
Problem 8.2. Given a square $A B C D$ of side length 1. Point $M \in B C$ and point $N \in C D$ are such that the perimeter of $\triangle M C N$ is 2 .

a) Find $\Varangle M A N$;

b) If $P$ is the foot of the perpendicular from $A$ to $M N$, find the locus of the point $P$.


Solution: a) Let $K$ be a point on the extension of $C B$ such that $B K=D N$. Then $M K=M B+B K=M B+D N=1-C M+$ $1-C N=2-(C M+C N)=M N$ (using that $C M+C N+M N=$ 2). Since $\triangle A B K \cong \triangle A D N$ we have that $A K=A N$. Therefore $\triangle A M N \cong \triangle A M K$ and it follows that $\Varangle M A N=\Varangle M A K$, i.e. $\Varangle M A N=\frac{1}{2} \Varangle K A N$. Furthermore $\Varangle K A B=\Varangle N A D$, which implies that $\Varangle K A N=\Varangle B A D=90^{\circ}$. Therefore $\Varangle M A N=45^{\circ}$.

b) It follows from $\triangle A M N \cong \triangle A M K$ that $\Varangle A M N=\Varangle A M K$. Thus, $\triangle A P M \cong \triangle A B M$ and $A P=A B=1$. Therefore $P$ lies on a circle of center $A$ and radius 1. Finally, the locus of $P$ is an arc of a circle with center $A$ and radius 1 excluding points $B$ and $D$.
