Problem 4

Let

ABCD

be a convex quadrilateral with

AB=AD+BC

. Suppose that an interior point

P

has distance

h

from the line

CD

and satisfies

AP=h+AD

and

BP=h+BC

. Prove that

\frac1{\sqrt h}\ge\frac1{\sqrt{AD}}+\frac1{\sqrt{BC}}

Step 2 of 5: Bound the possible height

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In plain words

The extremal circle is the largest small circle that can fit between two tangent circles and a line.

\Gamma_P\text{ lies inside the curvilinear triangle bounded by }\Gamma_A,\Gamma_B,t

Detailed analysis

Let $t$ be the common external tangent of $\Gamma_A$ and $\Gamma_B$ on the same side as $C,D$ . Because $P$ is inside $ABCD$ and $\Gamma_P$ is tangent to $CD$ , the circle $\Gamma_P$ is confined to the curvilinear triangle bounded by arcs of $\Gamma_A,\Gamma_B$ and $t$ . Moving it outward shows its radius is largest when it touches $t$ as well.