按两页源 PDF 的 16 题编号编排。第 10 题原问“最大值”但函数无上界，当前教学版保留此判断并增加最小值问题。
答案速览
题号
答案
题号
答案
1
4 4 4 9
2 2 2
2
3 − 2 2 3-2\sqrt2 3 − 2 2 10
无最大值；最小值 − 4 -4 − 4
3
16 16 16 11
9 / 16 9/16 9/16
4
2021 2021 2021 12
B、C
5
1898 1898 1898 13
2 − 1 \sqrt2-1 2 − 1
6
6 6 6 14
1 / 3 1/\sqrt3 1/ 3
7
n ≡ 1 ( m o d 3 ) n\equiv1\pmod3 n ≡ 1 ( mod 3 ) 15
− 4 -4 − 4
8
127 , 376 127,376 127 , 376 16
最小数字和 7 7 7 n = 21301 n=21301 n = 21301
逐题解析
第 1 题
题目
求方程 ∣ 2 x 3 y − 2 u 5 v ∣ = 2 |2^x3^y-2^u5^v|=2 ∣ 2 x 3 y − 2 u 5 v ∣ = 2 ( x , y , u , v ) (x,y,u,v) ( x , y , u , v )
解答
若 x = u x=u x = u 2 x ( 3 y − 5 v ) 2^x(3^y-5^v) 2 x ( 3 y − 5 v ) 4 4 4 ± 2 \pm2 ± 2 x ≠ u x\ne u x = u min ( x , u ) = 1 \min(x,u)=1 min ( x , u ) = 1
先准备一个初等整除事实。若奇素数 p p p p ∣ A + 1 , p 2 ∤ A + 1 p\mid A+1,p^2\nmid A+1 p ∣ A + 1 , p 2 ∤ A + 1 m m m A m + 1 A^m+1 A m + 1 p p p 1 + v p ( m ) 1+v_p(m) 1 + v p ( m ) v p ( m ) v_p(m) v p ( m ) p p p m m m m = p e h m=p^eh m = p e h p ∤ h p\nmid h p ∤ h ( A h + 1 ) / ( A + 1 ) ≡ h ( m o d p ) (A^h+1)/(A+1)\equiv h\pmod p ( A h + 1 ) / ( A + 1 ) ≡ h ( mod p ) B = − 1 + p r t B=-1+p^rt B = − 1 + p r t B + 1 B+1 B + 1 B p + 1 B^p+1 B p + 1 p p p e e e
当 x = 1 , u = k + 1 ≥ 2 x=1,u=k+1\ge2 x = 1 , u = k + 1 ≥ 2 ∣ 3 y − 2 k 5 v ∣ = 1 |3^y-2^k5^v|=1 ∣ 3 y − 2 k 5 v ∣ = 1 5 5 5 y y y y = 2 t y=2t y = 2 t
若 3 2 t − 1 = 2 k 5 v 3^{2t}-1=2^k5^v 3 2 t − 1 = 2 k 5 v 3 t − 1 , 3 t + 1 3^t-1,3^t+1 3 t − 1 , 3 t + 1 2 2 2 2 2 2 3 t + 1 = 2 j 3^t+1=2^j 3 t + 1 = 2 j t = 1 t=1 t = 1 t t t 8 8 8 2 2 2 t t t 4 4 4 2 2 2 4 4 4 3 t − 1 = 2 j 3^t-1=2^j 3 t − 1 = 2 j t t t t = 1 t=1 t = 1 t t t ( 3 t / 2 − 1 ) ( 3 t / 2 + 1 ) (3^{t/2}-1)(3^{t/2}+1) ( 3 t /2 − 1 ) ( 3 t /2 + 1 ) 2 2 2 2 2 2 2 , 4 2,4 2 , 4 t = 2 t=2 t = 2 t = 1 t=1 t = 1 5 5 5 t = 2 t=2 t = 2 80 = 2 4 ⋅ 5 80=2^4\cdot5 80 = 2 4 ⋅ 5 ( x , y , u , v ) = ( 1 , 4 , 5 , 1 ) (x,y,u,v)=(1,4,5,1) ( x , y , u , v ) = ( 1 , 4 , 5 , 1 )
若 3 2 t + 1 = 2 k 5 v 3^{2t}+1=2^k5^v 3 2 t + 1 = 2 k 5 v 8 8 8 k = 1 k=1 k = 1 5 5 5 t t t A = 9 , p = 5 A=9,p=5 A = 9 , p = 5 v = 1 + v 5 ( t ) v=1+v_5(t) v = 1 + v 5 ( t )
9 t + 1 = 2 ⋅ 5 v = 10 ⋅ 5 v 5 ( t ) ≤ 10 t . 9^t+1=2\cdot5^v=10\cdot5^{v_5(t)}\le10t. 9 t + 1 = 2 ⋅ 5 v = 10 ⋅ 5 v 5 ( t ) ≤ 10 t . t ≥ 2 t\ge2 t ≥ 2 9 t + 1 > 10 t 9^t+1>10t 9 t + 1 > 10 t t = 2 t=2 t = 2 t = 1 t=1 t = 1 ( 1 , 2 , 2 , 1 ) (1,2,2,1) ( 1 , 2 , 2 , 1 )
