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0|0|0|||1||A few suggestions.|sfwc||13:04:15|12/18/2006|%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FRIsoscelesOnQuadri.shtml%7CRight isosceles triangles on sides of a quadrilateral%5D%3A%0D%0AThe properties mentioned as equivalent appear not to be. The last three are true for any quadrilateral%2C whilst the first is equivalent to P %3D R.%0D%0AProperty 1 is%2C as you mention%2C equivalent to i.%5Bu%5Dac%5B%2Fu%5D %3D %5Bu%5Dbd%5B%2Fu%5D %28or%2C if you wish to be pedantic%2C to %5Bu%5Dac%5B%2Fu%5D%5Bsup%5D2%5B%2Fsup%5D %2B %5Bu%5Dbd%5B%2Fu%5D%5Bsup%5D2%5B%2Fsup%5D %3D 0%29.%0D%0AProperty 2 is true of any quadrilateral at all. Substituting %281 - %26mu%3B%29 %3D -i.%26mu%3B into %285%29 gives %5Bu%5Dpr%5B%2Fu%5D %3D -i%26mu%3B%5Bu%5Dac%5B%2Fu%5D %2B %26mu%3B%5Bu%5Dbd%5B%2Fu%5D %3D i%28-%26mu%3B%5Bu%5Dac%5B%2Fu%5D - i%26mu%3B%5Bu%5Dbd%5B%2Fu%5D%29 %3D i%5Bu%5Dqs%5B%2Fu%5D%2C so PR is orthogonal and equal to QS. This also shows that property 3 is true of any quadrilateral. This seems to fit better with the comment in the remark at the end of the page. The statement %27The two are orthogonal iff i%5Bu%5Dpr%5B%2Fu%5D %3D %5Bu%5Dqs%5B%2Fu%5D%27 is false%2C though i%5Bu%5Dpr%5B%2Fu%5D %3D %5Bu%5Dqs%5B%2Fu%5D implies P%3DR and Q%3DS%2C which in turn implies property 1. %0D%0AFor property 4%2C consider the new construction of q and s by triangles with opposite orientation. As you mention%2C %28p%2Br%29%2F2 %3D %28q%2Bs%29%2F2%2C so we have the slightly more general result that PQRS is always a parallelogram. To deal with the comment regarding the intersection of the diagonals%2C it is helpful to take the origin there. So c %3D ta%2C d %3D ub with t and u real. If p %3D r%2C we have -i%26mu%3Ba %2B %26mu%3Bb %3D -i%26mu%3Bta %2B %26mu%3Bub%2C so ia%281-t%29 %3D b%281-u%29%2C so ia is proportional to b. It follows that both q and s are proportional to %26mu%3Bb%2C so the line joining them passes through the origin%2C as required.%0D%0A%0D%0A%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FUnexpectedAngleBisector.shtml%7CAngle bisector in touching circles%5D%3A%0D%0AThis result could also be proved by considering the second point of intersection K of TP with the larger circle. By considering the scaling about T taking the smaller to the larger circle%2C The tangent to the larger circle at K is parallel to AB%2C so K is the midpoint of the arc AB%2C and so TK bisects ATB.%0D%0AThe %5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FEqualAnglesInTwoCircles.shtml%7Cgeneralisation%5D%0D%0Acan actually be extended even further. Given two circles that cross at S and T%2C let a chord AB of one intersect the other in P and Q. Then ASP %3D AST - PST %3D %28%26pi%3B - ABT%29 - %28%26pi%3B - PQT%29 %3D QTB.  %0D%0A%0D%0A%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FThreeParallelsInTriangle.shtml%7CThree parallels in a triangle%5D%3A%0D%0AThis can be proved in a slightly simpler manner by considering the symmetry of the figure.%0D%0AEK is the reflection of BC in the bisector of angle A%2C so the point X where these lines meet is the intersection of BC with that bisector%3A It satisfies XB%3AXC %3D c%3Ab %3D BD%3ACL%2C so XD%3AXL %3D %28BD - XB%29%3A%28CL - XC%29 %3D c%3Ab %3D XE%3AXK%2C so DE and KL are parallel. %0D%0A%0D%0A%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FInAndCircumcenter.shtml%7CA property of the line IO%5D%3A%0D%0AThis has a neat proof by transformation geometry. D is the reflection of A in BI%2C so DI %3D AI and AID %3D 2 AIB %3D 2%28%26pi%3B - A%2F2 - B%2F2%29 %3D 2%26pi%3B - %28%26pi%3B - C%29 %3D %26pi%3B %2B C. So letting f be the operation of rotation about I by %26pi%3B %2B C%2C D %3D f%28A%29. Similarly%2C B %3D f%28E%29. Let g be the operation of rotation about O by -2C. Then g%28B%29 %3D A%2C so fgf%28E%29 %3D D. But fgf rotates the plane by a total angle of %28%26pi %2B C%29 -2C %2B %28%26pi%3B %2B C%29 %3D 2%26pi%3B%2C so fgf is a translation by the vector D - E. Let E%27 %3D f%5Bsup%5D-1%5B%2Fsup%5D%28O%29%2C and let D%27 %3D E%27 %2B D - E %3D fgf%28E%27%29 %3D fg%28O%29 %3D f%28O%29. Then DE is parallel and equal in length to D%27E%27%2C which is perpendicular to IO and of length %7C2 IO sin%28%26pi%3B %2B C%29%7C %3D 2 IO sin%28C%29 %3D IO when C %3D 30 degrees.%0D%0AThis also throws some light on the %5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FInAndCircumcenterD.shtml%7Crelated theorem%5D mentioned elsewhere. Let O%27 be the reflection of O in CI. Then OO%27 is at right angles to the bisector CI of ACB%2C so to the internal bisector of NOM%2C so it is the external bisector of that angle. So O%27 is the midpoint of the longer arc MN and MIO%27 %3D O%27IN %3D MIN%2F2 %3D %282%26pi%3B - 2C%29%2F2 %3D %26pi%3B - C. So O%27IM %3D NIO%27 %3D %26pi%3B %2B C%2C and so M and N are the reflections of D%27 and E%27 in CI. Thus MN is equal and antiparallel to D%27E%27%2C which is equal and parallel to DE. Thus MN%2C NO and OM are antiparallel to ED%2C DC and CE respectively%2C so MNO and EDC are similar. As also MN %3D ED%2C they are congruent.%0D%0A%0D%0A%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FMixtilinearLike.shtml%7CA sangaku%3A Two unrelated circles%5D%3A%0D%0AAnother approach is to start from the smaller circle. Take coordinates such that it is the unit circle%2C with Q on the x axis. Let G %3D %28g%2C h%29%2C W %3D %28u%2C v%29. Then%3A%0D%0AQV has equation xg %2B yh %3D 1.%0D%0AQ is at %281%2Fg%2C 0%29.%0D%0AO is at %281%2Fg%2C v%2Fug%29 %3D W%2Fug.%0D%0AV satisfies xg %2B yh %3D 1 and %28x - 1%2Fg%29%5E2 %2B %28y - v%2Fgu%29%5E2 %3D %281 - 1%2Fug%29%5E2%0D%0ASo y%5E2 - 2ygv%2Fu %2B 2g%2Fu - g%5E2 - 1 %3D 0%0D%0ASo QR %3D -y %3D -gv%2Fu - sqrt%28g%5E2%2Fu%5E2 - 2g%2Fu %2B 1%29 %3D -gv%2Fu %2B 1 - g%2Fu %3D 1 - g%281%2Bv%29%2Fu%0D%0Aand PQ %3D 1 - 1%2Fug %2B v%2Fug %3D 1 - %281-v%29%2Fug.%0D%0ASo %28PQ - 1%29%28QR - 1%29 %3D %281-v%29%281%2Bv%29%2Fu%5E2 %3D 1.%0D%0AEquivalently%2C 1%2FPQ %2B 1%2FQR %3D 1 %3D 1%2Fq.%0D%0A%0D%0A%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FUnexpectedMedian.shtml%7CA median in touching circles%5D%3A%0D%0AApply Menelaus with respect to triangle ABC.%0D%0A%0D%0A%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FDissectCyclicQuadri.shtml%7CDissection of a cyclic quadrilateral%5D%0D%0A%22for any P%2C points M%2C N%2C Q%2C R always exist on the side lines of ABCD such the quadrilaterals AQPR%2C DNPQ%2C CMPN%2C and BRPM are cyclic.%22 Just let M%2C N%2C Q%2C R be the bases of the perpendiculars from P to the side lines.%0D%0A%22And for which P the set of N%27s that answers that question is not empty%3F%22%0D%0AThe rays PM%2C PN%2C PQ and PR are at fixed angles to one another%2C and must lie inside the angles CPB%2C DPC%2C APD and BPA respectively. Say AD and BC intersect at X%2C and AB and CD intersect at Y. Without loss of generality%2C A lies between X and D and B lies between A and Y. We must have BPA %3C%3D MPQ %3D %26pi%3B - AXB%2C so P lies outside the circumcircle of ABX. Similarly P lies outside the circumcircle of BCY. Suppose we are given any P lying outside these two circles. %0D%0AFor any choices of M on BC and R on AB%2C