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Question 1210201: Let ABCD be a square with side length 1. A laser is located at vertex A, which fires a laser beam at point X on side BC, such that BX = 2/3. The beam reflects off the sides of the square, until it ends up at another vertex; at this point, the beam will stop. Find the length of the total path of the laser beam. The diagram is linked below
https://artofproblemsolving.com/texer/zqcbfanp
Found 5 solutions by CPhill, greenestamps, ikleyn, AnlytcPhil, mccravyedwin:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let the square \( ABCD \) be oriented in the coordinate plane with points defined as follows:
- \( A = (0, 1) \)
- \( B = (0, 0) \)
- \( C = (1, 0) \)
- \( D = (1, 1) \)
The laser beam is fired from point \( A \) at an angle towards point \( X \), where \( X \) is the point on side \( BC \) such that \( BX = \frac{2}{3} \). The coordinates of point \( X \) are given by:
\[
X = (0, \frac{2}{3})
\]
To find the direction of the laser beam from \( A \) to \( X \), we calculate the vector from \( A \) to \( X \):
\[
\overrightarrow{AX} = X - A = (0, \frac{2}{3}) - (0, 1) = (0, -\frac{1}{3})
\]
The slope of this line is
\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{2}{3} - 1}{0 - 0} \quad \text{(vertical line)}.
\]
For the moment, let us compute the direct distance from \( A \) to \( X \):
\[
d_{AX} = \sqrt{(0 - 0)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{0 + \left(-\frac{1}{3}\right)^2} = \frac{1}{3}.
\]
The laser beam will reflect off of side \( BC \) and into side \( CD \). The point X is at a vertical location of \( \frac{2}{3} \) from the bottom of the square.
Now let's consider what happens when it reflects off side \( BC \). The line from \( A \) to \( X \) intersects \( BC \) at \( X \), and then heads towards side \( CD \).
The angle of incidence equals the angle of reflection. After reflecting off side \( BC \), the beam will change its direction but maintain its angle of slope towards the other vertical wall, \( CD \).
The horizontal direction can be preserved by extending or 'unfolding' the square. The strategy involves reflecting the square:
1. From \( A \) to \( X \), the laser hits \( BC \).
2. It then reflects and heads towards \( D \), calculating where it hits before potentially hitting at \( D \).
With this reflection, the new direction of travel can be calculated using the symmetry of the reflections:
- Upon reflecting, after reaching \( BC \) it would travel back upwards towards \( CD \) (at point \( D \)), moving rightward.
The distance until it would again meet the line \( AD \):
1. From \( X \) to \( D \) directly makes up the next segment. The coordinates for \( D \) are \( (1, 1) \) when revisiting from \( (0, \frac{2}{3}) \):
This results in:
\[
d_{XD} = \sqrt{(1 - 0)^2 + (1 - \frac{2}{3})^2} = \sqrt{1^2 + \left(\frac{1}{3}\right)^2} = \sqrt{1 + \frac{1}{9}} = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3}.
\]
Total distance:
\[
d_{AX} + d_{XD} = \frac{1}{3} + \frac{\sqrt{10}}{3} = \frac{1 + \sqrt{10}}{3}.
\]
Thus, the total path of the laser beam from A to D via reflections is
1 + \sqrt{10}.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Your link does not show a diagram....
The AI solution from the other tutor is wrong. That solution says the ray hits vertex D after the single reflection off side BC. That will not happen if the length of BC is 2/3. (Note that, by symmetry, the ray would hit vertex D after the single reflection off side BC if the length of BX were 1/2 instead of 2/3.)
Let Y be the point on CD where the ray hits after reflecting off side BC.
Since the angle of reflection is the same as the angle of incidence, triangles ABX and YCX are similar. Since BX is 2/3 and CX is 1/3, the ratio of similarity between the two triangles is 2:1.
But then, by that ratio of similarity, CY is half the length of AB, which is a side of the square. So Y is the midpoint of CD.
From there, we can see by symmetry that the ray will reflect off side CD and continue to vertex B.
Here is a diagram, with the path of the ray in red, starting from vertex A and ending at vertex B.
From the Pythagorean Theorem with AB=1 and BX=2/3, the length of AX is  .
Because of the similarity of triangles ABX and YCX, the length of XY is  .
That makes the length of the path of the ray from A to X to Y  .
And then by symmetry the path from Y to side AD and then on to vertex B is again  .
So the total length of the path from vertex A to vertex B, reflecting off sides BC, CD, and DA, is  .
ANSWER:
Answer by ikleyn(52847)  (Show Source):
You can put this solution on YOUR website! .
Let ABCD be a square with side length 1. A laser is located at vertex A, which fires a laser beam
at point X on side BC, such that BX = 2/3. The beam reflects off the sides of the square,
until it ends up at another vertex; at this point, the beam will stop.
Find the length of the total path of the laser beam. The diagram is linked below
https://artofproblemsolving.com/texer/zqcbfanp
~~~~~~~~~~~~~~~~~~~~~~~~
Couple of notices before to start.
(a) The link is EMPTY and does not show/(does not contain) any picture/diagram.
(b) The solution and all calculations in the post by @CPhill are WRONG.
I came to bring a correct solution.
(1) First beam goes from A to X on BC.
One leg is 1 unit, other leg is 2/3 of the unit.
Hence, AX =  =  =
(2) Then the beam reflects at X on BC and goes to point Y on DC.
The angle of incidence at X is equal to the angle of reflection at X.
Hence, right-angled triangles ABX and XCY are similar.
The similarity coefficient is BX/XC =  = 2.
Hence, XY is half of AX. i.e. .
(3) Next, the beam reflects at Y on DC and goes to point Z on AD.
Again, the angle of incidence at Y is equal to the angle of reflection at Y.
Hence, right-angled triangles XCY and YDZ are similar.
The similarity coefficient is CY/YD =  = 1.
So, the triangles XCY and YDZ are not only similar - - - they are CONGRUENT.
It implies that DZ = 1/3 and ZY = XY = .
(4) Now point Z is symmetric to point X.
It means that after next reflection at Z, the beam will go directly to vertex B.
and the journey will be complete.
The last beam's interval YB is the same long as AZ, i.e. .
(5) The last step is to find the total length of the journey
AX + XY + YZ + ZB = + + + = , or about 3.60555.
It is the ANSWER to the problem's question.
Solved.
The work and the calculations in the post by @CPhill
amaze me with their low level.
His work/technique is at the level of a student,
whose scores are between 2 and 3 in the five-score scale.
Closer to 2 than to 3.
The true AI level should be 6 + + +.
Answer by AnlytcPhil(1807)  (Show Source):
You can put this solution on YOUR website!
I think this particular AI system is getting better! That's the way AI works.
AI systems are programmed to learn from their mistakes.
When AI chess programs started out, anybody could beat them who knew the rules of
chess. But now no human chess master is able to beat the most sophisticated
AI chess programs.
Making mistakes and correcting them is essential for learning. If you only learn
the best ways first, you never learn why the other ways are inferior.
Edwin
Answer by mccravyedwin(408)  (Show Source):
You can put this solution on YOUR website!
I think this particular AI system is getting better! That's the way AI works.
AI systems are programmed to learn from their mistakes.
When AI chess programs started out, anybody could beat them who knew the rules of
chess. But now no human chess master is able to beat the most sophisticated
AI chess programs.
Making mistakes and correcting them is essential for learning. If you only learn
the best ways first, you never learn why the other ways are inferior.
Edwin
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