MathDB

5C

Part of 2023 LMT Fall

Problems(1)

2023 Fall Theme p5C

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12/23/2023
In equilateral triangle ABCABC, AB=2AB=2 and MM is the midpoint of ABAB. A laser is shot from MM in a certain direction, and whenever it collides with a side of ABCABC 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. 21\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 33. Consider one of these equilateral triangles ABCABC with MM being the midpoint of AB=2AB=2. Find the sum of the three minimum integer distances from MM 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("AA",A+(0,0.25),N); label("BB",B-(0.25,0),SW); label("CC",C+(0.25,0),SE); label("MM",M,S); [/asy] It is trivial to see that the vertical distance between MM and a given vertex is n3n\sqrt{3} for n∈N0n \in \mathbb{N}^{0}. If nn is even, the horizontal distance between OO and a given vertex is 1+2m1+2m for m∈N0m \in \mathbb{N}^{0}. If nn is odd, the horizontal distance is 2m2m for m∈N0m \in \mathbb{N}^{0}. We consider two separate cases:
1.1. nn is even. We thus want to find l∈Nl \in \mathbb{N} such that (n3)2+(1+2m)2=l2.\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.Make the substitution 1+2m=k1+2m=k to get that 3n2+k2=l2.3n^2+k^2=l^2.Notice that these equations form a family of generalized Pell equations y2−3x2=Ny^2-3x^2=N with N=k2N=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 ll values, and that gives us an upper bound on what the three ll values can be. From there, a simple bash of lower ll 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 (xn,yn)(x_n,y_n) to the above equation with sufficiently small xnx_n follow the formulaxn3+yn=(x03+y0)(un3+vn),x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),where (x0,y0)\left(x_0,y_0\right) is a solution to the generalized Pell equation and (un,vn)\left(u_n,v_n\right) are solutions to the Pell equation y2−3x2=1y^2-3x^2=1. Remember that the solutions to this last Pell equation satisfyun3+vn=(u03+v0)ku_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^kwhere the trivial positive integer solution (u0,v0)=(1,2)\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 3\sqrt{3}).
We thus get that(u1,v1)=(4,7),(u2,v2)=(15,26),(u2,v2)=(56,97)…\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)(u,v)=(0,1) is another solution).
From here, note that kk must be odd since k=1+2mk=1+2m for m∈N0m \in \mathbb{N}^{0}. For k=1k=1, the smallest three solutions to the Pell equation with nn 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 ll are thus 1,7,971,7,97. 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=3k=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)(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)(n,m,l)=(0,0,1) found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of ll are 1,7,211,7,21.
For k≥5k \ge 5, it is evident that there are no more smaller integral values of ll that can be found using the multiplicative principle: the solution set (n,m,l)=(0,k−12,k)(n,m,l)=\left(0,\dfrac{k-1}{2},k\right) is always extraneous for k>1k > 1 since it is equivalent to the path traced out by (0,0,1)(0,0,1) as described above, and any other solutions will give larger values of ll.
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=3,5,9,11l=3,5,9,11 gives no solutions for any odd kk and even nn, however n=13n=13 gives k=11k=11 and n=4n=4, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest ll values are 1,7,131,7,13.
22. nn is odd. We thus want to find l∈Nl \in \mathbb{N} such that (n3)2+(2m)2=l2.\left(n\sqrt{3}\right)^2+(2m)^2=l^2.Make the substitution 2m=k2m=k to get that 3n2+k2=l2.3n^2+k^2=l^2.This is once again a family of generalized Pell equations with N=k2N=k^2, however this time we must have kk even instead of kk odd. However, note that there are no solutions to this family of Pell equation with nn odd: k2≡0 (mod 4)k^2 \equiv 0 \text{ (mod }4) since kk is even, and 3n2≡3 (mod 4)3n^2 \equiv 3 \text{ (mod }4) since nn is odd, however 0+3≡3 (mod 4)0+3 \equiv 3 \text{ (mod }4) is not a possible quadratic residue mod 44. Thus, this case gives no solutions.
Our final answer is thus 1+7+13=211+7+13=\boxed{21}.
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