当 u = 1 , x = k + 1 ≥ 2 u=1,x=k+1\ge2 u = 1 , x = k + 1 ≥ 2 ∣ 2 k 3 y − 5 v ∣ = 1 |2^k3^y-5^v|=1 ∣ 2 k 3 y − 5 v ∣ = 1 2 k 3 y = 5 v + 1 2^k3^y=5^v+1 2 k 3 y = 5 v + 1 3 3 3 v v v 8 8 8 k = 1 k=1 k = 1 A = 5 , p = 3 A=5,p=3 A = 5 , p = 3 y = 1 + v 3 ( v ) y=1+v_3(v) y = 1 + v 3 ( v ) 5 v + 1 ≤ 6 v 5^v+1\le6v 5 v + 1 ≤ 6 v v = 1 v=1 v = 1 ( 2 , 1 , 1 , 1 ) (2,1,1,1) ( 2 , 1 , 1 , 1 )
若 2 k 3 y = 5 v − 1 2^k3^y=5^v-1 2 k 3 y = 5 v − 1 3 3 3 v = 2 t v=2t v = 2 t ( 5 t − 1 ) ( 5 t + 1 ) (5^t-1)(5^t+1) ( 5 t − 1 ) ( 5 t + 1 ) 2 2 2 5 t + 1 5^t+1 5 t + 1 t t t 8 8 8 6 6 6 t t t 2 2 2 5 t − 1 5^t-1 5 t − 1 2 2 2 t t t 3 3 3 t t t 8 8 8 4 4 4 4 4 4 t = 1 t=1 t = 1 25 − 1 = 2 3 ⋅ 3 25-1=2^3\cdot3 25 − 1 = 2 3 ⋅ 3 ( 4 , 1 , 1 , 2 ) (4,1,1,2) ( 4 , 1 , 1 , 2 )
四组均代入成立，且上述分类穷尽所有情形，故答案为 4 4 4
第 2 题
题目
设 α , β ∈ ( 0 , π / 2 ) \alpha,\beta\in(0,\pi/2) α , β ∈ ( 0 , π /2 )
( 1 − tan α 2 tan β 2 ) 2 cot α + cot β \frac{\left(1-\sqrt{\tan\frac\alpha2\tan\frac\beta2}\right)^2}{\cot\alpha+\cot\beta} cot α + cot β ( 1 − tan 2 α tan 2 β ) 2 的最大值。
解答
设 u = tan ( α / 2 ) , v = tan ( β / 2 ) u=\tan(\alpha/2),v=\tan(\beta/2) u = tan ( α /2 ) , v = tan ( β /2 ) 0 < u , v < 1 0<u,v<1 0 < u , v < 1 t = u v t=\sqrt{uv} t = uv
2 u v ( 1 − u v ) 2 ( u + v ) ( 1 − u v ) = 2 t 2 ( 1 − t ) ( u + v ) ( 1 + t ) ≤ t ( 1 − t ) 1 + t . \frac{2uv(1-\sqrt{uv})^2}{(u+v)(1-uv)}
=\frac{2t^2(1-t)}{(u+v)(1+t)}
\le\frac{t(1-t)}{1+t}. ( u + v ) ( 1 − uv ) 2 uv ( 1 − uv ) 2 = ( u + v ) ( 1 + t ) 2 t 2 ( 1 − t ) ≤ 1 + t t ( 1 − t ) . 等号要求 u = v u=v u = v ( 1 − 2 t − t 2 ) / ( 1 + t ) 2 (1-2t-t^2)/(1+t)^2 ( 1 − 2 t − t 2 ) / ( 1 + t ) 2 t = 2 − 1 t=\sqrt2-1 t = 2 − 1 3 − 2 2 3-2\sqrt2 3 − 2 2 α = β = π / 4 \alpha=\beta=\pi/4 α = β = π /4
第 3 题
题目
满足 A ∪ B ∪ C = { 1 , 2 , … , 2023 } A\cup B\cup C=\{1,2,\ldots,2023\} A ∪ B ∪ C = { 1 , 2 , … , 2023 } A ∩ B ∩ C = ∅ A\cap B\cap C=\varnothing A ∩ B ∩ C = ∅ ( A , B , C ) (A,B,C) ( A , B , C ) n n n n n n
解答
每个元素可以属于三集合中的任意一个或任意两个，共 3 + 3 = 6 3+3=6 3 + 3 = 6 n = 6 2023 n=6^{2023} n = 6 2023
模 100 100 100 6 2 6^2 6 2 36 , 16 , 96 , 76 , 56 36,16,96,76,56 36 , 16 , 96 , 76 , 56 5 5 5 2023 ≡ 3 ( m o d 5 ) 2023\equiv3\pmod5 2023 ≡ 3 ( mod 5 ) 16 16 16
第 4 题
题目
2023 2023 2023 3 3 3 0 0 0 1 1 1 1 1 1
解答
设最低分为 s s s 2023 ( s + 1011 ) 2023(s+1011) 2023 ( s + 1011 ) 2023 ⋅ 1011 2023\cdot1011 2023 ⋅ 1011 3 3 3 s ≤ 2022 s\le2022 s ≤ 2022 3 3 3 1 1 1 s ≤ 2021 s\le2021 s ≤ 2021
