1973 AMC 12 真题（P版）

1 答案 D

设 \(M\) 既是弦的中点，也是该半径的中点。圆心 \(O\) 到 \(M\) 的距离为 \(6\)，而圆的半径为 \(12\)。若 \(A\) 是弦的一个端点，则 \[ AM=\sqrt{12^2-6^2}=6\sqrt3\text{。}\]整条弦的长度为 \(2AM=12\sqrt3\)。

所以正确答案是 D。

English

Let \(M\) be the midpoint of the chord and also the midpoint of the radius. The distance from the center \(O\) to \(M\) is \(6,\) while the circle’s radius is \(12.\) If \(A\) is one endpoint of the chord, then \[ AM=\sqrt{12^2-6^2}=6\sqrt3. \] The full chord has length \(2AM=12\sqrt3.\)

Therefore, the correct answer is D.

2 答案 C

没有被涂色的小立方体恰好组成内部的 \(8\times8\times8\) 立方体。因此至少有一个面被涂色的小立方体个数为 \[ 10^3-8^3=1000-512=488\text{。}\]

所以正确答案是 C。

English

The unpainted cubes are precisely the interior \(8\times8\times8\) cube. Therefore the number having at least one painted face is \[ 10^3-8^3=1000-512=488. \]

Therefore, the correct answer is C.

3 答案 B

对满足 \(p\lt q\) 且 \(p+q=126\) 的质数，差 \(q-p=126-2p\) 在 \(p\) 尽可能小时最大。\(3\)、\(5\)、\(7\)、\(11\) 相应的补数分别为 \(123\)、\(121\)、\(119\)、\(115\)，它们都不是质数。当 \(p=13\) 时，另一个数为 \(113\)，它是质数。因此最大差为 \[ 113-13=100\text{。}\]

所以正确答案是 B。

English

For primes \(p\lt q\) with \(p+q=126,\) the difference \(q-p=126-2p\) is largest when \(p\) is as small as possible. The complements of \(3,\) \(5,\) \(7,\) and \(11\) are \(123,\) \(121,\) \(119,\) and \(115,\) none prime. For \(p=13,\) the complement is \(113,\) which is prime. Thus the largest difference is \[ 113-13=100. \]

Therefore, the correct answer is B.

4 答案 D

将公共斜边取为从 \(A=(0,0)\) 到 \(B=(12,0)\)。两个三角形的第三个顶点分别为 \((9,3\sqrt3)\) 和 \((3,3\sqrt3)\)。它们的公共区域是以 \(AB\) 为底的三角形，其顶点为 \(y=\frac{x}{\sqrt3}\) 与 \(y=\frac{12-x}{\sqrt3}\) 的交点。交点横坐标为 \(x=6\)，高度为 \(2\sqrt3\)。因此公共面积为 \[ \frac12(12)(2\sqrt3)=12\sqrt3\text{。}\]

所以正确答案是 D。

English

Put the common hypotenuse from \(A=(0,0)\) to \(B=(12,0).\) The two triangles have their third vertices at \((9,3\sqrt3)\) and \((3,3\sqrt3).\) Their common region is the triangle with base \(AB\) and apex where \(y=\frac{x}{\sqrt3}\) and \(y=\frac{12-x}{\sqrt3}\) meet. This occurs at \(x=6,\) with height \(2\sqrt3.\) Hence the common area is \[ \frac12(12)(2\sqrt3)=12\sqrt3. \]

Therefore, the correct answer is D.

5 答案 D

令 \(a*b=\frac{a+b}{2}\)。交换律显然成立。此外， \[ \begin{aligned} a+(b*c) &=\frac{2a+b+c}{2}\\ &=(a+b)*(a+c) \end{aligned}\text{。}\]所以加法对取平均数满足分配律。

结合律不成立，因为 \[ \begin{aligned} (a*b)*c&=\frac{a+b+2c}{4},\\ a*(b*c)&=\frac{2a+b+c}{4} \end{aligned}\text{。}\]取平均数对加法不满足分配律，因为一般有 \[ a*(b+c)\ne(a*b)+(a*c)\text{。}\]最后，若单位元为 \(e\)，则对每个 \(a\) 都必须有 \(\frac{e+a}{2}=a\)，即 \(e=a\)，这不可能由一个固定的 \(e\) 满足。因此只有 \(\mathrm{II}\) 和 \(\mathrm{IV}\) 恒成立。

所以正确答案是 D。

English

Let \(a*b=\frac{a+b}{2}.\) Commutativity is immediate. Also \[ \begin{aligned} a+(b*c) &=\frac{2a+b+c}{2}\\ &=(a+b)*(a+c). \end{aligned} \] so addition distributes over averaging.

