Maths Olympiad Prep

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, 2010

Combinatorics Difficulty 6.5 National olympiad Prove it Austria

We consider points with integer coordinates in the rectangle with corners in (0,0)(0,0), (n,0)(n,0), (n,2)(n,2) and (0,2)(0,2). It is possible to move from a point (a,b)(a,b) in the rectangle to either points (a+1,b)(a+1,b), (a+1,b+1)(a+1,b+1) or (a,b1)(a,b-1) if the second point is also in the given rectangle.
How many possible paths are there from (0,0)(0,0) to (n,2)(n,2) under these rules?

Solution

Let aka_k, bkb_k and ckc_k be the number of possible paths leading from (0,0)(0,0) to (k,0)(k,0), (k,1)(k,1) and (k,2)(k,2) respectively. It is obvious that a0=1a_0 = 1, b0=0b_0 = 0 and c0=0c_0 = 0 hold. Furthermore, for k1k \ge 1 we have the recursive equations
ck=bk1+ck1,bk=ak1+bk1+ck andak=ak1+bk. \begin{aligned} c_k &= b_{k-1} + c_{k-1}, \\ b_k &= a_{k-1} + b_{k-1} + c_k \text{ and} \\ a_k &= a_{k-1} + b_k. \end{aligned}
From the first equation, we obtain bm=cm+1cmb_m = c_{m+1} - c_m for m0m \ge 0. Substituting in the second equation therefore yields
am=cm+2cm+1cm+1+cmcm+1=cm+23cm+1+cm a_m = c_{m+2} - c_{m+1} - c_{m+1} + c_m - c_{m+1} = c_{m+2} - 3c_{m+1} + c_m
for m0m \ge 0, and substitution in the third equation finally yields
cm+23cm+1+cm(cm+13cm+cm1)(cm+1cm)=cm+25cm+1+5cmcm1=0 c_{m+2} - 3c_{m+1} + c_m - (c_{m+1} - 3c_m + c_{m-1}) - (c_{m+1} - c_m) = c_{m+2} - 5c_{m+1} + 5c_m - c_{m-1} = 0
for m1m \ge 1. The characteristic equation of the recursion is q35q2+5q1=0q^3 - 5q^2 + 5q - 1 = 0, and since q35q2+5q1=(q1)(q24q+1)q^3 - 5q^2 + 5q - 1 = (q-1)(q^2 - 4q + 1), the roots of the characteristic equation are q1=1q_1 = 1, q2=2+3q_2 = 2 + \sqrt{3} and q3=23q_3 = 2 - \sqrt{3}.
It follows that the required values are given by expressions of the form cn=A+B(2+3)n+C(23)nc_n = A + B \cdot (2+\sqrt{3})^n + C \cdot (2-\sqrt{3})^n. Since we know c0=c1=0c_0 = c_1 = 0 and c2=1c_2 = 1, we obtain the system of equations
A+B+C=0A+(2+3)B+(23)C=0A+(7+43)B+(743)C=1. \begin{aligned} A + B + C &= 0 \\ A + (2 + \sqrt{3})B + (2 - \sqrt{3})C &= 0 \\ A + (7 + 4\sqrt{3})B + (7 - 4\sqrt{3})C &= 1. \end{aligned}
Solving this system of equations yields A=12A = -\frac{1}{2}, B=14312B = \frac{1}{4} - \frac{\sqrt{3}}{12} and C=14+312C = \frac{1}{4} + \frac{\sqrt{3}}{12}, and the number of possible paths is therefore given by the expression
cn=12+(33)(2+3)n12+(3+3)(23)n12. c_n = -\frac{1}{2} + \frac{(3 - \sqrt{3})(2 + \sqrt{3})^n}{12} + \frac{(3 + \sqrt{3})(2 - \sqrt{3})^n}{12}.

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