1977 AMC 12 真题（P版）

1 答案 D

因为 \(y=2x\)，所以 \(z=2y=4x\)。因此 \(x+y+z=x+2x+4x=7x\)。

因此，正确答案是 D。

English

Since \(y=2x,\) we have \(z=2y=4x.\) Therefore \(x+y+z=x+2x+4x=7x.\)

Therefore, the correct answer is D.

2 答案 D

每个等边三角形都是等角三角形、等腰三角形和正多边形，并且与任意其他等边三角形相似。然而，边长不同的两个等边三角形并不全等。

因此，正确答案是 D。

English

Every equilateral triangle is equiangular, isosceles, regular, and similar to every other equilateral triangle. Two equilateral triangles with different side lengths are not congruent, however.

Therefore, the correct answer is D.

3 答案 E

每种硬币各取一枚，共值 \(1+5+10+25+50=91\) 美分。由于 \(\frac{273}{91}=3\)，五种硬币各有三枚，所以共有 \(5\cdot3=15\) 枚硬币。

因此，正确答案是 E。

English

One coin of each kind is worth \(1+5+10+25+50=91\) cents. Since \(\frac{273}{91}=3,\) there are three coins of each of the five kinds, or \(5\cdot3=15\) coins.

Therefore, the correct answer is E.

4 答案 C

\(\triangle ABC\) 的两个底角都等于 \(50^\circ\)。因为 \(CE=CD\)，所以三角形 \(CED\) 的顶角为 \(50^\circ\)，从而 \(\angle EDC=65^\circ\)。同理，\(BF=BD\) 推出 \(\angle BDF=65^\circ\)。直线 \(CB\) 上方的三个角之和为 \(180^\circ\)，所以 \[ \begin{aligned} \angle EDF&=180^\circ-65^\circ-65^\circ\\ &=50^\circ \end{aligned}\text{。}\]

因此，正确答案是 C。

English

The base angles of \(\triangle ABC\) are each \(50^\circ.\) Since \(CE=CD,\) triangle \(CED\) has vertex angle \(50^\circ,\) so \(\angle EDC=65^\circ.\) Similarly, \(BF=BD\) gives \(\angle BDF=65^\circ.\) The three angles above the straight line \(CB\) sum to \(180^\circ,\) so \[ \begin{aligned} \angle EDF&=180^\circ-65^\circ-65^\circ\\ &=50^\circ. \end{aligned} \]

Therefore, the correct answer is C.

5 答案 A

由三角不等式可得 \(AP+PB\ge AB\)。等号成立当且仅当 \(A\)、\(P\)、\(B\) 共线，且 \(P\) 位于 \(A\) 与 \(B\) 之间（包括两个端点）。因此，所求轨迹是从 \(A\) 到 \(B\) 的线段。

因此，正确答案是 A。

English

The triangle inequality gives \(AP+PB\ge AB.\) Equality holds exactly when \(A,\) \(P,\) \(B\) are collinear with \(P\) between \(A\) and \(B,\) including the endpoints. Thus the locus is the segment from \(A\) to \(B.\)

Therefore, the correct answer is A.

6 答案 D

有 \[ \begin{aligned} \left(2x+\frac y2\right)^{-1} &=\frac2{4x+y},\\ (2x)^{-1}+\left(\frac y2\right)^{-1} &=\frac1{2x}+\frac2y\\ &=\frac{y+4x}{2xy} \end{aligned}\text{。}\]两者的乘积为 \(\frac{1}{xy}=(xy)^{-1}\)。

因此，正确答案是 D。

English

We have \[ \begin{aligned} \left(2x+\frac y2\right)^{-1} &=\frac2{4x+y},\\ (2x)^{-1}+\left(\frac y2\right)^{-1} &=\frac1{2x}+\frac2y\\ &=\frac{y+4x}{2xy}. \end{aligned} \] Their product is \(\frac{1}{xy}=(xy)^{-1}.\)

Therefore, the correct answer is D.

