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7/18/2019 CRUXv33n5
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7/18/2019 CRUXv33n5
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8
9
?
N
25
x
2 ≤ N
x ≤ 5
15
1
3
x + x3 + x9 + x27 + x81 + x243
x2 − 1
H
A
n
3
1
n + 1
2
1
n − 1
6
6
14
9
20
35
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1
9
1233
2005
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9
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16
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..................................................................................................................................................................................................................................................................................................................................................................1
2
34
5
6
8
×9
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8
9
?
N
25
x
2 ≤ N
x ≤ 5
15
1
3
x+x3+x9+x27+x81+x243
x2 − 1
H
A
n
3
1
n + 1
2
1
n − 1
6
7/18/2019 CRUXv33n5
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x
12
x x
12 × 60
60
x/60
x
x
60 =
x
720 +
1
2
11x = 360
x = 32 811
32 811
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........................................
3.14
3
0.14
a1
a2
a3
a4
. . .
d
a2 − a1 = a3 − a2 = a4 − a3 = . . . = d
r = 0
a2
/a1
= a3
/a2
= a4
/a3
= . . . = r
x
n
y
x = n + y
n
0 ≤ y < 1
x
y
n
x
0 < x < 1
n = 0
x = y
y
n
x
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x ≥ 1
n ≥ 1
y < n ≤ x
y
n
x
n − y = x − n = d
d
x = n + y
x
n − y = y = d
2y = n
n
0 ≤ y < 1
n = 1
y = 12
x
x = n + y = 1 + 12
= 32
x
y
n
x
0 < x < 1
x ≥ 1
n ≥ 1
y < n ≤ x
y
n
x
n
y =
x
n = r
r = 0
x = n + y
x
n
y = 1 +
y
n = r
y
0
n/y
0 < y < 1
n/y > n
1 + (y/n) < 1 + (1/n) ≤ 2
n ≥ 1
n < n
y
= 1 + y
n ≤ 2
n
n < 2
n = 1
n = 1
1/y = 1 + y
y2 + y − 1 = 0
y = −1 ±
12 − 4(1)(−1)
2(1) =
−1 ±√ 5
2
y > 0
y = −1 +
√ 5
2
x
x = n + y = 1 + −1 +
√ 5
2 =
1 +√
5
2
60
60
40
15
A
B
C
D
BC
AD
AB
60
BC
7/18/2019 CRUXv33n5
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AB
60
EF
E
A
F
AB
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........... 60
A
B
E F
45 |AF |
|AE | = 40 |AB| = 60
F
AB
EF
∠EF A = 60
|AE | = |AF | tan 60 = |AF |√ 3
|AF | = |AE |/√ 3 = 40/√ 3
AEF
60
AEF 1
2 × 40 × 40√
3 × 60 = 48000√
3
3
15 × 60 × 60 = 54000 3
48000√ 3
3
48000√ 3
3
d
d × 60 × 60 = 48000√ 3
d = 40
3√ 3
40
3√ 3
1
2006
10
1
1
9
25
. . .
2006
452 = 2025
45 = 2 × 22 + 1
2025
22
22
1
2006
2025− 2006 = 19
2025
2006
22
3
1
(x, y)
x+y
xy
(5, 20)
5 + 20 = 25
5 × 20 = 100
3
3
x
3 + x = a2
3x = b2
a
b
3x
x
7/18/2019 CRUXv33n5
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3c2
c
x = 3c2
3 + x = a2
3(1 + c2) = a2
1 + c2
3
c
3
0
1
2
0
c = 3k
k
1 + c2 = 1 + 9k2
3
3
9k2
1
c = 3k + 1
k
1 + c2 = 9k2 + 6k + 2
3
3
9k2 + 6k
2
c = 3k + 2
k
1 + c2 = (9k2 + 12k + 3 ) + 2
3
3
9k2 + 12k + 3
x
y
2(x + y − 2) = y(x − y + 2)
x2(y − 1) + y2(x − 1) = xy − 1
a = x − 1
b = y − 1
2(a + b) = (b + 1)(a − b + 2)
(a + 1)2b + (b + 1)2a = (a + 1)(b + 1) − 1
ab(a + b + 3) = 0
a = 0
b = 0
a + b = −3
a = 0
2b = (b + 1)(−b + 2)
b2 + b −
2 = 0
(b + 2)(b −
1) = 0
b = 1
b = −2
b = 0
2a = a + 2
a = 2
a + b = −3
a = −b − 3
2(−3) = (b + 1)(−b − 3 − b + 2)
2b2 + 3b − 5 = 0
(2b+5)(b−1) = 0
b = −52
b = 1
a = −b−3
a = −12
a = −4
a b x y0 1 1 20 −2 1 −12 0 3 1
−12 −
52
12 −
32−4 1 −3 2
7/18/2019 CRUXv33n5
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tn = n2 + 20
n ≥ 1
n ≥ 1
tn
tn+1
81
ABC
∠ABC = ∠ACB = 40
P
∠P BC = 20
∠P CB = 30
BP = B A
n + 1
n(2n + 1)
n
n = 1
32 + 42 = 52
n = 2
102 + 112 + 122 = 132 + 142
n ≥ 1
7/18/2019 CRUXv33n5
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a
b
c
a +b+c
ab+ bc+ ca
a4 + b4 + c4
abc
√ x +
√ y +
√ z = 3
x√
x + y√
y + z√
z = 3
x2
√ x + y2
√ y + z2
√ z = 3
+
.................................................................................................................................................................................................................................................................................................................................................................
tn = n2 + 20
n ≥ 1
n ≥ 1
tn
tn+1
81
ABC
∠ABC = ∠ACB = 40
P
∠P BC = 20
∠P CB = 30
BP = B A
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n + 1
n(2n + 1)
n
n = 1
32 + 42 = 52
n = 2
102 + 112 + 122 = 132 + 142
n ≥ 1
a
b
c
a + b + c
ab + bc + ca
a4 + b4 + c4
abc
√ x +
√ y +
√ z = 3
x√
x + y√
y + z√
z = 3
x2√
x + y2√ y + z2
√ z = 3
+
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α
β
γ
A
B
C ABC
ABC
DBC
EAC
F BA
AD = EF
α = π/2
.
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A
B C
D
E
F
αβ
γ
β
β
β γ
γ
γ
α
α
α
ABC
DBC
AB = DB
AC = DC
AG
DG
BC
ABC
DBC
AG = AB sin β
GD = CD sin γ = AC sin γ
AD = AG + GD = AB sin β + AC sin γ
ACE
AC/ sin α = E A/ sin γ
EA = AC sin γ/ sin α AF B
AF = AB sin β/ sin α
α + β + γ = π
EF = E A + AF
EF = AC sin γ
sin α+
AB sin β
sin α
EF = AD/ sin α
α = π/2
EF = AD
EF = AD
sin α = 1
α = π/2
0 < α < π
EF = AD
α = π/2
x
y
z
x
y+
z3√
xyz
2+
y
z+
x3√
xyz
2+
z
x+
y3√
xyz
2≥ 12
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x
y +
z3√
xyz ≥ 2
xz
y 3√
xyz
y
z +
x3√
xyz ≥ 2
yx
z 3√
xyz
z
x +
y3√
xyz ≥ 2
zy
x 3√
xyz
x
y +
z3√
xyz
2+
y
z +
x3√
xyz
2+
z
x +
y3√
xyz
2≥ 4
xz
y 3√
xyz +
yx
z 3√
xyz +
zy
x 3√
xyz
4 xz
y 3√ xyz + yx
z 3√ xyz + zy
x 3√ xyz ≥ 4·3 3 xz
y 3√ xyz · yx
z 3√ xyz · zy
x 3√ xyz = 12
x
y +
z3√
xyz
2+
y
z +
x3√
xyz
2+
z
x +
y3√
xyz
2≥ 12
x = y = z
x1
x2
. . .
