Estonian Math Competitions

Answer: $3+\sqrt{2}$.

Denote $\angle EDB=\alpha$ (Fig. 32). The similarity of $ADB$ and $BEC$ yields $\angle DAB=\angle EBC$ and $\angle BDA=\angle CEB$; the second of which gives $\angle DEB=180^{\circ }-\angle CEB=180^{\circ }-\angle BDA=\angle EDB=\alpha$. Therefore $DB=BE$. Hence the similarity of $ADB$ and $BEC$ yields $\frac{BE}{AD}=\frac{EC}{DB}=\frac{EC}{BE}$; thus ${\left(\frac{BE}{AD}\right)}^2=\frac{BE}{AD}\cdot \frac{EC}{BE}=\frac{EC}{AD}=2$, which gives $\frac{BE}{AD}=\sqrt{2}$ or $BE=\sqrt{2}AD$.

The isosceles triangle $BDE$ gives $\angle DBE=180^{\circ }-2\alpha$. On the other hand, $$\begin{array}{rl}\angle DBE& =120^{\circ }-\angle ABD-\angle EBC=120^{\circ }-\angle ABD-\angle DAB\\ & =120^{\circ }-(\angle ABD+\angle DAB)=120^{\circ }-\alpha .\end{array}$$ The equation $180^{\circ }-2\alpha =120^{\circ }-\alpha$ yields $\alpha =60^{\circ }$. So the triangle $BDE$ is equilateral, which yields $DE=BE=\sqrt{2}AD$.

Therefore $\frac{AC}{AD}=\frac{AD+DE+EC}{AD}=\frac{AD+\sqrt{2}AD+2AD}{AD}=1+\sqrt{2}+2=3+\sqrt{2}$.



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Alternative solution.

The similarity of $ADB$ and $BEC$ yields $\angle DAB=\angle EBC$ and $\angle ABD=\angle BCE$. So the triangles $ADB$ and $BEC$ are also similar to $ABC$. Therefore $\angle ADB=\angle BEC=\angle ABC=120^{\circ }$. But this yields $\angle BDE=\angle DEB=180^{\circ }-120^{\circ }=60^{\circ }$, meaning that the triangle $BDE$ is equilateral.

The similarity of $ADB$ and $ABC$ yields $\frac{AD}{AB}=\frac{AB}{AC}$. The similarity of $BEC$ and $ABC$ yields $\frac{EC}{BC}=\frac{BC}{AC}$. Combining these with $2AD=EC$, we see that



Fig. 32

$$B{C}^2=EC\cdot AC=2AD\cdot AC=2A{B}^2.$$ So $BC=\sqrt{2}AB$, meaning that the scale factor between triangles $ADB$ and $BEC$ is $\sqrt{2}$. Thus $DE=BD=\sqrt{2}AD$.

$$\text{Therefore }\frac{AC}{AD}=\frac{AD+DE+EC}{AD}=\frac{AD+\sqrt{2}AD+2AD}{AD}=1+\sqrt{2}+2=3+\sqrt{2}.$$