Butterflies in Similar Co-axial Conics

Sidney Kung
May 14, 2012

Better butterfly theorem in similar conics

Let there be two co-axial similar conic sections \(C_1\) and \(C_2\). A line crosses them at \(P\), \(P'\) and \(Q\), \(Q'\), \(M\) being a point of \(PQ\) and \(P'Q'.\) Through \(M\), draw two lines \(AA'BB'\) and \(CC'DD'\) and connect \(AD',\) \(A'D,\) \(A'D',\) \(AD,\) \(BC',\) \(B'C,\) \(B'C',\) and \(BC.\) Let \(X,\) \(Z,\) \(U,\) \(V,\) \(Y,\) \(W,\) \(U',\) \(V'\) be the points of intersection of \(PP'Q'Q\) with the eight line segments, respectively. Then

(1) \(\frac{1}{MX} + \frac{1}{MZ} - \frac{1}{MP'} - \frac{1}{MP}= \frac{1}{MY} + \frac{1}{MW} - \frac{1}{MQ'} - \frac{1}{MQ}\).

(For convenience, we use the figure from the Better Butterfly ttheorem, so that in the diagram the two circles represent similar conics.)

For a proof, we apply the lemma from the Better Butterfly theorem page to triangles \(MA'D',\) \(MAD,\) \(MA'D,\) and \(MAD'\):

(2) \(\frac{\text{sin}(\alpha )}{MD'} + \frac{\text{sin}(\beta )}{MA'} = \frac{\text{sin}(\alpha +\beta )}{MU}\)

\(\frac{\text{sin}(\alpha )}{MD} + \frac{\text{sin}(\beta )}{MA} = \frac{\text{sin}(\alpha +\beta )}{MV}\)

\(\frac{\text{sin}(\alpha )}{MD} + \frac{\text{sin}(\beta )}{MA'} = \frac{\text{sin}(\alpha +\beta )}{MZ}\)

\(\frac{\text{sin}(\alpha )}{MD'} + \frac{\text{sin}(\beta )}{MA} = \frac{\text{sin}(\alpha +\beta )}{MX}\).

So

\( \begin{align} \frac{\text{sin}(\alpha +\beta )}{MU} + \frac{\text{sin}(\alpha +\beta )}{MV} &= \left(\frac{\text{sin}(\alpha )}{MD'} + \frac{\text{sin}(\beta )}{MA'}\right) + \left(\frac{\text{sin}(\alpha )}{MD} + \frac{\text{sin}(\beta )}{MA}\right) \\ &= \left(\frac{\text{sin}(\alpha )}{MD'} + \frac{\text{sin}(\beta )}{MA}\right) + \left(\frac{\text{sin}(\alpha )}{MD} + \frac{\text{sin}(\beta )}{MA'}\right) \\ &= \frac{\text{sin}(\alpha +\beta )}{MX} + \frac{\text{sin}(\alpha +\beta )}{MZ}. \end{align} \)

Hence,

(3) \(\frac{1}{MU} + \frac{1}{MV} = \frac{1}{MX} + \frac{1}{MZ}\).

Similarly,

(4) \(\frac{1}{MU'} + \frac{1}{MV'} = \frac{1}{MW} + \frac{1}{MY}\).

Now, consider the inner circle as a conic section to which we apply the result of E. J. Atzema:

(5) \(\frac{1}{MU} - \frac{1}{MP'} = \frac{1}{MU'} - \frac{1}{MQ'}\).

Likewise, for the outer circle, we have

(6) \(\frac{1}{MV} - \frac{1}{MP} = \frac{1}{MV'} - \frac{1}{MQ}\).

Adding (5) and (6) and taking into account (3) and( 4), we get (1).

Butterfly Theorem and Variants

  1. Butterfly theorem
  2. 2N-Wing Butterfly Theorem
  3. Better Butterfly Theorem
  4. The Lepidoptera of the Circles
  5. The Lepidoptera of the Quadrilateral
  6. The Lepidoptera of the Quadrilateral II
  7. Butterflies in Ellipse
  8. Butterflies in Hyperbola
  9. Butterflies in Quadrilaterals and Elsewhere
  10. Pinning Butterfly on Radical Axes
  11. Shearing Butterflies in Quadrilaterals
  12. The Plain Butterfly Theorem
  13. Two Butterflies Theorem
  14. Two Butterflies Theorem II
  15. Two Butterflies Theorem III
  16. Algebraic proof of the theorem of butterflies in quadrilaterals
  17. William Wallace's Proof of the Butterfly Theorem
  18. Butterfly theorem, a Projective Proof
  19. Areal Butterflies
  20. Butterflies in Similar Co-axial Conics
  21. Butterfly Trigonometry
  22. Butterfly in Kite
  23. Butterfly with Menelaus
  24. William Wallace's 1803 Statement of the Butterfly Theorem
  25. Butterfly in Inscriptible Quadrilateral

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