Associativity fails because \[ \begin{aligned} (a*b)*c&=\frac{a+b+2c}{4},\\ a*(b*c)&=\frac{2a+b+c}{4}. \end{aligned} \] Averaging does not distribute over addition since generally \[ a*(b+c)\ne(a*b)+(a*c). \] Finally, an identity \(e\) would require \(\frac{e+a}{2}=a\) for every \(a,\) or \(e=a,\) which is impossible for a fixed \(e.\) Thus only \(\mathrm{II}\) and \(\mathrm{IV}\) always hold.

Therefore, the correct answer is D.

6 答案 C

在十进制中， \[ \begin{aligned} 24_b&=2b+4,\\ 554_b&=5b^2+5b+4 \end{aligned}\text{。}\]因此 \[ \begin{aligned} (2b+4)^2&=5b^2+5b+4,\\ b^2-11b-12&=0,\\ (b-12)(b+1)&=0 \end{aligned}\text{。}\]底数为正且必须大于 \(5\)，所以 \(b=12\)。

所以正确答案是 C。

English

In base ten, \[ \begin{aligned} 24_b&=2b+4,\\ 554_b&=5b^2+5b+4. \end{aligned} \] Therefore \[ \begin{aligned} (2b+4)^2&=5b^2+5b+4,\\ b^2-11b-12&=0,\\ (b-12)(b+1)&=0. \end{aligned} \] A base is positive and must exceed \(5,\) so \(b=12.\)

Therefore, the correct answer is C.

7 答案 A

该数列为 \[ 51,61,\ldots,341\text{。}\]它共有 \[ \frac{341-51}{10}+1=30 \] 项。因此它的和为 \[ \begin{aligned} \frac{30(51+341)}2 &=15\cdot392\\ &=5880 \end{aligned}\text{。}\]

所以正确答案是 A。

English

The sequence is \[ 51,61,\ldots,341. \] It has \[ \frac{341-51}{10}+1=30 \] terms. Its sum is therefore \[ \begin{aligned} \frac{30(51+341)}2 &=15\cdot392\\ &=5880. \end{aligned} \]

Therefore, the correct answer is A.

8 答案 E

一座 \(1\) 英尺高的雕像相对于原雕像的线性比例为 \(\frac{1}{6}\)，所以它的表面积与油漆需求都是原来的 \(\frac{1}{36}\)。因此 \(540\) 座小雕像需要 \[ 540\cdot\frac1{36}=15 \] 品脱油漆。

所以正确答案是 E。

English

A \(1\)-ft. statue has linear scale \(\frac{1}{6}\) relative to the original, so its surface area and paint requirement are \(\frac{1}{36}\) as large. The \(540\) small statues therefore require \[ 540\cdot\frac1{36}=15 \] pints.

Therefore, the correct answer is E.

9 答案 E

由于 \(CH\perp AB\)，\(\triangle CHM\) 与 \(\triangle CHB\) 都在 \(H\) 处为直角。三等分使它们在 \(C\) 处的锐角相等，并且两三角形共用边 \(CH\)，所以它们全等。因此 \(HM=HB\)。

因为 \(M\) 是 \(AB\) 的中点，所以 \(AM=MB\)。沿斜边各点的顺序为 \(A,M,H,B\)，并且 \(MB=MH+HB=2MH\)。因此 \[ AB=2MB=4MH\text{。}\]三角形 \(ABC\) 与 \(CHM\) 的底边在同一直线上，并且从 \(C\) 到这条直线的高相同，所以面积比为 \(AB:HM=4:1\)。因此 \(\triangle ABC\) 的面积为 \(4K\)。

所以正确答案是 E。

English

Since \(CH\perp AB,\) both \(\triangle CHM\) and \(\triangle CHB\) are right at \(H.\) The trisection gives equal acute angles at \(C,\) and the triangles share side \(CH,\) so they are congruent. Hence \(HM=HB.\)

Because \(M\) is the midpoint of \(AB,\) \(AM=MB.\) Along the hypotenuse the order is \(A,M,H,B,\) and \(MB=MH+HB=2MH.\) Thus \[ AB=2MB=4MH. \] Triangles \(ABC\) and \(CHM\) have bases on the same line and the same altitude from \(C,\) so their areas are in the ratio \(AB:HM=4:1.\) Therefore the area of \(\triangle ABC\) is \(4K.\)

Therefore, the correct answer is E.

