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| Circumcircle Fixed Point (Posted on 2010-12-17) |
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Let B and C be points on two given non-collinear rays from the same point A
such that |AB| + |AC| is constant.
Prove that there exists a point D, distinct from A, such that the circumcircles
of triangles ABC pass through D for all choices of B and C.
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Submitted by Bractals
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Rating: 4.0000 (1 votes)
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Solution:
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Let 2k be the constant and let B' and C' be points on the rays
such that |AB'| = |AC'| = k. Construct the circumcircle of ΔAB'C'
and let D be the intersection, distinct from A, of this circumcircle
and the bisector of ∠B'AC'.
Let B and C be points on rays AB' and AC' respectively such that
|AB| + |AC| = 2k. WOLOG assume |AC| < k < |AB|. We want to
prove that the circumcircle of ΔABC passes through D.
Clearly, AB'D and AC'D are congruent right triangles. Therefore,
|DB'| = |DC'|.
|B'B| = |AB| - |AB'| = |AC'| - |AC| = |CC'|.
Therefore, BB'D and CC'D are congruent right triangles. Therefore,
∠BDB' = ∠CDC'.
∠BDC = ∠BDB' + ∠B'DC = ∠B'DC + ∠CDC' = ∠B'DC'.
Since angles B'AC' and B'DC' are supplementary, angles BAC and
BDC are supplementary.
Hence, B, A, C, and D are concylic. Therefore, the circumcircle
of ΔABC passes through D.
Note: See Harry's post for an alternate solution.
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Subject |
Author |
Date |
 | Solution | Harry | 2010-12-19 23:17:42 |
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