let Q be the point on DA with MPQ %3D %26pi%3B - AXB%2C and N the unique point on CD with NPR %3D %26pi%3B - BYC. We say a choice of M is suitable if both of M and Q lie in the interiors of the respective sides of the quadrilateral ABCD%3B the definition of suitability for R is similar. Now choose M%27 and R%27 as close to B as possible so that both choices are suitable. At most one of M%27%2C R%27 is not B%3B without loss of generality M%27 Then M%27PR%27 %3C%3D M%27PQ%27 - BPD %3C%3D %26pi%3B - CXD - XCD %3D CDA. On the other hand%2C choose M%27%27 and R%27%27 as far from B as possible so that both choices are suitable. We have either M%27%27 %3D C or Q%27%27 %3D A%2C and also R%27%27 %3D A or N%27%27 %3D C. This gives 4 possible values for M%27%27PR%27%27%3A CPA%2C M%27%27PQ%27%27%2C N%27%27PR%27%27 or M%27%27PQ%27%27 %2B N%27%27PR%27%27 - CPA%2C and in all of these cases M%27%27PR%27%27 %3E%3D CDA %28The last case is least obvious. We have M%27%27PQ%27%27 %2B N%27%27PR%27%27 - CPA %3E%3D %28%26pi%3B - BXD%29 %2B %28%26pi%3B - XBY%29 - %28%26pi%3B - XBY%29 %3D XDY %3D CDA%29. So we may choose M between M%27 and M%27%27 and R between R%27 and R%27%27 so that MPR %3D CDA%2C and then M%2C N%2C Q and R meet all of the conditions in the problem.%0D%0ATo summarise%2C a necessary and sufficient condition is that P should lie outside the circumcircles of ABX and BXY.%0D%0A%0D%0A%5Blink%3Awww.cut-the-knot.org%2FCurriculum%2FGeometry%2FFourFeetOnBisectors.shtml%7CProjections on internal and external angle bisectors%5D%3A%0D%0AP%2C N and M are collinear since they lie on the Simson line of A with respect to BII%5Bsub%5DC%5B%2Fsub%5D.%0D%0A%0D%0AThankyou%0D%0A%0D%0Asfwc%0D%0A%3C%3E%3C
1|1|0|||||RE%3A A few suggestions.|alexb||13:38:44|12/18/2006|I hope I meant %22related%22 not %22equivalent%22. I do not think that I tried to show the equivalence%2C but have to have a closer look.%0D%0A%0D%0AThank you on the whole%3B but where you%27ve been all this time%3F There is so much to do now.%0D%0A%0D%0AAlex
2|2|1|||1||RE%3A A few suggestions.|sfwc||14:54:15|12/18/2006|%3EThank you on the whole%3B but where you%27ve been all this time%3F %0D%0AYou are right%2C of course%2C that it would have been more convenient if I had commented on those pages as they were added to the website. However%2C during the last term I was rather busy%2C and certainly had a lot of other mathematics to do%3A I just about had time to keep an eye on the forum. I am now on holiday%2C and have used some of my free time to think about what has been added to the site since I last looked.%0D%0A %0D%0A%3EThere is so much to do now. %0D%0AThat%27s fine%3B I won%27t mind if you take a while to look at the suggestions as I know that you are busy. The main joy is in the mathematics itself%3B sharing it%2C though still a joy%2C is secondary.%0D%0A%0D%0AThankyou%0D%0A%0D%0Asfwc%0D%0A%3C%3E%3C
3|3|2|||||RE%3A A few suggestions.|alexb||14:59:09|12/18/2006|1.%0D%0A%3EHowever%2C during the last term I was %0D%0A%3Erather busy%0D%0A2.%0D%0A%3EThe main joy is in %0D%0A%3Ethe mathematics itself%3B sharing it%2C though still %0D%0A%3Ea joy%2C is secondary. %0D%0A%0D%0AAll the more I appreciate your making an effort.%0D%0A%0D%0AThank you%2C%0D%0AAlex
4|1|0|||1||RE%3A A few suggestions.|sfwc||18:48:44|12/18/2006|%3ETo summarise%2C a necessary and sufficient condition is that P %0D%0A%3Eshould lie outside the circumcircles of ABX and BXY. %0D%0AThat was a mistake%3A I mean ABX and BCY.%0D%0A%0D%0AThankyou%0D%0A%0D%0Asfwc%0D%0A%3C%3E%3C
5|2|4|||||RE%3A A few suggestions.|alexb||18:50:07|12/18/2006|OK. Thank you.