以下证明这个上界确实能达到。给球队指定目标分数 s i = 2021 + i s_i=2021+i s i = 2021 + i i = 0 , … , 2022 i=0,\ldots,2022 i = 0 , … , 2022 3 3 3 0 , 1 , 2 0,1,2 0 , 1 , 2 d i = 3 , 1 , 2 d_i=3,1,2 d i = 3 , 1 , 2 674 , 674 , 675 674,674,675 674 , 674 , 675 1349 1349 1349 1349 + 674 = 2023 1349+674=2023 1349 + 674 = 2023
剩余比赛都决出胜负。所需胜场数为整数 w i = ( s i − d i ) / 3 w_i=(s_i-d_i)/3 w i = ( s i − d i ) /3 673 ≤ w i ≤ 1347 673\le w_i\le1347 673 ≤ w i ≤ 1347
使用一个图的定向引理：若无向图 G G G w i w_i w i ∑ w i = ∣ E ( G ) ∣ \sum w_i=|E(G)| ∑ w i = ∣ E ( G ) ∣ S S S ∑ i ∈ S w i ≥ ∣ E ( S ) ∣ \sum_{i\in S}w_i\ge |E(S)| ∑ i ∈ S w i ≥ ∣ E ( S ) ∣ i i i w i w_i w i S S S ∣ E ( S ) ∣ |E(S)| ∣ E ( S ) ∣
现在 G G G 2022 − d i 2022-d_i 2022 − d i
∑ w i = 2023 ⋅ 3032 − 2 ⋅ 2023 3 = 2023 ⋅ 1010 = ∣ E ( G ) ∣ . \sum w_i=\frac{2023\cdot3032-2\cdot2023}{3}=2023\cdot1010=|E(G)|. ∑ w i = 3 2023 ⋅ 3032 − 2 ⋅ 2023 = 2023 ⋅ 1010 = ∣ E ( G ) ∣. 对 ∣ S ∣ = k ≤ 1012 |S|=k\le1012 ∣ S ∣ = k ≤ 1012 ∑ i ∈ S w i ≥ 673 k ≥ ( k 2 ) ≥ ∣ E ( S ) ∣ \sum_{i\in S}w_i\ge673k\ge\binom{k}{2}\ge|E(S)| ∑ i ∈ S w i ≥ 673 k ≥ ( 2 k ) ≥ ∣ E ( S ) ∣ k > 1012 k>1012 k > 1012 T T T t ≤ 1010 t\le1010 t ≤ 1010 2022 − d i − w i ≥ 672 2022-d_i-w_i\ge672 2022 − d i − w i ≥ 672
∑ i ∈ S w i − ∣ E ( S ) ∣ = ∑ i ∈ T ( deg G ( i ) − w i ) − ∣ E ( T ) ∣ ≥ 672 t − ( t 2 ) ≥ 0. \sum_{i\in S}w_i-|E(S)|
=\sum_{i\in T}(\deg_G(i)-w_i)-|E(T)|
\ge672t-\binom t2\ge0. i ∈ S ∑ w i − ∣ E ( S ) ∣ = i ∈ T ∑ ( deg G ( i ) − w i ) − ∣ E ( T ) ∣ ≥ 672 t − ( 2 t ) ≥ 0. 引理适用，把有向边的起点规定为胜者，即实现全部目标分数。因此最低分的最大值确为 2021 2021 2021
第 5 题
题目
已知 x , y ∈ N + x,y\in\mathbb N^+ x , y ∈ N + x , y ∈ [ 1 , 1897 ] x,y\in[1,1897] x , y ∈ [ 1 , 1897 ] ⌊ x y ⌋ + 1 \left\lfloor\dfrac xy\right\rfloor+1 ⌊ y x ⌋ + 1 x x x ( x , y ) (x,y) ( x , y )
解答
若 x = 1 x=1 x = 1 y = 1 , … , 1897 y=1,\ldots,1897 y = 1 , … , 1897 1897 1897 1897 x ≥ 2 x\ge2 x ≥ 2 y = 1 y=1 y = 1 x + 1 x+1 x + 1 x x x y ≥ 2 y\ge2 y ≥ 2
1 ≤ ⌊ x / y ⌋ + 1 ≤ ⌊ x / 2 ⌋ + 1 ≤ x . 1\le\lfloor x/y\rfloor+1\le\lfloor x/2\rfloor+1\le x. 1 ≤ ⌊ x / y ⌋ + 1 ≤ ⌊ x /2 ⌋ + 1 ≤ x . 要为 x x x x x x x ≥ 2 x\ge2 x ≥ 2 x = 2 x=2 x = 2 y = 2 y=2 y = 2 1897 + 1 = 1898 1897+1=1898 1897 + 1 = 1898
第 6 题
题目
【明确非退化配置】凸四边形 A B C D ABCD A B C D O O O O O O A B , C D AB,CD A B , C D K , L K,L K , L A C , B D AC,BD A C , B D M , N M,N M , N
