First round – City competition

3.1. Let $D$ be the point on the side $\overline{BC}$ such that $\angle DAC=90^{\circ }$. Let us denote $\phi =\angle CDA$ and $x=|CD|$. Then $\cos \angle ACB=\sin \phi$. Then we have $|AC|=x\sin \phi$ and $|BD|=|AD|=x\cos \phi$. We also have $\angle BAD=\frac{\phi }{2}$ and $|AB|=2x\cos \phi \cos \frac{\phi }{2}$. Since $|BC|+|AC|=2|AB|$, we get $$1+\cos \phi +\sin \phi =4\cos \phi \cos \frac{\phi }{2}.$$ By squaring both sides of the equation we get $$1+{\cos }^2\phi +{\sin }^2\phi +2\cos \phi +2\sin \phi +2\sin \phi \cos \phi =16{\cos }^2\phi {\cos }^2\frac{\phi }{2},$$ and further on $$\begin{array}{rl}2(1+\cos \phi )(1+\sin \phi )& =8{\cos }^2\phi (1+\cos \phi ),\\ 1+\sin \phi & =4(1-{\sin }^2\phi ),\\ (4\sin \phi -3)(\sin \phi +1)& =0,\\ \sin \phi & =\frac{3}{4},\end{array}$$ where we used $\cos \phi \ne -1$ and $\sin \phi \ne -1$, which is valid because the angle $\phi$ is acute. Hence, $\cos \angle ACB=\frac{3}{4}$.