xn
x1
x2+
x3n√ x1x2 · · ·xn
α≥ 2αn
α ∈ (−∞,0)∪ (1,∞)
n = 3
α = 2
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(x, y)
x
y
0 ≤ x ≤ 7
0 ≤ y ≤ 7
(0, 0)
0
XY Z
[XY Z ]
A(x1, y1)
B(x2, y2)
C (x3, y3)
[ABC ] = 1
2
det
x1 y1 1
x2 y2 1x3 y3 1
= 1
2
x1y2 + x2y3 + x3y1 − x2y1 − x3y2 − x1y3
P (a, b)
Q(c, d)
O
[P QO] = 12|ad − bc|
[P QO] |ad − bc|
ad
bc
ad
bc
a
b
c
d
a
b
c
d 0
1
2
3
4
5
6
7
12 · 1
2 · 1
2 · 1
2 = 1
16
ad
bc
a
d
b
c
a
d 3
4
b
c
ad
bc
34 · 3
4 = 9
16
116
+ 916
= 58
S 2006 =2006k=1
(−1)k k2 − 3
(k + 1)!
n! = n · (n − 1) · · · 3 · 2 · 1
6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
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k ≥ 1
k2 − 3
(k + 1)! =
k(k + 1)
(k + 1)! − k + 1
(k + 1)! − 2
(k + 1)! =
1
(k − 1)! − 1
k! − 2
(k + 1)!
=
1
(k − 1)! − 1
(k + 1)!
−
1
k! +
1
(k + 1)!
S 2006 =
2006
k=1
(
−1)k
k2 − 3
(k + 1)!
=
2006k=1
(−1)k
(k − 1)! −
2006k=1
(−1)k
(k + 1)!
−2006k=1
(−1)k
k! +
2006k=1
(−1)k
(k + 1)!
=
−1 + 1 +
2006k=3
(−1)k
(k − 1)! −
2004k=1
(−1)k
(k + 1)! +
1
2006! − 1
2007!
−
−1 +
2006k=2
(−1)k
k! −
2005k=1
(−1)k+1
(k + 1)! +
1
2007!
=
1
2006! − 1
2007!
−−1 +
1
2007!
=
1
2006! − 2
2007! + 1 =
2007
−2
2007! + 1 =
2005
2007! + 1
λ > 0
x
..................................................................................................................................................................................................................................................................................................
A
x
B
x
C
θ
λ = tan θ
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A
BC
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θ
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. θ .
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2θ
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∠BAC =
∠ABC = θ
ABC
C
2θ
tan2θ = 2tan θ
1 − tan2 θ=
2λ
1 − λ2
1
f (x)
n
5
k
f (n)
5
2k
f (555) = 555555
f (x) = ax2 + bx + c
f (5) = 25a + 5b + c = 55
f (55) = 3025a + 55b + c = 5555
f (555) = 308025a + 555b + c = 555555
a = 95
b = 2
c = 0
f (x) = 95
x2 + 2x = x95
x + 2
55 · · · 55 k
= 5
1 + 101 + 102 + · · · + 10k−1
= 5
10k − 1
9
= 59
10k − 1
f
59
(10k − 1)
=
59
(10k − 1)
95
59
(10k − 1)
+ 2
= 59(10
k
− 1)(10k
+ 1) = 59(10
2k
− 1 ) = 5 5 · · · 55 2k
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2
2
2 · 2 = 4
4
16 · 9 · 4
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2
1
16 · 15 · 14
16·9·416·15·14 = 6
35
16·916·15 = 3
5
414 = 27
4
14
35 · 2
7 = 6
35
3!
3
43
= 4
3
4
3
2
4 · 4 · 3 · 2 = 96
96560
= 635
3
5×5
11
12
13
14
21
22
23
24
31
32
33
34
41
42
43
44
k
n × n
k ≤ n
7/18/2019 CRUXv33n5
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gcd
gcd
a
b
0
a
b
gcd(a, b)
a
b
gcd(a, b)
a
b
d
a
d
a
a = qd
q
gcd(0, 0) = 0
0
0
0
gcd
gcd(2, −4) = 2
gcd(3, 5) = 1
gcd(−13, 1) = 1
gcd(a, 0) = a
a
gcd(a, 0) = −a
a
gcd(a, 0) = |a|
gcd(b, 1) = 1
b
gcd(1977, 2007)
1977
2007
1977 = 3 × 659
659
659
1977
1
3
659
1977
2007 = 3 × 669 = 3 × 3 × 223
223
2007
1
3
9
223
669
2007
1977
2007
1
3
gcd(1977, 2007) = 3
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1977
2007
a
b
q
r
a = qb + r
gcd(a, b) = gcd(b, r)
2007 = 1(1977) + 30 ==⇒ gcd(2007, 1977) = gcd(1977, 30)1977 = 65(30) + 27 ==⇒ gcd(1977, 30) = gcd(30, 27)
30 = 1(27) + 3 ==⇒ gcd(30, 27) = gcd(27, 3)27 = 9(3) + 0 ==⇒ gcd(27, 3) = gcd(3, 0)
gcd(2007, 1977) = gcd(3, 0)
3
gcd
0
gcd
6540
1236
12
gcd
gcd
gcd
gcd
21n + 4
14n + 3
n
gcd(21n + 4, 14n + 3) = 1
7/18/2019 CRUXv33n5
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n
21n + 4 = 1(14n + 3) + (7n + 1)
14n + 3 = 2(7n + 1) + 1
7n + 1 = (7n + 1)(1) + 0
gcd(21n + 4, 14n + 3) = gcd(14n + 3, 7n + 1) = gcd(7n + 1, 1) = 1
gcd(c, b) = 1
gcd(ac,b) = gcd(a, b)
gcd(a, b)
gcd(ac, b)
gcd(ac, b)
gcd(a, b)
gcd(n2, 2n + 1) = 1
n
2n + 1
n2
2n + 1
gcd(2n + 1, 2) = 1
gcd(n2, 2n + 1) = gcd(2n2, 2n + 1)
gcd(2n + 1, 2) = 1
= gcd(−n, 2n + 1)
2n2 = n(2n + 1) + (−n)
= gcd(n, 2n + 1)
gcd(−1, 2n + 1) = 1
= gcd(n, 1)
2n + 1 = 2(n) + 1
= 1
gcd
7/18/2019 CRUXv33n5
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ABC
∠A = 70
CA + AI = BC
I
ABC
∠B
f : →
f (x + y) = f (x) + f (y) + 2547
x
y ∈
f (2004) = 2547
f (2547)
a
b
c
a + b + c ≥ 1
a +
1
b +
1
c
a3 + b3 + c3 ≥ a + b + c
ABC
A
B
C
BC
CA
AB |AC | = 2|CB |
|BA| = 2|AC | |CB | = 2|BA|
[ABC ] = 126
AA
BB
CC
(x, y)
xx+y = yxy
x2y = 1
ABCD
[ABCD] ≤ 14
(AB2 + BC 2 + CD2 + DA2)
f
[0, 1]
f (0) = 0
f (1) = 1
f (x) =
f (2x)
4
0 < x < 1
2
3
4 +
f (2x − 1)
4
1
2 ≤ x < 1
x
2
x = (0.b1b2b3 . . .)2
f (x)
2
7/18/2019 CRUXv33n5
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p
p2
+ 2543
16
ABC
∠B = 90
P
∠A
ABC
M
AB
A = M = B
AP
CP
M P
BC
AB
AC
D
E
N
∠M P B = ∠P CN
∠N P C = ∠M BP
[AP C ]/[ACDE ]
20
14
M
N
P
BC
CA
AB
ABC
G
BN =√ 32
AB
BMGP
ABC
xk
yk
k = 1
2
. . .
