A lot of reflections
Triangle A B C has vertices A = ( 0 , 1 ) , B = ( 1 , 0 ) , and C = ( 0 , 0 ) in the coordinate plane.
The insides of the sides are lined with mirrors, and then a laser beam is fired from the origin with a slope of 3 7 9 3 1 4 .
Which corner of the triangle will the laser beam hit first?
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3 solutions
An error in your proof (that doesn't affect the solution to this problem). If the parity is (even,even) then the slope is not a fraction in lowest terms. Reduce it to find a closer point.
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Ah I see. Although, with the way I explained it (looking for the first time the line crosses a lattice point), that issue will never come up anyways.
I'll add a note to my solution about the ( even , even ) parity.
This is reminiscent of the Advanced billiards problem from a while back.
what does undefined slope mean?
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Undefined slope means vertical (up and down) slope.
Man, this is so awesome. I was all over the place on this problem.
I just don't understand why even or odd even effects it could you please explain
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The simplest way would probably be to look at some slopes that actually hit a lattice point on the small grid above.
Slope 1/1 hits C (odd, odd) at the point (1,1)
Slope 1/2 hits A (even, odd) at the point (2,1)
Slope 2/1 hits B (odd, even) at the point (1,2)
Slope 2/2 reduces to 1/1 so it does not actually hit point (2,2)
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Ok thanks is the rule that can be applied to other mirror problems
What is this field of maths? I haven't heard of lattice points, parity and I've no idea what even,odd you're referring to? Where can I learn more about this?
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Lattice points are simply the points in the coordinate plane with integer coordinates.
Parity simply refers to whether an integer is even or odd.
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Thanks! What if the coordinates were (0,0),(2,0),(0,2)? By the logic of your solution the line will not hit any vertex (but that seems wrong; the line will still hit (2,0)). I didn't get why odd/even parity matters.. it's like a formula you used but where did you get that from?
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@Yathish Dhavala – Then the first lattice point hit is ( 2 p , 2 q ) and look at the parity of ( p , q ) . The rest of the logic is the same.
@Yathish Dhavala – Observe the tiling of triangle A B C in the coordinate plane as I have shown above. You will notice that point A always has an even x -coordinate and an odd y -coordinate. Point B always has an odd x -coordinate and an even y -coordinate. And so on.
How did you prove this??
Could you explain the parity clearly?Thank you.
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Parity simply refers to whether an integer is even or odd. If you look at the tiling in my image, you see:
- A is on points ( 0 , 1 ) and ( 2 , 1 ) .
- B is on points ( 1 , 0 ) and ( 1 , 2 ) .
- C is on points ( 0 , 0 ) , ( 0 , 2 ) , ( 2 , 2 ) , ( 2 , 0 ) , and ( 1 , 1 ) .
You can then notice a pattern with the parity of the x - and y - coordinates, and this pattern continues as the tiling is extended outward.
If the slope is irrational, the beam will continue forever without hitting a corner.
If the slope is a rational number write it in lowest terms as q p
If p and q are both odd it will hit corner C.
If p < q and p is even it will hit corner B, if q is even it will hit corner A.
If p > q the opposite holds.
This can easily be seen by reflecting the triangle over its sides to fill the 1st quadrant. The images of A, B, and C end up as the above rules indicate.
Since 3 1 4 < 3 7 9 and 3 1 4 is even the answer is B.
I'm not seeing how p < q or p > q matters for the solution. Did I miss something with my solution?
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Our solutions are mostly the same. A slope of 2/3 has P even and < Q in my notation and hits (3,2) which is (odd,even) in yours so it hits corner B. A slope of 3/2 has P odd and > Q in my notation and hits (2,3) which is (even,odd) in yours so it hits corner A.
You missed what happens if the fraction can be reduced. There's is no such thing as (even,even). A slope of 2/4 is the same as 1/2 so it will hit the corner A at (2,1) before coming back to C.
I wonder if I should have made this harder by making the slope something like 314/378. Would a lot of people have picked C?
It doesn't seem as though it matters if p<q or q<p. If the slope was 380/379, the first lattice point the laser intersects would be at (379,380), which corresponds to corner B, but by your solution would result in corner A.
by reflectiong sides and points following red line ( y = 3 7 9 3 1 4 x )
(odd, odd). (even, even) : C
(odd, even) : B
(even, odd) : A
and last point is (379,314) : B
so
ans) B
There's no such thing as (even,even) because the slope would reduce until at least one of them is odd.
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yes, that's right. But those points in my solution are all lattice points Andrew Hayes's sketch can help you to understand my intention
Tile the coordinate plane with the triangle A B C like so:
Now to find the reflections of the laser, simply draw a straight line through this tiling. It can be seen that the first time the laser strikes A , B , or C is when the laser strikes a lattice point. We see that A has parity ( even , odd ) , B has parity ( odd , even ) , and C has parity ( even , even ) or ( odd , odd ) .
The line y = 3 7 9 3 1 4 x will first cross a lattice point at the point ( 3 7 9 , 3 1 4 ) . Since this point has parity ( odd , even ) , it corresponds to point B .
We can generalize in the following way:
Otherwise, put the slope into reduced fraction form, q p (as in @Jeremy Galvagni 's solution)
Note that it's not possible for the parity to be ( even , even ) , because then q p is not in reduced terms.