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).

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