10 答案 A

将三个方程相加，得到 \[ (n+1)(x+y+z)=3\text{。}\]当 \(n=-1\) 时，这变为 \(0=3\)，所以无解。若 \(n\ne-1\)，对称取值 \[ x=y=z=\frac1{n+1} \] 满足全部三个方程。因此方程组恰在 \(n=-1\) 时无解。

所以正确答案是 A。

English

Adding the equations gives \[ (n+1)(x+y+z)=3. \] When \(n=-1,\) this becomes \(0=3,\) so no solution exists. If \(n\ne-1,\) the symmetric assignment \[ x=y=z=\frac1{n+1} \] satisfies all three equations. Hence the system has no solution exactly when \(n=-1.\)

Therefore, the correct answer is A.

11 答案 B

对固定的正数 \(a\)，集合 \[ 2\operatorname{Max}(|x|,|y|)\le a \] 是边与坐标轴平行的正方形。集合 \[ \sqrt{2(x^2+y^2)}\le a \] 是它的外接圆，而 \[ |x|+|y|\le a \] 是该圆的外接菱形。因此所需的嵌套顺序是内部的轴向正方形、圆、外部菱形，对应图 \(\mathrm{II}\)。

所以正确答案是 B。

English

For a fixed positive \(a,\) the set \[ 2\operatorname{Max}(|x|,|y|)\le a \] is an axis-aligned square. The set \[ \sqrt{2(x^2+y^2)}\le a \] is its circumscribed circle, and \[ |x|+|y|\le a \] is the diamond circumscribed about that circle. Thus the required nesting is an inner axis-aligned square, then a circle, then an outer diamond, which is figure \(\mathrm{II}.\)

Therefore, the correct answer is B.

12 答案 D

若有 \(d\) 名医生和 \(\ell\) 名律师，则 \[ 35d+50\ell=40(d+\ell)\text{。}\]因此 \(5d=10\ell\)，所以 \[ d:\ell=2:1\text{。}\]

所以正确答案是 D。

English

If there are \(d\) doctors and \(\ell\) lawyers, then \[ 35d+50\ell=40(d+\ell). \] Thus \(5d=10\ell,\) so \[ d:\ell=2:1. \]

Therefore, the correct answer is D.

13 答案 D

该分数为正，其平方为 \[ \begin{aligned} &\frac{4(\sqrt2+\sqrt6)^2} {9(2+\sqrt3)}\\ &\quad=\frac{4(8+4\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16(2+\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16}{9} \end{aligned}\text{。}\]因此原分数为 \(\frac{4}{3}\)。

所以正确答案是 D。

English

The fraction is positive, and its square is \[ \begin{aligned} &\frac{4(\sqrt2+\sqrt6)^2} {9(2+\sqrt3)}\\ &\quad=\frac{4(8+4\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16(2+\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16}{9}. \end{aligned} \] Therefore the original fraction is \(\frac{4}{3}.\)

Therefore, the correct answer is D.

14 答案 C

设 \(a\)、\(b\)、\(c\) 为三个阀门每小时注满的水箱数。则 \[ \begin{aligned} a+b+c&=1,\\ a+c&=\frac23,\\ b+c&=\frac12 \end{aligned}\text{。}\]第一式的两倍减去另外两式，得到 \[ a+b=2-\frac23-\frac12=\frac56\text{。}\]因此阀门 \(A\) 与 \(B\) 注满水箱需要 \[ \frac1{a+b}=\frac65=1.2 \] 小时。

所以正确答案是 C。

English

Let \(a,\) \(b,\) \(c\) be the hourly rates in tankfuls. Then \[ \begin{aligned} a+b+c&=1,\\ a+c&=\frac23,\\ b+c&=\frac12. \end{aligned} \] Twice the first equation minus the other two gives \[ a+b=2-\frac23-\frac12=\frac56. \] Thus valves \(A\) and \(B\) fill the tank in \[ \frac1{a+b}=\frac65=1.2 \] hours.

Therefore, the correct answer is C.

15 答案 D

两条半径与扇形的弦组成一个腰长为 \(6\)、顶角为 \(\theta\) 的等腰三角形。其底边长为 \(12\sin(\frac{\theta}{2})\)。若 \(R\) 为该三角形的外接圆半径，则扩展正弦定理给出 \[ 2R=\frac{12\sin(\frac{\theta}{2})}{\sin\theta}\text{。}\]由于 \(\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\)， \[ R=3\sec\frac{\theta}{2}\text{。}\]

所以正确答案是 D。

English

The two radii and the sector’s chord form an isosceles triangle with equal sides \(6\) and vertex angle \(\theta.\) Its base has length \(12\sin(\frac{\theta}{2}).\) If \(R\) is the triangle’s circumradius, the extended sine rule gives \[ 2R=\frac{12\sin(\frac{\theta}{2})}{\sin\theta}. \] Since \(\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}),\) \[ R=3\sec\frac{\theta}{2}. \]

Therefore, the correct answer is D.