n)
kxkyk ≥ 1
nk=1
xk − yk
x2k
+ y2k
≤ 14
n√
n + 1
p
q
p2 − p + 1 = q3
(x, n)
xn+1 + 2n+1 + 1
xn + 2n + 1
7/18/2019 CRUXv33n5
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1 a +
1
b + 0.64
+ 1
b + 1
c + 0.64
+ 1
c + 1
a + 0.64
≥ 1.2
a > 0
b > 0
c > 0
abc = 1
y2 = 1 + x + x2 + x3 + x4
n
≥ 2
2n
−1
n
ABC
H
da
db
dc
H
BC
CA
AB
da + db + dc ≤ 3r
r
ABC
arcsinsin x = arccoscos x
a
a
ABC
O
AC
AB
BC
P
Q
∠P QB = ∠P BO
SA
SABC
ABC
σ1
σ2
A
B
C
σ
S
r1
r2
σ1
σ2
R
σ
n > 1
n
3n − 2n
7/18/2019 CRUXv33n5
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2004
2001
x + x√ x2 − 1
= 2004
f :
→
f x2y + f (x + y2) = x3 + y3 + f (xy)
x
y ∈
a
b
c
abc ≥ 1
a3 + b3 + c3 ≥ ab + bc + ca
2004
A1
A2
. . .
A2004
2003
1002
ω
ABC
L
N
E
ω
AB
BC
CA
LE
BC
H
LN
AC
J
H
J
N
E
AB
O
P
EJ
N H
S (HJNE )
S (ABOP ) = u2
S (COP ) = v2
S (F ) F
http://www.mathlinks.ro/Forum/viewtopic.php?t-15624
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f
x
y
f (xy)
f (x) − f (y)
= (x − y)f (x)f (y)
Φa,K
Φa,K (x) =
ax
x ∈ K
0
x /∈ K
K
∗ =
\0
a ∈
∗
a ∈
∗
K
∗
f = Φa,K
f (xy)
f (x) − f (y)
= (x − y)f (x)f (y)
x
y ∈
x
y ∈ K
xy ∈ K
axy(ax − ay) = (x − y)ax · ay
x
y /∈ K
Φ(x) = Φ(y) = 0
x ∈ K
y /∈ K
xy /∈ K
y = xy · 1
x
K
Φ(xy) = Φ(y) = 0
f
f
f (x0) = 0
x0 ∈
x = 1
y = 0
f (0) = 0
x0 = 0
y = 1
f (x)
f (x) − ax
= 0
a = f (1)
x = x0
a ∈
∗
f (x) = ax
f (x) = 0
K = x ∈
∗ | f (x) = ax = x ∈
| f (x) = 0
1 ∈
K
x1
x2 ∈
K
x1
= x2
f (x1
) = ax1
f (x2) = ax2
x = x1
y = x2
f (x1x2) = ax1x2
x1x2 ∈ K
x = x1
y = 1/x1
1/x1 ∈ K
x = x21
y = 1/x1
x21 ∈ K
K
∗
f = Φa,K
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7/18/2019 CRUXv33n5
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B
e2i(α
−γ )/3
C
e2πi/3 · e2i(α−γ )/3
B
C
A
eiθ
eiθ · e2πi/3
eiθ · e4πi/3 ABC
a = BC
b = CA
c = AB
α = ∠CAB
β = ∠ABC
γ = ∠BC A
β > α > γ
BA = 12
a
BD = c cos β
DA = BA − BD = 12
a − c cos β = R sin(β − γ )
R ABC
EB = R sin(α − γ )
N ABC
∠
DN A = 2(β − γ )
∠
EN B = 2(α − γ )
N AB
∠AN B = 2γ
∠AN B = 2γ + 23
(β − γ ) + 23
(α − γ ) = 23
(α + β + γ ) = 1 2 0
∠BN C = ∠C N A = 120 ABC
(a,b,c)
a ≤ b ≤ c
gcd(a,b,c) = 1
a3 + b3 + c3
a2b
b2c
c2a
(a,b,c)
p
a
b
a2b
a3 + b3 + c3
a3 + b3
p
c3
c
gcd(a, b) = 1
gcd(b, c) = gcd(c, a) = 1
a2
b2
c2
a3 + b3 + c3
a2b2c2
a3 + b3 + c3
3c3 ≥ a3 + b3 + c3 ≥ a2b2c2
3c ≥ a2b2
c2
a3 + b3
c2
≤ a3
+ b3
≤ 2b3
108c3 ≥ 4a6b6 ≥ a6c4
a ≤ c ≤ 108/a6
a7 ≤ 108
a = 1
b = 1
c2
a3 + b3 = 2
c = 1
(1, 1, 1)
b > 1
c > b > 1
c = b
gcd(b, c) = 1
c3 > b3 + 1
2c3 > 1 + b3 + c3 = a3 + b3 + c3 ≥ b2c2
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c >
1
2b2
2b3
>
1
4b4
b < 8
b = 2
c = 3
(1, 1, 1)
(1, 2, 3)
f :
→
x
y
f (x + y) + f (x)f (y) = f (x) + f (y) + f (xy)
f ≡ 0
f ≡ 2
f (x) = x
x = 0 = y
f (0)
2= 2f (0)
f (0) = 2
y = 0
f (x) = 2
x ∈
f ≡ 2
f (0) = 0
a = f (1)
x = 1
y = −1
af (−1) = a + 2f (−1)
f (−1) = a/(a − 2)
(x − 1, 1)
(−x + 1, −1)
(−x, 1)
(x, y)
f (x) + (a − 2)f (x − 1) = a
f (−x) + 2
a − 2f (−x + 1) − f (x − 1) =
a
a − 2
f (−x + 1) + (a − 2)f (−x) = a
f (x − 1)
f (−x + 1)
f (x) − (a − 2)f (−x) = 0
x −x
f (−x) − (a − 2)f (x) = 0
a /∈ 1
3
f (−x)
f (x) = 0
x ∈
f ≡ 0
a = 3
f (x) = 3 − f (x − 1)
x ∈
f (2) = 3 − f (1) = 0
f 52 = 3 − f
32 = f
12
2, 12
(x, y)
f 52 = f 12 + f (1) = f 12 + 3
a = 1
f (x) = f (x − 1) + 1
x ∈
f (x + n) = f (x) + n
x ∈
n ∈
f (n) = f (0 + n) = f (0) + n = n
n ∈
n
y
nf (x) = f (nx)
x ∈
n ∈
r = m/n ∈
x ∈
mf (x) = f (mx) = f (n · rx) = nf (rx)
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f (rx) = rf (x)
f (r) = f (r · 1) = rf (1) = r
y = r
f (x + r) = f (x) + r
y = x
f (x)2
= f (x2)
f (x) ≥ 0
x ≥ 0
f (−x) = −f (x)
f (x) ≤ 0
x ≤ 0
x ∈
r ∈
r < x
f (x) = f (x − r + r) = f (x − r) + r ≥ r
f (x − r) ≥ 0
f (x) ≥ x
f (x) ≤ r
r ∈
r > x
f (x) ≤ x
f (x) = x
P (x)
P (n) > n
n
m
P (1)
P P (1)
P P P (1)
. . .
m
P (x) = x + 1
P (i)(1)
i
P (1)
P
P (1)
. . .