7 答案 E

令 \(u=\sqrt[4]{2}\)。由于 \[ \begin{aligned} &(1-u)\bigl(-(1+u)(1+u^2)\bigr)\\ &\qquad=-(1-u^2)(1+u^2)\\ &\qquad=u^4-1=1 \end{aligned}\text{，}\]所以 \[ t=-(1+\sqrt[4]{2})(1+\sqrt2)\text{。}\]

因此，正确答案是 E。

English

Let \(u=\sqrt[4]{2}.\) Since \[ \begin{aligned} &(1-u)\bigl(-(1+u)(1+u^2)\bigr)\\ &\qquad=-(1-u^2)(1+u^2)\\ &\qquad=u^4-1=1, \end{aligned} \] it follows that \[ t=-(1+\sqrt[4]{2})(1+\sqrt2). \]

Therefore, the correct answer is E.

8 答案 B

设 \(r\)、\(s\)、\(t\in\{-1,1\}\) 分别为 \(a\)、\(b\)、\(c\) 的符号，则该式为 \(r+s+t+rst\)。三个符号全为正时，式子的值为 \(4\)；全为负时，值为 \(-4\)。若正负号混合，直接相消可得 \(0\)。因此所求集合为 \(\{-4,0,4\}\)。

因此，正确答案是 B。

English

Let \(r,\) \(s,\) \(t\in\{-1,1\}\) be the signs of \(a,\) \(b,\) \(c.\) The expression is \(r+s+t+rst.\) If all three signs are positive it is \(4,\) and if all are negative it is \(-4.\) If the signs are mixed, direct cancellation gives \(0.\) Hence the set is \(\{-4,0,4\}.\)

Therefore, the correct answer is B.

9 答案 B

设三条弧 \(AB\)、\(BC\)、\(CD\) 的度数均为 \(x^\circ\)，弧 \(AD\) 的度数为 \(y^\circ\)。于是 \[ \begin{aligned} 3x+y&=360,\\ \frac{x-y}{2}&=40 \end{aligned}\text{。}\]解得 \(y=30\)。圆周角 \(\angle ACD\) 所对的弧是 \(AD\)，所以其度数为 \(\frac{y}{2}=15^\circ\)。

因此，正确答案是 B。

English

Let each of arcs \(AB,\) \(BC,\) \(CD\) measure \(x^\circ\) and let arc \(AD\) measure \(y^\circ.\) Then \[ \begin{aligned} 3x+y&=360,\\ \frac{x-y}{2}&=40. \end{aligned} \] Solving gives \(y=30.\) The inscribed angle \(\angle ACD\) subtends arc \(AD,\) so it measures \(\frac{y}{2}=15^\circ.\)

Therefore, the correct answer is B.

10 答案 E

令 \(x=1\)，右边便成为所求的系数之和。因此 \[ \begin{aligned} a_7+a_6+\cdots+a_0 &=(3\cdot1-1)^7\\ &=2^7=128 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Setting \(x=1\) makes the right side the desired coefficient sum. Thus \[ \begin{aligned} a_7+a_6+\cdots+a_0 &=(3\cdot1-1)^7\\ &=2^7=128. \end{aligned} \]

Therefore, the correct answer is E.

11 答案 B

若 \(n\le x\lt n+1\)，则 \(n+1\le x+1\lt n+2\)，所以 \(\mathrm{I}\) 正确。取 \(x=y=\frac{3}{2}\)，可得 \([x+y]=3\ne2=[x]+[y]\) 且 \([xy]=2\ne1=[x][y]\)。因此 \(\mathrm{II}\) 和 \(\mathrm{III}\) 错误。

因此，正确答案是 B。

English

If \(n\le x\lt n+1,\) then \(n+1\le x+1\lt n+2,\) so \(\mathrm{I}\) is true. Taking \(x=y=\frac{3}{2}\) gives \([x+y]=3\ne2=[x]+[y]\) and \([xy]=2\ne1=[x][y].\) Thus \(\mathrm{II}\) and \(\mathrm{III}\) are false.

