Witness Minnesota Fats' Amazing Trick Shot: Angle MAZ

[Wilmer]

(U)     A            (Z)       D            (Y) 

                     [M] 
 B                                           E 
(V)          C       (W)            F       (X)

That there is an 8 by 4 pool table, with pockets U,V,W,X,Y,Z and center M.

Of course: UV = VW = WX = XY = YZ = ZU = 4.

The cue ball is at M.

Minnesota Fats performs the following "trick shot":
shoots cue ball at point A:
ball continues in order to points B,C,D,E,F, then from F directly to side pocket Z.

What is size of angle MAZ?

 

[caffeinemachine]

(U)     A            (Z)       D            (Y) 

                     [M] 
 B                                           E 
(V)          C       (W)            F       (X)

That there is an 8 by 4 pool table, with pockets U,V,W,X,Y,Z and center M.

Of course: UV = VW = WX = XY = YZ = ZU = 4.

The cue ball is at M.

Minnesota Fats performs the following "trick shot":
shoots cue ball at point A:
ball continues in order to points B,C,D,E,F, then from F directly to side pocket Z.

What is size of angle MAZ?

Interesting problem.
Here's a hint for you.
Reflect M about the upper side of the table to get a new point M'.
In out present scenario the player shoots the ball from M to A.
Conceive of a scenario where the player shoots the cue ball from M' to A and the upper side is "permeable" to the cue ball for the first time it crosses the upper side and after that it becomes impermeable.
Note that this does not change the fate of the cue ball.

You need to just keep reflecting this way--well almost. (How many times?)

 

[Wilmer]

Thanks CM; I've already solved this.
Was looking for confirmation: guess I should have said so!

I got angle MAZ = ATAN(9/8) ; ~48.37 degrees

I think the way I did it is a bit like what you're suggesting:
u = angle MAZ: v = TAN(u)
AZ = 2 / v
a = AU = 4 - AZ = 4 - 2 / v = (4v - 2) / v

BU = av
b = BV = 4 - BU = 4 - bv

...and similarly "around the table"...

 

[caffeinemachine]

Thanks CM; I've already solved this.
Was looking for confirmation: guess I should have said so!

I got angle MAZ = ATAN(9/8) ; ~48.37 degrees

I think the way I did it is a bit like what you're suggesting:
u = angle MAZ: v = TAN(u)
AZ = 2 / v
a = AU = 4 - AZ = 4 - 2 / v = (4v - 2) / v

BU = av
b = BV = 4 - BU = 4 - bv

...and similarly "around the table"...

Your method's good. But the two methods we have are different. The one I suggested is "purely geometric". That is, say you draw the table to scale on a (large) piece of paper and pose the problem as such. Now it is asked to "construct" geometrically the angle to be found. No trig and no coordinate geometry and no calculations. Just plain ruler and compass constructions. Using my method this can be done.

Your method is "computational". It certainly works and has wider scope, of course.

 

[Wilmer]

Gotcha! Merci beaucoup.
Play you a game of "9ball" for $1 ? (Bandit)

 

[caffeinemachine]

Gotcha! Merci beaucoup.
Play you a game of "9ball" for $1 ? (Bandit)

?? What? :)

 

[Wilmer]

Nine-ball - Wikipedia, the free encyclopedia

 

[caffeinemachine]

Nine-ball - Wikipedia, the free encyclopedia

Haha.. not really my game. Thanks anyway.

 

[Wilmer]

Sudharaka, you want to play?