How do I calculate the slope of a line knowing one intersection point and the intersection point of that line after two reflection angles?

Question

When a point starting at $O(0,0)$ intersects a side of the square $ABCO$, it reflects so the original angle and reflecting angle are equal. In the bottom side of the square ($AO$), we know the point intersects at $M\left(\dfrac12, 0\right)$.

What is the value of $m$ so that the line $y = mx$ intersects with the top of the square ($BC$) and right side of the square ($AB$) before intersecting with the bottom of the square ($AO$) at $M$?

I'm able to prove that the angle between each line and the sides of the square is $45^\circ$ and that the angle between two reflecting lines is $90^\circ$. But I haven't been able to use that to find the value of $m$.

I also see that if I continue drawing the line from $M$ so it reflects off the left and top of the square, it will intersect with $A(1,0)$, making a symmetric diagram if I draw a line from $M\left(\dfrac12, 0\right)$ to $\left(\dfrac12, 1\right)$. I believe there must be a way to prove that the two intersections with the top of the square must be at $\left(\dfrac12,1\right)$ and $\left(\dfrac34,1\right)$ (which would make the value of $m = \frac34$). But I don't see how to prove that.

Answer 1

Let $P$ and $Q$ be the points of intersection of the trajectory with $BC$ and $AB$, respectively. It's not hard to see that $\triangle OCP \sim \triangle PBQ \sim \triangle MAQ$.

From $\triangle PBQ \sim \triangle MAQ$, we get: $$\frac{BP}{BQ} = \frac{AM}{AQ} \Longrightarrow BP = BQ \cdot \frac{AM}{AQ} = (1-AQ)\frac{AM}{AQ} = \frac{AM}{AQ}-AM$$

But from similarity $\triangle OCP\sim \triangle MAQ$, we get $\displaystyle{\frac{AM}{AQ} = \frac{CP}{OC} = CP}$.

So we have that $BP = CP - AM$. But $AM = \frac{1}{2}$, and $BP = 1 - CP$, so $CP = \frac{3}{4}$. Hence, $P = \left(\frac{3}{4},1\right)$, and $m = \frac{4}{3}$.

Edit:

To prove that $\triangle OCP \sim \triangle MAQ$, we just need to do some angle chasing. Since $AO \parallel BC$, we have that $\angle AOP = \angle CPO$. Due to the reflection of the trajectory conditions, we also have that $\angle CPO = \angle BPQ$. Hence $\triangle OCP \sim \triangle PBQ$ by the $AA$ criterion (they are also both right triangles). Again by the reflection condition, $\angle BQP = \angle AQM$, and since they are both right triangles, $\triangle PBQ \sim \triangle MAQ$, which proves that the three triangles are similar.

Answer 2

I think I came up with an answer.

As I wrote in the question, I continued drawing the line from $M$ so it reflects off the left and top of the square. It must intersect with $A(1,0)$ because all the angles are the same as on the other side of the square.

I then drew a horizontal (dashed) line intersecting with the intersection of the angled lines touching the left and right of the square and vertical lines down from the intersections at the top of the square and a vertical line up from the intersection at $M$.

That vertical line must intersect with the crossing point of the two diagonal lines that intersect the top of the square, because the the angles of the other two angles within the triangles they form are the same. In fact, all 8 triangles formed along the top of the square must be identical because the angles must be all the same and the lengths of two of the sides are the same. So the lengths of the sides of the four triangles with sides along the top of the square must be the same. So each of those four sides must be $\frac14$.

So $y = mx$ becomes $1 = m\frac34$ and so $m = \frac43$.

Answer 3

Let $D$ and $E$ be the reflection points at $BC$ and $AB$, respectively. Then: $$CD=\frac{1}{m}, AE=\frac{m}{2},$$ because: $$m=\tan \measuredangle CDO=\tan \measuredangle AME=\tan \measuredangle BDE.$$ From triangle $BDE$: $$m=\tan \measuredangle BDE=\frac{BE}{BD}=\frac{1-\frac{m}{2}}{1-\frac{1}{m}} \Rightarrow \\ m=\frac43.$$