Therefore, the correct answer is B.

12 答案 D

设 \(a\) 为 Al 的年龄，\(s\) 为另外两人的年龄之和。于是 \(a-s=16\)，且 \[ \begin{aligned} 1632&=a^2-s^2\\ &=(a-s)(a+s)\\ &=16(a+s) \end{aligned}\text{。}\]因此，所求的总年龄 \(a+s\) 为 \(\frac{1632}{16}=102\)。

因此，正确答案是 D。

English

Let \(a\) be Al’s age and \(s\) the sum of the other two ages. Then \(a-s=16\) and \[ \begin{aligned} 1632&=a^2-s^2\\ &=(a-s)(a+s)\\ &=16(a+s). \end{aligned} \] Hence the requested total \(a+s\) is \(\frac{1632}{16}=102.\)

Therefore, the correct answer is D.

13 答案 E

接下来的各项为 \(a_3=a_1a_2\)、\(a_4=a_1a_2^2\) 和 \(a_5=a_1^2a_2^3\)。若该数列为等比数列，则 \[ \begin{aligned} \frac{a_2}{a_1}&=\frac{a_3}{a_2}=a_1,\\ \frac{a_4}{a_3}&=a_2=a_1 \end{aligned}\text{。}\]因此 \(a_2=a_1^2\)，且 \(a_2=a_1\)。由各项为正可知 \(a_1=a_2=1\)。反之，这两个初值会产生常数等比数列 \(1,1,1,\ldots\)。

因此，正确答案是 E。

English

The next terms are \(a_3=a_1a_2,\) \(a_4=a_1a_2^2,\) and \(a_5=a_1^2a_2^3.\) If the sequence is geometric, then \[ \begin{aligned} \frac{a_2}{a_1}&=\frac{a_3}{a_2}=a_1,\\ \frac{a_4}{a_3}&=a_2=a_1. \end{aligned} \] Thus \(a_2=a_1^2\) and \(a_2=a_1.\) Positivity forces \(a_1=a_2=1.\) Conversely, these initial values produce the constant geometric sequence \(1,1,1,\ldots.\)

Therefore, the correct answer is E.

14 答案 B

移项并配成乘积，得到 \[ \begin{aligned} mn-m-n&=0,\\ (m-1)(n-1)&=1 \end{aligned}\text{。}\]\(1\) 的整数因数对为 \((1,1)\) 和 \((-1,-1)\)，分别得到 \((m,n)=(2,2)\) 和 \((0,0)\)。因此共有两个有序整数对。

因此，正确答案是 B。

English

Rearranging and completing the product gives \[ \begin{aligned} mn-m-n&=0,\\ (m-1)(n-1)&=1. \end{aligned} \] The integer factor pairs of \(1\) are \((1,1)\) and \((-1,-1),\) producing \((m,n)=(2,2)\) and \((0,0).\) Thus there are two ordered pairs.

Therefore, the correct answer is B.

15 答案 D

三个圆心构成边长为 \(6\) 的等边三角形，其内切圆半径为 \(\frac{6\sqrt3}{6}=\sqrt3\)。外部三角形的每一边都是向外平移三单位后的平行切线，所以外部三角形的内切圆半径为 \(3+\sqrt3\)。内切圆半径为 \(r\) 的等边三角形边长为 \(2\sqrt3r\)，因此外部三角形的每条边长为 \[ 2\sqrt3(3+\sqrt3)=6+6\sqrt3\text{。}\]周长为 \(3(6+6\sqrt3)=18+18\sqrt3\)。

因此，正确答案是 D。

English

The circle centers form an equilateral triangle of side \(6,\) whose inradius is \(\frac{6\sqrt3}{6}=\sqrt3.\) Each side of the outer triangle is a parallel tangent line three units farther out, so the outer triangle has inradius \(3+\sqrt3.\) An equilateral triangle of inradius \(r\) has side \(2\sqrt3r,\) hence each outer side is \[ 2\sqrt3(3+\sqrt3)=6+6\sqrt3. \] The perimeter is \(3(6+6\sqrt3)=18+18\sqrt3.\)

Therefore, the correct answer is D.

