Is it possible to simplify the following relatively simple expression  (10*(5+sqrt(41)))/(sqrt(70+10*sqrt(41))*sqrt(130+10*sqrt(41)))  using 1-2 standard commands  simplify , combine, radnormal  and so on?   I was unable to do this in Maple 2018. Maybe newer versions of Maple will be able to handle this. I managed to simplify it in 3 steps:

expr:=(10*(5+sqrt(41)))/(sqrt(70+10*sqrt(41))*sqrt(130+10*sqrt(41)));
sqrt(simplify(expr^2));

If we solve the equation  x^x=1/sqrt(2)  in Maple 2018.2, then Maple returns only one solution  x=1/2 , although this equation has 2 solutions  x=1/2  or  x=1/4 . This can be seen if you plot graphs or , for example, solve by the  Student:-Calculus1:-Roots  command. In this case, the root  x=1/4   is returned only as a numerical approximation. I wonder if this bug has been fixed in the latest versions of Maple?

restart;
solve(x^x=1/sqrt(2));
Student:-Calculus1:-Roots(x^x=1/sqrt(2));

When we specify a set (a sequence of objects enclosed in curly braces), Maple removes duplicates, since the elements of the set must be unique, that is, they cannot be repeated. See below for 2 examples. With the first example  {a<=b  and  b>=a}, everything is in order, since they are one and the same. But Maple treats the same equality, written in two ways  {a=b, b=a} , as different objects. It seems to me that this is not very convenient:

restart;
{a<=b, b>=a}; # OK
{a=b, b=a}; # not OK
is((a=b)=(b=a)); # not OK

Can Maple prove this simple identity  binomial(2*n, n)/2 = binomial(2*n-1, n-1) ,where n is integer and positive. Doing it manually is very easy. My attempt was unsuccessful:

is(binomial(2*n, n)/2=binomial(2*n-1, n-1)) assuming n::posint;

                                                       FAIL

Here is a problem that I recently had to solve and I really liked it. I recommend it to all Maple fans:

"The contour of the Christmas tree on the occasion of the New Year 2022 is a triangle, which is divided into 7 small triangles, the bases of which are parallel. The areas of triangles and the star are measured in snowflakes, the values of which are presented on the balls (of course, the number of snowflakes must be integer). It is necessary to determine the areas of triangles, on the balls of which there are no numbers".

At first glance it seems that the area of the star is not related to the problem, but it is not. This will become clear only after the problem is solved.