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CTK Exchange
alexb
Charter Member
1897 posts |
Aug-04-06, 08:43 AM (EST) |
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"Property of the line through incenter and circumcenter"
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R. Honsberger in From Erdos to Kiev, pp 199-201 treats a problem from Crux Mathematicorum, 1987, 160: "As usual, let I and O be the incenter and circumcenter, respectively, of ÄABC. Suppose angle C is 30°, and that the side AB is laid off along each of the other two sides to give points D and E so that EA = AB = BD. Prove that the segment DE is both equal and perpendicular to IO." Now, I think that the problem is a mixture of two:
- DE is perpendicular to IO,
- DE = IO.
The first one is independent of the magnitude of angle C: DE is perpendicular to IO anyway. The second one is the consequence of the angle requirement. As a preliminary, the construction of points D and E can be repeated for the other two sides AC and BC, giving three lines. Perhaps, curiously, the three are parallel, see https://www.cut-the-knot.org/Curriculum/Geometry/ThreeParallelsInTriangle.shtml
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me
guest
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Oct-13-06, 11:12 AM (EST) |
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2. "RE: Property of the line through incenter and circumcenter"
In response to message #0
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If K, L, M in BC, CA, AB are feet of the external bisectors of the angles A, B, C, then KB/KC*LC/LA*MA/MB = b/c*c/a*a/b = 1 and K, L, M are collinear by Menelaus theorem. AL = bc/(c - a), AM = bc/(b - a) and AM/AL = (c - a)/(b - a) = AD/AE, triangles ADE, AML are centrally similar and DE parallel to LM, same as KLM. If I_a, I_b, I_c are excenters of the triangle ABC opposite to A, B, C, then I_aA, I_bB, I_cC are altitudes of the excentral triangle I_aI_bI_c, I its orthocenter, O its 9-point circle center, and the reflection O' of I in O its circumcenter, O'OI its Euler line. Quadrilaterals I_bI_cBC, I_cI_aCA, I_aI_bAB are all cyclic because of the right angles, KB*KC = KI_b*KI_c, LC*LA = LI_c*LI_a, MA*MB = MI_a*MI_b. Therefore KLM is radical axis of the circumcircles (O), (O') of the triangles ABC, I_aI_bI_c, perpendicular to the center line O'O, same as O'OI. In conclusion, DE is parallel to KLM, which is perpendicular to O'OI. |
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