∠ B K L = ∠ C L K , A M = 1 , M C = 2 , B N = 3. \angle BKL=\angle CLK,
\qquad AM=1,\quad MC=2,\quad BN=3. ∠ B K L = ∠ C L K , A M = 1 , M C = 2 , B N = 3. 求 N D ND N D
解答
题面校注： 按凸外切四边形、所述交点在线段内部理解；教学版明确这一非退化配置。
设内切圆半径为 r r r O O O K L KL K L O K sin ∠ B K L = O L sin ∠ C L K = r OK\sin\angle BKL=OL\sin\angle CLK=r O K sin ∠ B K L = O L sin ∠ C L K = r O K = O L = h OK=OL=h O K = O L = h K = ( 0 , − h ) , L = ( 0 , h ) K=(0,-h),L=(0,h) K = ( 0 , − h ) , L = ( 0 , h ) B , C B,C B , C A B : y = s x − h AB:y=sx-h A B : y = s x − h C D : y = − s x + h CD:y=-sx+h C D : y = − s x + h h 2 = r 2 ( 1 + s 2 ) h^2=r^2(1+s^2) h 2 = r 2 ( 1 + s 2 )
任一另外的切线写为 u x + v y = r ux+vy=r ux + v y = r u 2 + v 2 = 1 u^2+v^2=1 u 2 + v 2 = 1
r + v h u + v s , r − v h u − v s . \frac{r+vh}{u+vs},\qquad\frac{r-vh}{u-vs}. u + v s r + v h , u − v s r − v h . 二者乘积等于 ( r 2 − v 2 h 2 ) / ( u 2 − v 2 s 2 ) = r 2 (r^2-v^2h^2)/(u^2-v^2s^2)=r^2 ( r 2 − v 2 h 2 ) / ( u 2 − v 2 s 2 ) = r 2 A D , B C AD,BC A D , B C x A x D = x B x C = r 2 x_Ax_D=x_Bx_C=r^2 x A x D = x B x C = r 2
A M M C = ∣ x A ∣ ∣ x C ∣ = ∣ x B ∣ ∣ x D ∣ = B N N D . \frac{AM}{MC}=\frac{|x_A|}{|x_C|}
=\frac{|x_B|}{|x_D|}=\frac{BN}{ND}. M C A M = ∣ x C ∣ ∣ x A ∣ = ∣ x D ∣ ∣ x B ∣ = N D B N . 代入 1 / 2 = 3 / N D 1/2=3/ND 1/2 = 3/ N D N D = 6 ND=6 N D = 6
配置也确实存在：取圆 x 2 + y 2 = 2 x^2+y^2=2 x 2 + y 2 = 2 A = ( − 1 / 3 , − 2 ) , B = ( 3 , − 2 ) , C = ( 2 / 3 , 2 ) , D = ( − 6 , 2 ) A=(-1/3,-\sqrt2),B=(3,-\sqrt2),C=(2/3,\sqrt2),D=(-6,\sqrt2) A = ( − 1/3 , − 2 ) , B = ( 3 , − 2 ) , C = ( 2/3 , 2 ) , D = ( − 6 , 2 ) 3 , 9 3,9 3 , 9 1 : 2 1:2 1 : 2
第 7 题
题目
求所有正整数 n n n a 1 , a 2 , … , a n a_1,a_2,\ldots,a_n a 1 , a 2 , … , a n a 1 + 2 a 2 + ⋯ + n a n a_1+2a_2+\cdots+na_n a 1 + 2 a 2 + ⋯ + n a n a 1 , a 2 , … , a n a_1,a_2,\ldots,a_n a 1 , a 2 , … , a n
解答
写 a k = a 1 + ( k − 1 ) d a_k=a_1+(k-1)d a k = a 1 + ( k − 1 ) d
∑ k = 1 n k a k = n ( n + 1 ) 2 ( a 1 + 2 ( n − 1 ) 3 d ) . \sum_{k=1}^nka_k=\frac{n(n+1)}2\left(a_1+\frac{2(n-1)}3d\right). k = 1 ∑ n k a k = 2 n ( n + 1 ) ( a 1 + 3 2 ( n − 1 ) d ) . 若 n ≡ 1 ( m o d 3 ) n\equiv1\pmod3 n ≡ 1 ( mod 3 ) ( 2 n + 1 ) / 3 (2n+1)/3 ( 2 n + 1 ) /3
若 n ≢ 1 ( m o d 3 ) n\not\equiv1\pmod3 n ≡ 1 ( mod 3 ) d = 2 , a 1 = − 2 ( n − 1 ) 2 / 3 d=\sqrt2,a_1=-2(n-1)\sqrt2/3 d = 2 , a 1 = − 2 ( n − 1 ) 2 /3 0 0 0 2 \sqrt2 2 n ≡ 1 ( m o d 3 ) n\equiv1\pmod3 n ≡ 1 ( mod 3 )
