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Berkeley Math Circle Monthly Contest 2

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a square. Consider four circles $k_{1}, k_{2}, k_{3}, k_{4}$ which pass respectively through $A$ and $B$ , $B$ and $C$ , $C$ and $D$ , $D$ and $A$ , and whose centers are outside the square. Circles $k_{4}$ and $k_{1}$ , $k_{1}$ and $k_{2}$ , $k_{2}$ and $k_{3}$ , $k_{3}$ and $k_{4}$ intersect respectively at $L$ , $M$ , $N$ , $P$ inside the square. Prove that quadrilateral $LMNP$ can be inscribed in a circle.

Solution

Solution:

We use a standard theorem that states that a convex quadrilateral $WXYZ$ can be inscribed in a circle if and only if $\angle W + \angle Y = 180$ . We have
$\begin{aligned} \angle PLM + \angle MNP & = (360 - \angle MLA - \angle ALP) + (360 - \angle PNC - \angle CNM) \\ & = (180 - \angle MLA) + (180 - \angle ALP) + (180 - \angle PNC) + (180 - \angle CNM) \\ & = \angle ABM + \angle PDA + \angle CDP + \angle MBC \\ & = (\angle ABM + \angle MBC) + (\angle PDA + \angle CDP) \\ & = 90 + 90 = 180, \end{aligned}$
as desired.

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