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3 2 ( 1.1 F (n + 1) = F (n) + F (n 1) (n 1), F (0) = 0, F (1) = 1 0, 1, 1, 2, 3, 5, 8, 13, 21, Wikipedia Wikipedia F (n) F (n 1) ϕ = lim n ϕ 2 = ϕ + 1 ϕ = 1 2 (1 + 5) F (n) = 1 [( ) n ( 1 5 ) n ] F (n) s(x) F (n)x n x = 1 x x 2 n=0 s(x) = F (0) + F (1)x + n=2 [F (n 1) + F (n 2)]x n n=2 = x + F (n 1)x n + F (n 2)x n = x + x F (n)x n + x 2 n=0 = x + xs(x) + x 2 s(x). n=2 n=0 F (n)x n

4 3 1.2 F (n + 1) 1 2 n n n 2 F (n) F (n 1) F (n + 1) = F (n) + F (n 1) s(x) s(x) = x[1 + (x + x 2 ) + (x + x 2 ) (x + x 2 ) n + ] x n+1 F (n + 1) 1 2 n n = 2m F (2m + 1) = m k=0 m+kc m k C x 2 F (2m + 1) = m k=0 (m + k)! (m k)!(2k)! F (2m) = m 1 k=0 (m + k)! (m k 1)!(2k + 1)! F (2m + 1) = F (2m) + F (2m 1) F (2m + 2) = F (2m + 1) + F (2m) T (n + 2) n N 1 F N (n + N 1) = N 1 k=0 n 0 F N (n + k 1) (n 1), F N (m) = 0 (0 m N 2), F N (N 1) = 1 1, 2,, N n F N (n + N 1) F 1 (n) 1, F 2 (n) = F (n), F 3 (n) = T (n) [ ] n n

5 4 n F N (n) s N (x) = x N 1 1 N j=1 xj N F N (n + N 1) = F N k (n + N k 1) (0 n N k; 1 k N 1) n N k + 1, N k + 2,, N 1.3 Φ(k) (k = 0, 1,, N 1) a k (k = 0, 1,, N 1) Φ(n + N 1) = N 1 k=0 s(x) = N 1 m=0 a k Φ(n + k 1) (n 1). [ 1 N m 1 j=1 a N j x j ] Φ(m)x m 1 N j=1 a N jx j [ ] s(x) = Φ(n)x n = n=0 N 1 n=0 Φ(k)x k + Φ(n)x n 2 Ψ(x) Φ(n)x n = n=n n N + k = m n=n [ N 1 k=0 n=n ] a k Φ(n N + k) x n. Ψ(x) = N 1 k=0 a k x N k m=k Φ(m)x m = N 1 k=0 a k x N k[ s(x) k 1 m=0 Φ(m)x m]. N k = j Ψ(x) = N j=1 a N j x j[ N j 1 s(x) Φ(m)x m]. m=0 [ N Ψ(x) = a N j x j] s(x) j=1 N 2 m=0 [ N m 1 j=1 a N j x j] Φ(m)x m

6 5 m N 2 N 1 Ψ(x) s(x) [ N 1 m=0 1 ] N m 1 j=1 a N j x j Φ(m)x m s(x) = 1 N j=1 a N jx j [ N 1 N 1 N ] s(x) = Φ(m)x m m=0 j=n m a N jx j Φ(m)x m + m=0 1 N j=1 a. N jx j j + m N = k s(x) = N 1 m=0 N 1 Φ(m)x m + x N m=0 Φ(m) m k=0 a m kx k 1 N j=1 a. N jx j x s(x)

7 [1] θ 2θ cos 2 θ + sin 2 θ = 1 (2.1.1) ABC AB = CA = 1, CAB = 2θ (2.2.1) 2.1 CAB BC D C D AB AB E, F CE D DG ABD ACD, AD BC, BAD = CAD = θ (2.2.2) BD = CD = sin θ, AD = cos θ (2.2.3) AEC CAE = 2θ AE = cos 2θ, AF = AD cos θ = cos 2 θ ABD CBE CE DF 1 BAD = BCE = BDF = θ

8 7 C θ A θ θ G D θ E F B 2.1: EF DG EF = GD = CD sin θ = sin 2 θ cos 2θ = AE = AF EF = cos 2 θ sin 2 θ cos 2θ = cos 2 θ sin 2 θ (2.2.4) AEC CE = sin 2θ = CG + GE = CD cos θ + DF = sin θ cos θ + BD cos θ = 2 sin θ cos θ sin 2θ = 2 sin θ cos θ (2.2.5) (2.2.4),(2.2.5) cos 2 2θ + sin 2 2θ = (cos 2 θ sin 2 θ) 2 + (2 sin θ cos θ) 2 = (cos 2 θ + sin 2 θ) 2 (cos 2 θ + sin 2 θ) 2 = cos 2 2θ + sin 2 2θ θ 2θ (cos 2 2θ + sin 2 2θ) 2 = cos 2 4θ + sin 2 4θ ( ) (cos 2 θ + sin 2 θ) 4 = cos 2 4θ + sin 2 4θ

