IMO 1996 SL G8

Origin: RUS | Category: Geometry

Problem

Let $\triangle ABCD$ be a convex quadrilateral, and let $R_A$, $R_B$, $R_C$, and $R_D$ denote the circumradii of the triangles $\triangle DAB$, $\triangle ABC$, $\triangle BCD$, and $\triangle CDA$ respectively. Prove that

$$R_A+R_C>R_B+R_D$$

if and only if

$$\angle A+\angle C>\angle B+\angle D.$$

Solution

Let the diagonals $AC$ and $BD$ meet at $X$. Either $\angle AXB$ or $\angle AXD$ is greater than or equal to $90^\circ$, so we assume without loss of generality that

$$\angle AXB\geq 90^\circ.$$

Let

$$\alpha=\angle CAB,\qquad \beta=\angle ABD,\qquad \alpha'=\angle BDC,\qquad \beta'=\angle DCA.$$

These angles are all acute and satisfy

$$\alpha+\beta=\alpha'+\beta'.$$

Furthermore,

$$R_A=\frac{AD}{2\sin\beta}, \qquad R_B=\frac{BC}{2\sin\alpha}, \qquad R_C=\frac{BC}{2\sin\alpha'}, \qquad R_D=\frac{AD}{2\sin\beta'}.$$

Let

$$\angle B+\angle D=180^\circ.$$

Then $A$, $B$, $C$, $D$ are concyclic, and trivially

$$R_A+R_C=R_B+R_D.$$

Let

$$\angle B+\angle D>180^\circ.$$

Then $D$ lies within the circumcircle of $\triangle ABC$, which implies that

$$\beta>\beta'.$$

Similarly,

$$\alpha<\alpha',$$

so we obtain

$$R_A<R_D \qquad\text{and}\qquad R_C<R_B.$$

Thus,

$$R_A+R_C<R_B+R_D.$$

Let

$$\angle B+\angle D<180^\circ.$$

As in the previous case, we deduce that

$$R_A>R_D \qquad\text{and}\qquad R_C>R_B,$$

so

$$R_A+R_C>R_B+R_D.$$