### 1996 IMO SL #G7

Let $ABCD$ be a convex quadrilateral, and let $R_A, R_B, R_C, R_D$ denote the circumradii of the triangles $DAB, ABC, BCD, CDA,$ respectively. Prove that $R_A + R_C > R_B + R_D$ if and only if $\angle A + \angle C > \angle B + \angle D.$

Let $AC \in BD = X$. Note that either $\angle AXB$ or $\angle AXD$ is greater than $90^\circ$, so WLOG assume $\angle AXB \geq 90^\circ$. Let $\alpha, \beta, \alpha', \beta'$ be $\angle CAB, ABD, BDC, DCA$. Note that these are all acute and that $\alpha+\beta=\alpha'+\beta'$. Particularly, $$R_A=\frac{AD}{2 \sin \beta}, R_B=\frac{BC}{2 \sin \alpha}, R_C=\frac{BC}{2 \sin \alpha'}, R_D=\frac{AD}{2 \sin \beta'}.$$

Now, if $\angle B + \angle D =180^\circ$, then $ABCD$ is cyclic and $R_A+R_C=R_B+R_D$.

Next, if $∠B +∠D > 180^\circ$, then $D$ lies within $(ABC)$, which implies $\beta > \beta'$ . Similarly $\alpha <\alpha'$ , so we obtain $R_A < R_D$ and $R_C < R_B$. Hence, $R_A + R_C < R_B + R_D$.

Lastly, if $\angle B + \angle D < 180^\circ$, similar to the second case, note $R_A>R_D$ and $R_C>R_B$. Then, we have $R_A + R_C > R_B + R_D$.

Hence we have shown the desired. $\square$