Crux Mathematicorum 2005

∗ AED BC AD
ABCD AED 4 ........................................................................................................................
∗ A B A
A B
1 1

(n + 2)2 − 100 = n2 + 92
n2 + 4n + 4 − 100 = n2 + 92
n = 47
a b 24
n b = n3
(n + 2)3 = n3 + 488
6n2 + 12n− 480 = 0
n2 + 2n − 80 = 0
(n− 8)(n + 10) = 0
a n k
a = (n + k)2 − 100
n2 + 92 = n2 + 2kn + k2 − 100
k2 + 2kn − 192 = 0
(m− n)(m + n) = 263
192 1 2 3 4 6 8 12 16 24 32 48 64 96 192
m− n m + n m n a 1 192 2 96 49 47 2301 3 64 4 48 26 22 576 6 32 19 13 261 8 24 16 8 156
12 16 14 2 96
5 a 2301 576 261 156 96
3
(1, 2, 3)
(0, 2, 3) (0, 2, 2) (1, 1, 3) (1, 1, 0) (1, 0, 3) (1, 0, 1) (1, 2, 2) (0, 2, 2) (1, 2, 1) (1, 0, 1) (1, 2, 0) (1, 1, 0)
(2, 4, 4) (0, 4, 4) (2, 4, 3) (2, 1, 3) (2, 4, 2) (2, 0, 2) (2, 4, 1) (2, 3, 1) (2, 4, 0) (2, 2, 0) (2, 3, 5) (2, 3, 1) (2, 2, 5) (2, 2, 0) (2, 1, 5) (2, 1, 3) (2, 0, 5) (2, 0, 2) (1, 4, 5) (0, 4, 5) (0, 4, 4)
(1, 4, 5) 10 (0, 4, 5) (1, 1, 5) (1, 0, 5) (1, 4, 4)
(1, 4, 1) (1, 4, 0) (1, 3, 5) (1, 2, 5) (1, 4, 3) (1, 4, 2)
(1, 2, 3)

d = 2R R R h
r1 r2 h ≤ R R
R
R =

h
ABCD A(1, 4) B(5, 4) C (5, 8) D(1, 8) P
P P
ABC A BD CE D AC E AB AB AC
Γ1 Γ2 M Γ1 CE N Γ2 BD
M N ABC AM = AN
10a + b 7 a − 2b 7
75
75
(x, y) x2 + y2 + 2xy − 2005x − 2005y − 2006 = 0
ABC A BD CE D AC E AB Γ1
Γ2 AB AC M Γ1 CE N Γ2 BD M N
ABC AM = AN
75
75
a b c
a + b + c
a a− 2b − 2c 3a− 6b − 6c 9a − 18b − 18c b 3b 7b − 2c− 2a 21b − 6c − 6a c 3c 9c 25c− 2a− 2b
9a − 18b − 18c = 27
21b − 6c− 6a = 27
25c− 2a − 2b = 27
An Bn C n n A3 = B3 = C 3 = 27
A2 = 27 3
A1 = 9 3
C 1 = 63 3
B0 = 57 3
21+3cosx − 10 × 2−1+2cosx + 22+cosx − 1 = 0
2X 3 − 5X 2 + 4X − 1 = 0 X = 2cosx X = 1
(X − 1)(2X 2 − 3X + 1) = 0 X = 1 (X − 1)2(2X − 1) = 0 X = 1
2
X 2cosx = 0 2cos x = 1 2
cos x = 0 cos x = −1 0 ≤ x ≤ 2π x = π
2 x = π x = 3π
2
4 6 8
6 61 67 4 41 43 47
9 89 2 5
43 61 207
{2 3 5 41 67 89} {2 5 7 43 61 89}
300 207
207
{1 2 3 4 5 6 7 8 9 10}
{} {1 2 4 5 8} {6 7} {2 3 4 5 7 8 9} {1 2 3}
S n S S 2n
S
S n = {1 2 3 . . . n} W n S n W 10
S 0 = {} W 0 = 1
S 1 = {1} {} {1} W 1 = 2
S 2 4 W 2 = 4
S 3 8 {1 2 3} W 3 = 7
S 4 16 3 {1 2 3 4} {1 2 3} {2 3 4} W 4 = 13
S 9 S 10 A S 9 A
S 10 W 9 S 9 S 10
S 9 S 10 10

7 A S 6 S 9 8 9
W 6 S 6 S 9 8 9 W 9 − W 6
S 10 10 S 9 10
10 S 9
W 10 = W 9 + (W 9 − W 6) = 2W 9 − W 6
W 9 = 2W 8−W 5 W 8 = 2W 7−W 4
W 10 = 2W 9 − W 6 = 2(2W 8 − W 5) − W 6 = 4W 8 − W 6 − 2W 5
= 4(2W 7 − W 4) − W 6 − 2W 5 = 8W 7 − W 6 − 2W 5 − 4W 4
= 8(2W 6 − W 3) − W 6 − 2W 5 − 4W 4
= 15W 6 − 2W 5 − 4W 4 − 8W 3
= 15(2W 5 − W 2) − 2W 5 − 4W 4 − 8W 3
= 28W 5 − 4W 4 − 8W 3 − 15W 2
= 28(2W 4 − W 1) − 4W 4 − 8W 3 − 15W 2
= 52W 4 − 8W 3 − 15W 2 − 28W 1
= 52(13)− 8(7)− 15(4) − 28(2) = 504
S 10 504
A S n A n A n
A S n−1 W n−1 A A n A n− 1
A n S n−2 W n−2 S n n n− 1
A n− 1 A n− 2 3 A n
n− 1 S n−3 W n−3 A A
W n = W n−1 + W n−2 + W n−3

3 2 6
5 6 5 6
2 × 3 3 × 2
d(n) n d(1) = 1 d(6) = 4 d(19) = 2 d(25) = 3 d(n)
1
12 12 = 2 × 2 × 3 = 22 × 3
1 6 2 × 3 1 × 2 × 3 1 × 1 × ×2 × 3
d(n) n
d(n)
n d(n) 2 2 2 3 3 2 4 22 3
12 22 × 3 6
1 + 2 + · · · + k = 1 2
k(k + 1) 1 + 3 + · · · + (2− 1) = 2
A1 = {1 5 7 11 13 17 . . . } B1 = {2 4 8 10 14 16 . . . } A1 3
B1 3
m A1
1 5 6 = 1 + 5 7 8 = 1 + 7 11 12 = 5 + 7 13 = 1 + 5 + 7
14 = 1 + 13 16 = 5 + 11 17 18 = 1 + 17 19 20 = 1 + 19
21 = 1 + 7 + 13 22 = 5 + 17 23 24 = 1 + 23 . . .
c
2 n(n+1) 3 n(n+1)
3 n 3 n + 1 3
n 3 |C | > 1 3 |S |
C n
n+1 3 n n n n = 6k + 2
n n + 1 3 n n = 6k + 5 n n + 1 3
S = {1 2 . . . n} A0 = { S 3} B0 = { S 3} C 0 = { S 3}
S A B C A0
B0 C 0 C C 0 A B B0 A0
n A0 B0 C 0 A B C
n = 6k + 2
S = {1 2 3 . . . 6k 6k + 1 6k + 2} A0 = {1 5 7 11 . . . 6k − 1 6k + 1} B0 = {2 4 8 10 . . . 6k − 2 6k + 2} C 0 = {3 6 9 . . . 6k − 3 6k}
1 3 |S | = 1
= 1 3

= 1 + 3 + 5 + 7 + · · · + (6k − 1) + (6k + 1)
−
|B0| = 2 + 4 + 8 + 10 + · · · + (6k − 2) + (6k + 2)
= 2 + 4 + 6 + 8 + 10 + · · · + (6k − 2) + 6k + (6k + 2)
− (6 + 12 + · · · + 6k)
− 6(1 + 2 + · · · + k)
|C 0| = 3 + 6 + 9 + · · · + 6k = 3(1 + 2 + 3 + · · · + 2k)
= 3
1 2
(2k)(2k + 1)
= 6k2 + 3k
A B C 1 3 |S | k A0 C 0 k + 1 B0 C 0 n = 6k + 5
n = 6k + 5
S = {1 2 3 . . . 6k + 3 6k + 4 6k + 5} A0 = {1 5 7 11 . . . 6k + 1 6k + 5} B0 = {2 4 8 10 . . . 6k + 2 6k + 4} C 0 = {3 6 9 . . . 6k 6k + 3}
1 3 |S | = 6k2 + 11k + 5 |A0| = 6k2 + 12k + 6
|B0| = 6k2 + 12k + 6 |C 0| = 6k2 + 9k + 3
A B C 1
3 |S |
k + 1 A0 C 0 k + 1 B0 C 0 n = 6k + 2
A0 k B0
k + 1 n = 6k + 5 A0 k + 1 B0 k + 1
A0