16 答案 B

若省略的角为 \(x^\circ\)，则由凸性可得 \(0\lt x\lt180\)。因此 \[ 180(n-2)=2190+x\text{，}\]所以 \[ \frac{2190}{180} \lt n-2 \lt\frac{2370}{180}\text{。}\]即 \[ 12\frac16\lt n-2\lt13\frac16\text{。}\]唯一可能的整数是 \(n-2=13\)，所以 \(n=15\)。

所以正确答案是 B。

English

If the omitted angle is \(x^\circ,\) then convexity gives \(0\lt x\lt180.\) Hence \[ 180(n-2)=2190+x, \] so \[ \frac{2190}{180} \lt n-2 \lt\frac{2370}{180}. \] This is \[ 12\frac16\lt n-2\lt13\frac16. \] The only possible integer is \(n-2=13,\) so \(n=15.\)

Therefore, the correct answer is B.

17 答案 E

半角恒等式给出 \[ \begin{aligned} \cos\theta &=1-2\sin^2\frac{\theta}{2}\\ &=1-\frac{x-1}{x}\\ &=\frac1x \end{aligned}\text{。}\]因此 \[ \tan^2\theta=\sec^2\theta-1=x^2-1\text{。}\]因为 \(\theta\) 为锐角，其正切值为正，所以 \[ \tan\theta=\sqrt{x^2-1}\text{。}\]

所以正确答案是 E。

English

The half-angle identity gives \[ \begin{aligned} \cos\theta &=1-2\sin^2\frac{\theta}{2}\\ &=1-\frac{x-1}{x}\\ &=\frac1x. \end{aligned} \] Therefore \[ \tan^2\theta=\sec^2\theta-1=x^2-1. \] Because \(\theta\) is acute, its tangent is positive, so \[ \tan\theta=\sqrt{x^2-1}. \]

Therefore, the correct answer is E.

18 答案 C

质数 \(p\ge5\) 为奇数，所以 \(p-1\) 与 \(p+1\) 是两个连续偶数。其中一个能被 \(4\) 整除，因此它们的乘积能被 \(8\) 整除。在三个连续整数 \(p-1\)、\(p\)、\(p+1\) 中，有一个能被 \(3\) 整除。它不可能是 \(p\)，因为 \(p\ge5\) 是质数，所以 \(3\) 也整除 \((p-1)(p+1)\)。由于 \(3\) 与 \(8\) 互质， \[ (p-1)(p+1)=p^2-1 \] 恒能被 \(24\) 整除。

所以正确答案是 C。

English

A prime \(p\ge5\) is odd, so \(p-1\) and \(p+1\) are consecutive even integers. One is divisible by \(4,\) making their product divisible by \(8.\) Among the three consecutive integers \(p-1,\) \(p,\) \(p+1,\) one is divisible by \(3.\) It cannot be \(p,\) since \(p\ge5\) is prime, so \(3\) also divides \((p-1)(p+1).\) Because \(3\) and \(8\) are relatively prime, \[ (p-1)(p+1)=p^2-1 \] is always divisible by \(24.\)

Therefore, the correct answer is C.

19 答案 D

两个乘积为 \[ \begin{aligned} 72_8!&=72\cdot64\cdots8 =8^9(9!),\\ 18_2!&=18\cdot16\cdots2 =2^9(9!) \end{aligned}\text{。}\]因此它们的商为 \[ \frac{8^9(9!)}{2^9(9!)}=4^9\text{。}\]

所以正确答案是 D。

English

The two products are \[ \begin{aligned} 72_8!&=72\cdot64\cdots8 =8^9(9!),\\ 18_2!&=18\cdot16\cdots2 =2^9(9!). \end{aligned} \] Their quotient is therefore \[ \frac{8^9(9!)}{2^9(9!)}=4^9. \]

Therefore, the correct answer is D.

20 答案 C

将河流取为 \(x\) 轴，并把牛仔置于 \(C=(0,-4)\)。则小屋位于 \(H=(8,-11)\)。将 \(C\) 关于河流反射到 \(D=(0,4)\)。对河流上的任意点 \(S\)，有 \(CS=DS\)，所以使 \(CS+SH\) 最小等价于使 \(DS+SH\) 最小。当 \(D\)、\(S\)、\(H\) 共线时取得最小值。最短路程为 \[ DH=\sqrt{8^2+15^2}=\sqrt{289}=17\text{。}\]

所以正确答案是 C。

English

Take the stream as the \(x\)-axis and put the cowboy at \(C=(0,-4).\) His cabin is then \(H=(8,-11).\) Reflect \(C\) across the stream to \(D=(0,4).\) For any point \(S\) on the stream, \(CS=DS,\) so minimizing \(CS+SH\) is the same as minimizing \(DS+SH.\) This occurs when \(D,\) \(S,\) \(H\) are collinear. The minimum distance is \[ DH=\sqrt{8^2+15^2}=\sqrt{289}=17. \]

Therefore, the correct answer is C.