P (n) > n
deg(P ) ≥ 1
P
P (x) = x + b
1 + b = P (1) > 1
b ≥ 1
P (x) = x + 1
b ≥ 2
P (1) ≡ 1 (mod b)
P (i)(1) ≡ 1 (mod b)
i ≥ 1
P (x) = 2x + b
2 + b = P (1) > 1
b
≥ 0
b = 0
P (i)(1) = 2i
i ≥ 1
P (x) = 2x + b
b ≥ 1
P (x) = ax + b
a ≥ 3
deg(P ) ≥ 2
N ∈
P (n) > 2n
n ≥ N
1 < P (1) < P (P (1)) < P (P (P (1))) < · · ·
k ∈
P (k)(1) ≥ N
r = P (k)(1)
m = P (k+1)(1) − P (k)(1)
r ≥ N
m = P (r) − r > r
1 ≤ i ≤ k
1 < P (i)(1) ≤ r < m
m
P (i)(1)
1 ≤ i ≤ k
P (k+1)(1) = m + r ≡ r (mod m)
P (i)(1) ≡ r (mod m)
i ≥ k + 1
P (i+1)(1) = P (P (i)(1)) ≡ P (r) = P (P (k)(1))
= P (k+1)(1) ≡ r (mod m)
P (i)(1) ≡ r (mod m)
i ≥ k + 1
ABC
r
r1
r2
r3
A
B
C
a > r1
b > r2
c > r3
ABC
a + b + c > r + r1 + r2 + r3
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s ABC
tan(A/2) = r1/s
a > r1
tan(A/2) < a/s < 1
tan(B/2) < 1
tan(C/2) < 1
A < π2
B < π2
C < π2
ABC
ABC
s > r + 2R
R ABC
r1 + r2 + r3 = r + 4R
r + r1 + r2 + r3 = 2r + 4R < 2s = a + b + c
n A
B
C 1
2
. . .
3n |A| = |B| = |C | = n
x ∈ A
y ∈ B
z ∈ C
x
y
z
(a,b,c)
a ∈ A
b ∈ B
c ∈ C
1 ∈ A
k
A
B
x ∈ C
x − 1 ∈ A
x ∈ C
x − 1 /∈ A
(1, x − 1, x)
x − 1 /∈ B
x − 1 ∈ C
x − 1 > k
(x − k , k , x)
(k − 1, x − k, x − 1)
x − k /∈ A
x − k /∈ B
x −
k ∈
C
(x −
k −
1, k , x −
1)
(1, x − k − 1, x − k)
x − k − 1 ∈ C
i ≥ 0
x − ik
x − ik − 1
C
i
k
A
B
C = c1
. . .
cn
|A| = |C | = n
A = c1 − 1
. . .
cn − 1
i
ci > k > 1
ci − 1 > 1
1 ∈ A
A
B
C
1
2
. . .
n |A| |B| |C | > 14
n
n
1
p
n
p − 1
p
n3 − 1
4 p − 3
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(x, y)
x
y
x3 + y3 = 7
xy(x + y) = −2
(x, y)
(x + y)3 = x3 + y3 + 3xy(x + y) = 1
x + y = 1
xy = −2
x
y
X 2 − X − 2 = 0
(x, y) = (−1, 2)
(2, −1)
ABC
K 1
K 2
K 3
K 4
K 1
K 2
K 3
K 4
K 1
K 2
K 3
K 4
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................................................................................................................................................................................................................................
...............................................................................................................
.
..
.
..
..
..
..
..
..
..
.
..
..
..
..
........................
............
..........................................................................................................................................
........................................................ .
..
.
..
..
..
..
..
..
..
.
..
..
..
..
..
..
....................
............
..........................................................................................................................................
........................................................
..
..
.
..
.
..
.
..
..
..
.
..
.
..
..
..
..
.
..
...................
.............
.............................................................................................................
.......................................
.............................................
.
..
..
.
..
..
..
.................
..
.....................
................................................
....................................................................................................
.........................................................
A B
C
K 1 K 2
K 3
K 4
Oi
K i
i = 1
2
3
4
ρ
γ
I
r
Γ
O
R ABC
K 1
γ
A
k = ρ/r
−−→AO1 = k−→AI
−−→IO1 = (1 − k)−→IA
−−→IO2 = (1 − k)
−→IB
−−→IO3 = (1 − k)
−→IC
O1O2O3
ABC
I
1 − k
O1O2O3
O4
O
O4
I
O
2ρ O1O2O3
2ρ =
1 − ρ
r
R
ρ = rR
2r + R
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N × N
N
N ≥ 3
n 1
n
2
N = 2003
N ≥ 3
0, 9
1
2
. . .
N 2
N = 2k + 1
I = −k −k + 1
. . .
k − 1
k
(0, 0)
(0, 0)
(x, y)
(x, y)
|x| + |y| > |x| + |y|
x
y > 0
(x + 1, y)
(x, y + 1)
•
(0, 0)
(−1, 0)
(1, 0)
(0, 1)
(0, −1)
•
(n, 0)
n ≥ 1
(n, 1)
(n, −1)
1
2(n + 1)
(n + 1, 0)
n
n + 1
•
(n − p, p)
n > p ≥ 1
(n − p + 1, p)
2(n − p) + 1
2(n + 1)
(n − p, p + 1)
2 p + 1
2(n + 1)
•
(0, n)
n ≥ 1
(1, n)
(−1, n)
1
2(n + 1)
(0, n + 1)
n
n + 1
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(0, 0)
(x, y)
x
y ≥ 0
p(x, y)
(x, y) = (0, 0)
p(x, y) = 1
4(|x| + |y|)
X
E (X )
(x, y) = (0, 0)
w(x, y)
(x, y)
(x, y)
n
w(x, y) = 1/n
w(x, y) = 0
(x, y)
(x, y)
|x
|+
|y
| ≤ 2k
E (X ) =
(x,y)=(0,0)
p(x, y)w(x, y)
=
2kn=1
|x|+|y|=n
p(x, y)w(x, y)
=
2kn=1
1
4n
|x|+|y|=n
w(x, y)
≤ 1
4
1 +
1
2+
1
3+
1
4
+
1
8
1
5+
1
6+ · · · +
1
12
+ 1
12 1
13+ · · · +
1
24 + · · ·
= 1
4
1 +
1
2+
1
3+
1
4
+
1
8
1
5+
1
6+ · · · +
1
12
+ R
R =
2kn=3
1
4n
4ni=1
1
2n2 − 2n + i
14
1 + 1
2 + 1
3 + 1
4
+ 1
8
15
+ 16
+ · · · + 112
= 143771
221760 < 0, 65
R ≤2k
n=3
1
4n
4n
i=1
1
2n2 − 2n =
2k
n=3
1
4n 4n
2n2 − 2n=
2kn=3
1
2n(n − 1)=
1
2
2kn=3
1
n − 1− 1
n
=
1
2
1
2− 1
2k
<
1
4
E (X ) < 0, 9
0, 9
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A1
B1
C 1
ABC
6
A2
B2
C 2
A1C 2B1A2C 1B2 ABC
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
..
