In Memoriam of Maryam Mirzakhani

Oh, billiards dynamics. This quite reminds me of playing the world's billiard game 8 Ball Pool by Miniclip . This really comes to my mind when I perform bank and kick shots for the tough matches. ;)

Here is my solution...

By the given, the ball is $\dfrac{7}{4}$ meters away from the east wall and $\dfrac{3}{4}$ meters away from the south wall.

Since the angle of reflection equals the angle of incidence, we can draw another billiard table to the right, where there is no change in angle of the ball's path after passing through the east wall.

In that case, the ball travels $\dfrac{11}{4}$ meters east and $\dfrac{3}{4}$ meters south. Thus, by similarity of triangles, $\begin{array}{rl} \dfrac{\text{top to bottom}}{\text{left to right}} = \dfrac{\frac{3}{4}}{\frac{11}{4}} &= \dfrac{x}{1}\\ x &= \dfrac{3}{11} \end{array}$ where $m + n = 3 + 11 = \boxed{14}$ .

I really appreciate Maryam Mirzakhani's work in billiards. She was a brilliant mathematician! :)

I = angle of incidence

R = angle of reflection

As from the figure we can see that both triangles are similar.

Hence,

$\frac{1}{0.75}$ = $\frac{x}{0.75-x}$

0.75 - x = 1.75x

2.75x = 0.75

x = $\frac{3}{11}$

Here's my solution-

Since angles of incidence and reflection are equal, their respective complementary pair angles will also be equal (as each angle to both sides of point of incidence with the normal inside the table is a RIGHT angle).

Therefore the sides (of two apparent triangles on both sides of normal inside the table) are proportional.

It follows that,

x/1 = (0.75 - x)/1.75

i.e. x = (3 - 4x)/7

i.e. 7x = 3 - 4x

i.e. 11x = 3

i.e. x = 3/11

Since the angles of the smaller and the bigger triangle (turned down to the left smaller triangle gave us a bigger triangle) are the same and both are right-angled, we can use the Tangent.

Dividing $\frac{0,75}{2,75}$ = Tan(15,2551187)

tan(15,2551187) * 1 m = 0,272727... = $\frac{3}{11}$

I found this one surprisingly tough until I looked at it the right way.

I drew a line up from the side pocket and realized, from the top down, that $\frac{1}{4}$ plus $\frac{3}{4}$ x plus 2 x equals 1.

Cue traveled total of 11/4m in x while traveling 3/4m in y (falling into the pocket). Therefore to travel 1m in x, it will have to travel 1*(3/4)/(11/4)=3/11m

Drawing a perpendicular line from the ball to the other side of the rectangle we get 3 segments x, y and .25 x+y=1-.25=.75 Also tan(f)=y/1.75 and also tan(f)=x/1 Solving we find that x=.27 or 3/11 f is the angle of the incidence of the ball when it hits the rectangle's short side.