1. If an equilateral triangle shares two vertices with ABCD, those two vertices form a side of the equilateral triangle. Given a side of a triangle, there
are exactly two possible positions
of the final point to form an equilateral
triangle. Hence there are 42 · 2 = 12 possible triangles.
It remains to show that no two of these triangles are the same (so there
is no overcounting), which is equivalent to showing that no three points
of ABCD form an equilateral triangle. However, as any three points of
ABCD form a 45 − 45 − 90 triangle, there is no overcounting and our
answer is 12 .
ab
a+b
=
⇐⇒ (a+b)2 = 4ab ⇐⇒ (a−b)2 =
a+b
4
0 ⇐⇒ a = b. Hence the only possible value of a is 42 .
2. If a?b = ab, we have
3. Since ∠BDE = 90◦ , we have ∠ADF = ∠AF D = 30◦ , and so ∠BAC =
s
120◦ . Let DE = EF = F D = s. Then AF = AD = √ , F C = DB =
3
√
√
4s 3
. Then
s 3, so AB = BC =
3
1 2
s sin 60◦
2
√ !2
4 3
sin 120◦
3
[DEF ]
=
[ABC]
1
2
=
=
1
√ !2
4 3
3
3
16
4. Since a < b < c, we have ab < ac < bc, hence ab − 1 < ca − 1 < bc − 1.
We can quickly discard ab − 1 = 22 , 42 , 62 , 102 as a, b > 1. If ab − 1 = 32 ,
we have a = 2, b = 5, and so both 2c − 1, 5c − 1 are perfect squares.
The smallest value of c for which this is possible is c = 13. This means
that we can dispense with ab − 1 = k 2 for k ≥ 13, as this would give
c > b > k ≥ 13, irrelevant to our investigation. Hence we only need to
look at ab − 1 = 52 , 72 , 82 , 92 , 112 , 122 .
• If ab−1 = 25, we have a = 2, b = 13, which doesn’t produce a smaller
value of c.
• If ab−1 = 49, we have a = 5, b = 10 (other possibilities don’t produce
a smaller value of c), so 5c − 1 and 10c − 1 are both perfect squares,
which doesn’t hold for any 10 < c < 13.
• If ab−1 = 64, we have a = 5, b = 13, which doesn’t produce a smaller
value of c.
1
• If ab−1 = 81, we have a = 2, b = 41, which doesn’t produce a smaller
value of c.
• If ab − 1 = 121, we have a = 2, b = 61, which doesn’t produce a
smaller value of c.
• If ab−1 = 144, we hae a = 5, b = 29, which doesn’t produce a smaller
value of c.
Hence, (a, b, c) = (2, 5, 13 ) produces the smallest value of c.
5. We want to find how many lucky numbers are less than 474747. More
generally, we want to find how many lucky numbers are less than an ndigit lucky number x. We’ll do this by building a lucky number from the
left, and count how many lucky numbers are less than x throughout the
process.
Let f (k) denote the number of lucky numbers less than the number formed
by removing k digits from the left of x. Then f (n) = 0, and we are looking
for f (0). If the first digit of x is a 4, we can make the first digit of the
number we’re building either 4 or 0 (which means we skip this digit and
build an n − 1 digit lucky number). In the latter case, there are 2n−1
possible n − 1 digit lucky numbers, and in the former case there are f (1)
lucky numbers. Hence we’d have f (0) = 2n−1 + f (1). If, instead, the
first digit were a 7, the first digit could be 4 or 0 (which is effectively
the same case), or 7 (in which case there are f (1) possible numbers), so
f (0) = 2n + f (1).
This gives us a general method: f (k) = 2n−1−k + f (k + 1) if the kth digit
is a 4 and f (k) = 2n−k + f (k + 1) if the kth digit is a 7. Our answer is
then f (0) = 25 + 25 + 23 + 23 + 21 + 21 = 84 .
Note that if we replace ’4’ with ’0’, ’7’ with ’1’, and allow leading zeros,
there is a natural connection to binary numbers that is worth exploring.
6. Since ∠AED = ∠AF D = 90◦ , quadrilateral AEDF is cyclic. Hence
AD · EF = AE · DF + ED · AF .
We can easily calculate the area of ABC to be 84, so AD = 12. Hence
BD = 5 and DC = 9. Furthermore, we have 4DF C ∼ 4AF C, meaning
DF
9
36
=
=⇒ DF =
. Similarly, we have 4DEB ∼ 4ADB, so
12
15
5
ED
5
60
=
=⇒ ED =
. From the Pythagorean triples (5, 12, 13) =⇒
12
13
13
36 48
144
60 144
, 12) and (3, 4, 5) =⇒ ( , , 12), we have AE =
and
( ,
13 13
5 5
13
48
48 · 60 + 36 · 144
4 · 60 + 3 · 144
AF =
, so 12EF =
. Then EF =
=
5
5 · 13
5 · 13
672
.
