# Where should the yellow arrow placed to get the red lines as short as possible?

### 3 Answers | Add Yours

The figure has two parallel line segments with ends A and B of length 100 m and 200 m that are 300 m away from each other. At the point where the yellow arrow is placed red lines are drawn from A and from B.

Let the yellow arrow point to X that is x m away from the top of the 100 m line. The length of the red line from A is equal to `sqrt(100^2 + x^2)` and the length of the red line from point B is equal to `sqrt(200^2 + (300 - x)^2)` .

To determine the appropriate placement of the yellow arrow, the sum `D = sqrt(100^2 + x^2) + sqrt(200^2 + (300 - x)^2)` has to be minimized.

D' = `(1/2)*(1/(sqrt(100^2 + x^2)))*2x + (1/2)*(1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)*-1`

Solving D' = 0 for x,

`(1/2)*(1/(sqrt(100^2 + x^2)))*2x + (1/2)*(1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)*-1 = 0`

=> `(1/2)*(1/(sqrt(100^2 + x^2)))*2x = (1/2)*(1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)`

=> `(1/(sqrt(100^2 + x^2)))*2x = (1/(sqrt(200^2 + (300 - x)^2)))*2*(300-x)`

=> `sqrt(200^2 + (300 - x)^2)*2x = sqrt(100^2 + x^2)*2*(300-x)`

=> `(200^2 + (300 - x)^2)*4x^2 = (100^2 + x^2)*4*(300-x)^2`

=> `(200^2 + (300 - x)^2)*x^2 = (100^2 + x^2)*(300-x)^2`

=> `x^4-600*x^3+130000*x^2 =`

` x^4-600*x^3+100000*x^2-6000000*x+900000000`

=> `13*x^2 = 10*x^2-600*x+90000`

=> `x^2 + 200*x - 30000 = 0`

=> `x^2 + 300x - 100x - 30000 = 0`

=> `x(x +300) - 100(x + 300) = 0`

=> `(x - 100)(x + 300) = 0`

=> x = 100 and x = -300

Ignore the negative root.

**The yellow arrow should be placed 100 m from the top end of the line that is 100 m long.**

There's a very clever and much simpler alternate solution (I wish I was clever enough to have figured it out myself, but alas, I read it in a book).

Reflect point `B` across the horizontal line above it--the line where the tip of the arrow and the two red line segments all meet--and reflect the red line segment joining the tip of the arrow and point `B` across this line as well. Call the reflected point `B'` . What you get is a mirror image*, and the reflected lines have the same length as the original ones.

Now, since the shortest distance between two points is a straight line, the line from `A` to `B'` is the shortest length of the red lines. We now have similar right triangles with vertices `A` and `B',` and since the rightmost triangle is twice the size of the leftmost one, we can see that the yellow arrow should be twice as far from the vertical leg of rightmost triangle, or **200m away starting from the right side (equivalently, 100m away starting from the left side).**

*You can think of the red lines as a ray of light reflected by a mirror. This shows that if light takes the shortest path, it makes the angles of incidence and reflection equal.