16 答案 D

令 \(u_n=i^n\cos(45+90n)^\circ\)。则 \[ \begin{aligned} u_{n+2} &=i^{n+2}\cos(225+90n)^\circ\\ &=(-i^n)(-\cos(45+90n)^\circ)\\ &=u_n \end{aligned}\text{。}\]每个偶数下标的项都等于 \(\frac{\sqrt2}{2}\)，每个奇数下标的项都等于 \(-\frac{i\sqrt2}{2}\)。共有 \(21\) 个偶数下标和 \(20\) 个奇数下标，所以总和为 \(\frac{\sqrt2}{2}(21-20i)\)。

因此，正确答案是 D。

English

Let \(u_n=i^n\cos(45+90n)^\circ.\) Then \[ \begin{aligned} u_{n+2} &=i^{n+2}\cos(225+90n)^\circ\\ &=(-i^n)(-\cos(45+90n)^\circ)\\ &=u_n. \end{aligned} \] Every even-indexed term equals \(\frac{\sqrt2}{2},\) and every odd-indexed term equals \(-\frac{i\sqrt2}{2}.\) There are \(21\) even indices and \(20\) odd indices, so the sum is \(\frac{\sqrt2}{2}(21-20i).\)

Therefore, the correct answer is D.

17 答案 B

可能的集合为 \(\{1,2,3\}\)、\(\{2,3,4\}\)、\(\{3,4,5\}\) 和 \(\{4,5,6\}\)。每个集合都有 \(3!=6\) 种排列，因此在 \(6^3=216\) 个结果中，有 \(24\) 个符合条件。所求概率为 \(\frac{24}{216}=\frac{1}{9}\)。

因此，正确答案是 B。

English

The possible sets are \(\{1,2,3\},\) \(\{2,3,4\},\) \(\{3,4,5\},\) and \(\{4,5,6\}.\) Each has \(3!=6\) orderings, so \(24\) of the \(6^3=216\) outcomes work. The probability is \(\frac{24}{216}=\frac{1}{9}.\)

Therefore, the correct answer is B.

18 答案 B

由换底公式，\[ \begin{aligned} y&=\frac{\log 3}{\log2}\cdot \frac{\log4}{\log3}\cdots \frac{\log32}{\log31}\\ &=\frac{\log32}{\log2}\\ &=\log_2 32=5 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

By change of base, \[ \begin{aligned} y&=\frac{\log 3}{\log2}\cdot \frac{\log4}{\log3}\cdots \frac{\log32}{\log31}\\ &=\frac{\log32}{\log2}\\ &=\log_2 32=5. \end{aligned} \]

Therefore, the correct answer is B.

19 答案 A

\(P\) 和 \(Q\) 都在线段 \(BE\) 的垂直平分线上，而 \(R\) 和 \(S\) 都在线段 \(DE\) 的垂直平分线上。由于 \(B\)、\(E\)、\(D\) 共线，这两条垂直平分线平行，所以 \(PQ\parallel RS\)。同理，\(Q\)、\(R\) 在线段 \(CE\) 的垂直平分线上，\(S\)、\(P\) 在线段 \(AE\) 的垂直平分线上。由于 \(A\)、\(E\)、\(C\) 共线，所以 \(QR\parallel SP\)。因此 \(PQRS\) 始终是平行四边形。

因此，正确答案是 A。

English

Both \(P\) and \(Q\) lie on the perpendicular bisector of \(BE,\) while both \(R\) and \(S\) lie on the perpendicular bisector of \(DE.\) Since \(B,\) \(E,\) \(D\) are collinear, these bisectors are parallel, so \(PQ\parallel RS.\) Likewise, \(Q,\) \(R\) lie on the perpendicular bisector of \(CE,\) and \(S,\) \(P\) lie on that of \(AE.\) Since \(A,\) \(E,\) \(C\) are collinear, \(QR\parallel SP.\) Therefore \(PQRS\) is always a parallelogram.