9 8 (cos 2 θ + sin 2 θ) m = cos 2 mθ + sin 2 mθ (m = 2 n, n = 0, 1, 2, 3, ) (2.2.6) (2.2.6) n m θ 1 cos θ 1, 1 sin θ 1 cos θ sin θ (2.2.6) 0 < cos 2 θ + sin 2 θ < cos 2 θ + sin 2 θ < 1 lim m (cos2 θ + sin 2 θ) m = 1 cos 2 θ + sin 2 θ = 1 (2.2.7) 1 < cos 2 θ + sin 2 θ (2.2.6) 0 < cos 2 mθ + sin 2 mθ < 2 (2.2.8) (2.2.6),(2.2.7),(2.2.8) cos 2 θ + sin 2 θ = 1 cos 2 mθ + sin 2 mθ = (2.2.6) (2.2.6) mθ m θ < π/4 π/4 θ < π/2 π/2 θ < π (2.2.4),(2.2.5) sin θ cos θ 2π (2.2.4),(2.2.5) < θ < 2.1 AF = AD cos θ = cos 2 θ, F B = DB sin θ = sin 2 θ 1 = AB = AF + F B = cos 2 θ + sin 2 θ [1] 1 5 (2010) 12-20

10 [1] x, y { xy + x y = 183 (3.1.1) x + y = 27 y = 27 x x 2 29x = 0 (3.1.2) (x 15)(x 14) = 0 (3.1.3) x = 15 x = 14 (3.1.4) (x, y) = (14, 13) (x, y) = (15, 12) (3.1.5) (3.1.5) x 14 y 13 S 182 x 15 y 12 S 180 [1] 2

11 a 27 b { xy + x y = a (3.2.1) x + y = b x, y { (x 1)(y + 1) = a 1 (3.2.2) (x 1) + (y + 1) = b x 1, y + 1 t t 2 bt + (a 1) = 0 (3.2.3) t = b ± b 2 4(a 1) 2 (3.2.4) 10 OPTION ANGLE DEGRESSS 30 SET WINDOW -16,40, -16,40 40 DRAW GRID 50 SET POINT STYLE 4 60 FOR B=1 to FOR A=1 TO 1+B^2/4 80 LET K=(B-SQR(B^2-4*A+4))/2 90 LET S=K-INT(K) 100 IF S=0 THEN PRINT A, 110 IF S=0 THEN PRINT B 120 IF S=0 THEN PLOT POINTS: A, B 130 NEXT A 140 NEXT B 150 SET LINE COLOR PLOT LINES: 0,0; 40, PLOT LINES: -3,0; 39, PLOT LINES: -8,0; 40, PLOT LINES: -15,0; 37, PLOT LINES: -24,0; 36, END 2

12 11 図 2 実行結果 点が直線上に並んでいるので その方程式を a = mb + n (3.2.5) とおいてみる このとき解 t の分子の根号の中の式は b2 4(a 1) であるから この式に a = mb + n を代入すると b2 4(a 1) = (b 2m)2 4(m2 + n 1) となり これが完全平方となるためには n = 1 m2 (3.2.6) (3.2.7) となる整数 m が存在することが十分条件である グラフからは必要条件でもあるらしい その後 中野 潤 東京都立神代高校 さんからつぎのような補足の連絡があった 記述の一部は中西 襄 氏 京都大学名誉教授 の示唆により変更した N を整数として b2 4(a 1) = N 2 (3.2.8) (b + N )(b N ) = 4(a 1) (3.2.9) (b + N ) (b N ) = 2N (3.2.10) とおく この等式を変形すれば となる であるので b + N と b N の2数の差は偶数である したがって b N が奇数ならば b + N は奇数もなるので b N が奇数の場合には明らかに条件を満た さないので考える必要がない b N が偶数ならば b + N も偶数となる いま b N =m 2 (3.2.11)

13 12 N = b 2m b + N = 2(b m) (b + N)(b N) a 1 = 4 = m(b m) (3.2.12) a = mb + (1 m 2 ) (3.2.13) m b 2 4(a 1) a 5 a = b a = 2b 3 a = 3b 8 O 5 b 3

