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CTK Exchange
Bui Quang Tuan
Member since Jun-23-07
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Feb-02-08, 08:27 AM (EST) |
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"Infinite Of Congruent Circles From Translations"
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Dear All My Friends,
Given triangle ABC and two different points P, Q as a vector v = PQ.
Oa = translation of A by vetor v. Similarly define Ob, Oc.
(Oa) = circle centered at Oa passing A. Other than A this circle intersects AB at Bc, AC at Cb.
(Ob) = circle centered at Ob passing B. Other than B this circle intersects BC at Ca, BA at Ac.
(Oc) = circle centered at Oc passing C. Other than C this circle intersects CA at Ab, CB at Ba.
(Of course three circles (Oa), (Ob), (Oc) are congruent and their radii = |v|)
Three lines BcCb, CaAc, AbBa bound one triangle A'B'C'
Results: circumcircle of A'B'C' is congruent with circles (Oa), (Ob), (Oc)
Proof:
If (Q) is a circle centered at Q passing P then (Oa), (Ob), (Oc) are translations of (Q) by vectors PA, PB, PC respectively.
A line through P parallel with BC interesects (Q) at X (other than P). Similarly define Y, Z.
Easy to show that ABcCa, BCaAc, CAbBa are translations of PZY, PXZ, PYX by vectors PA, PB, PC respectively. (1)
Therefore A'B'C' is similar with ABC
La = line from A parallel with BcCb
Lb = line from B parallel with CaAc
Lc = line from C parallel with BcCb
Because ABC and A'B'C' are similar so these three lines are concurrent at D on circumcircle of ABC.
Easy to show that A'CaBa is translation of DBC by vector PX. Similarly with B', C'. So at the end: DA'B'C' is translation of PXYZ by vector PD or A'B'C' is translation XYZ by vector PD. From this circumcircle of A'B'C', say (Od) centered at Od passing D, is congruent with (Oa), (Ob), (Oc). Moreover OaObOcOd is translation of ABCD by v and therefore OaObOcOd are inscribed in circle (O'), translation of circumcircle (O) of ABC.
Now we can see three other triangles DAB, DBC, DCA each with "three translation by v congruent circles" at vertices. We can apply our result for these triangle and each can generated "fourth congruent circle" and "fourth D point". Repeate these step, we can generate infinite of congruent circles with center on circle (O').
We can apply this result for "three congruent circles by reflection" (orthocenter and angle bisector cases).
Best regards,
Bui Quang Tuan
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