Therefore, the correct answer is A.

20 答案 E

把路径反向，从底部中央的 \(T\) 出发拼出 TSETNOC。对于水平移动均向左的一类路径，六步中的每一步都有向上或向左两种选择，因此共有 \(2^6=64\) 条路径。由对称性，水平移动向右的路径也有 \(64\) 条。完全竖直的中央路径同时属于两类，所以总数为 \[ 64+64-1=127\text{。}\]这个数不在前四个选项中。

因此，正确答案是 E。

English

Reverse the paths and spell TSETNOC from the central bottom \(T.\) In the family whose horizontal moves go left, each of the six steps has two choices: up or left. This gives \(2^6=64\) paths. By symmetry, \(64\) paths have horizontal moves going right. The all-vertical central path belongs to both families, so the total is \[ 64+64-1=127. \] This number is not among the first four choices.

Therefore, the correct answer is E.

21 答案 B

用第一个方程减去第二个方程，得到 \[ \begin{gathered} (a+1)x+(a+1)\\ =(a+1)(x+1)=0 \end{gathered}\text{。}\]若 \(a=-1\)，两个方程均化为 \(x^2-x+1=0\)，它没有实根。否则 \(x=-1\)，代入 \(x^2-x-a=0\) 得 \(a=2\)。这个值确实满足条件，所以恰有一个 \(a\) 的取值符合要求。

因此，正确答案是 B。

English

Subtracting the second equation from the first gives \[ \begin{gathered} (a+1)x+(a+1)\\ =(a+1)(x+1)=0. \end{gathered} \] If \(a=-1,\) the common equation is \(x^2-x+1=0,\) which has no real roots. Otherwise \(x=-1,\) and substitution into \(x^2-x-a=0\) gives \(a=2.\) This value works, so exactly one value of \(a\) qualifies.

Therefore, the correct answer is B.

22 答案 C

令 \(a=b=0\)，得到 \(2f(0)=4f(0)\)，所以 \(f(0)=0\)。现在令 \(a=0\) 且 \(b=x\)，方程变为 \[ \begin{aligned} f(x)+f(-x) &=2f(0)+2f(x)\\ &=2f(x) \end{aligned}\text{，}\]因此对每个实数 \(x\)，都有 \(f(-x)=f(x)\)。

因此，正确答案是 C。

English

Setting \(a=b=0\) gives \(2f(0)=4f(0),\) so \(f(0)=0.\) Now set \(a=0\) and \(b=x.\) The equation becomes \[ \begin{aligned} f(x)+f(-x) &=2f(0)+2f(x)\\ &=2f(x), \end{aligned} \] and hence \(f(-x)=f(x)\) for every real \(x.\)

Therefore, the correct answer is C.

23 答案 B

设 \(x^2+mx+n\) 的两个根为 \(u\)、\(v\)。于是 \(u+v=-m\)，且 \(uv=n\)，而第一个方程的两个根为 \(u^3\)、\(v^3\)。因此 \[ \begin{aligned} -p&=u^3+v^3\\ &=(u+v)^3-3uv(u+v)\\ &=-m^3+3mn \end{aligned}\text{。}\]所以 \(p=m^3-3mn\)。

因此，正确答案是 B。

English

Let the roots of \(x^2+mx+n\) be \(u,\) \(v.\) Then \(u+v=-m\) and \(uv=n,\) while the roots of the first equation are \(u^3,\) \(v^3.\) Thus \[ \begin{aligned} -p&=u^3+v^3\\ &=(u+v)^3-3uv(u+v)\\ &=-m^3+3mn. \end{aligned} \] Therefore \(p=m^3-3mn.\)

Therefore, the correct answer is B.