..
..
.
...............
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.
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.
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.
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.
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.
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.
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.
..
.
..
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.
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.
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.
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................................................
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A
A1
A2
B
B1
B2
C
C 1
C 2
XY Z
[XY Z ]
A1B1C 1
ABC −1
2
[A1B1C 1] = 14
[ABC ]
h
A 1
2
[AB1C 1] = 14
[ABC ]
O
H ABC
ABC
ha
A
BC ABC
[BH C ] = 12 · BC · (ha − AH ) = [ABC ] − 1
2BC · AH
AH = 2OA1
[BH C ] = [ABC ] − BC · OA1 = [ABC ] − 2[OBC ]
h(H ) = A2
h
[B1A2C 1] = 14
[BH C ] = 14
[ABC ] − 12
[OBC ]
[C 1B2A1] = 14
[ABC ] − 12
[OCA]
[A1C 2B1] = 14
[ABC ] − 12
[OAB]
[A1C 2B1A2C 1B2] = [A1B1C 1] + [B1A2C 1] + [C 1B2A1] + [A1C 2B1]
= 14
[ABC ] + 34
[ABC ]
− 12
([OBC ] + [OCA] + [OAB])
= [ABC ] − 12 [ABC ] = 12 [ABC ]
n
a(n)
a(n)
!
n
n
a(n)
n=
2
3
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n = 9
3a(n) = 2n
3 | n
n = 3k
a(3k) = 2k
k = 1
a( 3 ) = 3
k ≥ 2
3k | (2k)!
(2k)! = (2k) · · · 2 · 1
k
3
k = 3
k > 3
(2k − 1)! = (2k − 1) · · · k · · · 2 · 1
k
3
a(3k) < 2k
k = 3
a(9) = 6
9 | 6!
9 5!
n = 9
(a, b)
a > b
a
b
1
a
b2 − 5
b
a2 − 5
(a, b)
a > b
(a, b)
a2 + b2 − 5
ab
b2 − 5 = λa
a2 − 5 = µb
λ
µ
ab2 − 5a = λa2 = 5λ + λµb
b(ab
−λµ) = 5(λ+a)
5
b
a2 = 5+µb
5
a
5
ab−λµ
ab−λµ = 5k
k
b(5k) = 5(λ + a)
λ = bk − a
b2 − 5 = (bk − a)a
a2 + b2 − 5 = kab
(a, b)
a2 + b2 − 5 = kab
k
d = gcd(a, b)
a = da
b = db
d2a2 + d2b2 − 5 = kd2ab
d2
5
d
1
(a, b)
a2 + b2 − 5 = kab
k > 1
(ak − b)2 + a2 − 5 = ka(ak − b)
(ak − b, a)
(4, 1)
k = 3
(a, b) → (ak − b, a)
a1 = 4
b1 = 1
an+1 = 3an − bn
bn+1 = an
n ∈
an < an+1
n
(an, bn)
an = L2n+1
bn = L2n−1 Ln
L0 = 2
L1 = 1
Ln+1 = Ln + Ln−1
n ∈
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f : →
f (nm) = f (n) + f (m)
n
m
f (nm) = f (n) + f (m)
n
m ∈
f
f
f (n + 1) ≥ f (n)
n ∈
f (n) = C log n
C ∈
f (
n+ 1)
≥ f (
n)
Ω(n)
Ω( pa11 pa22 · · · pakk ) = a1 + a2 + · · · + ak
n = pa11 pa22 · · · pakk
n
§
f
f (n+1) ≥ f (n)
n ∈
C ≥ 0
f (n) = C log n
n ∈
f
g
f (n)g(2) = f (2)g(n) ∀n ∈
g(n) = log n
f (n) = C log n
C = f (2)/ log 2
c
7/18/2019 CRUXv33n5
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n ∈
k ∈
l ∈
2l
−1
≤ nk
< 2l
(l − 1)f (2) ≤ kf (n) ≤ lf (2)
g −lg(2) ≤ −kg(n) ≤ −(l−1)g(2)
g(2)
f (2)
k
− 1
k f (2)g(2) ≤ f (n)g(2) − f (2)g(n) ≤ 1
k f (2)g(2)
k ∈
f :
→
∪ 0
f
f
C > 0
f (n) = C log n
n ∈
f
f (n)/f (m) = (log n)/(log m)
n
m ≥ 2
(log n)/(log m) = a/b
a
b ∈
nb = ma
n
m
f
f (1) = 0
f
f (n) > 0
n > 1
f
f (n) = f (1 · n) = f (1) + f (n)
f (1) = 0
f
a ∈
f (a) = 0
a ≥ 1
f (a) ≥ f (1) = 0
f (a) > 0
n > 1
k
nk > a
kf (n) = f (nk) ≥ f (a) > 0
f (n) > 0
f (n) = f (m)
n < m
n = 1
f (1) = 0
f (m) > 0
1 < n < m
k
nk+1 < mk
k > (log n)/(log m − log n)
f (nk) = kf (n) = kf (m) = f (mk)
f (r) = f (nk) = f (mk)
r
nk ≤ r ≤ mk
f (nk+1) = f (nk)
(k + 1)f (n) = kf (n)
f (n) > 0
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n ∈
k
2k+1 − 2k = 2k > n
n
ri
2k
2k+1
2k < r1 < r2 < · · · < rn < 2k+1
f
f
kf (2) = f (2k) < f (r1) < f (r2) < · · · < f (rn) < f (2k+1)
= (k + 1)f (2) = kf (2) + f (2)
f (2) > n
n ∈
[email protected] [email protected]
http://www.unirioja.es/dptos/dmc/jvarona/welcome.html
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()
ABC
a ≥ b ≥ c
A
B
C
AH
BC
H
BC
m = B H
n = C H
a(bm + cn) − bc(b + c)
∠A
u1
u2
u3
1
6
3i=1
cos2(ui − ui+1) + cos2(ui + ui+1)
≥ (cos u1 cos u2 cos u3)2 + (sin u1 sin u2 sin u3)2
3
S
2 × 2
A
B
C ∈ S
ABCAB = C
(ABC )n = AnBnC n
n
A
B
C ∈ S
S
loge(eπ − 1) loge(eπ + 1) + logπ(πe − 1) logπ(πe + 1) < e2 + π2
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C
AB C
−→v C
−→v
C C
C
A
B
C
O
OA
OB
OC
ABC
ABCD
ABCD
O
I
AC
BD
E
T
O
E
T
I
E
T
(A, B) 1
2
. . .
13 |A ∪ B|
ABC
∠ABC = 80
BD
∠ABC
D
AC
AD = DB + B C
∠A
P
n
P P · · · P n
3n
a
b
(a − 1)(b − 1) ≥ 0
ab + ba ≥ 1 + ab + (1 − a)(1 − b) · min1
ab
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F n
Ln
F 0 = 0
F 1 = 1
F n+1 = F n + F n−1
n ≥ 1
L0 = 2
L1 = 1
Ln+1 = Ln + Ln−1
n ≥ 1
∞
n=1
arctan
1
L2n
arctan
1
L2n+2
arctan 1
F 2n+1
≤ 4
π
arctan(β) arctan(β) + 1
3
β = 12
√ 5 − 1
m
m ≥ 2
a1
a2
. . .
am
Lm = limn→∞
1
nm e
1
m
k=1
ln
1 + akxn
dx
ABC
a ≥ b ≥ c
A
B
C
AH
BC
H
BC
m = BH
n = CH
a(bm + cn) − bc(b + c)
A
u1
u2
u3
1
6
3i=1
cos2(ui − ui+1) + cos2(ui + ui+1)
≥ (cos u1 cos u2 cos u3)2 + (sin u1 sin u2 sin u3)2
3
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S
2 × 2
A
B
C ∈ S
ABCAB = C
(ABC )n = AnBnC n
n
A
B
C ∈ S
S
loge(eπ − 1) loge(eπ + 1) + logπ(πe − 1) logπ(πe + 1) < e2 + π2
C
AB C
−→v C −→v
C
C C
A
B
C
O
OA
OB
OC
ABC
ABCD
I
O
E
AC
BD
T
O
E
T
I
E
T
(A, B)
1
2
. . .