21 答案 B

对从 \(a\) 开始的 \(k\ge2\) 个连续正整数， \[ 200=k(2a+k-1)\text{。}\]由正性可得 \(k\lt\sqrt{200}\)，所以 \(200\) 的可能约数 \(k\) 为 \[ 2,4,5,8,10\text{。}\]由 \[ 2a=\frac{200}{k}-k+1\text{，}\]只有 \(k=5\) 与 \(k=8\) 使右边为正偶数。它们分别给出集合 \(18,19,20,21,22\) 和 \(9,10,\ldots,16\)。因此共有 \(2\) 个集合。

所以正确答案是 B。

English

For \(k\ge2\) consecutive positive integers starting at \(a,\) \[ 200=k(2a+k-1). \] Positivity gives \(k\lt\sqrt{200},\) so the possible divisors \(k\) of \(200\) are \[ 2,4,5,8,10. \] From \[ 2a=\frac{200}{k}-k+1, \] only \(k=5\) and \(k=8\) give positive even right-hand sides. They yield the sets \(18,19,20,21,22\) and \(9,10,\ldots,16.\) Hence there are \(2\) sets.

Therefore, the correct answer is B.

22 答案 A

当 \(-2\le x\le1\) 时，\(x\) 到 \(-2\) 与 \(1\) 的距离之和为 \(3\)。若 \(x\) 位于该区间外，且到区间的距离为 \(u\)，则距离和为 \(3+2u\)。因此 \[ 3+2u\lt5 \quad\Longleftrightarrow\quad u\lt1\text{。}\]将区间 \([-2,1]\) 的两端各延伸 \(1\)，得到 \[ -3\lt x\lt2\text{。}\]

所以正确答案是 A。

English

For \(-2\le x\le1,\) the sum of the distances from \(x\) to \(-2\) and \(1\) is \(3.\) If \(x\) lies a distance \(u\) outside this interval, the sum is \(3+2u.\) Thus \[ 3+2u\lt5 \quad\Longleftrightarrow\quad u\lt1. \] Extending the interval \([-2,1]\) by \(1\) at each end gives \[ -3\lt x\lt2. \]

Therefore, the correct answer is A.

23 答案 D

在朝上一面为红色的结果中，可能朝上的是双红卡的两个红面之一，或红蓝卡的红面。这三个可见的红面等可能出现。前两种情况下背面为红色，第三种情况下背面为蓝色，所以条件概率为 \[ \frac23\text{。}\]

所以正确答案是 D。

English

Among outcomes having a red upper side, either of the two red faces of the red-red card or the red face of the red-blue card can be uppermost. These three visible red faces are equally likely. The underside is red in the first two cases and blue in the third, so the conditional probability is \[ \frac23. \]

Therefore, the correct answer is D.

24 答案 D

设一份三明治、一杯咖啡和一块派的价格分别为 \(s\)、\(c\)、\(p\)。两张账单给出 \[ \begin{aligned} 3s+7c+p&=3.15,\\ 4s+10c+p&=4.20 \end{aligned}\text{。}\]第一式的三倍减去第二式的两倍，得到 \[ \begin{aligned} s+c+p &=3(3.15)-2(4.20)\\ &=1.05 \end{aligned}\text{。}\]所求午餐费用为 \(\$1.05\)。

所以正确答案是 D。

English

Let \(s,\) \(c,\) \(p\) be the prices of a sandwich, coffee, and pie. The checks give \[ \begin{aligned} 3s+7c+p&=3.15,\\ 4s+10c+p&=4.20. \end{aligned} \] Three times the first equation minus twice the second gives \[ \begin{aligned} s+c+p &=3(3.15)-2(4.20)\\ &=1.05. \end{aligned} \] The requested luncheon costs \(\$1.05.\)

Therefore, the correct answer is D.

