## Tangent Lines and Circles in Convex Quadrilateral

Let A_{0}, A_{1}, A_{2}, and A_{3} list in order the vertices of a convex quadrilateral Q. Treat all indices as integers modulo 4. Let L_{k} denote the line through A_{k} and A_{k+1}, and let C_{k} be the circle tangent to L_{k-1}, L_{k}, and L_{k+1} outside Q. Let M_{k} be the line through the points where C_{k} is tangent to L_{k-1} and to L_{k+1}. Let E_{k} be the intersection of M_{k} with M_{k+1}. Prove that E_{1}E_{3} bisects A_{1}A_{3}.

What if applet does not run? |

(The problem was proposed by Chu Cheng, Yan An Middle School, Shanghai, China. The solution is by Marius Stefan, Los Angeles, CA. American Mathematical Monthly, v 114 (December 2007), 926-927).

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For distinct points A, B,C, write d(A, BC) for the distance from point A to the line through B and C. Denote by T_{lk} the point where the line L_{k} touches the circle C_{l}. Let M be the midpoint of the line segment A_{1}A_{3}. Let α be half the angle between L_{1} and L_{3}, lines that enclose C_{0} and C_{2}. Let β be half the angle between L_{0} and L_{2}, the lines enclosing C_{1} and C_{3}.

What if applet does not run? |

With these definitions, we have

d(M, E_{3}E_{0}) | = (d(A_{1}, E_{3}E_{0}) + d(A_{3}, E_{3}E_{0})) / 2 | |

= (A_{1}T_{01} cos(α) + A_{3}T_{03} cos(α)) / 2 | ||

= (A_{1}T_{01} + A_{3}T_{03}) cos(α) / 2 | ||

= (A_{1}T_{00} + A_{3}A_{0} + A_{0}T_{00}) cos(α) / 2 | ||

= (A_{1}A_{0} + A_{3}A_{0}) cos(α) / 2 |

Similarly,

d(M, E_{2}E_{3}) | = (d(A_{3}, E_{2}E_{3}) + d(A_{2}, E_{2}E_{3})) / 2 | |

= (A_{3}T_{32} cos(β) + A_{1}T_{30} cos(β)) / 2 | ||

= (A_{3}T_{32} + A_{1}T_{30}) cos(β) / 2 | ||

= (A_{3}T_{33} + A_{1}A_{0} + A_{0}T_{33}) cos(β) / 2 | ||

= (A_{1}A_{0} + A_{3}A_{0}) cos(β) / 2 |

It follows that

d(M, E_{3}E_{0}) / d(M, E_{2}E_{3}) | = cos(α) / cos(β). |

The lines E_{3}E_{0} and E_{1}E_{2} are parallel because they are both perpendicular to the bisector of 2α. The lines E_{0}E_{1} and E_{2}E_{3} are also parallel, since they are perpendicular to the bisector of 2β. Hence E_{0}E_{1}E_{2}E_{3} is a parallelogram. The distance d_{0} between E_{3}E_{0} and E_{1}E_{2} is

d_{0} | = T_{01}T_{21} cos(α) = T_{03}T_{23} cos(α) | |

= (T_{01}T_{21} + T_{03}T_{23}) cos(α)) / 2 | ||

= [(A_{1}T_{00} + A_{1}A_{2} + A_{2}T_{22}) + (A_{0}T_{00} + A_{0}A_{3} + A_{3}T_{22})] cos(α) / 2 | ||

= (A_{0}A_{1} + A_{1}A_{2} + A_{2}A_{3} + A_{3}A_{0}) cos(α) / 2. |

Similarly, the distance d_{1} between E_{0}E_{1} and E_{2}E_{3} is

d_{1} | = (A_{0}A_{1} + A_{1}A_{2} + A_{2}A_{3} + A_{3}A_{0}) cos(β) / 2. |

Therefore,

d_{0} / d_{1} | = cos(α) / cos(β) | = d(M, E_{3}E_{0}) / d(M, E_{2}E_{3}), |

which implies that M ∈ E_{1}E_{3}.

The configuration has additional properties:

The centers of the four circles C

_{0}, ..., C_{3}are concyclic.If s is the semiperimeter of the quadrilateral A

_{0}A_{1}A_{2}A_{3}thenT _{01}T_{21}= T_{03}T_{23}= T_{10}T_{30}= T_{12}T_{32}= s.

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Copyright © 1996-2018 Alexander Bogomolny