13 |A ∪ B|
ABC
ABC
80
BD
ABC
D
AC
AD = DB + BC
A
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P
n
P P · · · P n
3n
a
b
(a − 1)(b − 1) ≥ 0
ab + ba ≥ 1 + ab + (1 − a)(1 − b) · min1
ab
F n
Ln
F 0 = 0
F 1 = 1
F n+1 = F n + F n−1
n ≥ 1
L0 = 2
L1 = 1
Ln+1 = Ln + Ln−1
n ≥ 1
∞n=1
arctan 1
L2n arctan 1
L2n+2arctan
1
F 2n+1
≤ 4π
arctan(β) arctan(β) + 13
β = 12
√ 5 − 1
m
m ≥ 2
a1
a2
. . .
am
Lm
= limn→∞
1
nm e
1
m
k=1
ln1 + ak
xn dx
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D
BC ABC
E
F
B
AC
C
AB
P
AD
EF
AD =√ 32
BC
P
AD
P S
P T
S
T
P
X
Y
P
Z
ST
XY
1
P X +
1
P Y =
2
P Z
.
.
..
..
..
..
.
..
..
.
..
..
..
..
..
..
..
..
..
..
..
.
..
.
..
..
..
..
..
.
..
..
..
..
..
..
.
..
..
..
..
.
....................................................................................
...........
.........
.............
.............................................................................................................................................
..
..............
...
..
.............................................................................................................................................................
.......................................................................................................................................
.............
..............................................................................................................................................................................
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............
..
.
..
..
..
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
..
..
..
.
..
.
..
..
..
..
.
..
.
..
..
..
..
....................................................................................................................................................................................................................................................................................................................
.
..
.
..
..
..
.
..
..
..
.
..
.
..
..
..
.
..
..
..
..
..
..
..
.
..
.
..
..
..
.
..
..
..
.
..
.
..
..
.
..
.
..
..
..
.
..
..
.
................................................................................................................................................................................................................................................................................................................................................................................................
..
.
..
.
..
.
..
...............
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
M
OP
S
T
X
Y
Z
M
XY
O
F
P O
ST
1
P X +
1
P Y =
P X + P Y
P X · P Y =
2P M
P X · P Y
2P M
P X · P Y =
2
P Z
P M · P Z = P X · P Y
Z
F
O
M
P M · P Z = P F · P O = P S 2
P SF P OS
P S 2 = P X · P Y
E
F
BC
AC
AB
∠ABC = 180 − ∠F EC = ∠AEF
BDF
∠ABC = ∠BF D
∠AEF = ∠BF D
∠AF E = ∠DEC
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n
5m + 3n ≡ 1m + (−1)n ≡ 2 (mod 4)
2k ≡ 2 (mod 4)
k = 1
n
m = n = 0
k = 1
n
m = 0
1 + 3n = 2k
n > 2
3n ± 1
1 + 3n = 2k
m = 0
n
n = 1
k = 2
m > 0
3
(−1)m
≡ (−1)k
(mod 3)
m
k
5
2k ≡ (−2)n ≡ −2n ≡ ±2 (mod 5)
n
k
m
m ≥ 3
n ≥ 3
k ≥ 7
A = 22276800 = 26 · 32 · 52 · 7 · 13 · 17
5m + 3n = 2k
A
m ≥ 3
n ≥ 3
k > 7
551 ≡ 53
(mod A)
3243 ≡ 33 (mod A)
2127 ≡ 27 (mod A)
m = 1
n = 1
n = 1
5m+3 = 2k
B = 65792 = 28 · 257
B
5256 ≡ 1 (mod B)
225 ≡ 29 (mod B)
9
2
B
n = 1
m > 0
(m, k) ∈ (1, 3)
(3, 7)
k
2
B
k = 1
k = 7
(m , n , k) = (1, 1, 3)
(m,n,k) = (3, 1, 7)
m ≡ 1 (mod B)
m ≡ 3 (mod B)
m
1
3
n = 1
m = 1
n ≥ 3
C = 26 ·34 · 17
321 ≡ 35 (mod C )
2223 ≡ 27 (mod C )
5
7
3
2
C
(n, k) = (3, 5)
(m,n,k) = (1, 3, 5)
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2
3
5m+ 3n = 2k
(m,n,k) ∈ (0, 0, 1)
(0, 1, 2)
(1, 1, 3)
(3, 1, 7)
(1, 3, 5)
ε
x2
a2+
y2
b2−1 = 0
ε
M 1(x1, y1)
M 2(x2, y2)
x21
a2
+ y21
b2 −1x2
2
a2
+ y22
b2 −1
M 1 = (x1, y1)
M 2 = (x2, y2)
M 1, M 2 = x1x2
a2 +
y1y2
b2 − 1
Li
= [ui, v
i, w
i]
uix + viy + wi = 0
i = 1
i = 2
[L1, L2] = a2u1u2 + b2v1v2 − w1w2
M 1
M 2
M 1, M 1M 2, M 2 − M 1, M 22 = 0
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M 1, M 1M 2, M 2
M 1, M 1M 2, M 2 = 1
M 1, M 22 = 1
y1x − x1y = 0
y2x − x2y = 0
x1x2
a2 +
y1y2
b2 = 0
r1 < 0 < r2 < r3
8x3 − 6x +√
3 = 0
r23 = 4r22 − 4r42
r21 = 4r23 − 4r43
s1 < 0 < s2 < s3
8x3 − 6x +1 = 0
r2
1 + s2
2 = 1
s2
1 + r2
2 = 1
r2
3 + s2
3 = 1
x = sin θ
8x3 − 6x +√
3 = 0
√ 3
2 = −4x3 + 3x = −4sin3 θ + 3 sin θ = sin(3θ)
3θ = π3
+ 2πk
3θ = 2π3
+ 2πk
k
θ
[0, 2π]
θ ∈
π
9
2π
9
7π
9
8π
9
13π
9
14π
9
r1 = sin 13π
9 = − sin
4π
9
r2 = sin π
9
r3 = sin 2π
9
sin2 2θ = 4 sin2 θ − 4sin4 θ
r23 = sin2 2π
9 = 4 sin2 π
9 − 4sin4 π
9 = 4r22 − 4r42
r21 = sin2 4π
9 = 4 sin2 2π
9 − 4sin4 2π
9 = 4r23 − 4r43
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x = cos θ
8x3
− 6x + 1 = 0
−1
2 = 4x3 − 3x = 4 c o s3 θ − 3cos θ = cos(3θ)
3θ = 2π3
+ 2πk
3θ = 4π3
+ 2πk
k
θ
[0, 2π]
θ ∈
2π
9
4π
9
8π
9
10π
9
14π
9
16π
9
s1 = cos 8π
9 = − cos
π
9
s2 = cos 4π
9
s3 = cos 2π
9
cos2 θ +sin2 θ = 1
ri
si
cos 3θ
x1
x2
. . .
