2022 AIME I 第 4 题

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4.

w=3+i2w = \frac{\sqrt{3} + \mathrm{i}}{2}z=1+i32z = \frac{-1 + \mathrm{i}\sqrt{3}}{2},其中 i=1\mathrm{i} = \sqrt{-1}。求满足方程 iwr=zs\mathrm{i} \cdot w^r = z^s 的有序数对 (r,s)(r, s) 的个数,其中两个分量都是不超过 100100 的正整数。

Let w=3+i2w = \frac{\sqrt{3} + \mathrm{i}}{2} and z=1+i32,z = \frac{-1 + \mathrm{i}\sqrt{3}}{2}, where i=1.\mathrm{i} = \sqrt{-1}. Find the number of ordered pairs (r,s)(r, s) of positive integers not exceeding 100100 that satisfy the equation iwr=zs.\mathrm{i} \cdot w^r = z^s.

答案:834
知识点:复数棣莫弗定理模运算
难度评级:2300
小提示:

两个数都在单位圆上:ww 的辐角是 3030^\circzz 的辐角是 120120^\circ,而 i\mathrm{i} 的辐角是 9090^\circ

Both numbers lie on the unit circle: ww has argument 3030^\circ and zz has argument 120,120^\circ, while i\mathrm{i} has argument 9090^\circ

大提示:

比较辐角得 90+30r120s(mod360)90 + 30r \equiv 120s \pmod{360},即 r+34s(mod12)r + 3 \equiv 4s \pmod{12};数出 r[1,100]r \in [1, 100] 在每个余数类中的个数

Matching arguments gives 90+30r120s(mod360),90 + 30r \equiv 120s \pmod{360}, i.e. r+34s(mod12);r + 3 \equiv 4s \pmod{12}; count how many r[1,100]r \in [1, 100] hit each residue

解答:

wwzz 的模都是 11:极坐标形式为 w=cis30w = \operatorname{cis} 30^\circz=cis120z = \operatorname{cis} 120^\circ,而 i=cis90\mathrm{i} = \operatorname{cis} 90^\circ。因此方程 iwr=zs\mathrm{i} \cdot w^r = z^s 等价于以下辐角条件:90+30r120s(mod360)90 + 30r \equiv 120s \pmod{360}\text{,}也就是 r+34s(mod12)r + 3 \equiv 4s \pmod{12}\text{。}

对每个 ss,这确定了 r(mod12)r \pmod{12}:当 s1s \equiv 122、或 0(mod3)0 \pmod 3 时,余数 4s34s - 3 分别为 115599(模 1212)。在 1r1001 \le r \le 100 中,满足 r1(mod12)r \equiv 1 \pmod{12} 的有 99 个,满足 r5r \equiv 5r9(mod12)r \equiv 9 \pmod{12} 的各有 88 个。在 1s1001 \le s \le 100 中,满足 s1(mod3)s \equiv 1 \pmod 3 的有 3434 个,另外两个余数类各有 3333 个。

总数为 34934 \cdot 9 +338+ 33 \cdot 8 +338+ 33 \cdot 8 =306+264+264=834= 306 + 264 + 264 = 834

Both ww and zz have modulus 1:1: in polar form w=cis30w = \operatorname{cis} 30^\circ and z=cis120,z = \operatorname{cis} 120^\circ, while i=cis90.\mathrm{i} = \operatorname{cis} 90^\circ. The equation iwr=zs\mathrm{i} \cdot w^r = z^s is therefore a statement about arguments: 90+30r120s(mod360),90 + 30r \equiv 120s \pmod{360}, i.e. r+34s(mod12).r + 3 \equiv 4s \pmod{12}.

For each s,s, this determines r(mod12):r \pmod{12}: the residue 4s34s - 3 is 1,1, 5,5, or 99 modulo 1212 according as s1,s \equiv 1, 2,2, or 0(mod3).0 \pmod 3. Among 1r1001 \le r \le 100 there are 99 values with r1(mod12)r \equiv 1 \pmod{12} and 88 values each with r5r \equiv 5 or r9(mod12).r \equiv 9 \pmod{12}. Among 1s1001 \le s \le 100 there are 3434 values with s1(mod3)s \equiv 1 \pmod 3 and 3333 values in each of the other two classes.

The count is 34934 \cdot 9 +338+ 33 \cdot 8 +338+ 33 \cdot 8 =306+264+264=834.= 306 + 264 + 264 = 834.

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