Cherchez le quadrilatère cyclique II
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A Mathematical Droodle
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Copyright © 1996-2018 Alexander Bogomolny
The applet provides an illustration to a generalization of the following problem from an outstanding collection by T. Andreescu and R. Gelca:
Consider a semicircle of center O and diameter AB. A line intersects AB at M and the semicircle at C and D in such a way that
(This is one of the problems from the 1996 Balkan Mathematical Olympiad.)
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The generalization is twofold:
First, AB need not be a diameter of the given circle. Suffice it for the quadrilateral ABCD to be cyclic.
Second, the inequalities
BM < AM andCM < DM can be done away with, although the proof then may have to be split into several cases.
The outline of the proof is as follows:
First we show that quadrilaterals ABLK and CDKL are cyclic. By this means we introduce two circles meeting at K and L and thus having KL as their radical axis. Adjoining the given circle gives a set of three circles with AB, CD, and KL pairwise radical axes implying M is a radical center of the three circles. Thus L lies on KM.
Due to the first step, we need now prove that KO is perpendicular to KL (instead of KM.)
First step: ABLK and CDKL are cyclic quadrilaterals. Indeed,
| ∠AKB | = ∠AKO + ∠BKO |
| = ∠ADO + ∠BCO | |
| = (180° - ∠AOD)/2 + (180° - ∠BOC)/2 | |
| = 180° - ∠AOD/2 - ∠BOC/2 | |
| = 180° - arcAD/2 - arcBC/2 | |
| = 180° - ∠ABD - ∠BAC | |
| = 180° - ∠ABL - ∠BAL | |
| = ∠ALB, |
where (and also later on) the arcs are those of the given circle. This says that quadrilateral ABLK is cyclic. Using the cyclicity of ABLK we proceed to establish the cyclicity of CDKL:
| ∠DKL | = 360° - ∠AKD - ∠AKL |
| = 360° - ∠AOD - (180° - ∠ABL) | |
| = 180° - ∠AOD + ∠ABL | |
| = 180° - ∠AOD + ∠ABD | |
| = 180° - arcAD + arcAD/2 | |
| = 180° - arcAD/2 | |
| = 180° - ∠ACD | |
| = 180° - ∠LCD. |
Wherefrom we obtain the identity
∠DKL + ∠DCL = 180°,
so that quadrilateral CDKL is also cyclic. Thus L does lie on KM.
Second step: To prove the orthogonality of KM and KO note that
∠AKL + ∠ABL = 180°
implying that orthogonality of KL and KO is equivalent to
| (1) | ∠AKO + ∠ABL = 90°. |
But quadrilateral ADKO is cyclic:
| ∠AKO + ∠ABL | = ∠ADO + ∠ABD |
| = (180° - ∠AOD)/2 + ∠ABD | |
| = 90° - arcAD/2 + arcAD/2 | |
| = 90°, |
as required.
References
- T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.3.12 (p. 13)
Radical Axis and Radical Center
- How to Construct a Radical Axis
- A Property of the Line IO: A Proof From The Book
- Cherchez le quadrilatere cyclique II
- Circles On Cevians
- Circles And Parallels
- Circles through the Orthocenter
- Coaxal Circles Theorem
- Isosceles on the Sides of a Triangle
- Properties of the Circle of Similitude
- Six Concyclic Points
- Radical Axis and Center, an Application
- Radical axis of two circles
- Radical Axis of Circles Inscribed in a Circular Segment
- Radical Center
- Radical center of three circles
- Steiner's porism
- Stereographic Projection and Inversion
- Stereographic Projection and Radical Axes
- Tangent as a Radical Axis
- Two Circles on a Side of a Triangle
- Pinning Butterfly on Radical Axes
- Two Lines - Two Circles
- Two Triples of Concurrent Circles
- Circle Centers on Radical Axes
- Collinearity with the Orthocenter
- Six Circles with Concurrent Pairwise Radical Axes
- Six Concyclic Points on Sides of a Triangle
- Line Through a Center of Similarity
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
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