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## Question

On a 10 x 10 grid table, a light beam enters from the middle of the top left cell at an angle of 45 degrees as indicated by the arrow. When the beam reaches a mirror it is reflected and its path ends when it reaches the edge of the table.

Place the mirrors on the table to create the maximum beam length. The mirrors may not touch each other, not even diagonally. The size and quantities of the mirrors you can use are shown below. A squares diagonal length will be 1 unit.

What is the longest beam you can make?

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I assume that all four edges of all the mirrors are reflective.

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I assume that all four edges of all the mirrors are reflective.

Yes

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If the smallest mirror were to be placed within the grid block at 2B,

where would the beam exit the grid (assuming no other mirrors were

used)?

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If the smallest mirror were to be placed within the grid block at 2B,

where would the beam exit the grid (assuming no other mirrors were

used)?

I'm guessing 3A based on the slight offset of the beam entrance (it's not exactly at the northwest corner).

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In answer to superprismatic's question, on 03 December 2011 - 04:49 PM, said:

If the smallest mirror were to be placed within the grid block at 2B,

where would the beam exit the grid (assuming no other mirrors were

used)?

MM said, "I'm guessing 3A based on the slight offset of the beam entrance (it's not exactly at the northwest corner)."

The actual answer is that It would end the beam at the edge of the table at exactly the mid point of the left edge of 3A. I am not usually this picky, but it is important to note that the entrance point is not a "slight offset" but the problems stated that the "light beam enters from the middle of the top left cell at an angle of 45 degrees..." If it entered at the top corner, it would not reflect the same as it does with this setup. If it enters somewhere other than the "middle" at a "45 degree" angle, it will also affect where it hits the mirror and where it goes. It also affects the measurement of the beams. In other words, it is extremely important to the entire problem!

Edited by princecharmthings
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Can the mirrors be rotated?

I can make the path length Very Long if they can be rotated by any angle, not restricted to multiples of 90 degrees, and the beam of light is narrow

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23.5 Units is the best I have been able to get, so far. I will let you know if I find a better solution...but I have done it over and over and I have done far worse...so I am hanging on to this answer! Big, Bad 23 and a half!!! (Can't forget that half!!!) ;D

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Another question. Are we allowed to use all the mirrors listed? I was interpreting it to mean we could choose 4 small or 3 larger, etc. If we are to use all of the mirrors listed, my answer will change dramatically!!!

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Another question: if a square's diagonal length is 1 unit, then I guess that means a mirror square is significantly smaller than a grid square. Is that true, or did I misinterpret this statement? If so, then we have to fastidiously clear about exactly where inside a grid square we are placing a mirror square, as Princecharmthings points out.

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the same question as of captained

Edited by Arjit Saraswat
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Oh boy! It is becoming increasingly clear that we are most unclear as to the facts of this exercise!!!

Capt. Ed, I was working under the assumption that one mirror square was the same size as one square on the table.

I had also begun working with the assumption that we were free to use all of the mirror pieces in either a horizontal or vertical alignment.

Looking back, I see I am assuming a lot and we all know what happens when we ASSUME anything!!!

I will say that it was a lot more fun working with all the mirrors instead of just a few. My units increased significantly, too.

I would say I wasted all that time working on a problem that I really didn't have all the facts on, but I didn't have anything better to do anyway, and it was fun! Hopefully the Prof. will make an appearance in the morning and clear up this mess so we can get down to some serious time wasting!!! LOL

I will refrain from posting any more "scores" until I find out the real rules. I bid you all Good Night!

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If the smallest mirror were to be placed within the grid block at 2B,

where would the beam exit the grid (assuming no other mirrors were

used)?

It would exit at the middle of 3A.

The actual answer is that It would end the beam at the edge of the table at exactly the mid point of the left edge of 3A. I am not usually this picky, but it is important to note that the entrance point is not a "slight offset" but the problems stated that the "light beam enters from the middle of the top left cell at an angle of 45 degrees..." If it entered at the top corner, it would not reflect the same as it does with this setup. If it enters somewhere other than the "middle" at a "45 degree" angle, it will also affect where it hits the mirror and where it goes. It also affects the measurement of the beams. In other words, it is extremely important to the entire problem!

Correctly stated

Can the mirrors be rotated?

I can make the path length Very Long if they can be rotated by any angle, not restricted to multiples of 90 degrees, and the beam of light is narrow

No rotation is allowed. The mirrored block should be placed exactly covering the cells on the grid.

Another question. Are we allowed to use all the mirrors listed? I was interpreting it to mean we could choose 4 small or 3 larger, etc. If we are to use all of the mirrors listed, my answer will change dramatically!!!

You can and should use all of the mirrored block available to you. (4-one cell blocks, 3-two cell blocks, 2-three cell blocks, 1-four cell block)

Another question: if a square's diagonal length is 1 unit, then I guess that means a mirror square is significantly smaller than a grid square. Is that true, or did I misinterpret this statement? If so, then we have to fastidiously clear about exactly where inside a grid square we are placing a mirror square, as Princecharmthings points out.

Mirror squares and grid squares are the same size. A square's diagonal length is stated at 1 unit to make the beam length easier to count since the beam always travels on the diagonal.

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Oh boy! It is becoming increasingly clear that we are most unclear as to the facts of this exercise!!!