24 答案 D

对 \(1\le n\le128\)，有 \[ \begin{gathered} \frac1{(2n-1)(2n+1)}\\ =\frac1{2(2n-1)}-\frac1{2(2n+1)} \end{gathered}\text{。}\]因此，逐项相消后，原和为 \[ \frac12\left(1-\frac1{257}\right)=\frac{128}{257}\text{。}\]

因此，正确答案是 D。

English

For \(1\le n\le128,\) \[ \begin{gathered} \frac1{(2n-1)(2n+1)}\\ =\frac1{2(2n-1)}-\frac1{2(2n+1)}. \end{gathered} \] Hence the sum telescopes to \[ \frac12\left(1-\frac1{257}\right)=\frac{128}{257}. \]

Therefore, the correct answer is D.

25 答案 E

因数 \(2\) 比因数 \(5\) 多，所以 \(10\) 的指数为 \[ \begin{aligned} n={}&\left\lfloor\frac{1005}{5}\right\rfloor+ \left\lfloor\frac{1005}{25}\right\rfloor\\ &+\left\lfloor\frac{1005}{125}\right\rfloor+ \left\lfloor\frac{1005}{625}\right\rfloor\\ ={}&201+40+8+1=250 \end{aligned}\text{。}\]由于选项中没有 \(250\)，正确选项是“以上均不正确”。

因此，正确答案是 E。

English

There are more factors of \(2\) than of \(5,\) so the exponent of \(10\) is \[ \begin{aligned} n={}&\left\lfloor\frac{1005}{5}\right\rfloor+ \left\lfloor\frac{1005}{25}\right\rfloor\\ &+\left\lfloor\frac{1005}{125}\right\rfloor+ \left\lfloor\frac{1005}{625}\right\rfloor\\ ={}&201+40+8+1=250. \end{aligned} \] Since \(250\) is not listed, the correct choice is “none of these.”

Therefore, the correct answer is E.

26 答案 B

沿对角线 \(MP\) 分割并利用 \(\sin\theta\le1\)，可得 \(A\le\frac{ab+cd}{2}\)。同样，沿 \(NQ\) 分割可得 \(A\le\frac{ad+bc}{2}\)。对于非凸四边形，取两个三角形面积的适当差，这些上界仍然成立。将两式相加，得到 \[ \begin{aligned} 2A&\le\frac{ab+ad+bc+cd}{2}\\ &=\frac{(a+c)(b+d)}2 \end{aligned}\text{，}\]即 \(A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\)。等号成立要求四个正弦上界全部取等号，因此四个角都是直角；反之，矩形可以取到等号。所以等号成立当且仅当四边形为矩形。

因此，正确答案是 B。

English

Splitting along diagonal \(MP\) and using \(\sin\theta\le1\) gives \(A\le\frac{ab+cd}{2}.\) Splitting along \(NQ\) similarly gives \(A\le\frac{ad+bc}{2}.\) These bounds remain valid for a nonconvex quadrilateral by taking the appropriate difference of triangle areas. Adding them yields \[ \begin{aligned} 2A&\le\frac{ab+ad+bc+cd}{2}\\ &=\frac{(a+c)(b+d)}2, \end{aligned} \] or \(A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right).\) Equality requires equality in all four sine bounds, so all four angles are right angles; conversely a rectangle gives equality. Thus the equality holds exactly for rectangles.

Therefore, the correct answer is B.