25 答案 E

草地半径为 \(6\)。在小路的另一条边上，圆心到该边的垂直距离为 \(3\)，所以通向交点的半径与经过圆心的路边成 \(30^\circ\) 角。小路的一半由半径为 \(6\) 的 \(30^\circ\) 扇形和直角边长为 \(3\) 与 \(3\sqrt3\) 的直角三角形组成。其面积为 \[ \begin{aligned} &\frac{30}{360}\pi(6^2) +\frac12(3)(3\sqrt3)\\ &\qquad=3\pi+\frac{9\sqrt3}{2} \end{aligned}\text{。}\]因此整条小路的面积为 \(6\pi+9\sqrt3\)，剩余草地面积为 \[ 36\pi-(6\pi+9\sqrt3) =30\pi-9\sqrt3\text{。}\]

所以正确答案是 E。

English

The plot has radius \(6.\) At the path’s other edge, the perpendicular distance from the center is \(3,\) so the radius to an intersection point makes a \(30^\circ\) angle with the edge through the center. Half of the path consists of a \(30^\circ\) sector of radius \(6\) and a right triangle with legs \(3\) and \(3\sqrt3.\) Its area is \[ \begin{aligned} &\frac{30}{360}\pi(6^2) +\frac12(3)(3\sqrt3)\\ &\qquad=3\pi+\frac{9\sqrt3}{2}. \end{aligned} \] Thus the whole path has area \(6\pi+9\sqrt3,\) and the remaining grass area is \[ 36\pi-(6\pi+9\sqrt3) =30\pi-9\sqrt3. \]

Therefore, the correct answer is E.

26 答案 E

设数列有 \(2n\) 项，公差为 \(d\)。将每个奇数编号项与后一项配对，可得 \[ nd=30-24=6\text{。}\]末项与首项之差为 \[ (2n-1)d=10.5\text{。}\]因为 \(2nd=12\)，相减得到 \(d=1.5\)。于是 \(n=\frac{6}{1.5}=4\)，该数列共有 \(2n=8\) 项。

所以正确答案是 E。

English

Let the progression have \(2n\) terms and common difference \(d.\) Pairing each odd-numbered term with its successor shows that \[ nd=30-24=6. \] The difference between the last and first terms is \[ (2n-1)d=10.5. \] Since \(2nd=12,\) subtraction gives \(d=1.5.\) Hence \(n=\frac{6}{1.5}=4,\) and the progression has \(2n=8\) terms.

Therefore, the correct answer is E.

27 答案 A

汽车 \(A\) 的等距离平均速度与汽车 \(B\) 的等时间平均速度分别为 \[ x=\frac{2uv}{u+v}, \qquad y=\frac{u+v}{2}\text{。}\]两者之差为 \[ \begin{aligned} y-x &=\frac{(u+v)^2-4uv}{2(u+v)}\\ &=\frac{(u-v)^2}{2(u+v)}\\ &\ge0 \end{aligned}\text{。}\]因此 \(x\le y\)，当 \(u=v\) 时可取等号。

所以正确答案是 A。

English

Car \(A\)’s equal-distance average and Car \(B\)’s equal-time average are \[ x=\frac{2uv}{u+v}, \qquad y=\frac{u+v}{2}. \] Their difference is \[ \begin{aligned} y-x &=\frac{(u+v)^2-4uv}{2(u+v)}\\ &=\frac{(u-v)^2}{2(u+v)}\\ &\ge0. \end{aligned} \] Thus \(x\le y,\) with equality possible when \(u=v.\)

Therefore, the correct answer is A.

28 答案 C

由换底公式， \[ \begin{aligned} \frac1{\log_a n}&=\log_n a,\\ \frac1{\log_b n}&=\log_n b,\\ \frac1{\log_c n}&=\log_n c \end{aligned}\text{。}\]因为 \(a\)、\(b\)、\(c\) 成等比数列，所以 \(b^2=ac\)。两边取以 \(n\) 为底的对数，得到 \[ 2\log_n b=\log_n a+\log_n c\text{。}\]因此所给三项的倒数组成等差数列。

所以正确答案是 C。

English

By change of base, \[ \begin{aligned} \frac1{\log_a n}&=\log_n a,\\ \frac1{\log_b n}&=\log_n b,\\ \frac1{\log_c n}&=\log_n c. \end{aligned} \] Since \(a,\) \(b,\) \(c\) are in geometric progression, \(b^2=ac.\) Taking logarithms to base \(n\) gives \[ 2\log_n b=\log_n a+\log_n c. \] Hence the reciprocals of the three given terms form an arithmetic progression.

Therefore, the correct answer is C.

29 答案 A

设跑道长度为 \(L\)。由于 \(\gcd(5,9)=1\)，两人首次同时回到 \(A\) 的正时刻为 \(t=L\)：他们分别完成了 \(5\) 圈与 \(9\) 圈。相对速度为 \(5+9=14\)，所以在时刻 \(L\) 之前，他们在 \[ t=\frac{kL}{14}, \qquad k=1,2,\ldots,13 \] 时相遇。因此除去起点与终点，共相遇 \(13\) 次。

所以正确答案是 A。

English

Let the track length be \(L.\) Since \(\gcd(5,9)=1,\) the first positive time when both boys are back at \(A\) is \(t=L\): they have completed \(5\) and \(9\) laps. Their relative speed is \(5+9=14,\) so before time \(L\) they meet at \[ t=\frac{kL}{14}, \qquad k=1,2,\ldots,13. \] Thus there are \(13\) meetings excluding the start and finish.

