1976 AMC 12 真题（P版）

1 答案 B

方程为 \[ 1-\frac1{1-x}=\frac1{1-x}\text{。}\]两边乘以 \(1-x\ne0\)，得 \(1-x-1=1\)，所以 \(x=-1\)。

所以正确答案是 B。

English

The equation is \[ 1-\frac1{1-x}=\frac1{1-x}. \] Multiplying by \(1-x\ne0\) gives \(1-x-1=1,\) so \(x=-1.\)

Therefore, the correct answer is B.

2 答案 B

因为 \((x+1)^2\ge0\)，所以它的相反数至多为 \(0\)。它只有在 \((x+1)^2=0\) 时非负，而这只对应一个值 \(x=-1\)。

所以正确答案是 B。

English

Since \((x+1)^2\ge0,\) its negative is at most \(0.\) It is nonnegative only when \((x+1)^2=0,\) which occurs for the single value \(x=-1.\)

Therefore, the correct answer is B.

3 答案 E

与所选顶点相邻的两条边的中点到该顶点的距离均为 \(1\)。另外两个中点的横、纵坐标差分别为 \(2\) 和 \(1\)，所以每个距离都是 \(\sqrt{2^2+1^2}=\sqrt5\)。总和为 \(2+2\sqrt5\)。

所以正确答案是 E。

English

The two midpoints on sides meeting the chosen vertex are each at distance \(1.\) The other two midpoints have coordinate differences \(2\) and \(1,\) so each is at distance \(\sqrt{2^2+1^2}=\sqrt5.\) The sum is \(2+2\sqrt5.\)

Therefore, the correct answer is E.

4 答案 C

设倒数数列的和为 \(S'\)。乘以 \(r^{n-1}\) 后，原数列各项的顺序恰好反转：\[ \begin{aligned} r^{n-1}S' &=r^{n-1}+r^{n-2}+\cdots+1\\ &=s \end{aligned}\text{。}\]因此 \(S'=\frac{s}{r^{n-1}}\)。当 \(r=1\) 时结论也成立。

所以正确答案是 C。

English

Let \(S'\) be the reciprocal sum. Multiplying it by \(r^{n-1}\) reverses the original terms: \[ \begin{aligned} r^{n-1}S' &=r^{n-1}+r^{n-2}+\cdots+1\\ &=s. \end{aligned} \] Hence \(S'=\frac{s}{r^{n-1}}.\) This also holds when \(r=1.\)

Therefore, the correct answer is C.

5 答案 C

条件为 \(9(u-t)=9\)，所以 \(u=t+1\)。可能的整数是 \(12\)、\(23\)、\(34\)、\(45\)、\(56\)、\(67\)、\(78\) 和 \(89\)，共 \(8\) 个。

所以正确答案是 C。

English

The condition is \(9(u-t)=9,\) so \(u=t+1.\) The possibilities are \(12,\) \(23,\) \(34,\) \(45,\) \(56,\) \(67,\) \(78,\) and \(89,\) giving \(8\) integers.

Therefore, the correct answer is C.

6 答案 C

对于这样的根 \(r\)，\[ \begin{aligned} r^2-3r+c&=0,\\ (-r)^2+3(-r)-c&=0 \end{aligned}\text{。}\]两式相减得 \(2c=0\)。于是第一个方程化为 \(x(x-3)=0\)，其根为 \(0\) 和 \(3\)。

所以正确答案是 C。

English

For such a root \(r,\) \[ \begin{aligned} r^2-3r+c&=0,\\ (-r)^2+3(-r)-c&=0. \end{aligned} \] Subtracting gives \(2c=0.\) Thus the first equation is \(x(x-3)=0,\) with roots \(0\) and \(3.\)

Therefore, the correct answer is C.

7 答案 E

两个因子同时为正恰好发生在 \(|x|\lt1\) 时，此时 \(-1\lt x\lt1\)。两个因子同时为负需要 \(|x|\gt1\) 且 \(x\lt-1\)，这化为 \(x\lt-1\)。两个端点都使乘积为零。

所以正确答案是 E。

English

Both factors are positive exactly when \(|x|\lt1,\) giving \(-1\lt x\lt1.\) Both are negative when \(|x|\gt1\) and \(x\lt-1,\) which reduces to \(x\lt-1.\) The endpoints make the product zero.