第 8 题
题目
任取正 n n n 93 125 \dfrac{93}{125} 125 93 n n n
解答
三角形为钝角，当且仅当三个顶点位于某段小于半圆的弧内。沿圆周固定方向，从这段弧的第一个顶点开始，其余两点都必须落在随后严格小于半周的位置中。钝角三角形有唯一这样的起点，因而不会重复计数。
若 n = 2 m + 1 n=2m+1 n = 2 m + 1 m m m n ( m 2 ) n\binom m2 n ( 2 m ) 3 ( n − 3 ) / ( 4 ( n − 2 ) ) 3(n-3)/(4(n-2)) 3 ( n − 3 ) / ( 4 ( n − 2 )) 93 / 125 93/125 93/125 n = 127 n=127 n = 127
若 n = 2 m n=2m n = 2 m m − 1 m-1 m − 1 n ( m − 1 2 ) / ( n 3 ) = 3 ( n − 4 ) / ( 4 ( n − 1 ) ) n\binom{m-1}2/\binom n3=3(n-4)/(4(n-1)) n ( 2 m − 1 ) / ( 3 n ) = 3 ( n − 4 ) / ( 4 ( n − 1 )) n = 376 n=376 n = 376 127 , 376 127,376 127 , 376
第 9 题
题目
过椭圆 x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 a 2 x 2 + b 2 y 2 = 1 a > b > 0 a>b>0 a > b > 0 A C , B D AC,BD A C , B D A B C D ABCD A B C D [ 8 , 25 / 2 ] [8,25/2] [ 8 , 25/2 ] a / b a/b a / b
解答
设右焦点 F = ( c , 0 ) F=(c,0) F = ( c , 0 ) c 2 = a 2 − b 2 c^2=a^2-b^2 c 2 = a 2 − b 2 ( cos θ , sin θ ) (\cos\theta,\sin\theta) ( cos θ , sin θ ) F + t ( cos θ , sin θ ) F+t(\cos\theta,\sin\theta) F + t ( cos θ , sin θ )
L θ = 2 a b 2 b 2 cos 2 θ + a 2 sin 2 θ . L_\theta=\frac{2ab^2}{b^2\cos^2\theta+a^2\sin^2\theta}. L θ = b 2 cos 2 θ + a 2 sin 2 θ 2 a b 2 . 具体地，二次项系数为 D / a 2 b 2 D/a^2b^2 D / a 2 b 2 2 c cos θ / a 2 2c\cos\theta/a^2 2 c cos θ / a 2 − b 2 / a 2 -b^2/a^2 − b 2 / a 2 4 / a 2 4/a^2 4/ a 2
两对角线垂直，四边形面积为其长度乘积的一半：
S = 2 a 2 b 4 a 2 b 2 + ( a 2 − b 2 ) 2 sin 2 θ cos 2 θ . S=\frac{2a^2b^4}{a^2b^2+(a^2-b^2)^2\sin^2\theta\cos^2\theta}. S = a 2 b 2 + ( a 2 − b 2 ) 2 sin 2 θ cos 2 θ 2 a 2 b 4 . 因 sin 2 θ cos 2 θ \sin^2\theta\cos^2\theta sin 2 θ cos 2 θ [ 0 , 1 / 4 ] [0,1/4] [ 0 , 1/4 ] S max = 2 b 2 = 25 / 2 S_{\max}=2b^2=25/2 S m a x = 2 b 2 = 25/2 S min = 8 a 2 b 4 / ( a 2 + b 2 ) 2 = 8 S_{\min}=8a^2b^4/(a^2+b^2)^2=8 S m i n = 8 a 2 b 4 / ( a 2 + b 2 ) 2 = 8 t = a / b > 1 t=a/b>1 t = a / b > 1 ( t + 1 / t ) 2 = b 2 = 25 / 4 (t+1/t)^2=b^2=25/4 ( t + 1/ t ) 2 = b 2 = 25/4 t + 1 / t = 5 / 2 t+1/t=5/2 t + 1/ t = 5/2 2 , 1 / 2 2,1/2 2 , 1/2 t > 1 t>1 t > 1 a / b = 2 a/b=2 a / b = 2
第 10 题
题目
【教学修订】设 f ( x ) = x 2 − 2 x − 14 x − 1 + x x 2 − 4 x − 28 x − 1 + 61 f(x)=x^2-2x-14\sqrt{x-1}+x\sqrt{x^2-4x-28\sqrt{x-1}+61} f ( x ) = x 2 − 2 x − 14 x − 1 + x x 2 − 4 x − 28 x − 1 + 61 x x x
解答
题面校注： 源 PDF 也写“最大值”，并非转写错误。函数在充分大的 x x x x 2 − 2 x − 14 x − 1 x^2-2x-14\sqrt{x-1} x 2 − 2 x − 14 x − 1