65
2
7. We have sin x = sin(4x)−sin(2x) = sin(3x+x)−sin(3x−x) = sin 3x cos x+
sin x cos 3x − (sin 3x cos x − sin x cos 3x) = 2 sin x cos 3x = sin x. Hence
π
1
either sin x = 0 or cos 3x = = cos( ). The first case gives us the solution
2
3
π
5π
x = π, while the second gives us the solutions 3x = + 2πk,
+ 2πk for
3
3
17
nonnegative integers k. The maximal solution is thus
π, making our
9
answer 17 + 9 = 26 .
8. Let p(n, k) denote the probability that exactly k switches are on after n
minutes. If there are k switches on, the probability of selecting a switch
k
that is already on is , and the probability of selecting a switch that is
3
3−k
4−k
k+1
off is
. Hence p(n, k) =
p(n − 1, k − 1) +
p(n − 1, k + 1)
3
3
3
(as we could have either turned a switch on from a position that had one
less switch on or turned a switch off from a position that had one more
switch on).
4
1
1
Hence, we have p(n, 3) = p(n − 1, 2) + p(n − 1, 4) = p(n − 1, 2)
3
3
3
1
(as p(n − 1, 4) = 0). Similarly, p(n, 0) = p(n − 1, 1). Then p(n, 2) =
3
2
2
1
p(n − 1, 1) + p(n − 1, 3) = p(n − 1, 1) + p(n − 2, 2), and similarly
3
3
3
2
1
p(n, 1) = p(n − 1, 2) + p(n − 2, 1).
3
3
1
2
Let s(n) = p(n, 1) + p(n, 2). Then s(n) = s(n − 1) + s(n − 2). The
3
3
characteristic equation ofthis recurrence is 3x2 − 2x − 1 = 0 =⇒ x =
n
1
1
1, − , so s(n) = c1 + c2 −
for some constants c1 , c2 . As s(0) = 1
3
3
3
1
3n+1 + (−1)n
2
.
and s(1) = , we have c1 = , c2 = , so s(n) =
3
4
4
4 · 3n
Furthermore, since the parity of the number of on switches changes every
minute, p(n, 2) = 0 when n is even and p(n, 1) = 0 when n is odd, hence
p(n, 1) = s(n) when n is even and p(n, 2) = s(n) when n is even. As a
result, f (2015, 2) =
32016 + 1
.
4 · 32015
2
3
1
9. We have f (0) = 1, f (1) = , f (2) = , f (3) = , and so on, so we
2
3
5
Fn
conjecture f (n) =
where Fn is the nth Fibbonacci number (with
Fn+1
3
F0 = F1 = 1). Checking this, we get
Fn
=
Fn+1
1
Fn−1
1+
Fn
⇐⇒ Fn (Fn + Fn−1 ) = Fn+1 Fn
⇐⇒ Fn2 + Fn Fn−1 = Fn+1 Fn
⇐⇒ Fn2 = Fn (Fn+1 − Fn−1 )
F1 2
F1
1
F1 F2
·
·...·
=
=
.
F2 F3
F1 3
F1 3
377
p
1 ± 1 − 4(a − 2a2 )
2
2
10. We have b − b + a − 2a = 0, so b =
, so since b is an
2
2
2
integer, 8a −4a+1 is a perfect square. Let 8a −4a+1 = x2 . Then 16a2 −
8a + 2 = 2x2 , so we have (4a − 1)2 + 1 = 2x2 =⇒ (4a − 1)2 − 2x2 = −1.
Denote 4a−1 = y, so that y 2 −2x2 = −1. This is a√negative Pell
√ equation,
so the solutions are (y, x) = (an , bn ) where an + bn 2 = (1 + 2)2n−1 . We
also have the additional condition y ≡ −1 (mod 4), so n must be even.
√
Therefore,
the smallest solution occurs when n = 2 =⇒ an + bn 2 =
√
7 + 5 2 =⇒ y = 7, x = y =⇒ a = 2√ =⇒ b =√3. The second smallest
√
solution occurs when n = 4, so an + bn 2 = (1 + 2)7 = 239 + 169 2, so
a = 60 =⇒ b = 85, and a + b = 145 .
which is true. Hence f (1)f (2) . . . f (12) =
4