A0
A0
k A0 = {1 5 7 11 . . . 6k− 1 6k + 1} k
k A1 = {1 5 7 11 13 17 . . . } A0 A1 B0 B1
n = 6k + 2 A0 k k 1
5 6 7 8 11 12 13 14 k ≥ 16 B0
k + 1 k + 1 k n n = 6k + 2
k 1 5 7 11 13 17
n = 6k + 5 k 5 7 11
k 1 k 2+ 1 0
n = 12 + 8 ≥ 0 = 1 = 4 = 7
n = 12 + 11 ≥ 2 = 4
2003 = 12(166) + 11 2003
B1 2 6 j−2 j ≥ 1 6 j + 2 j ≥ 1
B1
B1
1 5 7 5 6 7 8 6 j − 1
6 j + 1 A1 j ≥ 2 11 13 17 19 5 6 7 8
6 j + 4 = 5 + (6 j − 1) 6 j + 5 = 6 + (6 j − 1) = 1 + 5 + (6 j − 1)
6 j + 6 = 7 + (6 j − 1) 6 j + 7 = 8 + (6 j − 1) = 1 + 7 + (6 j − 1)
6 j + 8 = 7 + (6 j + 1) 6 j + 9 = 8 + (6 j + 1) = 1 + 7 + (6 j + 1)
6 A1

n 3 n = 3k k S = {1 2 . . . 3k} |S | = 1
2 (3k)(3k + 1) = 1
3 + 6 + · · · + 3k = 3 ( 1 + 2 + · · · + k) = 3
1 2
k(k + 1)
= 1 2
(3k2 + 3k)
1 n 3 1 3
C C 1 3
= 1 3
|A0| = 1 + 5 + 7 + 11 + · · · + (6k + 1) + (6k + 5)
−
|B0| = 2 + 4 + 8 + 10 + · · · + (6k + 2) + (6k + 4)
= 2 + 4 + 6 + 8 + 10 + · · · + (6k + 2) + (6k + 4)
− (6 + 12 + · · · + 6k)
− 6(1 + 2 + · · · + k)
1 + 2 + · · · + (2k + 1)
n = 6k + 5 A0
k + 1 k + 1 1 5 6 7 8 11 12 13 14 k ≥ 16 B0 k + 1
k + 1 k n = 6k + 5 k 5 7 11
[email protected]
y + y
x ≥ 2
∠AP B = ∠CQD
10
2n−1 32n−1 43n−1 54n−1 65n−1 76n−1 87n−1 98n−1
n
BD = BE = y AE CD P AP C x y
2001 2001
y3 + 1
a
M N CAN BC M D CD
ABC AB K
a
ABCD a √
4 − a2 √
2 < a < 2 LA LB LC LD Γ A B C D
LA LB LB LC LC LD LD LA
A B C D
ABC D ABCD
X = {1 2 . . . 2001} m m W X u v ∈ W u v
u + v = 2k k
n 3
a b c a + b− c a + c− b b + c− a a + b + c 7 a b c 800 d
7 d
m < 2a 2n | (2am− m2 + n2)
n2 − m2 + 2mn ≤ 2a(n − m)
min (m,n)∈A
n
2 + x3
6 + · · · + xn
P n(x) ≥ 1 x ≥ 0 P n ρ ρ < 0
F n(x) = P n(x)e−x F n P n
F n(x) =
n! e−x
n F n F n(0) = P n(0) = 1 F n(x) ≥ 1 x ≤ 0 F n
P n P n n P n(x)
F n (−∞, 0] F n
P n P n ρ ρ
1 P n ρ k > 1 P n(x) = (x − ρ)kQn(x) Qn(x)
P n(x) = k(x− ρ)k−1Qn(x) + (x− ρ)kQ n(x)
P n(ρ) = P n(ρ) = 0 0 = P n(ρ) − P n(ρ) = ρn
n! ρ = 0 ρ < 0 ρ P n
n k k · 2s + 1 (1 ≤ s ≤ n
n k
k = (21 + 1)(22 + 1)(23 + 1) · · · (2n + 1) + 1

∠ABC 1 . . . n−1 n ∩ (AB) = A1 ∩ (BC ) = An+1
∩ i = Ai+1 1 ≤ i < n

|BA1| |BAk+1|
= sin(∠A1Ak+1B)
{an}∞n=1
k ≥ 1 S k k sk
S k k ≥ 2 sk = uk + vk + wk uk S k 1 vk
S k x1 x = 1 wk S k
1
wk+3 = vk+2 = uk+1 = 9sk
sk+4 = uk+4 + vk+4 + wk+4 = 9sk+3 + uk+3 + vk+3
= 10sk+3 − wk+3 = 10sk+3 − 9sk
{sk}∞k=2 4 P (x) = 0
P (x) = x4 − 10x3 + 9 = (x− 1)(x3 − 9x2 − 9x− 9)
Q(x) = x3 − 9x2 − 9x− 9 r 9 < r < 10
λ λ rλλ = 9 r > 9 |λ| < 1 sk
sk = a + bλk + cλ k
+ drk
sk = ∞ lim k→∞
|λ|k = 0
tn sk tn ≤ sk
ln10 + 1 k ≤ lnn
rk = ek ln r ≤ e lnn ln r ln(r+) = nα
α = ln r
ln(r + ) tn = O(nα)
tn
( pn − 1, p) = 1
pn − 1 | pφ(pn−1) − 1
a m n an − 1 | am − 1 n | m
m = qn + r q r ∈ ∪ {0} 0 ≤ r < n
am − 1 = aqn+r − 1 = ar
(an)q − 1
+ ar − 1
an − 1 | am − 1 an − 1 | (an)q − 1 an − 1 | ar − 1 r = 0 n | m
p > 1 (pn−1, p) = 1 p
X |X | = n Ai ⊂ X (1 ≤ i ≤ n) |Ai| = 3 |Ai∩A j | ≤ 1 (i = j) A ⊆ X
Ai |A| ≥ 2 √
∠HAA = ∠HBA = ∠HBQ = ∠HP Q
HAA ∼ HP Q T AA ∼ RP Q
∠T AA = ∠T AA = ∠ABA = ∠P BQ = ∠RP Q = ∠RQP
HAT ∼ HQR HAQ ∼ HT R ∠HQA = ∠HR T
∠HCC = ∠HQP ∠HC C = ∠HP Q HC C ∼ HP Q SC C ∼ RP Q
HCS ∼ HQR HQC ∼ HR S ∠HQC = ∠HR S ∠HR T + ∠HR S = ∠HQA + ∠HQC = 180
R T S HT R ∼ HAQ HR S ∼ HQC
T R : AQ = HR : HQ = RS : QC
T R : RS = AQ : QC AA ⊥ BC M Q ⊥ BC AA M Q AQ : QC = AM : M C T R : RS = AM : M C
M AC A C R T S T S T R : RS = AM : M C R T S
f A B 1 2
n |A−B| = n |f (A)−f (B)| = n
f E n E n E n n ≥ 2 L > 0
d
f (A), f (B)

n P (x) 2n P (0) = 1 P (k) = 2k−1 k = 1 2 . . . 2n
2P (2n + 1) − P (2n + 2) = 1
k = 0 1 . . . 2n
P (x) = 2n
a2n
Q(x) = a2n(x− 1)(x − 2) · · ·
(2n)!
2k + 1 k + 1 2k
k = 2 U n U 0 = 2 U n+1 = 2U n + 1
U n = 3 × 2n − 1 n ≥ 0 1999 3 × 2n− 1 1999+1 3 1999
U 9 = 3 × 29 − 1 = 1535
U 8 = 3 × 28 − 1 = 7 6 7
U 7 = 3 × 27 − 1 = 3 8 3
U 6 = 3 × 26 − 1 = 1 9 1
U 5 = 3 × 25 − 1 = 9 5
U 4 = 3 × 24 − 1 = 4 7
U 3 = 3 × 23 − 1 = 2 3
U 2 = 3 × 22 − 1 = 1 1
U 1 = 3 × 2 − 1 = 5
U 0 = 2
3k − 1 = xn
x k 3k− 1 = x3 x 3k − 1 x ≥ 2
3k = x3 + 1 = (x + 1)(x2 − x + 1)
x2 − x + 1 = 3s s ≥ 1 s = 0 x ≥ 2 x2−x + 1 0 9
x ∈ {0 1 . . . 8} s = 1 x2 − x + 1 = 3 x = 2 k = 2
n 1 x k 3k − 1 = xn n = 3
n n = 2r (xr)2 = 3k − 1 ≡ −1 (mod 3)
0 1 3 n = 2r + 1
3k = xn + 1 = x2r+1 + 1 = (x + 1)a
a = 1 − x + x2 − · · · + x2r x + 1 > 1 a > 1 n > 1 x + 1 = 3u a = 3v
u v x ≡ −1 (mod 3)
3ν = a ≡ 1 + 1 + · · · + 1 ≡ 2r + 1 (mod 3)
2r + 1 ≡ 0 (mod 3) r ≡ 1 (mod 3) r = 1 + 3w w 3k = x2r+1 + 1 = (x1+2w)3 + 1
x1+2w = 2 w = 0 r = 1 n = 3

Ax2 + Bx + C A B C
A x2 + B/Ax + C/A

()
a b c d e > 0 a2 + b2 + c2 + d2 + e2 ≥ 1
a2
r s ∈ 0 < r < s a b c ∈ (r, s)
a
ABC AB = AC AD A BC BE CF ∠B ∠C
E AC F AB B C
AD BE CF A
QC = AC P BC B C P B = AB
ABC AQP
R r ABC ha hb hc ABC A B C
wa wb wc
A B C
N u v w
(u , v , w , N )
BA1
A1C =
CB1
B1A =
P (T k) < P (ABC ) P (T ) T
[T k] = k2 + k + 1
(k + 1)2 [ABC ] [T ] T
R k = k √ k P (ABC )
(k + 1)(k2 + k + 1)
rk = k2 + k + 1

