13.440 Given two mutually perpendicular axes $O x$ and $O y$, and a point $A(a, a)$, where $a>0$. It is required to find the coordinates of a point $M$ on the axis $O x$ and a point $P$ on the axis $O y$, such that the triangle $A M P$ is equilateral.
Solution. Let $(o, y)$ and ( $x, o$ ) be the coordinates of points $P$ and $M$ respectively. Since $\triangle A M P$ is equilateral, then

$$
x^{2}+y^{2}=(a-x)^{2}+a^{2}=(a-y)^{2}+a^{2} \Rightarrow(a-x)^{2}=(a-y)^{2} \Leftrightarrow\left[\begin{array}{l}
x=y \\
2 a=x+y
\end{array}\right.
$$

If $x=y$, then $2 x^{2}=(a-x)^{2}+a^{2} \Rightarrow x^{2}+2 a x-2 a^{2}=0 \Rightarrow x=y=-a+\sqrt{3 a^{2}}$

If $y=2 a-x$, then $x^{2}+(2 a-x)^{2}=(a-x)^{2}+a^{2} \Rightarrow$

$\Rightarrow x^{2}-2 a x+2 a^{2}=0 \Rightarrow(x-a)^{2}+a^{2}=0$, which is impossible.

Answer: $M(a(\sqrt{3}-1) ; 0), P(0 ; a(\sqrt{3}-1))$.