14 13 m 1 m 2 b a = mb + (1 m 2 ) (b, a) (b, a) m a = b a = 2b 3 a = 3b 8 a = 13b 168 a = 14b 195 m = 1 m = 2 m = 3 m = 13 m = m a b m b a = mb + (1 m 2 ) a b a m b = 27 a = 182 b2 4(a 1) = 5 m a = 182 b 2 4(a 1) > 0 2 m = 13, b = 27 a = 183 a = 13b 168 b = 27 a = = 183 m = 14, b = 27 a = 14b 195 b = 27 a = = 183 (b, a) = (27, 183) m m 2 bm + (a 1) = 0 m = 13, 14 ( ) [1] W.S. J. 1997) 27-28

15 (2010 ) 5/6 6/ (Landoldt) mm 1.5 mm 5 m

16 rad = rad rad 1/2= /10= mm 15 mm [2] , 1.5, I θ I θ I 1 θ (4.2.1) k I = k θ (4.2.2) θ 1 θ 1 l r θ = l r (4.2.3) (rad) 7 (4.2.2) I = k r l (4.2.4) 5 [1] rad 7 l r 1 rad = r/r

17 16 θ r l θ = l r r = 5 m, l = 1.5 mm = m I = 1 k (4.2.4) k = I l r = m 5 m = = (4.2.5) k rad rad rad 5 m = 360 (4.2.6) 1 = 60 (4.2.7) 1 = 60 (4.2.8),, , 1.5, 2.0 5:1: mm 0.1, 0.2, = 75 mm (4.3.1) = 37.5 mm (4.3.2) = 25 mm (4.3.3) = 6.25 mm (4.3.4) 1.2

18 17 5 m (4.2.4) I r r 1 I 1 r 2 I 2 I 2 I 1 I 2 I 1 = k r2 l k r1 l = r 2 r 1 (4.3.5) 5 m m 4 m 0.1 I 1 = 0.1, r 1 = 5 m, r 2 = 4 m I 2 I 2 = r 2 I 1 = = 0.08 (4.3.6) r 1 5 3m I 2 = r 2 I 1 = = 0.06 (4.3.7) r m [3] 1909 (Snellen) 8 [1] 6/6, 6/12, 6/60, 20/20, 20/40, 20/200 = / 1 (4.4.1) 6/12(=0.5 ) 6 m m 1 = 1/ (4.4.2) 5 m 6 m 20 ft (=6.096 m) ft= cm

19 18 log MAR m ft 10 6/60 20/ /48 20/ (=0.9031) 6.25 * 20/ (=0.7959) 5 6/30 20/ (=0.6990) 4 6/24 20/ (=0.6021) 3 6/18 20/ (=0.4949) 2.5 6/15 20/ (=0.3979) 2 6/12 20/ (=0.3010) /10 * (=0.2014) 1.5 6/9 20/ * 1.25 * 20/ (=0.0969) * * 0.9 * 1 6/6 20/ * 20/ (= ) /4 * 1.5 * 0.5 6/3 20/ (= ) E log MAR 4.5 (4.4.2) (4.2.2) (4.2.2) (4.4.2) (4.2.2) k k d I θ d (4.2.2) I = k d θ d (4.5.1) k d (4.5.1) k d = Iθ d = = 1 (4.5.2) k d = 1 k d = 1 (4.4.2)

20 19 (4.4.2) I = 1 θ d (4.5.3) = rad rad k k d k = rad 1 = = k d (4.5.4) (4.5.2) k d = 1 (4.2.2 (4.6.2) 1 1 rad = ( ) θ θ d = θ (4.5.5) I = k θ = k ( ) θ ( ) = k d θ d (4.5.6) (4.5.4), (4.5.5) k d (4.5.4) k d = r l 2πr θ = 2πr = 2π 360 2π rad = 360 π rad = rad = 180 π r

21 20 π rad = = (4.6.1) 1 = = rad (4.6.2) 1 = rad [4] θ r = 20 ft θ 0 = 1 1 r 0 = 20 ft θ θ 0 = 1 r 0 r 0 = 1 θ = 1 r 0 r θ θ 0 = r r 0 r = θ θ 0 r 0 θ 0 = 1 θ r

22 21 θ( ) r(ft) [5] AGO v v 1 v 1 = 50 log 10 v + 100, v 1 = 4 log v, AGO v = 1.0 v 1 = 100 AGO v 1 = 40 log MAR log MAR logarithmic minimum angle of resolution log MAR = log 10 1 r h l 3

23 22 θ l r h h l θ 0 l h h = r sin θ l = rθ sin θ = θ θ3 O(θ 5 ) Maclaurin θ = rad E E r l 1 3! + [6] [1] 1982) 13 [2] id-28 [3] 1, [4] [5] [6] detail/q

24 NHK latex 4

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

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,

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