MathDB

In equilateral triangle

ABC

AB=2

and

M

is the midpoint of

AB

. A laser is shot from

M

in a certain direction, and whenever it collides with a side of

ABC

it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled.
Proposed by Jerry Xu
Solution.

\boxed{21}

Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows:
Tessellate the plane with equilateral triangles of side length

3

. Consider one of these equilateral triangles

ABC

with

M

being the midpoint of

AB=2

. Find the sum of the three minimum integer distances from

M

to any vertex in the plane.
[asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("

A

",A+(0,0.25),N); label("

B

",B-(0.25,0),SW); label("

C

",C+(0.25,0),SE); label("

M

",M,S); [/asy] It is trivial to see that the vertical distance between

M

and a given vertex is

n\sqrt{3}

for

n \in \mathbb{N}^{0}

. If

n

is even, the horizontal distance between

O

and a given vertex is

1+2m

for

m \in \mathbb{N}^{0}

. If

n

is odd, the horizontal distance is

2m

for

m \in \mathbb{N}^{0}

. We consider two separate cases:

1.

n

is even. We thus want to find

l \in \mathbb{N}

such that

\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.

Make the substitution

1+2m=k

to get that

3n^2+k^2=l^2.

Notice that these equations form a family of generalized Pell equations

y^2-3x^2=N

with

N=k^2

. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small

l

values, and that gives us an upper bound on what the three

l

values can be. From there, a simple bash of lower

l

values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case.
By the multiplicative principle some set of solutions

(x_n,y_n)

to the above equation with sufficiently small

x_n

follow the formula

x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),

where

\left(x_0,y_0\right)

is a solution to the generalized Pell equation and

\left(u_n,v_n\right)

are solutions to the Pell equation

y^2-3x^2=1

. Remember that the solutions to this last Pell equation satisfy

u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k

where the trivial positive integer solution

\left(u_0, v_0\right)=(1,2)

(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of

\sqrt{3}

).
We thus get that

\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots

(also don't forget that

(u,v)=(0,1)

is another solution).
From here, note that

k

must be odd since

k=1+2m

for

m \in \mathbb{N}^{0}

. For

k=1

, the smallest three solutions to the Pell equation with

n

even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of

l

are thus

1,7,97

. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex).
For

k=3

, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that

(n,m,l)=(0,1,3)

is extraneous since is equivalent to the path that is traced out by the solution

(n,m,l)=(0,0,1)

found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of

l

are

1,7,21

.
For

k \ge 5

, it is evident that there are no more smaller integral values of

l

that can be found using the multiplicative principle: the solution set

(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)

is always extraneous for

k > 1

since it is equivalent to the path traced out by

(0,0,1)

as described above, and any other solutions will give larger values of

l

.
Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that

l=3,5,9,11

gives no solutions for any odd

k

and even

n

, however

n=13

gives

k=11

and

n=4

, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest

l

values are

1,7,13

2

n

is odd. We thus want to find

l \in \mathbb{N}

such that

\left(n\sqrt{3}\right)^2+(2m)^2=l^2.

Make the substitution

2m=k

to get that

3n^2+k^2=l^2.

This is once again a family of generalized Pell equations with

N=k^2

, however this time we must have

k

even instead of

k

odd. However, note that there are no solutions to this family of Pell equation with

n

odd:

k^2 \equiv 0 \text{ (mod }4)

since

k

is even, and

3n^2 \equiv 3 \text{ (mod }4)

since

n

is odd, however

0+3 \equiv 3 \text{ (mod }4)

is not a possible quadratic residue mod

4

. Thus, this case gives no solutions.
Our final answer is thus

1+7+13=\boxed{21}

Problem Statement