We construct two triangles: ABC and CDE, where
A - position of of white ball
B - position on bumper directly above white ball
C - position on bumper where the 2 red arrows meet
D - position on bumper directly above red ball
E - position of red ball
AB = 3 ft
BC = x
CD = 5-x
DE = 1.8 ft
△ABC and △EDC are both right angled (at ∠ABC and ∠EDC respectively)
△ABC and △EDC have congruent angles (∠ACB and ∠ECD)
Therefore, △ABC is similar to △EDC
By similar triangles:
BC/DC = AB/ED
x/(5-x) = 3/1.8
1.8x = 3(5-x)
1.8x = 15 - 3x
4.8x = 15
x = 3.125
So player should hit the white ball 3.125 ft to right of point B.
Point B is 1 ft to right of upper left corner.
Therefore, player should hit white ball 4.125 ft to right of upper left corner.