Therefore, the correct answer is E.

8 答案 A

共有 \(9^2=81\) 个可能的点。到原点距离不超过 \(2\) 的点包括原点；四个点 \((\pm1,0)\)、\((0,\pm1)\)；四个点 \((\pm1,\pm1)\)；以及四个点 \((\pm2,0)\)、\((0,\pm2)\)。因此共有 \(13\) 个点符合条件，概率为 \(\frac{13}{81}\)。

所以正确答案是 A。

English

There are \(9^2=81\) possible points. Within distance \(2\) are the origin; the four points \((\pm1,0),\) \((0,\pm1);\) the four points \((\pm1,\pm1);\) and the four points \((\pm2,0),\) \((0,\pm2).\) Thus \(13\) points qualify, for probability \(\frac{13}{81}.\)

Therefore, the correct answer is A.

9 答案 D

因为 \(D\) 是 \(AB\) 的中点，而 \(E\) 是 \(DB\) 的中点，所以 \(AE=\frac34AB\)。又因为 \(F\) 是 \(BC\) 的中点，所以它到直线 \(AB\) 的距离是 \(C\) 到该直线距离的一半。因此 \[ \begin{aligned} [AEF]&=\frac34\cdot\frac12[ABC]\\ &=\frac38(96)=36 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Since \(D\) is the midpoint of \(AB\) and \(E\) is the midpoint of \(DB,\) \(AE=\frac34AB.\) Since \(F\) is the midpoint of \(BC,\) its distance from line \(AB\) is half that of \(C.\) Therefore \[ \begin{aligned} [AEF]&=\frac34\cdot\frac12[ABC]\\ &=\frac38(96)=36. \end{aligned} \]

Therefore, the correct answer is D.

10 答案 D

有 \[ \begin{aligned} f(g(x))&=mpx+mq+n,\\ g(f(x))&=pmx+pn+q \end{aligned}\text{。}\]对任意 \(x\)，两边的 \(x\) 项都会消去，所以方程有解当且仅当 \(mq+n=pn+q\)。整理得 \(n(1-p)-q(1-m)=0\)。

所以正确答案是 D。

English

We have \[ \begin{aligned} f(g(x))&=mpx+mq+n,\\ g(f(x))&=pmx+pn+q. \end{aligned} \] The \(x\)-terms cancel for every \(x,\) so a solution exists exactly when \(mq+n=pn+q.\) Rearranging gives \(n(1-p)-q(1-m)=0.\)

Therefore, the correct answer is D.

11 答案 B

令 \(P\) 表示大象长着紫色眼睛，令 \(Q\) 表示野猪长着长鼻子。原命题 \(P\Rightarrow\neg Q\) 等价于其逆否命题 \(Q\Rightarrow\neg P\)，即 \(\mathrm{III}\)；也等价于 \(\neg P\lor\neg Q\)，即 \(\mathrm{IV}\)。陈述 \(\mathrm{I}\) 和 \(\mathrm{II}\) 都是原命题的逆命题，不一定成立。

所以正确答案是 B。

English

Let \(P\) mean the elephant has purple eyes and \(Q\) mean the pig has a long nose. The original statement \(P\Rightarrow\neg Q\) is equivalent to its contrapositive \(Q\Rightarrow\neg P,\) which is \(\mathrm{III},\) and to \(\neg P\lor\neg Q,\) which is \(\mathrm{IV}.\) Statements \(\mathrm{I}\) and \(\mathrm{II}\) are converses and need not follow.

Therefore, the correct answer is B.

