66th Belarusian Mathematical Olympiad

Answer: 2.

Show that any two intersecting diagonals of the pentagon cannot be good at the same time. Suppose, contrary to our claim, that there are two good intersecting diagonals. Without loss of generality, we assume that $AD$ and $BE$ are good diagonals of the pentagon $ABCDE$ (see Fig. 1). Then $BCDE$ and $ACDE$ are circumscribed quadrilaterals, therefore the sums of their opposite sides are equal. Hence, $$AC+DE=AE+CD\text{and}BE+CD=BC+DE,$$ which gives $$AC+BE=AE+BC.(1)$$ Let $M$ be the intersection point of $AC$ and $BE$. Then, by the triangle inequality, we have $$\begin{gathered} AC+BE=(AM+MC)+(BM+ME)= \\ =(AM+ME)+(BM+MC)>AE+BC, \end{gathered}$$ contrary to (1). Similarly, the crossing diagonals $AC$ and $BD$ of the pentagon $ABCDE$ cannot be good at the same time. Since there are no more than two noncrossing diagonals in the pentagon, the number of good diagonals in the pentagon is less than or equal to 2.

Рис. 1 Рис. 2

Show that there exist a pentagon with two good diagonals. We mark five points $A,B,C,D,E$ in the plain such that $AB=AC=AD=AE$ and $\angle BAC=\angle CAD=\angle DAE=\alpha <60^{\circ }$ (see Fig. 2). Then it is evident that the triangles $BAC,CAD,DAE$ are equal. Therefore, $BC=CD=DE$ and $\angle ABC=\angle ACB=\angle ACD=\angle ADC=\angle ADE=\angle AED=(180^{\circ }-\alpha )/2=\beta <90^{\circ }$. The constructed pentagon is convex since, by construction, $\angle A=3\alpha <180^{\circ }$, $\angle B=\angle E=\beta <90^{\circ }$, $\angle C=\angle D=2\beta <180^{\circ }$. Two diagonals $AC$ and $AD$ are good since the quadrilaterals $ACDE$ and $ABCD$ are circumscribed quadrilaterals because $AC+DE=AE+CD$ and $AB+CD=AD+BC$.