AkG Γ
AI 2 ∠A 90
A1AB
CAB =
B1BC
ABC =
r1 s1
s1 ≥ s
r1 ≥ r
ABC B D B AC E
∠BDC BC M N BE DC F M N BD AD = 2BF
DBC EC A F AB ABC ∠DBC = ∠EC A = ∠F AB ∠DCB = ∠EAC = ∠F BA

a b c d e > 0 a2 + b2 + c2 + d2 + e2 ≥ 1
a2
r s ∈ 0 < r < s a b c ∈]r, s[
a
ABC AB = AC AD A BC BE CF B C
E AC F AB B C
AD BE CF A BE CF
Q BC C B QC = AC P BC B C
P B = AB
A B C wa wb wc
A B C
(u , v , w , N )
BA1
A1C =
CB1
B1A =
P (T k) < P (ABC ) P (T ) T
[T k] = k2 + k + 1
(k + 1)2 [ABC ] [T ] T
R k = k √ k P (ABC )
(k + 1)(k2 + k + 1)
rk = k2 + k + 1
A1A2 · · · An Γ G k = 1 2 . . . n Bk
AkG Γ
A 90
A1AB
CAB =
B1BC
ABC =
0 < λ < 1 r s ABC r1 s1 A1B1C 1
s1 ≥ s
r1 ≥ r
ABC B D B AC E
BDC BC M N BE DC F M N BD AD = 2BF
DBC EC A F AB ABC DBC = EC A = F AB
DCB = EAC = F BA

N = 0 1 2 . . . T N N T N T N (cos θ) = cos(Nθ) T N
N xN T N (x) 2N −1 y = T N (x) (x2 − 1)y + xy = N 2y
T N
cos(Nθ) + i sin(Nθ) = (cos θ + i sin θ)N
T N (x) = N/2
T N (x) =
S N (n) = N · 2N −2n−1
N − n

I ABC d e f IAC IBC IC A AB BC CA D E F
IA d f
AEPF B F P D CDPE P ABC
∠P EA = ∠P F A = 90
P A2 = P F 2 + F A2 = P E 2 + EA2
(AF + P E )(AF − P E ) = (AE + P F )(AE − P F )
A F P E
AF + P E = AE + P F
AF −P E = AE −P F P E = P F P D = P E P D = P E = P F P
ABC
|JM −J M | < M M
t A3
4 (1 + 2 ln2)− 1
2
n
n

sin A
sin B =

A + B = 135
A = B A + B = 135 B = C B + C = 135
C = A C + A = 135 A + B + C = 180
A = B = C = 60
http://www.cut-the-knot.org/ctk/BridesChair.shtml
m + m2
k − 2
m(m + 1)(2m + 1)
ABCD Γ O Γ AB AD M N Γ
AM AN E F EF BC CD P Q
[AMON ] [AEF ] [CP Q] [Z 1Z 1 . . . Z n] n Z 1Z 2 . . . Z n
σ = 1 2
(CP + P Q + QC ) Γ A AEF AN = AM = σ

AD BC P ∠AP C = α α < π/2 B C E F
B C E F AD A = P D φ θ B C
[ABDE ] = [ABD] + [ADE ] = AD · (BB + EE )
2
2
N CN θ M BM θ
G A CN ∠N AG +∠CAE = π 2
AGC EE A
EE = AG · AE
sin(θ + φ)
sin(θ + φ)
BC BDF CDE
= sinα
= sinα(BP −CP ) = BB − CC
ABCD AD AB AC BD CD E F G H
[EBCF ] : [GBCH ] = EF : GH [Z 1Z 1 . . . Z n] n Z 1Z 2 . . . Z n
C AD B AD C AD AB BD X Y
EF
XC =
AE
AX =
DH
DC =
HG
CY
EF
GH =
CX
CY
EF
GH
FS

2 , x
1 + α + α2
2 , x
1 + α + α2

z w ∈ |z| = 1 j k
|1 − zk| = |z j| · |1 − zk| = |z j − z j+k| ≤ |z j + w| + |z j+k + w|
n k k < n
(n − k)|1 − zk| =
=
n−1
k=1
k=1
n−k
j=1
j=1
n−1
k=1
k=1
lim n→∞
1 n √
n! loga
n
loga
a √

2 +
a √
2 +loga
2
√ abc
2
√ abc

∗
∗ 72 73 76 77 79 80 81 83 84 87
∗ 60 1 2 3 4 5 6 . . . 383940
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15
∗ 72 73 76 77 79 80 81 83 84 87
∗ 60 1 2 3 4 5 6 . . . 383940

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15
.................................................................................................................................................................................. .. .. .. .. . .. . .. .. .. .. . .. . .. .. .. .. . .. .. .. . .. .. .. . .. .. .. .. . .. . .. .. .. .. . .. . .. .. .. .. . .. .. .. .. .. . .. . .. .. .. .. . .. . .. .. .. .. . .. . .. .. .. .. . .. .. .. .. . .. . .. .. .. .. .. . .. . .. .. .. .. . .. .. .. .. . .. . .....................................................................................................................................................................................
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... .. .. .. .. .. .. .. .. .. ..................
........................... .. .. .. .. .. .. .. .. .. .
x2 + y2 = 25
(m, n) x2 + y2 = 25 m2 + n2 = 25 m n
m2 ≥ 0 n2 {0 1 4 9 16 25} m2
• n2 = 1 n2 = 4 m2 m2 = 24 m2 = 21
• n2 = 0 n = 0 m2 = 25 m = ±5 (−5, 0) (5, 0)
• n2 = 9 n = ±3 m2 = 25− 9 = 16 m = ±4 (−4,−3) (−4, 3) (4,−3)
(4, 3)
• n2 = 16 n = ±4 m2 = 25−16 = 9 m = ±3 (−3,−4) (−3, 4) (3,−4)
(3, 4)
• n2 = 25 n = ±5 m2 = 0 m = 0 (0,−5) (0, 5)
12
(k − 1)
8
A abba a b a = 0 B cddc c d c = 0
A B A + B = C 19998 C 1 C
1f gf 1 f g
C 1 a + c = 1 a + c = 11 a + c = 11 a = 0 c = 0
f = 1 f = 2
f = 1 f = 2
f = 1
a + c = 11 f = 1 b + d 0 b+d = 10 a+c = 11

f = 2 a + c = 11 f = 2 b + d 1 b + d = 1
b + d = 11 b + d = 1 a + c = 11 f C f 1 b + d = 11
g = 2 C = 12221
∠AOB
11
√ 2
11
by = 1 b = x2 − 3x + 1 y = x + 1 by = 1 b y
b = 1 x2 − 3x = 0 x = 0 x = 3
b = −1 y (x−2)(x−1) = 0 x + 1 x = 1
y = 0 b = 0 x = −1
x = −1 x = 0 x = 1 x = 3
8
f (a, b) a b f (1, 5) = 1 + 2 + 3 + 4 + 5 = 15 f (3, 6) = 3+ 4+ 5+ 6 = 18
f (133333, 533333)

11
(765x − 43) x = 43/765
x = 56/1234 x 3
11
p = x + y q = xy 6 p = 5q pq = 30 p = 5q/6
q2 = 36 q = ±6 q = 6 p = 5 q = −6 p = −5 x y (x, y) = ( 3, 2)
(2, 3) (−6, 1) (1,−6)
8
x = 0 sin x = 0 = x/315 (0, 0)
y = sin x y = x/315 sin x 2π −1 ≤ sin x ≤ 1 y = x/315
y = sin x −1 ≤ x/315 ≤ 1 −315 ≤ x ≤ 315 315 ≈ 100.27π −100.27π ≤ x ≤ 100.27π
−100.27π ≤ x ≤ −100π −1 ≤ x/315 ≤ −0.997 −0.75 ≤ sin x ≤ 0
100π ≤ x ≤ 100.27π 2 100 −100π ≤ x ≤ −98π −98π ≤ x ≤ −96π . . .
−2π ≤ x ≤ 0 0 ≤ x ≤ 2π . . . 98π ≤ x ≤ 100π x = 0 2 × 100 − 1 = 199
11
(1 + a)(1 + b)(1 + c) ≥ 8(1 − a)(1 − b)(1 − c)

1 2 26
10× $11
26× $27
$5.50 × 3 + $13.50 × 3 = $57 263 × 103 = 17 576 000
17 576 000 × $57 = $1 001 832 000
$100 $26 3 $78
$(70 + k) $(30− k)
3 3 k ∈ {0 1 . . . 8} 3n − 8 3n − 7 . . . 3n
n = 26 n = 10 3n −
1 n 1 2
= 0 1 2 3
3n n, n, n 1 3n − 1 n,n,n − 1 3 3n − 2 n,n,n − 2 3 3n − 3 n,n,n − 3 3
n, n − 1, n − 1 3 n, n − 1, n − 2 6 n− 1, n − 1, n − 1 1
6 10

1215 a a3 + 2a2 + a + 5 221 a 2a2 + 2a + 1
a3 + 2a2 + a + 5 = (2a2 + 2a + 1)
1 2
2003
2√ 2003 ≈ 44.75 2 3 5 7 11 13 17 19 23 29 31 37 41 43
2003
122003
56 = 15625 66 = 46656 76 = 117649
A B C D E F
ABC × DEF = 232323
ABC DEF 3
851 273 232323
3
a2 + ab + b2 ≥ 1 4
a a = 1 + b
(1 + 2b + b2) + (b + b2) + b2 ≥ 1 4
3b2 + 3b + 3 4
3 P (A) = 1 3
P (B) = 1 6
P (C ) = 1 2
1
1
8
P (N H )
= P (N H | A) · P (A) + P (N H | B) · P (B) + P (N H | C ) · P (C )
= 1
P (A | N H ) = P (N H | A) · P (A)
P (N H ) =
45 45
45 3 × 3 × 3 33 = 27 4 × 4 × 4 43 = 64
4 × 4 × 4
4 × 4 × 4 5 × 5 × 5 4 × 4 × 4 64
45 19 16
5 × 5 × 5 45 80 53 = 125
3 × 3 × 3 27 80 18
5 × 5 4
45
360 = 23 × 32 × 5 d(360) = 24 = 4 × 3 × 2
40 = 8 × 5 d(8) = 4 d(5) = 2 d(40) = 8
12 = 4 × 3 d(4) = 3 d(3) = 2 d(12) = 6
450 = 18 × 25 d(18) = 6 d(25) = 3 d(450) = 18
120 = 12 × 10 d(12) = 6 d(10) = 4 d(120) = 12
500 = 25 × 20 d(25) = 3 d(20) = 6 d(500) = 12
n
d(n) = (e1 + 1) × (e2 + 1) × (e3 + 1) × · · · × (en + 1)

n = 24 × 32 × 55 = 450000 d(450000) = (4 + 1) × ( 2 + 1 ) × (5 + 1) = 5 × 3 × 6 = 90
450000 90 1 450000 450000
2a × 3b × 5c 0 ≤ a ≤ 4 0 ≤ b ≤ 2 0 ≤ c ≤ 5
450000 2 3 5 2 0 1 2 3 4
5 = 4 + 1 3 = 2 + 1 3 6 = 5 + 1 5
2 3 3 6 5
5 × 3 × 6 = 90 450000
d(n) a b
d(a×b) = d(a)×d(b) a b 1 f

