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[Physicsguy51]

According to the condition of the problem, the ball rolls without slipping, therefore, the speeds of those points of the ball that at a given moment of time touch the metal at AB (Fig.) are equal to zero. Considering the ball to be an absolutely rigid body (that is, the distance between any two points of the ball is unchanged), we conclude that at a given moment in time all points of the ball lying on the segment AB are motionless. And this means that at each moment of time the movement of the ball is a rotation about the axis AB. (It is clear that points A and B - the points where the ball touches the metal - are moving with the speed .)

The instantaneous velocity of any point of the ball is , where is the angular velocity of rotation, is the distance from the point to the axis AB. The speed of the center of the ball (point O in the figure) is equal to ; the distance from point O to axis AB is . Hence,


It is clear that the points of the ball most distant from the axis AB have the maximum speed. From geometric considerations it is clear that at any moment there is only one point that is maximally distant from the axis - in the figure this is point Q. The distance from point Q to the axis of rotation is , and speed of Q is:

Sorry, but I disagree with this solution. In the diagrams attached by @Pigkappa, which coincides with that of AAPT in this document it is not entirely clear where points and are. Are they respectively the lowest and highest points of the sphere, or the point nearest where the plates meet and the point furthest from it?
The right-hand diagram appears to show the plates end-on, so is looking up the slope. That means is the correct interpretation. This makes the answer '' correct. The text is wrong to refer to them as .
Under interpretation , the speed of is multiplied by the ratio of distances from to and to , namely, , so it is .
Under interpretation , is not the fastest point, and its speed is indeterminate because we do not know the angle of the slope. As to how the AAPT answer came to be wrong, it was a rehash of an old question and they forgot to change everything consistently. It happens, and to note, this is not the first time this has occurred in the USAPhO.
It sometimes helps to consider an extreme case. Suppose the plates are vertical, which they could be, as far as we know. The point at the top of the sphere is from , so moves at speed . Meanwhile, the point furthest from the plate join still moves at . I have tried to verify your solution, graphically, using all kinds of software and graphing calculator, but it is in no way possible to obtain an equilateral triangle , with all three angles of 60 degrees, while keeping the two metal planes orthogonal to each other (as explicitly required by the text). It is not really possible. There is a huge discrepancy between the text and the proposed graphical scheme.

As I already mentioned this is the official solution for the original problem I had solved last year. As I said this problem was proposed in 1987 and has been verified countless times. So it is not possible that the solution is incorrect. However, you may refer here: https://earthz.ru/solves/Zadacha-po-fizike-6534
This is the diagram which I am referring to.

[Physicsguy51]

I agree completely with Pigkappa. I dont understand why a difference of level between A and B changes the angle O. As if we rotate with an axis for Z

The problem which I had solved had explicitly mentioned that AB = radius of the circle.

[Physicsguy51]

According to the condition of the problem, the ball rolls without slipping, therefore, the speeds of those points of the ball that at a given moment of time touch the metal at AB (Fig.) are equal to zero. Considering the ball to be an absolutely rigid body (that is, the distance between any two points of the ball is unchanged), we conclude that at a given moment in time all points of the ball lying on the segment AB are motionless. And this means that at each moment of time the movement of the ball is a rotation about the axis AB. (It is clear that points A and B - the points where the ball touches the metal - are moving with the speed .)

The instantaneous velocity of any point of the ball is , where is the angular velocity of rotation, is the distance from the point to the axis AB. The speed of the center of the ball (point O in the figure) is equal to ; the distance from point O to axis AB is . Hence,


It is clear that the points of the ball most distant from the axis AB have the maximum speed. From geometric considerations it is clear that at any moment there is only one point that is maximally distant from the axis - in the figure this is point Q. The distance from point Q to the axis of rotation is , and speed of Q is:

Sorry, but I disagree with this solution. In the diagrams attached by @Pigkappa, which coincides with that of AAPT in this document it is not entirely clear where points and are. Are they respectively the lowest and highest points of the sphere, or the point nearest where the plates meet and the point furthest from it?
The right-hand diagram appears to show the plates end-on, so is looking up the slope. That means is the correct interpretation. This makes the answer '' correct. The text is wrong to refer to them as .
Under interpretation , the speed of is multiplied by the ratio of distances from to and to , namely, , so it is .
Under interpretation , is not the fastest point, and its speed is indeterminate because we do not know the angle of the slope. As to how the AAPT answer came to be wrong, it was a rehash of an old question and they forgot to change everything consistently. It happens, and to note, this is not the first time this has occurred in the USAPhO.
It sometimes helps to consider an extreme case. Suppose the plates are vertical, which they could be, as far as we know. The point at the top of the sphere is from , so moves at speed . Meanwhile, the point furthest from the plate join still moves at . I have tried to verify your solution, graphically, using all kinds of software and graphing calculator, but it is in no way possible to obtain an equilateral triangle , with all three angles of 60 degrees, while keeping the two metal planes orthogonal to each other (as explicitly required by the text). It is not really possible. There is a huge discrepancy between the text and the proposed graphical scheme. I cannot understand why in your opinion the triangle is equilateral in the case and are not at the same level.

