In a convex quadrilateral, two sides are equal to 1, while the other sides and both diagonals are no more than 1. What is the maximum value that the perimeter of the quadrilateral can take?

#
Answer: $2+4 \sin 15^{\circ}$. Note that opposite sides cannot be equal to 1, because the sum of the diagonals of a quadrilateral is greater than the sum of the opposite sides, so the adjacent sides are equal to 1. Let the quadrilateral be $A B C D$, and let $A B=B C=1$.

If $B D<1$, then we can increase $B D$ by moving point $D$ further away until one of the conditions is violated. Which condition will this be? This condition is that $B D \leq 1$, because if $A D=1$ or $C D=1$ occurs earlier, the sum of the diagonals will be less than the sum of the opposite sides.

Now let $A B=B C=B D=1, \angle A B D=\alpha, \angle C B D=\beta$. Then $A D=2 \sin \frac{\alpha}{2}, C D=2 \sin \frac{\beta}{2}$,

$$
A D+C D=2 \sin \frac{\alpha}{2}+2 \sin \frac{\beta}{2}=4 \sin \frac{\alpha+\beta}{4} \cos \frac{\alpha-\beta}{4}
$$

Note that $\frac{\alpha+\beta}{4} \leq 15^{\circ}$, since $\alpha+\beta \leq 60^{\circ}$ (because $A C \leq 1$); $\cos \frac{\alpha-\beta}{4} \leq 1$, so $A D+C D \leq 4 \sin 15^{\circ}$.

Equality is achieved when $\alpha=\beta=\frac{\pi}{6}$, and the perimeter of $A B C D$ in this case is $2+4 \sin 15^{\circ}$.

Submit a comment