A =
f (z) = z
2 × 2 I =
g(z)
c
f {f k}∞k=0
f 0 f 1 = f f k = f f k−1
k ≥ 2 f k Ak
A f {f k} n
f n = f 0 n n m < n
f m = f 0 m n f = f 0
n f
f (z) = z 1

A2 I a + d = 0 f (z) = −z f (z) = 1/z
c = 0 d = 0 f
2 c = 0 c = 1
g(z) = z− d g g−1(z) = z + d h(z) = g−1
f (g(z))
I Ak {f k} {hk}
f h c = 1 d = 0
A =
I a2 +b = 0 A3 = −a3I a = 1 {f k}
3
a(a2 + 2b) b(a2 + b)
I a(a2 + 2b) = 0 a = 0 2 a2 + 2b = 0
A4 = −a4
a4 + 3a2b + b2 ab(a2 + 2b)

n = 6
a(a2 + b)(a2 + 3b) a4b + 3a2b2 + b3

I a(a2 + b)(a2 + 3b) = 0 a = 0 2 a2 + b = 0
3 a2 + 3b = 0 A4 = −a4
4 I a = 1
a(a2 + 2b)(a4 + 4a2b + 2b2) a6b + 5a4b2 + 6a2b3 + b4

a(a2 + 2b)(a4 + 4a2b + 2b2) = 0
a = 0 2 a2 + 2b = 0 4 a4 + 4a2b + 2b2 = 0
a2/b n = 8

a12 + 11a10b + 45a8b2 + 84a6b3 ab(a2 + b)(a2 + 2b)(a2 + 3b) +70a4b4 + 19a2b5 + b6 (a4 + 4a2b + b2)
a(a2 + b)(a2 + 2b)(a2 + 3b) a10b + 9a8b2 + 28a6b3
(a4 + 4a2b + b2) +35a4b4 + 15a2b5 + b6

a(a2 + b)(a2 + 2b)(a2 + 3b)(a4 + 4a2b + b2) = 0
a = 0 2 a2 + b = 0 3 a2 + 2b = 0 4
a2 + 3b = 0 6 a4 + 4a2b + b2 = 0 a2/b
{f k} 1 2 3 4 6
[email protected]
(0, −→ I , −→ J , −→ K )
P x + y + z = 1
(a , b , c) a + b + c = 0
C O
L L L
1 −1
H 1 (x,y, 1) (x,y, 1) 1
(−1,−1, 1) H 1 H 1
h Z h (a , b , c) c = h Z h h = 1 h = 2
Z h H (h) h2 4
N (h) − 2
N (h) N (h) h
(a,b,c) (r,s,t) r s s
a = r(r + s)t b = s(r + s)t c = −rst
(a,b,c) (r,s,t)
h P (h) (a,b,c) c = h P (h) 2
h P (h) = N (h) (hn)
P (hn)/N (hn)

(xn, yn, zn)
3 2
T = (a,b,c) T = (a,c,b) S (T ) = a + b + c z(T ) = a + bj + cj2
z(T ) S (T ) z(T ) = 0 θ z(T ) a b
c
z0 T = (a,b,c) z(T ) = z0
T 1 T 2 T 1 ∗ T 2 S (T 1 ∗ T 2) = S (T 1)S (T 2) z(T 1 ∗ T 2) = z(T 1)z(T 2)
T 1 ∗ T 2 T 1 T 2 z(T 1 ∗ T 2) z(T 1 ∗ T 1)
T 1 T 2 T 1 ∗ T 2 T 1 T 2 T 1 ∗ T 2 T 1 T 2
T 1 ∗ T 2
T 1 ∗ T 2 T 2 ∗ T 1 (T 1 ∗ T 2) ∗ T 3 T 1 ∗ (T 2 ∗ T 3) T 1 T 1 ∗ (1, 0, 0)
T 1 T 2 T 1 ∗T = T 2 T
T (T n) T 0 = (1, 0, 0) T n+1 = T ∗ T n S (T n) p
T 1 ∗ T = T 2 T
A m u v m = u2 + 3v2 A
z u v z = u+ iv √
3 |z|2 = u2 + 3v2 B n
r s n = r2 + rs + s2
A A
A A

4 A A A
A n0
u v
p/2 p u − u v − v p u2 + 3v2 n0 < p
u0 v0
0
u2 1 + 3v2
A m = C 2 · p1 · · · pk C pi A
p 3 p−1 K (x , y , z) x y z 0
p p (xyz− 1) K ( p−1)2
3 K x = y = z
x 1 p p x2 + x + 1 p A
A
D A

ABC a ≤ b ≤ c a = BC b = CA c = AB R r ∠C
a + b − 2R − 2r
an = n
an = (−1)n + nan−1 n ≥ 2 n = 2 n ≥ 3
an−1 = (−1)n−1 + (n − 1)an−2
an = n

an = n! − n!
3
n n1 n2 . . . nk

2k
ni ≥ 5 i
j ≥ 10 2 j − 1 > j3
j j = 10 210 − 1 = 1023 > 103 = 1000 2 j − 1 > j3
j ≥ 10 2 j+1 − 1 > ( j + 1)3 j ≥ 10
j + 1
5
4
3
< 2
2 j+1 − 1 > 2 j+1 − 2 = 2(2 j − 1) > 2 j3 > ( j + 1)3
m ≥ 10
n = 7
(x − a1)(x − a2) · · · (x − a10) = (x + a1)(x + a2) · · · (x + a10)
5
a1 a2 . . . a10 a1 . . . a5
a6 = 0 a7 + a8 = 0 a9 + a10 = 0 k ∈ {6 7 8 9 10} x − ak ak
x /∈ {a6 a7 a8 a9 a10} (x − a1)(x − a2) . . . (x − a5) = (x + a1)(x + a2) . . . (x + a5)
x > 0 k ∈ {1 2 3 4 5} |x − ak| = max{x − ak, ak − x} < x + ak = |x + ak|
x < 0 x = 0
a1 a2 . . . a2000
a3 1 + a3
n 1 ≤ n ≤ 2000
a3 1 + a3
2 + · · · + a3 n =
ai = i i = 1 2 . . . n n+1 i=1
a3 i =
an+1 ∈ {n + 1 −n 0} n+1 i=1
a3 i =
n(n + 1)
a3 i =
1 = a2 1 a1 ∈ {0 1}
k ≥ 1 P i i = 1 2 . . . k
a1 a2 . . . ak+1
n 1 ≤ n ≤ k + 1
i ∈ {1 2 . . . k + 1} ai = 0 ai a1 a2 . . . ak+1 k
P k P k a1 a2 . . . ak+1
i ∈ {1 2 . . . k + 1} ai+1 = −ai k ≥ 2 P i a1 a2 . . . ai
ai+1 = −ai ai ai+1
a1 a2 . . . ak+1 k − 1 P k−1 P k−1
a1 a2 . . . ak+1
i ≤ k + 1 ai = 0 ai = −ai−1
ai = i i ∈ {1 2 . . . k + 1} a1 a2 . . . ak+1
1 √
x = 0 1 + 1
1 + y2 ≤ 2
y > 0 y = 0 0 < x ≤ 1 0 < y ≤ 1 u ≥ 0 v ≥ 0
x = e−u y = e−v
1 √

p x y z w x2 + y2 + z2 − wp = 0 0 < w < p
p = 2 x = y = w = 1 z = 0 p
1 + 02 1 + 12 . . . 1 + p − 1
2
2
2
2
0 ≤ a b ≤ ( p − 1)/2 p a2 − b2 = (a − b)(a + b) p a + b a − b a = b
0 ≤ x y ≤ ( p − 1)/2
1 + x2 ≡ −y2 (mod p)
x2 + y2 + 1 = wp w
0 < x2 + y2 + 1 ≤ 2 p − 1
2
2
f (x) + f (y)

f (x) + f (y)
f (x2 − y2) = (x + y)
f (x) − f (y)
f (x) − f (y)
xf (y) = yf (x) y = 1 f (x) = xf (1) f
p p ≡ 1 (mod 4)
p−1
k=1
2k2
p
− 2
k2
p
8 | mn m × n
m n
t1 t2
3t1 + t2
2 | m 4 | n 8 | n mn
8
a b ∈ (a, b) = 1 m ≥ (a − 1)(b − 1) m = ra + sb r s ∈ ∪ {0}
a = 2 b = 3 m 1 m = 2r + 3s r s
nr
4
ns
8
a b c x y z a > b > c > 0 x > y > z > 0
a2x2
a ≥ b ≥ c > 0 x ≥ y ≥ z > 0
a2x2 ≥ b2y2 ≥ c2z2
1
S ≥ 3(a2x2 + b2y2 + c2z2)
S = (by + cz)(bz + cy) + (cz + ax)(cx + az) + (ax + by)(ay + bx)
= a2(xy + xz) + b2(yz + yx) + c2(zx + zy)
+ (ab + ac)x2 + (bc + ba)y2 + (ca + cb)z2
xy + xz ≤ 1 2
(x2 + y2) + 1 2
S ≤ a2
x = y = z a = b = c
S ≤ 3(a2x2 + b2y2 + c2z2) + a2x2 + b2y2 + c2z2
= 4(a2x2 + b2y2 + c2z2)
S ≥ 3 4
x = y = z a = b = c
n ≥ 3 α (0, π) P n(x) = xn sinα− x sin nα + sin(n − 1)α
f (x) = x2 + ax + b n ≥ 3 P n(x) f (x)
g(x) g(x) = x+c n ≥ 3 P n(x) g(x)