12 答案 C

可能的苹果数共有 \(144-120+1=25\) 种。因为 \(128>25\cdot5\)，所以某一种数量至少出现 \(6\) 次。这个界可以达到：让三种数量各出现 \(6\) 次，其余二十二种各出现 \(5\) 次，恰好分配 \(128\) 箱。因此能保证的最大 \(n\) 是 \(6\)。

所以正确答案是 C。

English

There are \(144-120+1=25\) possible apple counts. Since \(128>25\cdot5,\) some count occurs at least \(6\) times. This is sharp: distribute \(128\) crates so that three counts occur \(6\) times and the other twenty-two occur \(5\) times. Thus the largest guaranteed \(n\) is \(6.\)

Therefore, the correct answer is C.

13 答案 A

产奶率为每头奶牛每天 \(\frac{x+1}{x(x+2)}\) 罐。设新的用时为 \(T\)，则 \[ \frac{x+5}{(x+3)T}=\frac{x+1}{x(x+2)}\text{。}\]因此 \(T=\frac{x(x+2)(x+5)}{(x+1)(x+3)}\)。

所以正确答案是 A。

English

The production rate is \(\frac{x+1}{x(x+2)}\) cans per cow-day. If the new time is \(T,\) then \[ \frac{x+5}{(x+3)T}=\frac{x+1}{x(x+2)}. \] Hence \(T=\frac{x(x+2)(x+5)}{(x+1)(x+3)}.\)

Therefore, the correct answer is A.

14 答案 A

平均角度为 \(\frac{100+140}{2}=120^\circ\)，所以内角和为 \(120n\)。凸 \(n\) 边形的内角和为 \(180(n-2)\)，故 \(120n=180(n-2)\)，解得 \(n=6\)。

所以正确答案是 A。

English

The average angle is \(\frac{100+140}{2}=120^\circ,\) so the angle sum is \(120n.\) A convex \(n\)-gon has angle sum \(180(n-2),\) hence \(120n=180(n-2)\) and \(n=6.\)

Therefore, the correct answer is A.

15 答案 B

除数 \(d\) 同时整除 \(358=2\cdot179\) 和 \(895=5\cdot179\)。这两个数的最大公因数为质数 \(179\)，所以 \(d=179\)。又因为 \(1059=5\cdot179+164\)，所以 \(r=164\)，从而 \(d-r=15\)。

所以正确答案是 B。

English

The divisor \(d\) divides both \(358=2\cdot179\) and \(895=5\cdot179.\) Their greatest common divisor is \(179,\) a prime, so \(d=179.\) Since \(1059=5\cdot179+164,\) we have \(r=164\) and \(d-r=15.\)

Therefore, the correct answer is B.

16 答案 E

设四条相等的边长均为 \(s\)，令 \(AB=2a\)，并设从 \(F\) 作出的高为 \(a\)。在两个等腰三角形中，从 \(C\) 作出的高平分 \(AB\)，从 \(F\) 作出的高平分 \(DE\)。\(ABC\) 的一半是斜边为 \(s\)、一条直角边为 \(a\) 的直角三角形；\(DEF\) 的一半也有相同的斜边和长度为 \(a\) 的高。两个直角三角形全等，只是两条直角边互换。因此相应的锐角互余，完整的顶角 \(\angle ACB\) 与 \(\angle DFE\) 之和为 \(180^\circ\)。相应的半三角形面积也相等，所以两个完整三角形的面积相等。陈述 \(\mathrm{II}\) 和 \(\mathrm{III}\) 正确。

所以正确答案是 E。

English

Let the equal sides have length \(s,\) let \(AB=2a,\) and let the altitude from \(F\) be \(a.\) In the isosceles triangles, the altitude from \(C\) bisects \(AB,\) while the altitude from \(F\) bisects \(DE.\) A half of \(ABC\) has hypotenuse \(s\) and leg \(a;\) a half of \(DEF\) has the same hypotenuse and altitude leg \(a.\) The two right triangles are congruent with their legs interchanged. Thus their relevant acute angles are complementary, so the full vertex angles \(\angle ACB\) and \(\angle DFE\) sum to \(180^\circ.\) The corresponding half-areas are equal as well, hence the full triangle areas are equal. Statements \(\mathrm{II}\) and \(\mathrm{III}\) hold.

