Problem:
Let △A0B0C0 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−1Cn−1. Likewise, define Bn to be the foot of the altitude from Bn−1 to line An−1Cn−1, and Cn to be the foot of the altitude from Cn−1 to line An−1Bn−1. What is the least positive integer n for which △AnBnCn is obtuse?
Answer Choices:
A. 10
B. 11
C. 13
D. 14
E. 15
Solution:
For all nonnegative integers n, let ∠CnAnBn=xn,∠AnBnCn=yn, and ∠BnCnAn=zn.
Note that quadrilateral A0B0A1B1 is cyclic since ∠A0A1B0=∠A0B1B0=90∘; thus, ∠A0A1B1= ∠A0B0B1=90∘−x0. By a similar argument, ∠A0A1C1=∠A0C0C1=90∘−x0. Thus, x1=∠A0A1B1+ ∠A0A1C1=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