= x2 sin nα− x
sin(n + 1)α + sin(n − 1)α
= (sin nα)(x2 − 2x cosα + 1)
f (x) = x2 + ax + b f (x) P n(x) n ≥ 3 f (x) P n+1(x) − P n(x) = (sin nα)(x2 − 2x cosα + 1)
n ≥ 3 n sin nα = 0 n = 3 n = 4 f (x) = x2 − 2x cosα + 1
P n(x) = xP n−1(x) +
sin(n − 1)α
sin(n − 2)α
f (x)
P n(x) = f (x)
f (x) P n(x) n ≥ 3
g(x) = x + c P n(x) n ≥ 3 g(x) P 3(x) P 3(x) = (sin α)f (x)(x + 2 cosα)
c g(x) f (x) f (x) eiα e−iα g(x) x + 2 cosα c = −2cosα
g(x) P 4(x) P 4(−2cosα) = 0 3 − 4sin2 α = 0 α = π
3 α = 2π
P n(x) c
100
100
ABCD AB = BC = CD Γ E AD Γ B C
F G EB AC EC BD
FBCG
ABCD O AC BD M N AC
BD [OAB] + [OCD] = [OBC ] [P QR ] P QR AN DM BC
ABC BC < AB BC < AC ABC ACB
AC AB D E I
BD
CE =
AP
ABC AL BM CN K AU BV CW T
M N V W P N L W U Q LM U V R
QAR RBP P CQ
s

A1A2 · · · An Γ P Γ P A1 P A2 . . . P An Γ
B1 B2 . . . Bn
n k=1
(P Ak)2 ≥ n
f (E ) E
E 1003 2003≤f (E )≤3003
ABC C AB M N BC AC C M AC C N BC
C CN AC N C BM
ac + bc − 2 = c3 − c
ba + ca − 2 = a3 − a
cb + ab − 2 = b3 − b
n 3
−bc b2 + bc c2 + bc a2 + ca −ca c2 + ca a2 + ab b2 + ab −ab
= (bc + ca + ab)3

ABCD AB = BC = CD Γ AD Γ B C E
F G EB AC EC BD
FBCG
ABCD O AC BD M N AC
BD [OAB] + [OCD] = [OBC ] [P QR ] P QR AN DM BC
ABC I BC < AB BC < AC ∠ABC ∠ACB AC AB D E
BD
CE =
AP

ABC AL BM CN K AU BV CW T M N
V W P N L W U Q LM U V R
QAR RBP P CQ
s

A1A2 · · · An Γ P Γ P A1 P A2 . . . P An Γ
B1 B2 . . . Bn
n k=1
(P Ak)2 ≥ n
E f (E ) E

ABC C AB M N BC AC C M AC
C N BC C CN
AC N C BM
n 3

−bc b2 + bc c2 + bc a2 + ca −ca c2 + ca a2 + ab b2 + ab −ab
= (bc + ca + ab)3

( )
k=0
ab−kbk > ab + bb
a > b > 1 d = gcd(a, b) a1 = a/d b1 = b/d a1 > b1 gcd(a1, b1) = 1
(a1 + b1)b = ab1 + da−bba1
b2 1
1
(a1 + 1)d = ad1 + dd(a1−1)
a1 = 2 3d = 2d + dd
3d > 2d + dd d = 1 d = 2 3d < 2d + dd
d ≥ 3

d > 1
gcd(a1, b1) = 1
a1 = 3 25d = 9d + 8ddd
d = 1 d = 2 d = 3 d > 3 25 < 8d d > 3
a1 = 5 49d = 25d + 32dd3d
49d < 25d + 32dd3d d ≥ 1
a1 > 6 (a1 + 2)2 < 2a1
b1 = 2 b1 = 2 > 1 a1 > 2
dab−1 1 + 2−1d(2d− 1)ab−2
1 +
dab−1 1 = da2d−1
1
2 a1 b1 gcd(a1, b1) = 1 a2d−1 1
d d = 2d1
b1
p a1 > b1 ≥ 3 gcd(a1, p) = 1 gcd(a1, b1) = 1 p b1
p dab−1 1 a1 p d p d = pkd1 b1 = pb2 d1 b2
p k b = pk+d1b2
pkd1ab−1 1 +
(b−3)bb−3 2
+ p(b−2)bb−2 2 = p(a−2)+k(a−b)ba−2
2 da−b1
2 m ≥ 2
b
m
p j · · · pk+d1b2 − m + 1
m − 1 · 1
1 · · · pk+d1b2 − p j + 1
p j − 1 · 1
pk+− j tm
pk+− j p(pj−2) = pk+− j+pj−2 = pk+(pj−1)− j
tm pk+1 ( p j − 1)− j ≥ 1
tm pk+1
ab2 + bc2 + ca2 ≥ ab + bc + ca
b
c ,
c
a ,
a
b
xpyqzr ≥
xuyvzw
3

= ∠F AB = ∠F BA = ∠BAC
M BC P Q DE AC DF AB
M P : M Q = AB : AC
N AC AN = F B AFBN AE : AN = AC : AB AEN ∼ ACB
EN = CD EN CD CN ENDC
P CN M P = 1 2
BN = 1 2
ABC ∠B ∠C ∠A
∠A D E F
E CI B AI DE AC
A = AI ∩BC B = BE ∩AC BC CA AB a b c
BB ⊥ AIA
AB = AB = c BC = b − c EC ∠BCB
BE
EB =
BC
CB =
a
b + c
S = A1A2A3A4 M
4
M A j j = 1 2 3 4
M = (x , y , z) A j = (x j , y j, z j)
F j =
(x − x j)2 + (y − y j)2 + (z − z j)2 grad(F j) = −→u j α1 α2 α3 α4
α1 + α2 + α3 + α4 = 1
α1A1 + α2A2 + α3A3 + α4A4 = M
4 j=1
−→ 0
(α1M A1 − α4M A4)−→u1 + (α2M A2 − α4M A4)−→u2
+ (α3M A3 − α4M A4)−→u3 = −→ 0
α1M A1 = α2M A2 = α3M A3 = α4M A4 c
4
j=1
(1 − α j) −−−→ M A
j = −α j −−−→ M A j j = 1 2 3 4
j = 1 M − α1A1
9
1
α1
+ 1
α2
+ 1
α3
+ 1
α4
i=1 AiM ≥ n

x1 x2 x3 x4 x5 ≥ 0 x1 + x2 + x3 + x4 + x5 = 1
x1
≥ 5
6
x1 = .5 x2 = .4 x3 = .05 x4 = .03 x5 = .02 x1 x2 x3 x4 x5 ≥ 0 x1 + x2 + x3 + x4 + x5 = 1
x1
3 + a2 4 + · · · + a2
a2(a3 + a4 + · · · + a100) ≥ a2 3 + a2
4 + · · · + a2 100 ≥ 200
a3 + a4 + · · · + a100 ≥ 200/a2
a2 ≥ 2a2 +
40 a1 = a2 = a3 = a4 = 10 a5 = a6 = · · · = a100 = 0
a1 a2 . . . an a1 + a2 + · · · + an
a1 ≥ a2 ≥ · · · ≥ an ≥ 0
a21 + a22 + · · · + a2m ≥ A

j=1
j=1
n j=1
λ ∈ [0, 1]
n

S n =
j=1 x2 j = 1

n ∈ n > 1 p q r ∈ S n q · r cos qr
pr < qr r p q min{ p j r j} ≤ s j ≤ max{ p j r j}
j s p r s p q q · s p · s q C 2
q λ ∈ (0, 1)
a = x b = x + p c = x + q
E = E (x) = ( p2 − pq + q2)x2 − ( p3 − 5 p2q + 4 pq2 + q3)x
+ ( p4 − 3 p3q + 2 p2q2 + q4)
E (x) x
− 4( p2 − pq + q2)( p4 − 3 p3q + 2 p2q2 + q4)
= −3( p6 − 2 p5q − 3 p4q2 + 6 p3q3 + 2 p2q4 − 4 pq5 + q6)
= −3( p3 − p2q − 2 pq2 + q3)2
≤ 0 p2 − pq + q2 ≥ 0 E (x) ≥ 0 x
a ≤ b ≤ c
( ) a = b = c
a b c
a = b = c k a = k(1 + z + z) b = k(1 + wz + w2z) c = k(1 + w2z + wz) ( ) w = e2πi/3
z = 1√ 7

α n
k=1
k=1
x3kxk+1
4(a2 + b2 + c2 − ab− bc− ca)
(a + b + c)2 − 3(a3b + b3c + c3a)
= (A− 5B + 4C )2 + 3(A−B− 2C + 2D)2
A = a3 + b3 + c3 B = a2b + b2c + c2a C = ab2 + bc2 + ca2 D = 3abc
( ) ( ) a b c
a2(a− b)(a− 2b) + b2(b− c)(b− 2c) + c2(c−a)(c− 2a) ≥ 0
A B A BC ≤ π
2
AX AZ AX B Z AZ AXB
Y W
K AZ µ K KZ = KA ν