Therefore, the correct answer is E.

17 答案 A

有 \[ \begin{aligned} (\sin\theta+\cos\theta)^2 &=\sin^2\theta+\cos^2\theta\\ &\quad+2\sin\theta\cos\theta\\ &=1+a \end{aligned}\text{。}\]当 \(\theta\) 为锐角时，正弦和余弦都为正，所以所求和为 \(\sqrt{a+1}\)。

所以正确答案是 A。

English

We have \[ \begin{aligned} (\sin\theta+\cos\theta)^2 &=\sin^2\theta+\cos^2\theta\\ &\quad+2\sin\theta\cos\theta\\ &=1+a. \end{aligned} \] Both sine and cosine are positive for acute \(\theta,\) so the sum is \(\sqrt{a+1}.\)

Therefore, the correct answer is A.

18 答案 E

延长 \(BD\)，使其越过 \(D\) 后与圆交于 \(E\)。因为 \(BC=3\)、\(BD=6\)，切割线定理给出 \[ 3\cdot BE=AB^2=36\text{，}\]所以 \(BE=12\)、\(DE=6\)。点 \(D\) 的幂满足 \[ \begin{aligned} DC\cdot DE&=(R-OD)(R+OD)\\ &=R^2-OD^2 \end{aligned}\text{，}\]因而 \(3\cdot6=(R-2)(R+2)=R^2-4\)。所以 \(R^2=22\)，且 \(R=\sqrt{22}\)。

所以正确答案是 E。

English

Extend \(BD\) through \(D\) to meet the circle at \(E.\) Since \(BC=3\) and \(BD=6,\) the tangent-secant theorem gives \[ 3\cdot BE=AB^2=36, \] so \(BE=12\) and \(DE=6.\) The power of \(D\) gives \[ \begin{aligned} DC\cdot DE&=(R-OD)(R+OD)\\ &=R^2-OD^2, \end{aligned} \] hence \(3\cdot6=(R-2)(R+2)=R^2-4.\) Thus \(R^2=22\) and \(R=\sqrt{22}.\)

Therefore, the correct answer is E.

19 答案 B

设余式为 \(r(x)=ax+b\)。因为 \(p(1)=3\)、\(p(3)=5\)，所以 \(a+b=3\) 和 \(3a+b=5\)。因此 \(a=1\)、\(b=2\)，余式为 \(x+2\)。

所以正确答案是 B。

English

Let the remainder be \(r(x)=ax+b.\) Since \(p(1)=3\) and \(p(3)=5,\) we have \(a+b=3,\) and \(3a+b=5.\) Thus \(a=1,\) and \(b=2,\) so the remainder is \(x+2.\)

Therefore, the correct answer is B.

20 答案 E

因式分解得 \[ \begin{aligned} &\bigl(2\log_a x-\log_b x\bigr)\\ &\quad\cdot \bigl(2\log_a x-3\log_b x\bigr)=0 \end{aligned}\text{。}\]因为 \(x\ne1\)，由换底公式可知，第一个因式在 \(a=b^2\) 时为零，第二个因式在 \(a^3=b^2\) 时为零。因此原方程在这两个条件中的任一个成立时成立，而选项中没有任何一个“当且仅当”陈述能描述它们的并集。

所以正确答案是 E。

English

Factoring gives \[ \begin{aligned} &\bigl(2\log_a x-\log_b x\bigr)\\ &\quad\cdot \bigl(2\log_a x-3\log_b x\bigr)=0. \end{aligned} \] Since \(x\ne1,\) change of base shows that the first factor vanishes when \(a=b^2,\) while the second vanishes when \(a^3=b^2.\) Thus the equation holds under either of two conditions, and no listed “if and only if” statement describes their union.

Therefore, the correct answer is E.

