How should I point my laser beam ?

Geometry Level pending

A laser light source is placed in a 10 × 10 × 10 10 \times 10 \times 10 room where the floor and the ceiling and the four walls are covered with mirrors. A coordinate reference frame is attached to the room, such that its origin is at one of the bottom corners, and its z z axis pointing vertically upward, its x y xy plane coincident with the floor, and the plane z = 10 z= 10 coincident with the ceiling, and the x = 0 , x = 10 , y = 0 , y = 10 x = 0, x = 10, y = 0, y=10 planes coincident with the four walls, so that the room extends between ( 0 , 0 , 0 ) (0, 0,0) and ( 10 , 10 , 10 ) (10,10,10) . The laser source is placed at A ( 3 , 2 , 2 ) A(3,2, 2) . You want to point the laser source such that its light beam reflects off the planes z = 0 , x = 10 , y = 10 , z = 10 z = 0, x = 10, y =10, z = 10 in that order then reaches the point B ( 3 , 4 , 6 ) B ( 3, 4,6 ) . What is the total distance travelled by the laser beam? If the distance can be expressed as a b a \sqrt{b} where a , b a, b are positive integers, and b b is square-free, find a + b a + b .


The answer is 20.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

David Vreken
Dec 22, 2020

Reflect the cube several times, first in z = 0 z = 0 , next in x = 10 x = 10 , next in y = 10 y = 10 , and finally in z = − 10 z = -10 , as shown below:

Since B B has coordinates ( 3 , 4 , 6 ) (3, 4, 6) , it is 0 − 6 = − 6 0 - 6 = -6 units away from z = 0 z = 0 , so when it is reflected to B ′ B' it has coordinates ( 3 , 4 , 0 − 6 ) = ( 3 , 4 , − 6 ) (3, 4, 0 - 6) = (3, 4, -6) .

Since B ′ B' has coordinates ( 3 , 4 , − 6 ) (3, 4, -6) , it is 10 − 3 = 7 10 - 3 = 7 units away from x = 10 x = 10 , so when it is reflected to B ′ ′ B'' it has coordinates ( 10 + 7 , 4 , − 6 ) = ( 17 , 4 , − 6 ) (10 + 7, 4, -6) = (17, 4, -6) .

Since B ′ ′ B'' has coordinates ( 17 , 4 , − 6 ) (17, 4, -6) , it is 10 − 4 = 6 10 - 4 = 6 units away from y = 10 y = 10 , so when it is reflected to B ′ ′ ′ B''' it has coordinates ( 17 , 10 + 6 , − 6 ) = ( 17 , 16 , − 6 ) (17, 10 + 6, -6) = (17, 16, -6) .

Since B ′ ′ ′ B''' has coordinates ( 17 , 16 , − 6 ) (17, 16, -6) , it is − 10 − − 6 = − 4 -10 - -6 = -4 units away from z = − 10 z = -10 , so when it is reflected to B ′ ′ ′ ′ B'''' it has coordinates ( 17 , 16 , − 10 − 4 ) = ( 17 , 16 , − 14 ) (17, 16, -10 - 4) = (17, 16, -14) .

The minimum distance would be equivalent to the length of A B ′ ′ ′ ′ AB'''' , which is A B ′ ′ ′ ′ = ( 17 − 3 ) 2 + ( 16 − 2 ) 2 + ( − 14 − 2 ) 2 = 18 2 AB'''' = \sqrt{(17 - 3)^2 + (16 - 2)^2 + (-14 - 2)^2} = 18\sqrt{2} .

Therefore, a = 18 a = 18 , b = 2 b = 2 , and a + b = 20 a + b = \boxed{20} .

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