Source: I.F.Sharygin contest 2009 - Correspondence round - Problem 11
Tags: geometry, circumcircle, trapezoid, number theory, least common multiple, geometry proposed
Toggle Solutions
Given quadrilateral $ ABCD$. The circumcircle of $ ABC$ is tangent to side $ CD$, and the circumcircle of $ ACD$ is tangent to side $ AB$. Prove that the length of diagonal $ AC$ is less than the distance between the midpoints of $ AB$ and $ CD$.
$ K, L$ are midpoints of $ AB, CD.$ From circumcircle tangencies, $ \angle ABC = \angle ACD,$ $ \angle CAB = \angle CDA$ $ \Longrightarrow$ $ \triangle ABC \sim \triangle DCA$ are oppositely similar $ \Longrightarrow$ $ \angle BCA = \angle CAD$ $ \Longrightarrow$ $ BC \parallel AD.$ Midline $ KL \parallel (BC \parallel AD)$ of trapezoid $ ABCD$ cuts its diagonal $ AC$ in half at $ M$ $ \Longrightarrow$ $ \triangle AKM \sim \triangle LCM$ are oppositely similar $ \Longrightarrow$ $ AKCL$ is cyclic with diagonal intersection $ M.$ Power of $ M$ to its circumcircle is $ - KM \cdot ML = - AM \cdot MC = - \frac {_1}{^4}AC^2$ and $ KL = KM + ML \ge 2 \sqrt {KM \cdot ML} = AC.$ Nothing prevents $ KL = AC$ when $ AKCL$ becomes a square.