### 1986 IMO SL #15

Let $ABCD$ be a convex quadrilateral whose vertices do not lie on a circle. Let $A'B'C'D'$ be a quadrilateral such that $A',B', C',D'$ are the centers of the circumcircles of triangles $BCD,ACD,ABD$, and $ABC$. We write $T (ABCD) = A'B'C'D'$. Let us define $A''B''C''D'' = T (A'B'C'D') = T (T (ABCD)).$

(a) Prove that $ABCD$ and $A''B''C''D''$ are similar.

(b) The ratio of similitude depends on the size of the angles of $ABCD$. Determine this ratio.

a) Notice $C'D'$ is the perpendicular bisector of $AB$ and similarly $A''B''$ is the perpendicular bisector of $C'D'$. Hence, $AB\parallel A''B''$. Similarly $BC\parallel B''C'', CD\parallel C''D'',DA\parallel D''A'',$ $AC\parallel A''C'', BD\parallel B''D''$. Hence, we have $ABCD\sim A''B''C''D''$, as desired. $\square$

b) Denote $S$ the midpoint of $AC$. Then, we get $$B'S=AC\frac{\cos \angle D}{2 \sin \angle D}$$ $$D'S=AC\frac{\cos \angle B}{2 \sin \angle B}$$ $$B'D'=AC|\frac{\cos(\angle B + \angle D)}{2 \sin \angle B \sin \angle D}|.$$ These also hold if $\angle B, \angle D > 90^\circ$. Similarly, we get $$A''C''=B'D'|\frac{\sin(\angle A' + \angle C')}{2 \sin \angle A' \angle C'}|.$$

Plugging in $B'D'$, we get $$A''C''=AC \frac{\sin^2(\angle A + \angle C)}{2 \sin A \sin B \sin C \sin D},$$ so we have the fixed ratio shown above. $\square$

## Comments

## Post a Comment