21 答案 B

这个乘积为 \[ 2^{\frac{1+3+\cdots+(2n+1)}{7}} =2^{\frac{(n+1)^2}{7}}\text{。}\]当 \(n=7\) 时，指数 \(\frac{64}{7}\) 小于 \(\frac{19}{2}\)，所以乘积小于 \(2^{\frac{19}{2}}=512\sqrt2\lt768\lt1000\)。所有更小的正奇数 \(n\) 也都不满足。当 \(n=9\) 时，指数为 \(\frac{100}{7}\gt10\)，所以乘积大于 \(2^{10}=1024\)。因此第一个符合条件的奇数 \(n\) 是 \(9\)。

所以正确答案是 B。

English

The product is \[ 2^{\frac{1+3+\cdots+(2n+1)}{7}} =2^{\frac{(n+1)^2}{7}}. \] For \(n=7,\) the exponent \(\frac{64}{7}\) is less than \(\frac{19}{2},\) so the product is less than \(2^{\frac{19}{2}}=512\sqrt2\lt768\lt1000.\) All smaller positive odd \(n\) also fail. For \(n=9,\) the exponent is \(\frac{100}{7}\gt10,\) so the product exceeds \(2^{10}=1024.\) Thus the first allowable odd \(n\) is \(9.\)

Therefore, the correct answer is B.

22 答案 A

设三个顶点为 \(A\)、\(B\)、\(C\)，重心为 \(G\)，并令 \(T=GA^2+GB^2+GC^2\)。重心恒等式给出 \[ PA^2+PB^2+PC^2=3PG^2+T\text{。}\]在等边三角形中，每个 \(GA=\frac{s}{\sqrt3}\)，所以 \(T=s^2\)。因此 \(3PG^2=a-s^2\)。当 \(a\lt s^2\) 时轨迹为空集；当 \(a=s^2\) 时只有一个点；当 \(a\gt s^2\) 时是半径为正的圆。

所以正确答案是 A。

English

For vertices \(A,\) \(B,\) and \(C\) with centroid \(G,\) put \(T=GA^2+GB^2+GC^2.\) The standard centroid identity gives \[ PA^2+PB^2+PC^2=3PG^2+T. \] In an equilateral triangle, each \(GA=\frac{s}{\sqrt3},\) so \(T=s^2.\) Hence \(3PG^2=a-s^2.\) The locus is empty for \(a\lt s^2,\) one point for \(a=s^2,\) and a circle of positive radius for \(a\gt s^2.\)

Therefore, the correct answer is A.

23 答案 A

利用 \[ \binom n{k+1}=\frac{n-k}{k+1}\binom nk\text{，}\]并写成 \[ \frac{n-2k-1}{k+1}=\frac{n-k}{k+1}-1\text{，}\]原式化为 \[ \binom n{k+1}-\binom nk\text{。}\]对每一组允许的 \(k\) 和 \(n\)，这都是整数。

所以正确答案是 A。

English

Using \[ \binom n{k+1}=\frac{n-k}{k+1}\binom nk, \] and writing \[ \frac{n-2k-1}{k+1}=\frac{n-k}{k+1}-1, \] the given expression becomes \[ \binom n{k+1}-\binom nk. \] This is an integer for every permitted \(k\) and \(n.\)

Therefore, the correct answer is A.

24 答案 C

令圆 \(K\) 的圆心为 \((0,0)\)、半径为 \(2\)，并令 \(AB\) 为 \(x\) 轴。则圆 \(L\) 的圆心为 \((0,1)\)、半径为 \(1\)。若圆 \(M\) 的圆心为 \((x,t)\)、半径为 \(t\)，它与圆 \(K\) 内切并与圆 \(L\) 外切，故 \[ \begin{aligned} x^2+t^2&=(2-t)^2,\\ x^2+(t-1)^2&=(1+t)^2 \end{aligned}\text{。}\]两式分别化为 \(x^2=4-4t\) 和 \(x^2=4t\)，所以 \(t=\frac12\)。半径之比为 \(2:\frac{1}{2}=4:1\)，因此面积之比为 \(16\)。

所以正确答案是 C。

English

Take \(K\) to have center \((0,0)\) and radius \(2,\) with \(AB\) the \(x\)-axis. Then \(L\) has center \((0,1)\) and radius \(1.\) If \(M\) has center \((x,t)\) and radius \(t,\) internal tangency to \(K\) and external tangency to \(L\) give \[ \begin{aligned} x^2+t^2&=(2-t)^2,\\ x^2+(t-1)^2&=(1+t)^2. \end{aligned} \] These simplify to \(x^2=4-4t\) and \(x^2=4t,\) so \(t=\frac12.\) The radius ratio is \(2:\frac{1}{2}=4:1,\) hence the area ratio is \(16.\)

Therefore, the correct answer is C.

