Iranian Mathematical Olympiad

Suppose ${\omega }_1$ is tangent to $AB$, $AD$ at $P_1$, $Q_1$ respectively and ${\omega }_2$ is tangent to $CD$, $BC$ at $P_2$, $Q_2$ respectively.

Lemma. Let a circle $\omega$ be tangent to the half-lines $BA$, $BC$ at $P$, $Q$ respectively. $\omega$ is tangent to ${\omega }_1$ if and only if $\sqrt{BP}=\sqrt{AB}\pm \sqrt{A{P}_1}$ for some choice of the sign.

Proof. $P{P}_1$ is the common external tangent of $\omega$ and ${\omega }_1$. So $\omega$, ${\omega }_1$ are tangent if and only if $P{P}_1=2\sqrt{r{r}_1}$, where $r$, $r_1$ are the radii of $\omega$, ${\omega }_1$ respectively. If we let $\angle ABC=2\alpha$, then $r=BP\cdot \tan \alpha$ and $r_1=AP\cdot \cot \alpha$. So $2\sqrt{r{r}_1}=2\sqrt{BP\cdot A{P}_1}$. So $\omega$, ${\omega }_1$ are tangent if and only if $AB-A{P}_1-BP=\pm 2\sqrt{BP\cdot A{P}_1}$ and the claim follows. $\square$

By assumption, there is a circle $\omega$ tangent to the lines $DA$, $DC$ at $Q$, $P$ respectively and externally tangent to ${\omega }_1$, ${\omega }_2$. It is easily seen that $\omega$ should be inside the angle $\angle ADC$. So, by the lemma we have: $$\begin{gathered} \sqrt{DQ}=\sqrt{AD}\pm \sqrt{A{Q}_1} \\ \sqrt{DP}=\sqrt{CD}\pm \sqrt{C{P}_2} \end{gathered}$$ for some choice of the signs. We have $DP=DQ$, $A{Q}_1=A{P}_1$, $C{P}_2=C{Q}_2$, $AD=BC$ and $CD=AB$. So by the above equations we get $$\sqrt{BC}\pm \sqrt{A{P}_1}=\sqrt{AB}\pm \sqrt{C{Q}_2}\Rightarrow \sqrt{BC}\pm \sqrt{C{Q}_2}=\sqrt{AB}\pm \sqrt{A{P}_1}.$$





Let ${\omega }^{\prime }$ be a circle tangent to the half-lines $BA$, $BC$ at $P^{\prime }$, $Q^{\prime }$ respectively, such that both sides of this equation are equal to $\sqrt{B{P}^{\prime }}$ (note that the value is positive). So ${\omega }^{\prime }$ is tangent to ${\omega }_1$, ${\omega }_2$ by the lemma and the assertion is proved. $\square$