APBQ M
M M A (= M B) APBQ P B : P A = (
√ 5 + 1)/2 A P B Q
∠AZP ∠P ZB

xkf (x) xk f (x) xn−kf (x) = xnxkf (x)
k
xk(1 − 2x)n = (−2)k

= yn(1 + y)2n n
= yn(1 − y2)n =
x3 + y3 ≤ 2
x3 + y3 = x3/2, y2 · x3/2, y
x3 + y3 ≤
x3 + y4 ·
x3 + y2
x3 + y3 ·
x + y
x2 + y2 · √
x + y ≤ 2 x3 + y3 ≤ 2
x y > 0 r xr−1 + yr ≥ xr + yr+1 xr + yr ≤ xr−3 + yr−3
r = 3 x = 0 y = 0
S = xr−2(1−x) + yr−1(1−y) T = xr−1(1−x) + yr(1−y) T ≥ 0
S − T = xr−2 − 2xr−1 + xr + yr+1 − 2yr + yr−1
= xr−2(1 − x)2 + yr−1(1 − y)2 ≥ 0
S ≥ T ≥ 0
xr−3 + yr−3 − xr − yr − xr−3(1 − x)2 − yr−3(1 − y2)2
= −xr − yr + 2xr−2 − xr−1 + 2yr−1 − yr+1
= 2xr−2(1 − x) + 2yr−1(1 − y) + xr−1 − xr + yr − yr+1
= 2S + T
= 2S + T + xr−3(1 − x)2 + yr−3(1 − y2)2 ≥ 0
xr + yr ≤ xr−3 + yr−3

≥ α
t1 + t2 + · · · + tn
= 1
(sin x)f n(x) =

P n−1(z) n−1 cot(kπ/n) k ∈ {1 2 . . . n − 1}
z + i
z − i
n = 1
P n−1(z) = (n − 1)P n−2(z) P n−1(z)
P n−2(z) cot
P n−1(z) P n−1(z)/P n−1(z)
f n(x) x
cot x = cot
mπ
f n(x) = n sin

B1(− cosβ, sinβ) B2(cosβ, sin β)
C 1(− cos γ, sin γ ) C 2(cos γ, sin γ )
(± cosα ± cosβ ± cos γ, sinα + sin β + sin γ )
P Q R O O M
(P + Q + R )/3 O H M OH = 3OM
y = sinα+sin β+sin γ

∗
b2
648 832 1168 1944
x + z > y + z x − z > y − z
xz > yz xz2 > yz2
∗ ax = cq = b cy = az = d
∗
∗ 0 < α β < π 2
α > β
sin(α− β) = sin α− sin β
∗ f (x) = 3x + 5 f −1
(0, +∞) (5, +∞) (8, +∞) (−∞, +∞)
51 3
b2
648 832 1168 1944
∗ x y > 0 x > y z = 0

∗ ax = cq = b cy = az = d
∗
∗ 0 < α β < π 2
α > β
sin(α− β) > sinα− sinβ sin(α− β) < sinα− sin β sin(α− β) = sin α− sin β
∗ f (x) = 3x + 5 f −1
(0, +∞) (5, +∞) (8, +∞) (−∞, +∞)
1 1 5
9 8 7 6 10 11 12 13
17 16 15 14

8 1 (1) 0 2 (2) 3 7 (3) 4 6 (4) 5
(5) 2004 ≡ 4 (mod 8) 1999 ≡ 7 (mod 8) 2004 1999 n2
8 0 1 4 n2 + 1 (1) (2) (5)
105
34 + 35 + 36
3 = 35 105
105 = 3 × 5 × 7 105 1 3 5 7 15 21 35 105
1
3 5 7
105 1 3 5 7 15 21 35 105
2 6 10 14 30 42 70 210 105
14
105
5 21 19 + 20 + 21 + 22 + 23
6 17.5 15 + 16 + 17 + 18 + 19 + 20
7 15 12 + 13 + 14 + 15 + 16 + 17 + 18
10 10.5 6 + 7 + · · · + 10 + 11 + · · · + 14 + 15

f n n f 4 = 3 f 7 = 13
f 38 f 51 f 150 f 200 f 300
3
f 48 f 75 f 196 f 379 f 1000
2 1 1 0 1 1 0 1 1 0 . . .
3 f 51 f 150 f 300 f 38 f 200
3 1 1 2 0 2 2 1 0 . . . 4 3 f 48 f 196 f 1000
α + β
α β
1/3 + 1/2
1 − 1/6 =
α β [0,π/2] tan [0,π] π/2 α + β π/4
O = (0, 0) A = (2, 1) B = (3,−1) AO = AB
2 × 1 90 A B O ∠OAB = 90 ∠AOB 45 AOB
α + β = 45
α + β = ∠AOB

20044 (4 + 1)(4 + 1)(8 + 1) = 225
225 2α × 3β × 167γ α β γ
20044 20 21 . . . 28
3 167 5 5 167 25
20044 2(36×25)
9 2 5 167 45 20044 3(10×45)
β = 10 × 45 = 450 γ = 450 20044 2900 × 345 × 167450 = 2004450
70 20 10
0.7 × 0.7 = 0.49

= 0.72 × (1 + 0.22 + 0.24 + · · · ) = 0.49 × 1
1 − 0.04 =
f (2) = 3 f (2003) = 2004
48
3 48(1 + 1/2 + 1/3) 4 48(1 + 1/2 + 1/3 + 1/4) 100
4
∠AM C = 150

A1B1C 1 A2B2C 2 A1C 1 = A2C 2 B1M 1 B2M 2 A1D1
A2D2
AMC 150
1254 1254 770 770 1995 1995 2
A1B1C 1 A2B2C 2 A1C 1 = A2C 2 B1M 1 B2M 2
A1D1 A2D2
N N
N N
10 19 1 8 1
21 31 . . . 91 19 100 1 100 199 1 100 199 100 1 200
299 1 99 19 1 300 399 400 499
. . . 900 999 19 1 N T
1 N T T = N/2 T /N = 1
2 T /N
0.192 N = 99 0.598 N = 199 0.462 N = 299 T
1 2
1 19 + 100 + 19 + 11 = 149 T /N = 149/319 < 1
2 T /N = 1
2 100 ≤ N ≤ 299
n
T
N =

n = 60 N = 100 + 60 = 160 T = 20 + 60 = 80 160
200 ≤ N ≤ 300 T N 200 ≤ N ≤ 209 T = 119
N = 200 T = 120 201 ≤ N ≤ 209 N T/N > 1
2 210 ≤ N ≤ 219
2 N ≥ 219 T ≥ 130 N ≥ 260
T /N = 1 2
270 N = 270 N N = 272 T = 136 N = 272
1 281 T = 137 T/N < 0.488 T /N
N {160 270 272} T N
N ∈ {1458 3398 13120 44686} 1
1 9 1 1 10 19 10 11 20 99 8 19
100 199 100 119 200 999 152 [= 8(19)] 271
1 000 1 999 1 000 1 271 2 000 9 999 2 168 [= 8(271)] 3 439
10 000 19 999 10 000 13 439 20 000 99 999 27 5 12 [= 8(3439)] 40 593
T 1
3 4 4 8 4 9 4
7 1 3 6 7 4 3 7 5 1 6
5 8 4 3 2 8 7 2 3 5 8 4 7 1 6
1 9
4 9 5 3 1 8 6
2 1 9 8 7 8 3 7 2 6 1
3 5 7 9 2 4 6 1 2 3 4 5 6 7 8
9 8 1
8 9 1 6 2 7 3
4 2 9 7 5 7 6 5 4 3 2
6 1 5 9 4 8 3 2 4 6 8 1 3 5 7
9
a1 + a8 = a2 + a7 = a3 + a6 = a4 + a5 = 9
a3 a5 ∈ {3 6} a2 + a3 + a4 = a5 + a6 + a7 = 9
a1 + a2 = a3 a7 + a8 = a6
9 4 5
4 9 5 6 7 2 3
8 7 9 2 1 2 6 1 8 3 7
6 5 1 9 8 4 3 7 8 6 4 5 3 1 2
9 8 1
8 9 1 3 5 4 6
7 5 9 4 2 4 3 2 7 6 5
3 1 2 9 7 8 6 5 7 3 8 1 6 2 4
9 2 7
2 9 7 6 5 4 3
1 5 9 4 8 4 6 8 1 2 3
6 7 8 9 1 2 3 5 1 6 2 7 3 8 4
9 2 7
2 9 7 3 8 1 6
4 8 9 1 5 1 3 5 4 6 8
3 7 5 9 4 2 6 8 4 3 2 7 6 5 1
B ......................................................................................................................................................................................................................................................................................................
E AC BD a = AE = EC b = BE = ED ABE
b
a =
B E
AC BD a = BE = EC b = BE = ED α = ∠BAC β = ∠BC A
ABE CB E
√ 2/2
1/2 =
√ 2
ABC α + β + 75 = 180 α = 105 − β sinα = sin(105 − β) = sin 105 cosβ − cos 106 sinβ
√ 2 =
sinβ
=
√
cot β
N T /N T /(N − 2)
T
3N 2 − 6N = 2T
6N 3N 2 N
N = 2n n n 9 N 18 2T = 3(2n)2 − 6(2n) = 12n2 − 12n
T = 6n2 − 6n
− 3n
n + 2 n + 2
18 n 9 n + 2 11 n + 2 18 18 n = 16 N = 32
T = 6n2 − 6n = 1440
30 $1440÷ 30 = $48
3 5 1
(7a + 3) + (7b + 5) = 7(a + b + 1) + 1
33 5
5 7
a + c ≡ b + d (mod m)
a − c ≡ b − d (mod m)
a × c ≡ b × d (mod m)
ak ≡ bk (mod m)
9 1 9
43 658 912 4 + 3 + 6 + 5 + 8 + 9 + 1 + 2 = 38 9 3 + 8 = 11
1 + 1 = 2 43 658 912 9
2
5 4 3 5 × 4 = 20 2
9
cz + d z a b c d
ad = bc f {f k}∞k=0 f 0 f 1 = f
f k = f f k−1 k ≥ 2 {f k} n f n = f 0
n
A =
A det A = ad− bc
A tr A = a + d A det(A − xI ) = 0 I 2 × 2
A
2 × 2 A B det(AB) = det(A) det(B) 2 × 2 A λ
µ λ + µ = tr(A) λµ = det(A) λ µ 2 × 2 A λk µk Ak
k
m m n
n = m An = αI A f α B = A Bm = αI B j = βI j < m β = 0 A j = βI
j < n n {f k} n f
f {f k} n f B
Bn = I
f f {f k} n A
f An = αI α = 0 A2n = α2I α2 = det(αI ) = det(An) = (det A)n
α2/n = det A g
β = 0 A2k =
θ/π cos θ cos θ ∈ {0 ±1 ±1 2 }
θ/π = m/n m n cos nθ = cos mπ = (−1)m
2cos nθ = (2 cos θ)n + a1(2 cos θ)n−1 + · · · + an−1(2 cos θ) + an
ai 2cos θ xn + a1xn−1 + · · · + an−1x + an
2cos θ cos θ ∈ {0 ±1 ±1 2 }
A f An = αI n α
α = 1 (det A)n = det(An) = det I = 1 n det A = 1 λ µ A λn
µn An = I λn = 1 µn = 1 λµ = det(A) = 1 λ µ n µ = λ
λ + µ = tr A λ = cos θ + i sin θ θ = 2tπ/n t
0 ≤ t < n cos θ = 1 2