25 答案 D

直接计算得 \[ \begin{aligned} \Delta u_n&=3n^2+3n+2,\\ \Delta^2u_n&=6n+6,\\ \Delta^3u_n&=6,\\ \Delta^4u_n&=0 \end{aligned}\text{。}\]因此四阶差分对所有 \(n\) 都为零，而三阶差分不为零。

所以正确答案是 D。

English

Directly, \[ \begin{aligned} \Delta u_n&=3n^2+3n+2,\\ \Delta^2u_n&=6n+6,\\ \Delta^3u_n&=6,\\ \Delta^4u_n&=0. \end{aligned} \] Thus the fourth differences vanish for all \(n,\) but the third differences do not.

Therefore, the correct answer is D.

26 答案 C

设内公切线与圆 \(O\) 和 \(O'\) 分别切于 \(R\) 和 \(S\)。设经过 \(P\) 的外公切线与两圆分别切于 \(X\) 和 \(Y\)，经过 \(Q\) 的外公切线与两圆分别切于 \(V\) 和 \(W\)。从同一点引出的切线段相等，所以 \[ \begin{gathered} PR=PX,\\ PS=PY,\\ QR=QV,\\ QS=QW \end{gathered}\text{。}\]在内公切线上，\(PR+QR=PQ\)，且 \(PS+QS=PQ\)。相加得 \[ \begin{aligned} PR+QR+PS+QS=2PQ \end{aligned}\text{。}\]在两条外公切线上，\(PX+PY=XY\)，且 \(QV+QW=VW\)。因此 \(2PQ=XY+VW\)。两条外公切线段长度相等，所以 \(XY=VW\)，从而 \(PQ=XY=VW\)。

所以正确答案是 C。

English

Let the internal tangent touch \(O\) and \(O'\) at \(R\) and \(S.\) Let the external tangent through \(P\) touch them at \(X\) and \(Y,\) and let the one through \(Q\) touch them at \(V\) and \(W.\) Equal tangent segments from a point give \[ \begin{gathered} PR=PX,\\ PS=PY,\\ QR=QV,\\ QS=QW. \end{gathered} \] Along the internal tangent, \(PR+QR=PQ\) and \(PS+QS=PQ.\) Adding, \[ \begin{aligned} PR+QR+PS+QS=2PQ. \end{aligned} \] Along the external tangents, \(PX+PY=XY\) and \(QV+QW=VW.\) Therefore \(2PQ=XY+VW.\) The two external common tangent segments have equal length, so \(XY=VW\) and hence \(PQ=XY=VW.\)

Therefore, the correct answer is C.

27 答案 A

设 \(T\) 为第一个分数。它的分子平方为 \[ \begin{aligned} &(\sqrt5+2)+(\sqrt5-2)\\ &\quad+2\sqrt{(\sqrt5+2)(\sqrt5-2)}\\ &=2\sqrt5+2 \end{aligned}\text{。}\]因此 \(T^2=\frac{2\sqrt5+2}{\sqrt5+1}=2\)，且 \(T=\sqrt2\)。另外，\(\sqrt{3-2\sqrt2}=\sqrt2-1\)。所以 \(N=\sqrt2-(\sqrt2-1)=1\)。

所以正确答案是 A。

English

Let \(T\) be the first fraction. Its numerator squared is \[ \begin{aligned} &(\sqrt5+2)+(\sqrt5-2)\\ &\quad+2\sqrt{(\sqrt5+2)(\sqrt5-2)}\\ &=2\sqrt5+2. \end{aligned} \] Thus \(T^2=\frac{2\sqrt5+2}{\sqrt5+1}=2,\) and \(T=\sqrt2.\) Also \(\sqrt{3-2\sqrt2}=\sqrt2-1.\) Hence \(N=\sqrt2-(\sqrt2-1)=1.\)

Therefore, the correct answer is A.

