A direct shot in the far right corner

Geometry Level 2

In a simulation of a pool game, the pool table is a rectangle with a horizontal width of 15 units and a vertical length of 30 units.

The left bottom corner of this rectangle is at the origin of an x y xy reference frame with standard orientation. You have two billiard balls of radius 1 unit, the first is the cue ball (drawn in light green) and is placed at ( 6 , 11 ) (6,11) , the second is the target ball and is placed at ( 9 , 19 ) (9, 19) . You want to shoot the cue ball (green) in a certain direction, such that it collides with the target ball, causing it to move to the target point which, in this problem, is the far right corner, whose coordinates are ( 15 , 30 ) (15, 30) . Find the angle θ \theta that the direction of your shot of the cue ball makes with the positive x x -axis, in radians , and report 1000 θ \lfloor 1000 \hspace{4pt} \theta \rfloor .


The answer is 1254.

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1 solution

Steven Chase
May 2, 2019
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import math

R = 1.0    # radius

xf = 15.0  # top right corner coordinates
yf = 30.0

x1 = 6.0   # shooting ball initial coordinates
y1 = 11.0

x2 = 9.0   # target ball coordinates
y2 = 19.0

####################

vx = xf - x2    # vector from target ball center to top right
vy = yf - y2

v = math.hypot(vx,vy)

ux = vx/v       # corresponding unit vector
uy = vy/v

x1c = x2 - 2.0*R*ux  # shooting ball center coordinates at collision
y1c = y2 - 2.0*R*uy  # co-linear with target ball center and corner
                     # center is a distance 2R away from target ball center

####################

Dx = x1c - x1        # vector from shooting ball starting location.........
Dy = y1c - y1        # ........to collision location

theta = math.atan(Dy/Dx)  # angle

####################

print (180.0 * theta / math.pi)
print math.floor(1000.0*theta)  # answer is 1254

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