Therefore, the correct answer is A.

30 答案 B

小数部分满足 \(0\le T\lt1\)。集合 \(S\) 是以 \((T,0)\) 为圆心、\(T\) 为半径的闭圆盘。其面积为 \[ \pi T^2\text{，}\]所以 \(0\le\operatorname{Area}S\lt\pi\)，从而特别有 \(0\le\operatorname{Area}S\le\pi\)。原点位于每个这样的圆盘上；当 \(T\gt0\) 时，圆盘延伸到 \(x\) 轴下方；而它的圆心一般不在 \(y=x\) 上。

所以正确答案是 B。

English

The fractional part satisfies \(0\le T\lt1.\) The set \(S\) is the closed disk centered at \((T,0)\) with radius \(T.\) Its area is \[ \pi T^2, \] so \(0\le\operatorname{Area}S\lt\pi,\) which in particular gives \(0\le\operatorname{Area}S\le\pi.\) The origin lies on every such disk, the disk extends below the \(x\)-axis when \(T\gt0,\) and its center is generally not on \(y=x.\)

Therefore, the correct answer is B.

31 答案 C

由于 \[ TTT=111T=3\cdot37\cdot T\text{，}\]质数 \(37\) 整除 \(YE\) 与 \(ME\) 中的一个。\(37\) 的两位数倍数为 \(37\) 或 \(74\)。\(74\) 不可能：另一个末位为 \(4\) 的两位数因子至少是 \(14\)，而 \(74\cdot14\gt999\)。因此一个因子为 \(37\)，所以 \(E=7\)。

乘积的个位数字等于 \(7^2\) 的个位数字，所以 \(T=9\)。因此乘积为 \(999\)，另一个因子为 \[ \frac{999}{37}=27\text{。}\]四个数字为 \(2\)、\(3\)、\(7\)、\(9\)，它们的和为 \(21\)。

所以正确答案是 C。

English

Since \[ TTT=111T=3\cdot37\cdot T, \] the prime \(37\) divides one of \(YE\) and \(ME.\) A two-digit multiple of \(37\) is \(37\) or \(74.\) The value \(74\) is impossible: the other two-digit factor ending in \(4\) is at least \(14,\) and \(74\cdot14\gt999.\) Hence one factor is \(37,\) so \(E=7.\)

The units digit of the product is the units digit of \(7^2,\) so \(T=9.\) Thus the product is \(999,\) and the other factor is \[ \frac{999}{37}=27. \] The four digits are \(2,\) \(3,\) \(7,\) \(9,\) whose sum is \(21.\)

Therefore, the correct answer is C.

32 答案 A

底面积为 \[ \frac{\sqrt3}{4}(6^2)=9\sqrt3\text{。}\]因为顶点到三个底面顶点的距离相等，所以其垂直投影是底面的外心。等边三角形底面的外接圆半径为 \(\frac{6}{\sqrt3}=2\sqrt3\)。若 \(h\) 为棱锥的高，则 \[ h^2+(2\sqrt3)^2=(\sqrt{15})^2\text{，}\]所以 \(h=\sqrt3\)。体积为 \[ \frac13(9\sqrt3)(\sqrt3)=9\text{。}\]

所以正确答案是 A。

English

The base area is \[ \frac{\sqrt3}{4}(6^2)=9\sqrt3. \] Because the apex is equally distant from all three base vertices, its perpendicular projection is the base circumcenter. The circumradius of the equilateral base is \(\frac{6}{\sqrt3}=2\sqrt3.\) If \(h\) is the pyramid’s height, then \[ h^2+(2\sqrt3)^2=(\sqrt{15})^2, \] so \(h=\sqrt3.\) The volume is \[ \frac13(9\sqrt3)(\sqrt3)=9. \]

Therefore, the correct answer is A.

