The Ramanujan Problems I originally thought of asking this question at the Mathematics Stackexchange, but then I decided that I'd have a better chance of a good discussion here.
In the Wikipedia page on Ramanujan (current revision), there is a link to a collection of problems posed by him. The page has a collection of about sixty problems which have appeared in the Journal of the Indian Mathematical Society.
While I was browsing through them at random, I came across this, which I recognized as the Brocard-Ramanujan problem. So,

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*Are all these problems solved? Or are there more unsolved problems among them?

*Why were the problems posed exactly? Was there some sort of contest in the journal back then?


Here is one problem from the list. (I just cannot resist posting it here!)

Let $AB$ be a diameter and $BC$ be a chord of a circle $ABC$. Bisect the minor arc $BC$ at $M$; and draw a chord $BN$ equal to half the length of the chord $BC$. Join $AM$. Describe two circles with $A$ and $B$ as centers and $AM$ and $BN$ as radii cutting each other at $S$ and $S'$ and cutting the given circle again at the points $M'$ and $N'$ respectively. Join $AN$ and $BM$ intersecting at $R$ and also join $AN'$ and $BM'$ intersecting at $R'$. Through $B$, draw a tangent to the given circle meeting $AM$ and $AM'$ produced at $Q$ and $Q'$ respectively. Produce $AN$ and $M'B$ to meet at $P$ and also produce $AN'$ and $MB$ to meet at $P'$. Show that the eight points $PQRSS'R'Q'P'$ are cyclic and that the circle passing through these points is orthogonal to the given circle $ABC$.

How does one even begin to prove this? I've tried to construct the entire thing using compass and straightedge, but the resulting mess only added to my confusion. Here is the link to the original problem.
Thanks a lot in advance!
EDIT: This problem is mentioned in the link provided in Andrey Rekalo's answer below in pp. 34-35. The solution is given in the second link in pp. 244-246.
 A: An unelegant solution (to the described problem) which works in principle: Transform everything in algebraic equations,
eg. by setting $A=(-1,0),B=(1,0),C=(x_C,y_C),M=(x_M,y_M),\dots$,
$x_C^2+y_C-1=0,x_M=1/2(1+x_C),\dots$ and compute a Groebner basis. Doing a few numerical examples proves then that there is at least a large enough set of real solutions.
This is of course ugly but has the advantage that you can give the heavy work to a machine.
A: You may want to see this paper as well by Berndt:

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*https://faculty.math.illinois.edu/~berndt/articles/galway.pdf
A: There is a survey article by Berndt, Choi, and Kang devoted to the set of 58 Ramanujan's problems. They indicate that the questions had originally appeared in the problems section of the Journal and apparently the editors published readers' solutions  in subsequent issues.
Concerning your question 1, let me just quote from the Introduction to the survey:

Several of the problems are elementary and can be attacked with a background of only high school mathematics. For others, significant amounts of hard analysis are necessary to effect solutions, and a few problems have not been completely solved.

An elementary solution to the specific geometric problem you've mentioned can be found in  Ramanujan's Notebooks, Part III by Berndt (Springer, 1991, pp. 244-246). The problem  stems from Ramanujan's work on modular equations of degree 3...
A: Concerning the planimetry problem: Having a suitable background, it is not hard to produce many puzzles like this. For example, here is one more pair of points on that mysterious circle: intersect a circle centered at $B$ with radius $BM$ and a circle centered at $A$ with radius $AB$, then the two intersection points are also on that 8-points circle.
This circle is a circle of Apollonius with foci at $A$ and $B$. More precisely, it is the locus of points $X$ such that $BX:AX=\sin\alpha$ where $\alpha=\angle BAM=\frac12\angle BAC$. The fact that this is a circle orthogonal to the original one is a general property of circles of Apollonius, and verifying that the ratio of distances equals $\sin\alpha$ for each point is straightforward.
