X2 (It is true because polygon is convex). X1X3X4...Xn is n1 side convex polygon and P is on arc X1X3 (It is true because polygon is convex) so: R13 = R34 + R45 + ... + Rn1 (our theorem for n1) In the triangle X1X2X3 R12 = R23 + R13 = R23 + R34 + R45 + ... + Rn1 So the theorem is true for any n>3 We can also know which ratio is sum of other: the ratio of side respective with the arc where P is. Please correct for me if I have made any mistake! Best regards, Bui Quang Tuan


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alexb
Charter Member
2031 posts 
Jun2207, 00:19 AM (EST) 

21. "RE: Challenge problems"
In response to message #20

Excellent result which appears to generalize van Schooten's theorem in two ways: 1. To nonequilateral triangles 2. To ngons with n > 3. Beautiful. I do not see any errors. P.S. For the next post, may I ask you to start a new thread. Thank you. 

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