Rioplatense Mathematical Olympiad

Name $\angle BAD=\alpha$, $\angle BCD=2\alpha$, $\angle ADC=\beta$, $\angle ABC=2\beta$. The interior angles of $ABCD$ add up to $3\alpha +3\beta =360^{\circ }$. Hence, $\alpha +\beta =120^{\circ }$. As $\alpha \ne \beta$, we may assume without loss of generality that $\beta <60^{\circ }<\alpha$.



Let $X=AB\cap CD$. Notice that $\angle AXD=60^{\circ }$. Let $E$ be the point on the same side as $A$ with respect to line $CD$ such that $CDE$ is equilateral. We have $\angle ECD=60^{\circ }=\angle AXD$ and $CE=CD=AB$. Hence, $EC\parallel AB$ and $ABCE$ is a parallelogram. This implies $\angle BAE=\angle BCE=2\alpha -60^{\circ }$ and $\angle EAD=(2\alpha -60^{\circ })-\alpha =\alpha -60^{\circ }$. On the other hand, $\angle EDA=60^{\circ }-\beta$. Since $60^{\circ }-\beta =\alpha -60^{\circ }>0$, triangle $AED$ is isosceles with $AE=ED$. Hence, $ABC$ and $BCD$ are also isosceles. In particular $\angle BAC=\frac{180^{\circ }-\angle ABC}{2}=90^{\circ }-\beta$, which implies $\angle CAD=\alpha -(90^{\circ }-\beta )=30^{\circ }$; and $\angle BDC=\frac{180^{\circ }-\angle BCD}{2}=90^{\circ }-\alpha$, which implies $\angle BDA=\beta -(90^{\circ }-\alpha )=30^{\circ }$. Therefore, the acute angle formed by $AC$ and $BD$ is $30^{\circ }+30^{\circ }=60^{\circ }$.