Butterflies in Similar Co-axial Conics
Sidney Kung
May 14, 2012
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
- Butterfly theorem
- 2N-Wing Butterfly Theorem
- Better Butterfly Theorem
- Butterflies in Ellipse
- Butterflies in Hyperbola
- Butterflies in Quadrilaterals and Elsewhere
- Pinning Butterfly on Radical Axes
- Shearing Butterflies in Quadrilaterals
- The Plain Butterfly Theorem
- Two Butterflies Theorem
- Two Butterflies Theorem II
- Two Butterflies Theorem III
- Algebraic proof of the theorem of butterflies in quadrilaterals
- William Wallace's Proof of the Butterfly Theorem
- Butterfly theorem, a Projective Proof
- Areal Butterflies
- Butterflies in Similar Co-axial Conics
- Butterfly Trigonometry
- Butterfly in Kite
- Butterfly with Menelaus
- William Wallace's 1803 Statement of the Butterfly Theorem
- Butterfly in Inscriptible Quadrilateral
- Camouflaged Butterfly
- General Butterfly in Pictures
- Butterfly via Ceva
- Butterfly via the Scale Factor of the Wings
- Butterfly by Midline
- Stathis Koutras' Butterfly
- The Lepidoptera of the Circles
- The Lepidoptera of the Quadrilateral
- The Lepidoptera of the Quadrilateral II
- The Lepidoptera of the Triangle
- Two Butterflies Theorem as a Porism of Cyclic Quadrilaterals
- Two Butterfly Theorems by Sidney Kung
- Butterfly in Complex Numbers
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