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1 ( 12 11 44 7 20 10 10 1 1 ( ( 2 10 46 11 10 10 5 8 3 2 6 9 47 2 3 48 4 2 2 ( 97 12 ) 97 12 -Spencer modulus moduli (modulus of elasticity) modulus (le) module modulus module 4 b θ a q φ p 1: 3 (le) module

a (a + ), a + a > (a + ), a + 4 a < a 4 a,,, y y = + a y = + a, y = a y = ( + a) ( x) + ( a) x, x y,y a y y y ( + a : a ) ( a : a > ) y = (a + ) y = a

[] a x f(x) = ( + a)( x) + ( a)x f(x) = ( a + ) x + a + () x f(x) a a + a > a + () x f(x) a (a + ) a x 4 f (x) = ( + a) ( x) + ( a) x = ( a + a) x + a + = ( a + ) x + a +, () a + a f(x) f(x) = f() = a

iii 1 1 1 1................................ 1 2.......................... 3 3.............................. 5 4................................ 7 5................................ 9 6............................

Case 1 a,b,α, β α α + β β α = ua + vb β = sa + tb α α + β β = (ua + vb (ua + vb + (sa + tb (sa + tb = (u a a + uva b + uvb a + v b b + (s a a + sta b

Bogoliubov H = a a + aa < 0 H 0 >, < 1 H 1 >, < H > < H 0 > a H = α α α α, α = 1, α, α = α, α = 0 α α α H = α α Postulate( a,b α, β a, a = b, b = 1, a, b = b, a = 0 α, α = β, β = 1, α, β = β, α = 0 α,

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

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OABC OA OC 4, OB, AOB BOC COA 60 OA a OB b OC c () AB AC () ABC D OD ABC OD OA + p AB + q AC p q () OABC 4 f(x) + x ( ), () y f(x) P l 4 () y f(x) l P

4 ( ) ( ) ( ) ( ) 4 5 5 II III A B (0 ) 4, 6, 7 II III A B (0 ) ( ),, 6, 8, 9 II III A B (0 ) ( [ ] ) 5, 0, II A B (90 ) log x x () (a) y x + x (b) y sin (x + ) () (a) (b) (c) (d) 0 e π 0 x x x + dx e

さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1

... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =

( ) ( ) 1729 (, 2016:17) = = (1) 1 1

1729 1 2016 10 28 1 1729 1111 1111 1729 (1887 1920) (1877 1947) 1729 (, 2016:17) 12 3 1728 9 3 729 1729 = 12 3 + 1 3 = 10 3 + 9 3 (1) 1 1 2 1729 1729 19 13 7 = 1729 = 12 3 + 1 3 = 10 3 + 9 3 13 7 = 91

学習の手順

NAVI 2 MAP 3 m 17 13 19 12 17 24 1 20 18 23 18 12 1 12 17 11 14 16 19 22 m 12 16 A 16 20 B 20 24 24 28 C 20 AC 40 cm AD A 0.20 12 0.300 B 0.200 0.12 12 C D 40 1.000 20 2 2 0 20 30 cm 14 1 1 160 160 16

7. 1 max max min f g h h(x) = max{f(x), g(x)} f g h l(x) l(x) = min{f(x), g(x)} f g 1 f g h(x) = max{f(x), g(x)} l(x) = min{f(x), g(x)} h(x) = 1 (f(x)

7. 1 ma ma min f g h h() = ma{f(), g()} f g h l() l() = min{f(), g()} f g 1 f g h() = ma{f(), g()} l() = min{f(), g()} h() = 1 (f() + g() + f() g() ) 2 1 1 l() = 1 (f() + g() f() g() ) 2 2 1 45 = 2 e 1

新・明解Ｃ言語で学ぶアルゴリズムとデータ構造

第 1 章 基本的 1 n 141 1-1 三値 最大値 algorithm List 1-1 a, b, c max /* */ #include int main(void) { int a, b, c; int max; /* */ List 1-1 printf("\n"); printf("a"); scanf("%d", &a); printf("b"); scanf("%d",

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3-1 ( sit ) (stead state vibratio) (trasiet vibratio) sit(a)w s ( W s ) W / g C (b) sit ( + s ) ( + s ) c + W + sit W s si t + s + c + si t (3.1) si t (3.1) a C W b sit(respose) () 3- acost+ bsit a sit+

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C

0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,

Taro10-名張１審無罪判決.PDF

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ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)

2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................