n 1 3 n
n 2u 2u3 u ≥ 1 8 12 2 4 6
n
[email protected]
ABC AB = AC M BC AB AC
P MPDC P D = M C P D M C AP D AP M
z
n ≥ 4 S = {P 1 P 2 . . . P n} n
at 1 ≤ t ≤ n P iP j P k P t
m(S ) = a1 + a2 + · · · + an
f (n) n S
b a = b2 − 1
n k nk = n
t 6t + 3
a b c a2+1 b2+1 (a2 + 1)(b2 + 1) = c2 + 1
AC BC ABC ACPQ BKCL
C ABC P L
n ≥ 3 x1 x2 . . . xn
n
2

ABC s ABC K L M P s AK BL CM
s AK
ABCP s
k {an} a1 = k + 1 an+1 = a2n − kan + k n ≥ 1
m = n am an
a b k m k + m = n ≥ 3 k ≤ 2m m ≤ 2k x1 . . . xn
• k xi a x1 = a
• m xi b xn = b
•
xnx1x2 + x1x2x3 + · · · + xn−1xnx1
ABC S BC = a CA = b AB = c a ≤ b ≤ c u
v P ABC u ≤ P D +P E +P F ≤ v D E F
AP BP CP u v
a b c S
n S = {1 2 . . . n} f : S → S x + f 4(x) = n + 1 x ∈ S
f 4 f 4(x) = f
f

n > 1 p n p − 1 p n3 − 1 4 p − 3
ABC O P Q AC BC
AP
a21 + · · · + a2n−1
n n − 2
BC BC B
AB
mb mc ABC b c ABC
ABC
P ≡ 0 P ≡ 1 P
P P (x)P (x+1) = P (x2) x P (z)P (z+1) = P (z2)
z ∈ z ∈ P P (z2) = 0 P
(z − 1)2
= 0
0 < |z| < 1 {zn}∞n=0 z0 = z zn+1 = z2n n ≥ 0 n ≥ 0 0 < |zn+1| < |zn| < 1
P (zn) = 0 P P ≡ 0
|z| > 1 |zn+1| > |zn| > 1 P (zn) = 0
|z| = 1 z ∈ {1 eiπ 3 e−iπ
3 } z = eiθ
|(z−1)2| = 2(1− cos θ) ∈ (0, 1)∪ (1, 4) z0 = (1− z)2 zn+1 = z2n n ≥ 0
P 0 1 eiπ3 e−iπ3 P P (eiπ3 ) = 0 P (e−iπ3 ) = 0 eiπ3
P (eiπ6 )P (eiπ6 + 1) = P (eiπ3 ) = 0 eiπ6 eiπ6 + 1 P {0 1 eiπ3 e−iπ3 }
0 1 P (x) = axp(x − 1)q
a p q P (x)P (x + 1) = P (x2)
a2xp+q(x − 1)p(x + 1)q = ax2p(x − 1)q(x + 1)q
a = 1 p = q P (x) = xp(x − 1)p p P
P (x) = 0 P (x) = xp(x − 1)p
p
D DO E BC B AC

AC
C
DE
L
M
O
P

A n 0 ∈ A c = c1 c2 . . . cn
a = a1 a2 . . . an b = b1 b2 . . . bn ci = 0 ai = bi ci = 1 ai = bi f : A → A
f (0) = 0 a b k f (a) f (b) k a b
c A a+ b + c = 0 f (a) + f (b) + f (c) = 0
A /2 0 · a = 0 1 · a = a n
B1 = (e1, . . . , en) e1 = (1, 0, . . . , 0) e2 = (0, 1, 0, . . . , 0)
en = (0, . . . , 0, 1) a = a1, a2, . . . , an a = n
i=1
f (a), f (b)
= d(a, b) f i ∈ {1 . . . n} f (0) = 0 d(0, ei) = 1
d
0, f (ei)

B2 = (f 1, . . . , f n) (e1, . . . , en) A a = (a1, a2, . . . , an)B1
d(0, a) = k d
f i, f (a)
• ai = 0 d(ei, a) = k d
f i, f (a)
a = n
i=1
f a + b + c = 0
f (a) + f (b) + f (c) = f (a + b + c) = f ( 0 ) = 0
a b c
x3 + ax2 + bx + c = 0 x3 + bx2 + cx + a = 0 x3 + cx2 + ax + b = 0
a = −1 b c = −b x3−x2+bx−b = 0
x3 + bx2 − bx− 1 = 0 x3− bx2− x + b = 0 x = 1
b = 0 r
r3 − r2 = 0 r = 0 1

2400 n q
0 < q < 2000 a b q = a/b gcd(a, b) = 1 b > 1 q (n, q)
{q2} =
n!
2000
a2
b2 −
a2
b2
b2
2000 b ∈ {2 4 5 10 20} b = 5 b = 2b b ∈ {1 2 5 10} a
gcd(a, b) = 1
a2 = 25q2 + n!
(n, q)
n, a
5 a

n = 7
(8, q)
n = 9
(9, q)
ei ∈ {1 −1} i = 1 2 . . . N

ei . . . ei+4
(∅) = 0 (e1, . . . , eN )
(e1, . . . , eN ) U = (U 1, . . . , U p) i = 1 . . . N ei ei
i−4 i−3 . . . i U
U U 1 < · · · a i ≥ 2 ea
U ei
i < a U 1 = a i ei = ei
ea = ea V V U V 1 = a = U 1
U V (U ) p = (U ) q = (V ) U = (U 1, . . . , U p, U 1)
V = (V 1, . . . , V q, V 1) U V (e1, . . . , eN ) U V
U 1 = V 1 U = (U 2, . . . , U p)
V = (V 2, . . . , V p) (U ) < (U ) (U )
U = ∅ V = ∅ U = V
∼ E
2N
AB
S AD BC AP
P S AC BD P Q = QR = RS
............................................................................................................................................................................................................................................................................................................................................................................................................................................................ .. .. .. . .. .. .. .. .. . .. .. .. .. .. .. . .. .. .. .. .. .. . .. .. .. .. . .. .. .. .. .. .. . .. .. .. .. .. . .. .. .. .. .. .. . .. .. .. .. .. .. . .. .. .. .. .. .. . .. .. .. . .. .. .. .. .. .. . .. . .. . .. .. .. . .. .. .. .. .. .. . .. .. .. .. .. .. . .. .. .. .. .. .. . .. .. .. . .. .. .. .. .. .. . .. .. .. . .. . .. . .. .. .. .. .. .. . .. .. .. .. .. . .. .. .. .. .. .. . .. . .. . .. .. .. . .. .. .. .. . .. .. .. .. .. . .. ...........................................................................................................................................................................................................................................................................
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.... .... ..... ..... .... ..... .... ..... ..... .... ..... ..... .... ..... ..... .... ..... ..... .... ..... ..... .... ..... .... ..... ..... .... ..... .... ..... ..... .... ..... .... ..... ..... .... ..... .... ..... ..... .... ..... .... ..... ..... .... ..... ..... .... ..... ..... .... ..... ..... .... ..... ..... .... ..... ..... .... ..... .... ..... ..... .... ..... .... ..... ..... .... ..... .... ..... ..... .. .. .. .. .. ... .. .. ... .. .. ... ... .. .. ... .. .. ... .. .. ... .. .. ... .. .. ... .. .. ... ... .. .. ... .. .. ... .. .. ... .. .. ... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. ... ... .. .. ... .. .. ... ..................................................................................................................................................................................................................................................................................
C T
T D =
X O BD P T AC Y T S AC
XR T S P T AC AB CD
P R
P Q
n n = 2 n = 1 a = b ≥ 2
n ≥ 3 a = (n − 1)n−1 b = (n − 1)n a b ≥ 2
(aa)n = aan = (n − 1)n(n−1)n =
(n − 1)n (n−1)n
(aa)2 = bb
b ≤ a bb ≤ aa < (aa)2 a > 1 2a ≤ b bb ≥ (2a)2a = 22a(aa)2 > (aa)2
a < b < 2a a b p a p b
α β p a b
2aα = bβ α
(aa)5 = bb
a < b < 2a a < b < 5a p a p b
α β p a b α < β
a b b = ka k ∈ {2 3 4} 1 < a 1 y > 1 x y
x =
x y pi
p2αii α j j p2αii
αi 9αi < p2αii ≤ αi
9n > n n x y 2 x = 2α y = 2β
α β ≥ 1 22α−1β = α(2β−1 + 1) β ≥ 2 2β−1 + 1 22α−1 α
22n−1 > n n ≥ 1 β = 1 22α−2 = α 22n−2 > n
n ≥ 2 α = 1 x = y = 2
ABC AC > BC M AB AP A BQ B AP
H AB P Q R RH CM
∠BQA = ∠BP A = 90 Γ AB P Q H
QPBA R Γ QP AB R H C CR
H M Γ CR HM H ABC CH ⊥ RM
H CRM H CRM RH ⊥ CM