27 答案 C

对于半径 \(r\)，球心为 \((r,r,r)\)，所给点为 \((5,5,10)\)。因此 \[ 2(5-r)^2+(10-r)^2=r^2\text{，}\]化简得 \(r^2-20r+75=0\)，即 \((r-5)(r-15)=0\)。两个半径分别为 \(5\) 和 \(15\)，所以直径之和为 \(2(5+15)=40\) 英寸。

因此，正确答案是 C。

English

For radius \(r,\) the center is \((r,r,r)\) and the given point is \((5,5,10).\) Thus \[ 2(5-r)^2+(10-r)^2=r^2, \] which simplifies to \(r^2-20r+75=0,\) or \((r-5)(r-15)=0.\) The two radii are \(5\) and \(15,\) so the sum of the diameters is \(2(5+15)=40\) inches.

Therefore, the correct answer is C.

28 答案 A

设 \(R(x)\) 为余式，则 \(\deg R\le4\)。\(g\) 的五个根 \(\alpha\) 满足 \(\alpha^6=1\) 且 \(\alpha\ne1\)。因此 \(\alpha^{12}=1\)，并且 \[ R(\alpha)=g(\alpha^{12})=g(1)=6\text{。}\]所以次数至多为 \(4\) 的多项式 \(R(x)-6\) 有五个不同的根。它只能恒为零，因此 \(R(x)=6\)。

因此，正确答案是 A。

English

Let \(R(x)\) be the remainder, so \(\deg R\le4.\) The five roots \(\alpha\) of \(g\) satisfy \(\alpha^6=1\) and \(\alpha\ne1.\) Hence \(\alpha^{12}=1\) and \[ R(\alpha)=g(\alpha^{12})=g(1)=6. \] Therefore \(R(x)-6,\) a polynomial of degree at most \(4,\) has five distinct roots. It must be identically zero, so \(R(x)=6.\)

Therefore, the correct answer is A.

29 答案 B

由柯西－施瓦茨不等式，\[ \begin{aligned} (x^2+y^2+z^2)^2&\le3Q,\\ Q&=x^4+y^4+z^4 \end{aligned}\text{。}\]当 \(x^2=y^2=z^2\ne0\) 时取等号，所以更小的值都不成立。因此最小整数为 \(3\)。

因此，正确答案是 B。

English

By Cauchy-Schwarz, \[ \begin{aligned} (x^2+y^2+z^2)^2&\le3Q,\\ Q&=x^4+y^4+z^4. \end{aligned} \] Equality occurs when \(x^2=y^2=z^2\ne0,\) so no smaller value can work. Thus the smallest integer is \(3.\)

Therefore, the correct answer is B.

30 答案 A

若外接圆半径为 \(r\)，则三条弦所对的圆心角分别为 \(40^\circ\)、\(80^\circ\)、\(160^\circ\)。因此 \[ \begin{aligned} a&=2r\sin20^\circ,\\ b&=2r\sin40^\circ,\\ d&=2r\sin80^\circ \end{aligned}\text{。}\]由和差化积公式，\[ \begin{gathered} \sin20^\circ+\sin40^\circ\\ =2\sin30^\circ\cos10^\circ\\ =\cos10^\circ=\sin80^\circ \end{gathered}\text{。}\]两边乘以 \(2r\)，得到 \(a+b=d\)。

因此，正确答案是 A。

English

If the circumradius is \(r,\) the three chords subtend central angles \(40^\circ,\) \(80^\circ,\) \(160^\circ,\) respectively. Thus \[ \begin{aligned} a&=2r\sin20^\circ,\\ b&=2r\sin40^\circ,\\ d&=2r\sin80^\circ. \end{aligned} \] The sum-to-product identity gives \[ \begin{gathered} \sin20^\circ+\sin40^\circ\\ =2\sin30^\circ\cos10^\circ\\ =\cos10^\circ=\sin80^\circ. \end{gathered} \] Multiplying by \(2r\) yields \(a+b=d.\)

Therefore, the correct answer is A.