Bouncing around

Geometry Level pending

A ball start at a pool table at the point $X$ (the table 1" wide , 2" long) , then a man shoot it to the right and then it bounce to the top and then left , finally , it stoped at the point $Y$ . And we know that $X$ is 0.6" to the left and 0.2" to the bottom . $Y$ is 0.2" from the left and 0.25" from the top . If the ball always rebounds at the same angle as it hits the side , then what is the distance the ball moves?

I don't know 3.84" None of these 3.82" 3.80" 3.87" 3.85"

2 solutions

Ricky Huang
May 22, 2020

Just unfold it! Using the Pythagorean theorem you can figure out the distance is √[(2 + 2 - 0.6 + 0.2) $^2$ + (1 - 0.2 + 0.25)] = 3.75".

A Former Brilliant Member
May 21, 2020

Let the angle of incidence in the case of first reflection be $α$ . Then, after some calculation work we get the equation :

$3.6\tan α=1.05\implies \tan α=\dfrac{1.05}{3.6}\implies \sec α=\sqrt {1+\tan^2 α}=\dfrac{3.75}{3.6}$ .

Hence the length of total path covered is

$1.4\sec α+(0.8-1.4\tan α)\cosec α+(3.4\tan α-0.8)\cosec α+0.2\sec α=3.6\sec α= \boxed {3.75''}$ .

Bouncing around

2 solutions

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