F = {M | M ⊂ S |M | = n} T ∈ P T ∪{(T )} F (T ) ∈ T M ∈ F
M = T ∪ {(T )} T ∈ P P

M ∈ F M = T ∪ {(T )} T ∈ P M x ∈ M M \ {x} x M = T ∪ {(T )}
T ∈ P
2n n
φ(k) n ≤ k gcd(n, k) = 1 φ(5m − 1 ) = 5n − 1
m n gcd(m, n) > 1
gcd(m, n) = 1 d = gcd(5m − 1, 5n − 1) d = 5gcd(m,n) − 1 = 4
5m − 1 2a pa1 1 . . . pak
k a ≥ 2
5n − 1 = (5m − 1 ) = 2a−1 pa1−1 1 · · · pak−1
k ( p1 − 1) · · · ( pk − 1)
5m − 1 2 a > 3 (5m − 1) ≡ 0 (mod 4) 5n − 1 ≡ 0 (mod 8)
d = 4 5m−1 d = 4
ai = 1 i = 1 . . . k
m 5m − 1 ≡ 0 (mod 8) a ≥ 3
2a−1( p1− 1) ≡ 0 (mod 8) 5n − 1 ≡ 0 (mod 8) d = 4
m = 2k + 1 a = 2 5m − 1 = 4 p1 · · · pk
pi 5 i 5 × 52k = 5m ≡ 1 (mod pi) 5 (mod pi)
pi (mod 5) x x2 ≡ 0 1 −1 (mod 5)
pi ≡ 1 −1 (mod 5) i i pi ≡ 1 (mod 5) 5

5n − 1 = 2( p1 − 1) · · · ( pk − 1) ≡ 2(−2)k (mod 5) 5n − 1 ≡ 2 −2 (mod 5)
gcd(m, n) > 1
f f (1) = 0
f (n) = m a x{f ( j) + f (n − j) + j} n ≥ 2 f (2000)
f (n) j 1 ≤ j ≤ n− 1 j f ( j) f (n− j)
f (n) = n(n − 1)
2 n ≥ 1
f (2) = f (1) + f (1) + 1 = 1 f (3) = max{f (1) + f (2) + 1 f (2) + f (1) + 2} = 3 f (4) = 6
n ≥ 5 f (k) = k(k − 1)
2 1 ≤ k < n
f (n − 1) + f (1) + n − 1 = (n − 2)(n − 1)
2 + 0 + n − 1 =
2 +
2 <
2
f (n) = max{f ( j)+ f (n− j)+ j} = n(n − 1)
2
1 + f (n + 1) = 1 + n+1
2
A B C D AC BD AC x DB y
A B C D F G F
ps( pr + qs)/(s2 − p2), ps( pq + rs)/(s2 − p2)
G
EF EG AC AC BD
A B F AC BD C D G AC BD
ABC D A EG ∠AF B = π
2 E ABF
AC BD A B C D
A D AC C D BD
XY Z XY |∠X −∠Y | = π
2 ABCD AC BD
= ∠BC A + ∠BAD − ∠F AC
= ∠BC A + ∠CAD = ∠BC A + ∠CB E = π 2
[email protected]
BA1
A1C =
CB1
B1A =
T k
[T k] = k2 + k + 1
(k + 1)2 [ABC ] [T ] T
R k ≥ k √
k P (ABC )
a > 0
a2 + 1
ABCD M AB CB DC AD CB E DCF ADG
N EF P F G
M N P
1 + (a1a22)
1 S2
a1 + 2a2 +
1
S n = 1 + 2 + · · · + n
(a,b,c,d) ab + 1 ac + 1 ad + 1 bc + 1 bd + 1
cd+1
(3, 5, 16, 1008) (3, 8, 21, 2080) (3, 16, 33, 6440) (3, 21, 40, 10208) (3, 33, 56, 22360) (3, 40, 65, 31416) (3, 56, 85, 57408) (3, 65, 96, 75208) (3, 85, 120, 122816)
(an, bn, cn, dn)
1
(bc + ca + ab)(yz + zx + xy)
≥ bcyz + cazx + abxy + 2
2
M B X A AX M C Y
XY D
2005 K
√ 2 −
BA1
A1C =
CB1
B1A =
P (T k) 0
a2 + 1
CB E DC F ADG N EF P F G
M N P
1 + (a1a22)
1 S2
a1 + 2a2 +
1
(a,b,c,d) ab + 1 ac + 1 ad + 1 bc + 1
bd+1 cd+1
(3, 5, 16, 1008) (3, 8, 21, 2080) (3, 16, 33, 6440) (3, 21, 40, 10208) (3, 33, 56, 22360) (3, 40, 65, 31416) (3, 56, 85, 57408) (3, 65, 96, 75208) (3, 85, 120, 122816)
(an, bn, cn, dn)
1
(bc + ca + ab)(yz + zx + xy)
≥ bcyz + cazx + abxy + 2
2
M B X A AX M C Y
XY D
2005 K
√ 2 −
a + b + c
a3
a3(b + c)

(b3 + c3)
(b3 + c3)
a = b = c

A B P C ∠P BF = ∠P CA ∠P Y X = ∠P CA
∠P Y X + ∠P Y Z = ∠P CA + ∠P CZ = 180
X Y Z
ABC 2 G
√ 3) B = (1, 0)
X =
u, √
P ABC u2 +
v − 1√ 3
ABC ∠A = 90 ∠B > ∠C H A BC B BC
B AH D B AC E D

DBC ∼ ABC ∠BB A = ∠BBA F BB ∼ ABC AH DE
F G ABC DBC F BB
DE
AH +
AB M N P Q
AM
AN =
AC + √
2AB
AB + √
2AC
ABC
AM
AN =
AC + √
2AB
AB + √
h
QMNP BU = CV = a
sin A − cos A = √
B + π
= c2 + (b2 + c2 − 2bc cos A) + 2bc sin A
= 2c2 + b2 + 2bc(sin A − cos A)
=
sin
( )
= 2 sin(60 − 40) = √
3 cos(40) − sin(40)
7sin2(40) < 3 sin(40) <
= sin(30) = 3 sin(10) − 4sin3(10) = 3c − 4c3
8c3− 6c + 1 = 0 8c3 > 0 −6c + 1 < 0 c > 1
6
ABC R AD BE CF P P
EF AC E 1 AB F 1 P F D AB F 2 BC D2
P DE BC D3 AC E 3
E 1F 1 cot A + F 2D2 cot B + D3E 3 cot C = 2R
L M N P BC CA AB
∠P F 1F 2 = ∠EF A = ∠C = ∠DF B = ∠P F 2F 1
P F 1 = P F 2 P D2 = P D3 P E 3 = P E 1
E 1F 1 cot A + F 2D2 cot B + D3E 3 cot C
= (P F 1 + P E 1) cot A + (P F 2 + P D2) cot B + (P E 3 + P D3) cot C
= (P F 1 + P E 3) cot A + (P F 1 + P D2) cot B + (P E 3 + P D2) cot C
= P F 1(cot A + cot B) + P D2(cot B + cot C ) + P E 3(cot C + cot A)
= P F 1 · AB
CF +
BE
a
3
p > 0 n
np/2 − 1
√ 3 ( )
n √
1
(n + 1)2/n = 1 (n + 1)2 = 3n
n = 2 3n > (n + 1)2 n≥ 3

= (x1 − x2)2 + (x2 − x3)2 + · · · + (xn − x1)2 ≥ 0
x2 1 + x2
2(x2 1 + x2
2 + · · · + x2 n)

·
·

≥ n2
n
≤ a n
k=1
m

·
≥ n2
x1 · · · xn
n
x1 + α . . . xn + α
4 1 0 −1
1 −1 3 4
f (x + y) = f (xy) f (7) = 7 f (49) =
49 14 7 1
9 5 3 0
−1 2 3 4
−35 4
35 4
−70 70
L −2 (r,−3) K L (a, b) (6, r)
a
100 231
115 231
1 2
118 231
4 1 0 −1
1 −1 3 4
f (x + y) = f (xy) f (7) = 7 f (49) =
49 14 7 1
9 5 3 0
−1 2 3 4
−35 4
35 4
−70 70
(6, r) a
100 231
115 231
1 2
118 231
∠ABC AC
n · (n− 1) · (n− 2) · (n− 3) · ·

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Crux Mathematicorum 2005