number z corresponding to a point on the segment
DE satisfies
z=+λλλ⎜⎛⎝^1 ab+ i⎞⎠⎟+ − ⎝⎜⎛ +bc+ i⎠⎟⎞ ≤≤
421 3
42() , 01Substitute the above expression into the equation
of the curve z = z 0 cos^4 t + 2z 1 cos^2 t · sin^2 t + z 2 sin^4 t,
and separate the real and imaginary parts from both
sides to give the following two equations,
3
42
1
2
1222 4422− =++ − =+ +λλλsin cos sin ,[ ( ) ] cos sin costt tab c a t b t t cctsin^4⎧
⎨⎪⎪⎩⎪
⎪Eliminating λ from the equations, we get
3
42
()ac− +bc+= acos^4 t + (2b + a – c) sin^2 tcos^2 t + asin^4 t
= a(1 – 2sin^2 t cos^2 t) + (2b + a – c) sin^2 t cos^2 t
= a + (2b – a – c)sin^2 t cos^2 t.
Then (2b – a – c) sin^22 cos^1
4
⎜⎝⎛ tt− ⎞⎠⎟=^0.
Since A, B, C are non-collinear, we know that
zzz 1021
2≠ ()+. So 2b – a – c^ ≠ 0.Then sin^2 t cos^2 t = sin^2 t(1 – sin^2 t) =^1
4
, that is,sin^2
1 2
2⎛⎝⎜ t− ⎞⎠⎟= 0. Then we have 3
421
41
21
22
−λ=+⎛⎝⎜ ⎠⎟⎞ = ,^soλ=^1 ∈[ , ].
2(^01) That means that the curve and
the line DE have one and only one common point
and the complex number corresponding to this
common point is
z=+^1 ⎛⎝⎜ ab+ i⎟⎞⎠++⎛⎝⎜ bc+ i⎞⎠⎟
2
1
42
1
2
3
42
=+
1 ++
2
2
4
ac bi
- Working out the first few terms gives us an idea
of how the given sequence develops:
ns 2 n – 1 s 2 n
1 ab
2 b – a 2 b – a
3 b 3 b – a
42 b – a 5 b –2a
53 b – a 8 b – 3a
65 b – 2a 13 b – 5a
78 b – 3a 21 b – 8a
It appears that the coefficients in the even terms
form a Fibonacci sequence and from the 5th term,
every odd term is a repeat of the third term before
it.
These observations are true for the entire sequence
since, for m ≥ 1, we have:
s 2 m + 2 = s 2 m + 1 + s 2 m
s 2 m + 3 = s 2 m + 2 – s 2 m + 1 = s 2 m
s 2 m + 4 = s 2 m + 3 + s 2 m + 2 = s 2 m + 2 + s 2 m
So, defining F 1 = 1, F 2 = 2 and Fn = Fn – 1 + Fn – 2
for n ≥ 3, we have s 2 n = bFn – aFn – 2 f o r
n ≥ 3. Since a = 3 and b < 1000, none of the first
five terms of the given sequence equal 2015. So we
are looking for integer solutions of
bFn – 3Fn – 2 = 2015 for n ≥ 3.
s 6 = 3b –3 = 2015, has no solution.
s 8 = 5b – 6 = 2015, has no solution.
s 10 = 8b – 9 = 2015 implies b = 253.
For n ≥ 6 we have b = 2015/Fn + 3Fn –2/Fn. Since
Fn increases, we have Fn ≥ 13 and Fn – 2/Fn < 1
for n ≥ 6.
Hence b < 2015/13 + 3 = 158. So the largest value
of b is 253.