I would also like to state that I do not know what AAPT had in mind while creating their problem, but I have not checked it. I just posted my solution for the problem which is the original one from 1987.

[Tarapìa Tapioco]

I agree completely with Pigkappa. I dont understand why a difference of level between A and B changes the angle O. As if we rotate with an axis for Z

The problem which I had solved had explicitly mentioned that AB = radius of the circle.

This confirms my theory. The solution you offered is correct for the problem you just sent me, but not for the one @Pigkappa proposed, where " radius of the circle" is not specified. As I suspected, AAPT recycled this result by adapting it to a similar but not the same problem. This is confirmed by the fact that a demonstration of it is not offered on their site. Differently from the AAPT exercise, in the 1987 problem the left rail is not tangent to the sphere at the point of contact.

[Physicsguy51]

Interestingly, this document from an American competition, in problem 2, reports your same result, without proof.

But... I am still confused in the problem I posted, and the result is , isn't it?

Here is why I think ... Adding a point Z in the figure at the bottom, OAZB is a square of radius R, and half of the diagonal is

Sono d'accordissimo. Più tardi posterò il mio procedimento completo (purtroppo non ho potuto farlo prima, in quanto in questi giorni sono stato poco bene), che per i punti 1.) e 2.) è identico a quello proposto da te e @Higgs. Questo problema è parecchio interessante, non solo per il procedimento da svolgere al fine di soddisfare le tre richieste, ma anche per qualcos'altro, su cui metterò l'accento nella parte conclusiva della mia risoluzione; ciononostante, è reso meno interessante da una soluzione ufficiale fuorviante, peraltro non dimostrata (già questo dovrebbe mettere in guardia dalla sua provenienza) e perciò - con buona probabilità - riciclata da qualche esercizio simile ma non identico (come ho già detto, non è veramente la prima volta, se si prova a guardare le soluzioni fornite dal sito ufficiale delle USAPhO, da me già visionate in passato). Come rovinare un'ottima idea, aggiungerei.

Yes, I would agree with your last few lines. As I stated before, I do not know what the authors had in mind while manipulating the problem, because they turned a simple problem into an ambiguous one. The original problem had a crucial statement which was: "The length AB = Radius of the sphere" and hence it was simple to get to the conclusion that the triangle is an equilateral triangle. However, by removing this statement they have not only changed their problem but also have changed the correct answer. As now, you cannot assume the triangle to be equilateral. I guess what I meant with A and B not being in the same level was that the distance between AB is equal to the radius of sphere. AB is the IAOR.

[Physicsguy51]

I agree completely with Pigkappa. I dont understand why a difference of level between A and B changes the angle O. As if we rotate with an axis for Z

The problem which I had solved had explicitly mentioned that AB = radius of the circle.

This confirms my theory. The solution you offered is correct for the problem you just sent me, but not for the one @Pigkappa proposed, where = radius of the circle is not specified. As I suspected, AAPT recycled this result by adapting it to a similar but not the same problem. This is confirmed by the fact that a demonstration of it is not offered on their site.

Yes, and this is why I wrote that this solution is for the original problem which appeared in 1987. I wrote this on purpose because I hadn't looked at the AAPT problem closely enough to figure out the dissimilarities, if any.

[Tarapìa Tapioco]

The problem which I had solved had explicitly mentioned that AB = radius of the circle.

This confirms my theory. The solution you offered is correct for the problem you just sent me, but not for the one @Pigkappa proposed, where = radius of the circle is not specified. As I suspected, AAPT recycled this result by adapting it to a similar but not the same problem. This is confirmed by the fact that a demonstration of it is not offered on their site.

Yes, and this is why I wrote that this solution is for the original problem which appeared in 1987. I wrote this on purpose because I hadn't looked at the AAPT problem closely enough to figure out the dissimilarities, if any.

Sorry for the misunderstanding, I thought your result was referring to the problem proposed by @Pigkappa. For the problem you sent, of course, everything is correct.

[Physicsguy51]

Alright!

[Pigkappa]

What's the problem statement for the 1987 problem?

[Physicsguy51]

What's the problem statement for the 1987 problem?

Here ya go: https://earthz.ru/solves/Zadacha-po-fizike-6534