# Butterflies in a Pencil of Conics

### Problem

Let $O$ be the midpoint of a given chord $RS$ of a proper (nondegenerate) conic $c_1,$ let two other chords $TU$ and $VW$ be drawn through $O,$ and let a conic $c_2$ through $T,$ $U,$ $V,$ $W$ cut the given chord in $E$ and $F.$ Then $O$ is the midpoint of $EF.$

### Solution 1

Let $MN,$ the polar of $O$ for coinc $c_1,$ cut $TU$ in $M$ and $VW$ in $N.$ Then, since $(VWON)+(TUOM)=-1,$ $MN$ is also the polar of $O$ for conic $c_2.$ Moreover, $RS$ is parallel to $MN$ since $(RSO\infty )=-1.$ Therefore, $(EF)\infty )=-1$ and $O$ is the midpoint of $EF.$

(This is practically the same argument that was used in one of the proofs of the original Butterfly theorem.)

### Solution 2

This solution is a straightforward adaptation of Hubert Shutrick's argument that was originally meant for the standard Butterfly theorem:

The pencil of conics that pass through $T,$ $U,$ $V,$ $W$ in the diagram include $c_1$ and the pair of lines $TU,$ $VW,$ whose intersection $O$ is a double point of the involution on $RS$ and the other double point must be the point at infinity since $M$ is the midpoint of $RS.$ Hence, it is also the midpoint of $EF$ because $c_2$ is another conic in the pencil.

### References

1. H. Eves, A Survey of Geometry, Allyn and Bacon, 1972
2. D. Pedoe, Geometry: A Comprehensive Course, Dover, 1988

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