令 t = x − 1 ≥ 0 t=\sqrt{x-1}\ge0 t = x − 1 ≥ 0 K = 53 K=\sqrt{53} K = 53 Q = x 2 − 4 x − 28 t + 61 Q=x^2-4x-28t+61 Q = x 2 − 4 x − 28 t + 61
Q = ( K − x ) 2 + 2 ( K − 2 ) ( t − K + 2 7 ) 2 ≥ 0 , Q=(K-x)^2+2(K-2)\left(t-\frac{K+2}{7}\right)^2\ge0, Q = ( K − x ) 2 + 2 ( K − 2 ) ( t − 7 K + 2 ) 2 ≥ 0 , 所以定义域恰为 x ≥ 1 x\ge1 x ≥ 1 u = Q u=\sqrt Q u = Q x + u ≥ x + ∣ K − x ∣ ≥ K x+u\ge x+|K-x|\ge K x + u ≥ x + ∣ K − x ∣ ≥ K
f ( x ) = x 2 + u 2 − 61 2 + x u = ( x + u ) 2 − 61 2 ≥ 53 − 61 2 = − 4. f(x)=\frac{x^2+u^2-61}{2}+xu
=\frac{(x+u)^2-61}{2}\ge\frac{53-61}{2}=-4. f ( x ) = 2 x 2 + u 2 − 61 + xu = 2 ( x + u ) 2 − 61 ≥ 2 53 − 61 = − 4. 等号要求 t = ( K + 2 ) / 7 t=(K+2)/7 t = ( K + 2 ) /7 x ≤ K x\le K x ≤ K
x = 1 + ( 53 + 2 7 ) 2 = 106 + 4 53 49 < 53 . x=1+\left(\frac{\sqrt{53}+2}{7}\right)^2
=\frac{106+4\sqrt{53}}{49}<\sqrt{53}. x = 1 + ( 7 53 + 2 ) 2 = 49 106 + 4 53 < 53 . 它确实达到等号，因此最小值为 − 4 -4 − 4
第 11 题
题目
复数 a , b , c a,b,c a , b , c ∣ a ∣ 2 + ∣ b ∣ 2 + ∣ c ∣ 2 = 1 |a|^2+|b|^2+|c|^2=1 ∣ a ∣ 2 + ∣ b ∣ 2 + ∣ c ∣ 2 = 1 ∣ a b ( a 2 − b 2 ) + b c ( b 2 − c 2 ) + c a ( c 2 − a 2 ) ∣ |ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)| ∣ ab ( a 2 − b 2 ) + b c ( b 2 − c 2 ) + c a ( c 2 − a 2 ) ∣
解答
原多项式因式分解为 − ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) -(a-b)(b-c)(c-a)(a+b+c) − ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) μ = ( a + b + c ) / 3 \mu=(a+b+c)/3 μ = ( a + b + c ) /3 t = 3 ∣ μ ∣ 2 ∈ [ 0 , 1 ] t=3|\mu|^2\in[0,1] t = 3∣ μ ∣ 2 ∈ [ 0 , 1 ]
∣ a − μ ∣ 2 + ∣ b − μ ∣ 2 + ∣ c − μ ∣ 2 = 1 − t , ∣ a − b ∣ 2 + ∣ b − c ∣ 2 + ∣ c − a ∣ 2 = 3 ( 1 − t ) . |a-\mu|^2+|b-\mu|^2+|c-\mu|^2=1-t,
\qquad |a-b|^2+|b-c|^2+|c-a|^2=3(1-t). ∣ a − μ ∣ 2 + ∣ b − μ ∣ 2 + ∣ c − μ ∣ 2 = 1 − t , ∣ a − b ∣ 2 + ∣ b − c ∣ 2 + ∣ c − a ∣ 2 = 3 ( 1 − t ) . 由三个非负数的算术几何平均不等式，三个差的模之积至多为 ( 1 − t ) 3 / 2 (1-t)^{3/2} ( 1 − t ) 3/2 ∣ a + b + c ∣ = 3 t |a+b+c|=\sqrt{3t} ∣ a + b + c ∣ = 3 t 3 t ( 1 − t ) 3 \sqrt{3t(1-t)^3} 3 t ( 1 − t ) 3 t ( 1 − t ) 3 t(1-t)^3 t ( 1 − t ) 3 ( 1 − t ) 2 ( 1 − 4 t ) (1-t)^2(1-4t) ( 1 − t ) 2 ( 1 − 4 t ) t = 1 / 4 t=1/4 t = 1/4 9 / 16 9/16 9/16
取 ω = e 2 π i / 3 \omega=e^{2\pi i/3} ω = e 2 π i /3 μ = 1 / ( 2 3 ) \mu=1/(2\sqrt3) μ = 1/ ( 2 3 ) a = μ + 1 / 2 , b = μ + ω / 2 , c = μ + ω 2 / 2 a=\mu+1/2,b=\mu+\omega/2,c=\mu+\omega^2/2 a = μ + 1/2 , b = μ + ω /2 , c = μ + ω 2 /2 1 1 1 t = 1 / 4 t=1/4 t = 1/4 9 / 16 9/16 9/16