xn
n ≥ 2
ni=1
xi = 0
ni=1
x2i = 1
ni=1
|xi|
ni=1
|xi|2
=
ni=1
x2i +
ni=1
n j=1 j=i
|xi||x j | ≥ 1 + ni=1
n j=1 j=i
xix j= 1 +
ni=1
xi
2−
ni=1
x2i
= 2
x1 = 1/√
2 = −x2
x3 = x4 = · · · = xn = 0
√ 2
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m
xi
n − m
1 ≤ m ≤ n
x1
. . .
xm
xm+1
. . .
xn
mi=1
x2i ≥ 1
m
mi=1
xi
2
ni=m+1
x2i ≥ 1
n − m
ni=m+1
xi
2 =
1
n − m
mi=1
xi
2
1 =ni=1
x2i ≥
1
m+
1
n − m
mi=1
xi
2
= n
m(n − m)
mi=1
xi
2
ni=1
|xi| = 2mi=1
xi ≤ 2√ n
m(n − m)
r
x(r − x)
r2/4
x = r/2
n = 2k
k
m(n − m) ≤ k2
ni=1
|xi| ≤ 2k√ n
=√
n
x1 = x2 = · · · = xk = −xk+1 = · · · = −xn = 1/√ n
n = 2k + 1
k
m(n − m) ≤ maxk(2k + 1 − k)
(k + 1)(2k + 1 − k − 1) = k(k + 1)
ni=1
|xi| ≤ 2
k(k + 1)√ n
=
n − 1
n
x1 = x2 = · · · = xk =
k + 1
k(2k + 1)
xk+1 = · · · = xn = − k(k + 1)(2k + 1)
n +
(−1)n − 1
2n
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n
3xy − 1
x + y= n
x
y
k
n = 3k
3xy − 3k(x + y) = 1
n = 3k + 1
x = k
y = −(3k2 + k + 1)
n = 3k − 1
x = k
y = 3k2 − k + 1
n
3
β > 1
x − β log2 x = β − β ln β
α1 < α2
x1
x2
α1 ≤ x1 < x2 ≤ α2
λ
0 < λ < 1
λ log2 x1 + (1 − λ)log2 x2 ≥ ln
λx1 + (1 − λ)x2
f (x) = x − β log2 x − β + β ln β
x > 0
f (x) = 1 − β/(x ln2)
x = β/(x ln2)
f (x) < 0
0 < x < β/ ln 2
f (x) > 0
x > β/ ln 2
f
(0, β/ ln2)
(β/ ln 2, ∞)
f
x = β/ ln 2
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limx→0+ f (x) = ∞
limx→∞
f (x) = −β + β ln β + limx→∞
x
1 − β log2 x
x
= ∞
limx→∞
β log2 x
x = 0
f (β/ ln2) < 0
f (β/ ln2) = β/ ln 2 − β log2(β/ ln2) − β + β ln β
1/ ln 2 − log2(β/ ln2) − 1 + ln β < 0
1 − (ln 2)
log2(β/ ln2) − ln 2 + (ln 2)(ln β) < 0
1 − ln 2 − ln(β/ ln 2) + (ln 2)(ln β) < 0
1 − ln 2 − ln β − ln(ln 2) + (ln 2)(ln β) < 0
1 − ln 2 + ln(ln 2) − (1 − ln 2)(ln β) < 0
(1−ln 2)(ln β) > 0
1−ln 2+ln(ln 2) < 0
f
α1
α2
0 < α1 < α2
f (x) ≤ 0
x ∈ [α1, α2]
0 < α1 < β/ ln 2 < α2
log2 x ≥ (x/β) − 1 + ln β
0 < λ < 1
λ log2 x1 + (1 − λ) log2 x2
≥ λx1
β− 1 + ln β + (1 − λ)
x2
β− 1 + ln β
= 1
β
λx1 + (1 − λ)x2
− 1 + ln β
t > 0
t − β + β ln β ≥ β ln t
g(t) = t − β + β ln β − β ln t
g(t) = 1 − β/t
t = β
g(t) = β /t2 > 0
g(β) = 0
g
g(t) ≥ 0
t > 0
t = λx1 + (1 − λ)x2
λx1 + (1 − λ)x2 − β + ln β ≥ β lnλx1 + (1 − λ)x2
λ log2 x1 + (1 − λ)log2 x2 ≥ ln
λx1 + (1 − λ)x2
x1 = α1
x2 = α2
λα1 + (1 − λ)α2 = β
λ = α2 − β
α2 − α1
f (β) = β(ln β − log2 β) < 0
α1 < β < β
ln 2 < α2
0 < λ < 1
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Γ ABC
M
AB
N
AC
D
M N
Γ
M B
M A· AC
DB− N C
N A· AB
DC
= BC
DA
P = AD∩BC
Q = M D∩BC
D
B
AC
B
Q
C
P
Q
QM D BAP
QN D CAP
BM
M A= −BQ
QP · P D
DA
CN
N A= −CQ
QP · P D
DA
P CA ∼ P DB
AC
BD=
P C
P D
P BA ∼ P DC
AB
CD=
P B
P D
BM
M A· AC
BD− CN
N A· AB
CD= −BQ
QP · P D
DA· AC
BD+
CQ
QP · P D
DA· AB
CD
= −BQ
QP · P D
DA· P C
P D+
CQ
QP · P D
DA· P B
P D
= −BQ · P C + CQ · P B
QP
·DA
= −BQ · P C + (CB + BQ) · (P C + CB)QP · DA
= CB · (P C + CB + BQ)
QP · DA
= CB · P Q
QP · DA=
BC
DA
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•
BN ∩CM = G
b
DB− c
DC
= a
DA
•
BN ∩CM = I
1
DB− 1
DC
= 1
DA
•
BN ∩CM = H
cosB
DB− cos C
DC
= | cosA|DA
p
α(n)
k
pk
11 · 22 · 33 · · · nn
limn→∞
α(n)
n2 =
1
2( p − 1)
α(n)
p
n!
α(n) = p
1 + 2 + · · · +
n
p
+ p2
1 + 2 + · · · +
n
p2
+ p3
1 + 2 + · · · +
n
p3
+ · · ·
α(n) =
N (n) j=1
p j
2
n
p j
n
p j
+ 1
N (n) = ln n/ ln p
n
p j − 1
n
p j
<
n
p j
n
p j
+ 1
≤
n
p j
n
p j + 1
p j
n
n
p j − 1
< p j
n
p j
n
p j
+ 1
≤ n
n
p j + 1
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1
2
N (n)
j=1
1
p j
− N (n)
n
<
α(n)
n2 ≤ 1
2
N (n)
j=1
1
p j
+
N (n)
n
n → ∞
limn→∞
N (n)
n = 0
limn→∞
α(n)
n2 =
1
2
∞ j=1
1
p j =
1
2( p − 1)
E = (x, y) ∈ ×
| x + y
N (n)
(x, y) ∈ E | x ≤ n
y ≤ n
n ∈
limn→∞
N (n)
n√
n=
4
3(√
2 − 1)
n ∈
k ∈
ϕ(n, k)
(x, y)
1 ≤ x
y ≤ n
x + y = k
N (n) =∞i=1
ϕ(n, i2)
ϕ(n, k)
1 ≤ x
y ≤ n
2 ≤ x + y ≤ 2n
ϕ(n, k) = 0
2 ≤ k ≤ 2n
2 ≤ k ≤ n +1
x+y = k
(1, k−1)
(2, k − 2)
. . .
