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CTK Exchange
Bui Quang Tuan
Member since Jun-23-07
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Mar-04-08, 11:54 AM (EST) |
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"Four, Five Then Nine Point Circles"
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Dear All My Friends,
In "Concyclic Points in a Triangle" there are four concyclic points related altitude and angle bisector from vertex A. It is interesting that we can generalize this fact for any two lines passing A and in this generalization there are five concyclic points. Moreover, there is triplet of five concyclic points and from this triplet we can construct one little nine point circle. When one of two lines is altitude so this little nine point circle is really great nine point circle. Detail as following:P and X is any two points on sideline BC of triangle ABC
(Ob), (Oc) are circumcircles of ABP, ACP respectively
B', C' are intersection of line AX with circles (Ob), (Oc) respectively (other than A)
Q = intersection of ObB', OcC'
R = intersection of line through B' parallel with AC and line through C' parallel with AB
(R is on sideline BC) Result 1: five points B', C', P, Q, R are concyclic on one circle, say (A'). Line ObB' intersect circle (Ob) at B'' (other than B')
Line OcC' intersect circle (Oc) at C'' (other than C')
S = intersection of line through B'' parallel with AC and line through C'' parallel with AB
(S is on sideline BC) Result 2: five points B'', C'', P, Q, S are concyclic on one circle, say (A''). Two circles (A') and (A'') are of same kind. In fact B'', C'', A are collinear on one line. This line is perpendicular with AX and intersect sideline BC at Y. We can construct circle (A'') by the same way in (1) but with two points P, Y rather than P, X. Line AOb intersect circle (Oc) at Ca (other than A)
Line AOc intersect circle (Ob) at Ba (other than A)
T = intersection of lines CCa, BBa
U = intersection of line through Ba parallel with AC and line through Ca parallel with AB Result 3: five points A, T, U, Ba, Ca are concyclic on one circle, say (A*). A*, T, U are collinear or TU is diameter of circle (A*) and A* is on AP. Now we can have nine point circle:
Result 4: nine points: P, Q, Ob, Oc, A', A'', A*, Ba, Ca are concyclic on one circle (N). Independing of X, this circle (N) is nine point circle of ABC if AP is altitude of ABC. Best regards,
Bui Quang Tuan
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Bui Quang Tuan
Member since Jun-23-07
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Mar-06-08, 01:53 PM (EST) |
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2. "RE: Four, Five Then Nine Point Circles"
In response to message #1
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Dear Alex,
Thank you for compliment but I think you over praised: I only try to combine some little results to get some more interesting. Of course combined configuration is very complicated. My idea is to extract they into some more simple facts as following:
a. In fact, intersections of parallel lines are not interesting: I added them into configuration only to get 4, 5 and 9 numbers. So we can remove R, S, U from configuration. Please note that they are not on our little nine point circle (N).
b. The circle in result (3) can be extract in one fact because it is independent enough
c. Two circles in results (1) and (2) can be combine in one fact because they are same essential.
d. For less complication we can use two intersecting circles instead of triangle ABC. In fact, B and C can be removed from our configuration.
I try to extract results (1) and (2) as following. I use here all notations as in my previous message. We can change if we want to make one independent fact.
Two circles (Ob), (Oc) intersect each other at A, P. Two perpendicular lines passing A: one intersects (Ob), (Oc) at B', C' respectively; other intersects (Ob), (Oc) at B'', C'' respectively. Two diameter lines B'B'' and C'C'' intersect each other at Q.
Results: P, Q, B', C' are concyclic on one circle, say A'. Similarly P, Q, B'', C'' are concyclic on one circle, say A''. Three points Q, A', A'' are on circumcircle (N) of PObOc taken A'A'' as diameter.
So, in this simple configuration we get already six points on (N)I think we can extract result (3) to get three points on the same circumcircle (N) of PObOc and total we can get nine points circle on (N).
From these extractions we can understand more about essence of this configuration.
You also can to relate these two extractions with other results in your sit's. I try to do it but my Internet is very slow and I already design to change my Internet in this weekend. Best regards,
Bui Quang Tuan
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Bui Quang Tuan
Member since Jun-23-07
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Mar-07-08, 09:45 AM (EST) |
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3. "RE: Four, Five Then Nine Point Circles"
In response to message #1
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Dear Alex,
We can extract result (3) as following:
Two circles (Ob), (Oc) intersect each other at A and P. Any line passing P intersects (Ob) at B, (Oc) at C.
Line AOb intersect circle (Oc) at Ca (other than A)
Line AOc intersect circle (Ob) at Ba (other than A)
T = intersection of lines CCa, BBa
then four points A, T, Ba, Ca are concyclic on one circle, say (A*). Three points A*, Ba, Ca are on the circle (N), circumcircle of PObOc.The circumcircle PObOc is common for two extracted facts.
If Ab = reflection of A in Ob and Ac = reflection of A in Oc then circumcircle PObOc is nine point circle of triangle AAbAc
Best regards,
Bui Quang Tuan
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