第 12 题
题目
【多选】判断下列命题中正确的选项，并说明理由：A. 自然数集与有理数集无双射；B. 有理数集与实数集无双射；C. 实数集与整数集无双射；D. 实数集与复数集无双射。
解答
自然数、整数、有理数均可数：整数按 0 , 1 , − 1 , 2 , − 2 , … 0,1,-1,2,-2,\ldots 0 , 1 , − 1 , 2 , − 2 , … ( 0 , 1 ) (0,1) ( 0 , 1 ) 1 1 1 2 2 2
复数等同于有序实数对，与实数等势。具体地，先将每个实数映入 ( 0 , 1 ) (0,1) ( 0 , 1 ) 1 1 1 0 , 2 0,2 0 , 2 x ↦ ( x , 0 ) x\mapsto(x,0) x ↦ ( x , 0 )
第 13 题
题目
数列 { a n } \{a_n\} { a n } a 1 = 1 2 a_1=\dfrac12 a 1 = 2 1
a n + 1 = a n ( 1 − 2 ) n + 1 a n + 2 + 1 , a_{n+1}=\frac{a_n}{(1-\sqrt2)^{n+1}a_n+\sqrt2+1}, a n + 1 = ( 1 − 2 ) n + 1 a n + 2 + 1 a n , 求 lim n → + ∞ a n n \displaystyle\lim_{n\to+\infty}\sqrt[n]{a_n} n → + ∞ lim n a n
解答
令 r = 1 + 2 , s = 1 − 2 = − 1 / r r=1+\sqrt2,s=1-\sqrt2=-1/r r = 1 + 2 , s = 1 − 2 = − 1/ r b n = 1 / a n b_n=1/a_n b n = 1/ a n b 1 = 2 b_1=2 b 1 = 2 b n + 1 = r b n + s n + 1 b_{n+1}=rb_n+s^{n+1} b n + 1 = r b n + s n + 1
b n r n − 1 = 2 + s ∑ j = 1 n − 1 ( s r ) j . \frac{b_n}{r^{n-1}}=2+s\sum_{j=1}^{n-1}\left(\frac sr\right)^j. r n − 1 b n = 2 + s j = 1 ∑ n − 1 ( r s ) j . 由于 s / r ∈ ( − 1 , 0 ) s/r\in(-1,0) s / r ∈ ( − 1 , 0 ) 2 2 2 3 3 3 n = 1 n=1 n = 1 2 2 2 b n > 0 b_n>0 b n > 0
a n n = r − ( n − 1 ) / n ( b n r n − 1 ) − 1 / n ⟶ 1 r = 2 − 1. \sqrt[n]{a_n}=r^{-(n-1)/n}
\left(\frac{b_n}{r^{n-1}}\right)^{-1/n}\longrightarrow\frac1r=\sqrt2-1. n a n = r − ( n − 1 ) / n ( r n − 1 b n ) − 1/ n ⟶ r 1 = 2 − 1. 第 14 题
题目
计算
tan 96 ∘ − tan 12 ∘ ( 1 + 1 sin 6 ∘ ) 1 + tan 96 ∘ tan 12 ∘ ( 1 + 1 sin 6 ∘ ) . \frac{\tan96^\circ-\tan12^\circ\left(1+\dfrac1{\sin6^\circ}\right)}{1+\tan96^\circ\tan12^\circ\left(1+\dfrac1{\sin6^\circ}\right)}. 1 + tan 9 6 ∘ tan 1 2 ∘ ( 1 + sin 6 ∘ 1 ) tan 9 6 ∘ − tan 1 2 ∘ ( 1 + sin 6 ∘ 1 ) . 解答
先证明 tan 12 ∘ ( 1 + 1 / sin 6 ∘ ) = cot 24 ∘ \tan12^\circ(1+1/\sin6^\circ)=\cot24^\circ tan 1 2 ∘ ( 1 + 1/ sin 6 ∘ ) = cot 2 4 ∘ cos 36 ∘ − sin 18 ∘ = 1 / 2 \cos36^\circ-\sin18^\circ=1/2 cos 3 6 ∘ − sin 1 8 ∘ = 1/2
cos 12 ∘ − cos 48 ∘ = 2 sin 30 ∘ sin 18 ∘ = sin 18 ∘ , \cos12^\circ-\cos48^\circ=2\sin30^\circ\sin18^\circ=\sin18^\circ, cos 1 2 ∘ − cos 4 8 ∘ = 2 sin 3 0 ∘ sin 1 8 ∘ = sin 1 8 ∘ , 得到 cos 12 ∘ − cos 36 ∘ = cos 48 ∘ − 1 / 2 = 2 sin 6 ∘ cos 36 ∘ \cos12^\circ-\cos36^\circ=\cos48^\circ-1/2=2\sin6^\circ\cos36^\circ cos 1 2 ∘ − cos 3 6 ∘ = cos 4 8 ∘ − 1/2 = 2 sin 6 ∘ cos 3 6 ∘ 2 sin 24 ∘ sin 12 ∘ = cos 12 ∘ − cos 36 ∘ 2\sin24^\circ\sin12^\circ=\cos12^\circ-\cos36^\circ 2 sin 2 4 ∘ sin 1 2 ∘ = cos 1 2 ∘ − cos 3 6 ∘