Capt. Ed, I was working under the assumption that one mirror square was the same size as one square on the table.

I had also begun working with the assumption that we were free to use all of the mirror pieces in either a horizontal or vertical alignment.

Looking back, I see I am assuming a lot and we all know what happens when we ASSUME anything!!!

I will say that it was a lot more fun working with all the mirrors instead of just a few. My units increased significantly, too.

I would say I wasted all that time working on a problem that I really didn't have all the facts on, but I didn't have anything better to do anyway, and it was fun! Hopefully the Prof. will make an appearance in the morning and clear up this mess so we can get down to some serious time wasting!!! LOL

I will refrain from posting any more "scores" until I find out the real rules. I bid you all Good Night!

I apologize for the delay in answering. My wife dragged my out of the house to go indoor rock climbing and Christmas shopping (I'm a bit of a homebody by nature). All your assumptions where correct, however I sometimes forget that I know the intended rules while others don't.

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Can I use only one mirror 4x at the position on the line between 7E and 7F? If possible then we get beam of infinite length

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Can I use only one mirror 4x at the position on the line between 7E and 7F? If possible then we get beam of infinite length

I'm not sure what you mean. The mirrors are actually square or rectangular blocks with four (or all, doesn't matter) mirrored surfaces. The OP says they cannot touch, plus if you place one at 7F the beam is reflected away from 7E.

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"If the smallest mirror were to be placed within the grid block at 2B,

where would the beam exit the grid (assuming no other mirrors were

used)?" with "It would exit at the middle of 3A." In this case, the

total path length would be 1 (4 diagonal bits each of length Â¼).

Is this correct?

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"If the smallest mirror were to be placed within the grid block at 2B,

where would the beam exit the grid (assuming no other mirrors were

used)?" with "It would exit at the middle of 3A." In this case, the

total path length would be 1 (4 diagonal bits each of length Â¼).

Is this correct?

The beam length would be 2, because it would traverse a total of two diagonals or four half diagonals.

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The beam length would be 2, because it would traverse a total of two diagonals or four half diagonals.

Thanks, I made a mental miscalculation. I think that I completely understand the problem now.

Time to attack it!

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Here's my pitiful first attempt.

Path length 51.5 using:

one 4-long at 9C-9D-9E-9F

two 3-longs at 6A-7A-8A and 5J-6J-7J

three 2-longs at 1C-1D, 1H-1I, and 3A-4A

three 1-longs at 1F, 3J, and 9H (could put another at 10J, but it doesn't change beam)

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41,5 only - I desperately tried to keep the mirrors horizontal. ( I thought they were not allowed to be rotated at all. )

anyway, just to keep the things moving -

1 cell blocks - A6, A8, D7, G10,

2 cell blocks - I4-J4, I6-J6, I8-J8,

3 cell blocks - C10-E10, E1-G1,

4 cell blocks - D4-G4.

Edited by = W =
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Here's my pitiful first attempt.

Path length 51.5 using:

one 4-long at 9C-9D-9E-9F

two 3-longs at 6A-7A-8A and 5J-6J-7J

three 2-longs at 1C-1D, 1H-1I, and 3A-4A

three 1-longs at 1F, 3J, and 9H (could put another at 10J, but it doesn't change beam)

I get 51.5 units as well, but my mirrors were placed a slightly differently (only affecting the exit location of the beam).

1x4u @ J7-J10

2x3u @ F1-H1, F9-H9

3x2u @ A3-A4, A8-A9, J2-J3

4x1u @ A6, D1, D9, J5

The beam exits from the middle of the outer edge (bottom) of D10 at a 45 degree angle to the right.

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41,5 only - I desperately tried to keep the mirrors horizontal. ( I thought they were not allowed to be rotated at all. )

anyway, just to keep the things moving -

1 cell blocks - A6, A8, D7, G10,

2 cell blocks - I4-J4, I6-J6, I8-J8,

3 cell blocks - C10-E10, E1-G1,

4 cell blocks - D4-G4.

I think that the Professor meant that rotations which allow mirrors not to be completely contained in consecutive cells are not allowed.

This makes the number of ways to place the mirrors finite, albeit very large. If my first attempt isn't acceptable, I trust Templeton will

post that.

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41,5 only - I desperately tried to keep the mirrors horizontal. ( I thought they were not allowed to be rotated at all. )

anyway, just to keep the things moving -

1 cell blocks - A6, A8, D7, G10,

2 cell blocks - I4-J4, I6-J6, I8-J8,

3 cell blocks - C10-E10, E1-G1,

4 cell blocks - D4-G4.

The mirrored blocks (other then square) may be placed horizontally or vertically, for example a mirrored block of length 3 may be placed in any three consecutive cells in a row or column.

I think that the Professor meant that rotations which allow mirrors not to be completely contained in consecutive cells are not allowed.

This makes the number of ways to place the mirrors finite, albeit very large. If my first attempt isn't acceptable, I trust Templeton will

post that.

Your attempt is quite acceptable, but much less then the longest length that I have. Although, to be honest, there may exist a better solution then that.

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51.5 total length

51.5 total length

Edited by Prof. Templeton
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A possible alternative problem would be finding the arrangement of mirrors that produces the longest finite arrangement if they can touch.

Edited by Morningstar

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