28 答案 B

能被 \(4\) 整除的下标有 \(25\) 个，与 \(1\pmod4\) 同余的下标也有 \(25\) 个。从 \(\binom{100}{2}=4950\) 个交点出发，平行线组不产生交点，因此减去 \(\binom{25}{2}=300\)。共点线组的 \(300\) 对直线只产生一个点，而不是 \(300\) 个不同的点，因此再减去 \(299\)。其余交点都可以选成互不相同，所以最大值为 \[ 4950-300-299=4351\text{。}\]

所以正确答案是 B。

English

There are \(25\) indices divisible by \(4\) and \(25\) congruent to \(1\pmod4.\) Starting from \(\binom{100}{2}=4950,\) the parallel group contributes no intersections, removing \(\binom{25}{2}=300.\) The concurrent group’s \(300\) pairs all give one point rather than \(300\) distinct points, removing another \(299.\) All remaining intersections can be chosen distinct, so the maximum is \[ 4950-300-299=4351. \]

Therefore, the correct answer is B.

29 答案 B

设安和芭芭拉现在的年龄分别为 \(x\) 和 \(y\)，并令她们恒定的年龄差为 \(d=x-y\)。在第一个所述时刻，安的年龄为 \(y\)，所以芭芭拉的年龄为 \(y-d\)。在更早的所述时刻，安的年龄为 \(y-d\)，所以芭芭拉的年龄为 \(y-2d\)。最后一句说明这个年龄等于 \(\frac{x}{2}\)。因此 \[ y-2(x-y)=\frac{x}{2}\text{，}\]从而 \(6y=5x\)。再结合 \(x+y=44\)，得 \(x=24\)、\(y=20\)。

所以正确答案是 B。

English

Let Ann’s and Barbara’s present ages be \(x\) and \(y,\) and let their constant age difference be \(d=x-y.\) At the first referenced time Ann was \(y,\) so Barbara was \(y-d.\) At the earlier referenced time Ann was that age \(y-d,\) so Barbara was \(y-2d.\) The final clause says this last age was \(\frac{x}{2}.\) Hence \[ y-2(x-y)=\frac{x}{2}, \] which gives \(6y=5x.\) Together with \(x+y=44,\) this yields \(x=24\) and \(y=20.\)

Therefore, the correct answer is B.

30 答案 E

令 \(x=2u\)、\(y=v\) 和 \(z=\frac{w}{2}\)。前两个方程都除以 \(2\)，得 \[ \begin{aligned} u+v+w&=6,\\ uv+vw+uw&=11 \end{aligned}\text{，}\]而 \(uvw=6\)。因此 \(u\)、\(v\)、\(w\) 是多项式 \(p(t)=t^3-6t^2+11t-6\) 的三个根。该多项式分解为 \[ p(t)=(t-1)(t-2)(t-3)\text{，}\]所以这三个数是按任意顺序排列的 \(1\)、\(2\)、\(3\)。它们的 \(3!=6\) 种排列产生 \(6\) 个不同的有序三元组 \((x,y,z)\)。

所以正确答案是 E。

English

Set \(x=2u,\) \(y=v,\) and \(z=\frac{w}{2}.\) Dividing the first two equations by \(2\) gives \[ \begin{aligned} u+v+w&=6,\\ uv+vw+uw&=11, \end{aligned} \] while \(uvw=6.\) Thus \(u,\) \(v,\) and \(w\) are the three roots of \(p(t)=t^3-6t^2+11t-6.\) This polynomial factors as \[ p(t)=(t-1)(t-2)(t-3), \] so they are \(1,\) \(2,\) and \(3\) in any order. Their \(3!=6\) permutations produce \(6\) distinct ordered triples \((x,y,z).\)

Therefore, the correct answer is E.