sin 24 ∘ sin 12 ∘ = sin 6 ∘ cos 36 ∘ . \sin24^\circ\sin12^\circ=\sin6^\circ\cos36^\circ. sin 2 4 ∘ sin 1 2 ∘ = sin 6 ∘ cos 3 6 ∘ . 将右边的 cos 36 ∘ \cos36^\circ cos 3 6 ∘ cos 24 ∘ cos 12 ∘ − sin 24 ∘ sin 12 ∘ \cos24^\circ\cos12^\circ-\sin24^\circ\sin12^\circ cos 2 4 ∘ cos 1 2 ∘ − sin 2 4 ∘ sin 1 2 ∘
于是原式是 tan ( 96 ∘ − 66 ∘ ) = tan 30 ∘ = 1 / 3 \tan(96^\circ-66^\circ)=\tan30^\circ=1/\sqrt3 tan ( 9 6 ∘ − 6 6 ∘ ) = tan 3 0 ∘ = 1/ 3 cos 30 ∘ / ( cos 96 ∘ cos 66 ∘ ) \cos30^\circ/(\cos96^\circ\cos66^\circ) cos 3 0 ∘ / ( cos 9 6 ∘ cos 6 6 ∘ )
第 15 题
题目
数列 { a n } \{a_n\} { a n } a 1 = 1 a_1=1 a 1 = 1 a n + 1 = a n 2 a n 2 + 1 a_{n+1}=\dfrac{a_n}{2a_n^2+1} a n + 1 = 2 a n 2 + 1 a n ⌊ 2 lg a 2023 ⌋ \lfloor2\lg a_{2023}\rfloor ⌊ 2 lg a 2023 ⌋
解答
令 b n = a n − 2 > 0 b_n=a_n^{-2}>0 b n = a n − 2 > 0 b 1 = 1 b_1=1 b 1 = 1 b n + 1 = b n + 4 + 4 / b n b_{n+1}=b_n+4+4/b_n b n + 1 = b n + 4 + 4/ b n b 2 = 9 b_2=9 b 2 = 9 b 2023 > 1 + 4 ⋅ 2022 = 8089 > 1000 b_{2023}>1+4\cdot2022=8089>1000 b 2023 > 1 + 4 ⋅ 2022 = 8089 > 1000 n ≥ 2 n\ge2 n ≥ 2 b n ≥ 9 b_n\ge9 b n ≥ 9 40 / 9 40/9 40/9
b 2023 ≤ 9 + 2021 ⋅ 40 9 < 10000. b_{2023}\le9+2021\cdot\frac{40}{9}<10000. b 2023 ≤ 9 + 2021 ⋅ 9 40 < 10000. 于是 2 lg a 2023 = − lg b 2023 ∈ ( − 4 , − 3 ) 2\lg a_{2023}=-\lg b_{2023}\in(-4,-3) 2 lg a 2023 = − lg b 2023 ∈ ( − 4 , − 3 ) − 4 -4 − 4
第 16 题
题目
求一个五位数 n n n 2556 ∣ n 3 − 1 2556\mid n^3-1 2556 ∣ n 3 − 1 n n n
解答
分解 2556 = 4 ⋅ 9 ⋅ 71 2556=4\cdot9\cdot71 2556 = 4 ⋅ 9 ⋅ 71 n 3 ≡ 1 ( m o d 4 ) n^3\equiv1\pmod4 n 3 ≡ 1 ( mod 4 ) n ≡ 1 ( m o d 4 ) n\equiv1\pmod4 n ≡ 1 ( mod 4 ) 9 9 9 n ≡ 1 ( m o d 3 ) n\equiv1\pmod3 n ≡ 1 ( mod 3 ) 71 71 71 gcd ( 3 , 70 ) = 1 \gcd(3,70)=1 g cd( 3 , 70 ) = 1 n 70 ≡ 1 n^{70}\equiv1 n 70 ≡ 1 n 3 ≡ 1 n^3\equiv1 n 3 ≡ 1 n ≡ 1 n\equiv1 n ≡ 1
n ≡ 1 ( m o d 852 ) . n\equiv1\pmod{852}. n ≡ 1 ( mod 852 ) . 数字和必模 3 3 3 1 1 1 7 7 7 1 1 1 4 4 4 1 1 1 10000 10000 10000 4 4 4
数字和为 4 4 4 4 4 4 1 1 1 01 01 01 21 21 21 4 4 4
30001 , 21001 , 20101 , 12001 , 11101 , 10201 , 10021. 30001,21001,20101,12001,11101,10201,10021. 30001 , 21001 , 20101 , 12001 , 11101 , 10201 , 10021. 它们模 71 71 71 39 , 56 , 8 , 2 , 25 , 48 , 10 39,56,8,2,25,48,10 39 , 56 , 8 , 2 , 25 , 48 , 10 1 1 1 7 7 7 n = 21301 = 1 + 25 ⋅ 852 n=21301=1+25\cdot852 n = 21301 = 1 + 25 ⋅ 852 7 7 7