Problem:
Let △A0​B0​C0​ be a triangle whose angle measures are exactly 59.999∘,60∘, and 60.001∘. For each positive integer n define An​ to be the foot of the altitude from An−1​ to line Bn−1​Cn−1​. Likewise, define Bn​ to be the foot of the altitude from Bn−1​ to line An−1​Cn−1​, and Cn​ to be the foot of the altitude from Cn−1​ to line An−1​Bn−1​. What is the least positive integer n for which △An​Bn​Cn​ is obtuse?
Answer Choices:
A. 10
B. 11
C. 13
D. 14
E. 15
Solution:
For all nonnegative integers n, let ∠Cn​An​Bn​=xn​,∠An​Bn​Cn​=yn​, and ∠Bn​Cn​An​=zn​.
Note that quadrilateral A0​B0​A1​B1​ is cyclic since ∠A0​A1​B0​=∠A0​B1​B0​=90∘; thus, ∠A0​A1​B1​= ∠A0​B0​B1​=90∘−x0​. By a similar argument, ∠A0​A1​C1​=∠A0​C0​C1​=90∘−x0​. Thus, x1​=∠A0​A1​B1​+ ∠A0​A1​C1​=180∘−2x0​. By a similar argument, y1​=180∘−2y0​ and z1​=180∘−2z0​.
Therefore, for any positive integer n, we have xn​=180∘−2xn−1​ (identical recurrence relations can be derived for yn​ and zn​ ). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to n (and the coefficient of x0​ is (−2)n ). Hence, we let xn​=pqn+r+(−2)nx0​. We will solve for p,q, and r by iterating the recurrence to obtain x1​=180∘−2x0​,x2​=4x0​−180∘, and x3​=540−8x0​. Letting n=1,2,3 respectively, we have
pq+r=180(1)
pq2+r=−180(2)
pq3+r=540(3)
Subtracting (1) from (3), we have pq(q2−1)=360, and subtracting (1) from (2) gives pq(q−1)=−360. Dividing these two equations gives q+1=−1, so q=−2. Substituting back, we get p=−60 and r=60.
We will now prove that for all positive integers n,xn​=−60(−2)n+60+(−2)nx0​=(−2)n(x0​−60)+60 via induction. Clearly the base case of n=1 holds, so it is left to prove that xn+1​=(−2)n+1(x0​−60)+60 assuming our inductive hypothesis holds for n. Using the recurrence relation, we have
xn+1​=180−2xn​
=180−2((−2)n(x0​−60)+60)
=(−2)n+1(x0​−60)+60
Our induction is complete, so for all positive integers n,xn​=(−2)n(x0​−60)+60. Identical equalities hold for yn​ and zn​.
The problem asks for the smallest n such that either xn​,yn​, or zn​ is greater than 90∘. WLOG, let x0​=60∘, y0​=59.999∘, and z0​=60.001∘. Thus, xn​=60∘ for all n,yn​=−(−2)n(0.001)+60, and zn​=(−2)n(0.001)+ 60. Solving for the smallest possible value of n in each sequence, we find that n=15 gives yn​>90∘. Therefore, the answer is (E)15​ .
OR
We start from Solution 1 until we reach the recurrence relation xn​=180−2xn−1​. Iterate this again, to get xn−1​=180−2xn−2​. Subtract the two, getting xn​=−xn−1​+2xn−2​. This recurrence has characteristic equation x2+x−2=0=(x+2)(x−1)=0⟺x=−2,1. Now, write
xn​=p+q.(−2)n
We obtain similar recursions for y,z that can be easily solved by getting x1​,y1​,z1​ from the original recursive formula and then using those two values to solve for p and q. Then proceed with the last paragraph of Solution 1 .
The problems on this page are the property of the MAA's American Mathematics Competitions