33 答案 C

设原混合物含 \(x\) 盎司水和 \(y\) 盎司酸。两次添加给出 \[ \begin{aligned} \frac{y}{x+y+1}&=\frac15,\\ \frac{y+1}{x+y+2}&=\frac13 \end{aligned}\text{。}\]化简得 \[ x+1=4y, \qquad x=2y+1\text{。}\]因此 \(y=1\)，\(x=3\)。原混合物中酸的百分比为 \[ 100\cdot\frac{y}{x+y} =100\cdot\frac14=25\%\text{。}\]

所以正确答案是 C。

English

Let the original mixture contain \(x\) ounces of water and \(y\) ounces of acid. The two additions give \[ \begin{aligned} \frac{y}{x+y+1}&=\frac15,\\ \frac{y+1}{x+y+2}&=\frac13. \end{aligned} \] These simplify to \[ x+1=4y, \qquad x=2y+1. \] Hence \(y=1\) and \(x=3.\) The original acid percentage was \[ 100\cdot\frac{y}{x+y} =100\cdot\frac14=25\%. \]

Therefore, the correct answer is C.

34 答案 C

设两城距离为 \(d\)，返程时间为 \(x\) 分钟。逆风与顺风速度分别为 \(\frac{d}{84}\) 与 \(\frac{d}{x}\)。它们的平均值是无风速度 \(\frac{d}{x+9}\)。因此 \[ \frac{2}{x+9}=\frac1{84}+\frac1x\text{。}\]清除分母，得到 \[ \begin{gathered} 168x=(x+9)(x+84),\\ x^2-75x+756=0,\\ (x-63)(x-12)=0 \end{gathered}\text{。}\]两个正值都符合题设条件，所以两个返程时间为 \(63\) 分钟与 \(12\) 分钟。

所以正确答案是 C。

English

Let the distance be \(d\) and the return time be \(x\) minutes. The against-wind and with-wind speeds are \(\frac{d}{84}\) and \(\frac{d}{x}.\) Their average is the still-air speed \(\frac{d}{x+9}.\) Therefore \[ \frac{2}{x+9}=\frac1{84}+\frac1x. \] Clearing denominators gives \[ \begin{gathered} 168x=(x+9)(x+84),\\ x^2-75x+756=0,\\ (x-63)(x-12)=0. \end{gathered} \] Both positive values are consistent with the stated conditions, so the two return times are \(63\) and \(12\) minutes.

Therefore, the correct answer is C.

35 答案 E

设 \(K\) 为水平直径的左端点。关于竖直直径的反射表明 \(KM=NR=s\)，且 \(QN=MP=s\)。结合题目给出的相等弦，上半圆被分成五段相等的弧。每段弧在 \(O\) 处所对的圆心角为 \(36^\circ\)。因此 \[ \begin{aligned} s&=2\sin18^\circ,\\ d&=2\sin54^\circ\\ &=2\cos36^\circ \end{aligned}\text{。}\]

利用倍角恒等式， \[ \begin{aligned} d&=2(1-2\sin^218^\circ)\\ &=2-s^2,\\ s&=2\cos72^\circ\\ &=4\cos^236^\circ-2\\ &=d^2-2 \end{aligned}\text{。}\]两式相加，得到 \[ d+s=d^2-s^2=(d-s)(d+s)\text{。}\]因为 \(d+s\gt0\)，所以 \(d-s=1\)。将 \(d=s+1\) 代入 \(d=2-s^2\)，得到 \[ s^2+s=1, \qquad s=\frac{\sqrt5-1}{2}\text{。}\]因此 \[ ds=s(s+1)=1 \] 且 \[ \begin{aligned} d^2-s^2 &=(d-s)(d+s)\\ &=2s+1\\ &=\sqrt5 \end{aligned}\text{。}\]三个等式都必然成立。

所以正确答案是 E。

English

Let \(K\) be the left endpoint of the horizontal diameter. Reflection across the vertical diameter shows that \(KM=NR=s\) and \(QN=MP=s.\) Together with the given equal chords, the upper semicircle is split into five equal arcs. Each subtends \(36^\circ\) at \(O.\) Thus \[ \begin{aligned} s&=2\sin18^\circ,\\ d&=2\sin54^\circ\\ &=2\cos36^\circ. \end{aligned} \]

Using the double-angle identities, \[ \begin{aligned} d&=2(1-2\sin^218^\circ)\\ &=2-s^2,\\ s&=2\cos72^\circ\\ &=4\cos^236^\circ-2\\ &=d^2-2. \end{aligned} \] Adding these equations gives \[ d+s=d^2-s^2=(d-s)(d+s). \] Since \(d+s\gt0,\) it follows that \(d-s=1.\) Substituting \(d=s+1\) into \(d=2-s^2\) yields \[ s^2+s=1, \qquad s=\frac{\sqrt5-1}{2}. \] Therefore \[ ds=s(s+1)=1 \] and \[ \begin{aligned} d^2-s^2 &=(d-s)(d+s)\\ &=2s+1\\ &=\sqrt5. \end{aligned} \] All three equations are necessarily true.

Therefore, the correct answer is E.