(k − 1, 1)
ϕ(n, k) = k − 1
n + 1 ≤ k ≤ 2n
x + y = k
(k − n, n)
(k − n + 1, n − 1)
. . .
(n, k − n)
ϕ(n, k) = 2n + 1 − k
ϕ(n, k) =
k − 1
2 ≤ k ≤ n + 1
2n + 1 − k
n + 1 ≤ k ≤ 2n
0
i ∈
ϕ(n, i2) =
i2 − 1
2 ≤ i ≤ I 1
2n + 1 − i2
I 1 ≤ k ≤ I 2
0
I 1 =√
n + 1
I 2 =√
2n
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N (n) =∞i=1
ϕ(n, i2) =
I 1i=2
(i2 − 1) +
I 2i=I 1+1
(2n + 1 − i2)
=
I 1i=1
i2 − I 1 + (2n + 1)(I 2 − I 1) −I 2
i=I 1+1
i2
= 2
I 1i=1
i2 −I 2i=1
i2 + 2n(I 2 − I 1) + I 2 − 2I 1
= I 1(I 1 + 1)(2I 1 + 1)
3− I 2(I 2 + 1)(2I 2 + 1)
6
+ 2n(I 2 − I 1) + I 2 − 2I 1
=
√ n · √
n · 2√
n
3−
√ 2n · √
2n · 2√
2n
6+ 2n
√ 2n − √
n
+ O(n)
= n√
n
2
3− 2
√ 2
3+ 2
√ 2 − 2
+ O(n)
= 4
3n
√ n√
2 − 1
+ O(n)
N (n)
n√
n=
4
3
√ 2 − 1
+ O
1√
n
limn→∞
N (n)
n√
n=
4
3
√ 2 − 1
n
γ
γ − 1
48n3 < 1 +
1
2 + · · · +
1
n − ln
n +
1
2 +
1
24n
< γ − 1
48(n + 1)3
n
xn = 1 + 1
2 + · · · +
1
n − ln
n +
1
2 +
1
24n
+
1
48n3
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xn+1 − xn = f (n)
f (x) = 1
x + 1 − ln
x +
3
2 +
1
24(x + 1)
+ ln
x +
1
2 +
1
24x
+
1
48(x + 1)3 − 1
48x3
f (x) > 0
x > 0
f (x) = 2656x6+10096x5+15008x4+10836x3+3870x2+652x+3716x4(x+1)4(24x2+12x+1)(24x2+60x+37)
limx→∞
f (x) = 0
f (x) < 0
x > 0
xn∞n=1
limn→∞
xn = γ
xn > γ
n
n
yn = 1 + 1
2 + · · · +
1
n − ln
n +
1
2 +
1
24n
+
1
48(n + 1)3
yn+1 − yn = g(n)
g(x) = 1
x + 1 − ln
x +
3
2 +
1
24(x + 1)
+ ln
x +
1
2 +
1
24x
+
1
48(x + 2)3 − 1
48(x + 1)3
g(x) < 0
x > 0
g(x) =− 8864x7+72336x6+247520x5+456204x4+483110x3+288492x2+86997x+947216x(x+1)4(x+2)4(24x2+12x+1)(24x2+60x+37)
limx→∞
g(x) = 0
g(x) > 0
x > 0
yn∞n=1
limn→∞
yn = γ
yn < γ
n
ABC
AD
H
E
AD
M
BC
AD = B C
HM = H E
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HM = HE
AD = BC
D(0, 0)
E (0, 1)
A(0, 2)
y
B(b, 0)
C (c, 0)
x
c > b
M
12
(b + c), 0
CH
C
AB
y = 12
bx − 12
bc
y
CH
H (0, −12
bc)
HM 2 − HE 2 =
b + c
2
2+
b2c2
4 −
1 +
bc
2
2=
b2 + c2
4 − bc
2 − 1
= 14
(b − c)2 − 4
= 1
4(BC 2 − AD2)
HM = H E
BC = AD
ABD ∼ CH D
∠BAD = 90−∠ABD = ∠HCD
AD
CD =
BD
HD
DA · HD = DB · DC
HM 2 − HE 2 = HD2 + DM 2 − HE 2
= HD2 + (DB + BM )2 − (HD + DE )2
= HD2 +
DB + 12
BC 2 −
HD + 12
DA2
= 14
(BC 2 − AD2) − DA · HD + DB(DB + BC )
= 14
(BC 2 − AD2) − DA · HD + DB · DC
= 14
(BC 2 − AD2)
HM = H E
AD = B C
(
)
(
)
ABC
BC
AD = BC
H
A
BC
AD = BC
M
BC
BC
E
AD
HM
HE
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D
BC ABC
P
AD
BP
AC
E
CP
AB
F
AD ⊥ B C
∠BDF = ∠CDE
A
BC
DE
DF
G
H
1
2
3
4
5
6
BDF
F DA
EDA
CDE
AHD
AGD
∠5 = ∠1
∠6 = ∠4
HA = AG
AHF
BF D
AF
F B =
HA
BD
.
.............................
..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.............................................................................................................................................................................................................................................................................................
..................................................................
....................................................................................................................................................................................................................................................................................................................................................................................................................................
.
..
..
..
..
..
.
..
..
..
.
..
..
..
..
.
..........................................................................................................................................................................................................................................................
......
...........................................
...................
.........
.........
............................
.........................
.....................................
.........
.....................................................................................................................
......
..
...............................................................................................................................................................................................................................................................
..
..................................................................................................................................................................................................................................................................................................................................................................................................................................................
A
B C D
E F
GH
P
1..............................................2
...............................
3.....
..............................
4 .
..
..
.
..
.
..
.
..
..
..
..
..
..
..
.
..
........................
....
5
.
..
..
.
..
..
..
..
..
...
..
........
................. 6
.
.
..
..
.
.
..
..
..
..
..
..
..
......
.............................
AGE
CDE CE
EA =
DC
AG
ABC
1 =
AF
F B · BD
DC · CE
EA =
HA
BD · BD
DC · DC
AG =
HA
AG
HA = AG
AD ⊥ BC
DAH
DAG
∠1 = ∠4
∠1 = ∠4
∠5 = ∠6 DGH
DG = DH
DA DGH
D
GH
AD ⊥
AD ⊥ B C
(
)
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(x, y)
x5 + y7 = 20041007
71
71
(16, 71) = 1
(1614)70 ≡ 1 (mod 71)
20041007
≡ 161007 = (1614)70(1627)
≡ 1627 = 40969
≡ 499 ≡ 1176493 ≡ 23 = 8 ( (mod 71)
71
0
1
20
23
26
30
32
34
37
39
41
45
48
51
70
71
0
1
5
14
17
25
46
54
57
66
70
x5 + y7 (mod 71)
k
0 ≤ k ≤ 70
k /∈ 8
10
11
60
61
63
8
71
limn→∞
ln
n
k=1
k2 + n2
n2
k1
n2
S n = ln
nk=1
k2 + n2
n2
k 1
n2
S n = 1
n2
nk=1
k ln
k2 + n2
n2
=
1
n
nk=1
k
nln
1 +
k2
n2
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f (x) = x ln(1+ x2
)
1
n
2
n
. . . n
n
[0, 1]
limn→∞
S n =
10
x ln(1 + x2) dx = 1
2
21
ln u du = 1
2(u ln u − u)
21
= 1
2
(2ln2 − 2) − (−1)
=
1
2(−1 + 2ln 2)