1957 AMC 12 真题（P版）

1 答案 B

从顶角顶点引出的高、中线和角平分线是同一条直线。在两个底角顶点中的每一个处，这三条直线互不相同。因此共有 \[ 1+3+3=7 \] 条不同的直线。

因此，正确答案是 B。

English

The altitude, median, and angle bisector from the apex are the same line. At each of the two base vertices, those three lines are distinct. Thus there are \[ 1+3+3=7 \] distinct lines.

Therefore, the correct answer is B.

2 答案 E

由韦达定理，根的和与积满足 \[ \begin{aligned} \frac h2&=4,\\ \frac{2k}{2}&=k=-3 \end{aligned}\text{。}\]因此 \(h=8\)，且 \(k=-3\)。

因此，正确答案是 E。

English

By Vieta’s formulas, the sum and product of the roots are \[ \begin{aligned} \frac h2&=4,\\ \frac{2k}{2}&=k=-3. \end{aligned} \] Hence \(h=8\) and \(k=-3.\)

Therefore, the correct answer is E.

3 答案 E

原式要求 \(a\ne1\)。对于这样的 \(a\)， \[ \begin{aligned} 1+\frac{a}{1-a} &=\frac{1-a+a}{1-a}\\ &=\frac1{1-a} \end{aligned}\text{。}\]因此原式等于 \(1-(1-a)=a\)。限制条件 \(a\ne1\) 仍须保留。

因此，正确答案是 E。

English

The original expression requires \(a\ne1.\) For such \(a,\) \[ \begin{aligned} 1+\frac{a}{1-a} &=\frac{1-a+a}{1-a}\\ &=\frac1{1-a}. \end{aligned} \] Therefore the expression is \(1-(1-a)=a.\) The restriction \(a\ne1\) remains.

Thus, the correct answer is E.

4 答案 C

严格按题中所给形式应用分配律，令 \(a=3x+2\)、\(b=x\)，且 \(c=-5\)。则 \[ \begin{gathered} (3x+2)(x-5)\\ =(3x+2)x+(3x+2)(-5) \end{gathered}\text{。}\]

因此，正确答案是 C。

English

Using the property in precisely the stated form, set \(a=3x+2,\) \(b=x,\) and \(c=-5.\) Then \[ \begin{gathered} (3x+2)(x-5)\\ =(3x+2)x+(3x+2)(-5). \end{gathered} \]

Thus, the correct answer is C.

5 答案 B

合并各对数，得到 \[ \begin{aligned} &\log\left(\frac ab\frac bc\frac cd\frac{dx}{ay}\right)\\ &\quad=\log\left(\frac ad\frac{dx}{ay}\right)\\ &\quad=\log\frac xy \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Combining the logarithms gives \[ \begin{aligned} &\log\left(\frac ab\frac bc\frac cd\frac{dx}{ay}\right)\\ &\quad=\log\left(\frac ad\frac{dx}{ay}\right)\\ &\quad=\log\frac xy. \end{aligned} \]

Therefore, the correct answer is B.

6 答案 A

盒高为 \(x\)，底面的两个边长分别为 \(10-2x\) 和 \(14-2x\)。因此 \[ \begin{aligned} V&=x(10-2x)(14-2x)\\ &=140x-48x^2+4x^3 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

The box has height \(x\) and base dimensions \(10-2x\) and \(14-2x.\) Hence \[ \begin{aligned} V&=x(10-2x)(14-2x)\\ &=140x-48x^2+4x^3. \end{aligned} \]

Thus, the correct answer is A.

7 答案 E

内切圆半径满足 \(r^2=48\)，所以 \(r=4\sqrt3\)。若边长为 \(s\)，则 \[ r=\frac{s\sqrt3}{6}\text{，}\]解得 \(s=24\)。周长为 \(3s=72\)。

因此，正确答案是 E。

English

The inradius satisfies \(r^2=48,\) so \(r=4\sqrt3.\) If \(s\) is the side length, then \[ r=\frac{s\sqrt3}{6}, \] which gives \(s=24.\) The perimeter is \(3s=72.\)

Thus, the correct answer is E.

8 答案 A

令 \(x=2t\)、\(y=3t\)，且 \(z=5t\)。因为 \(10t=100\)，所以 \(t=10\)，从而 \(x=20\) 且 \(y=30\)。因此 \[ 30=20a-10\text{，}\]解得 \(a=2\)。

因此，正确答案是 A。

English

Let \(x=2t,\) \(y=3t,\) and \(z=5t.\) Since \(10t=100,\) we get \(t=10,\) so \(x=20\) and \(y=30.\) Thus \[ 30=20a-10, \] and \(a=2.\)

Therefore, the correct answer is A.

9 答案 B

此时 \(x-y=2-(-2)=4\)。因此 \[ \begin{aligned} x-y^{x-y} &=2-(-2)^4\\ &=2-16=-14 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Here \(x-y=2-(-2)=4.\) Therefore \[ \begin{aligned} x-y^{x-y} &=2-(-2)^4\\ &=2-16=-14. \end{aligned} \]

Thus, the correct answer is B.

10 答案 C

配方得到 \[ y=2(x+1)^2+1\text{。}\]平方项非负，所以图像的最低点为 \((-1,1)\)。

因此，正确答案是 C。

English

Completing the square, \[ y=2(x+1)^2+1. \] The squared term is nonnegative, so the graph has its lowest point at \((-1,1).\)

Thus, the correct answer is C.

11 答案 E

分针从十二点方向顺时针转过 \(90^\circ\)。时针从十二点方向转过 \[ 2(30^\circ)+\frac14(30^\circ)=67.5^\circ\text{。}\]两针所成的较小角为 \(90^\circ-67.5^\circ=22.5^\circ\)，不在选项中。

因此，正确答案是 E。

English

The minute hand is \(90^\circ\) clockwise from twelve. The hour hand is \[ 2(30^\circ)+\frac14(30^\circ)=67.5^\circ \] from twelve. Their smaller angle is \(90^\circ-67.5^\circ=22.5^\circ,\) which is not listed.

Therefore, the correct answer is E.

12 答案 C

使用相同的十的幂，得到 \[ \begin{gathered} 10^{-49}-2\cdot10^{-50}\\ =(10-2)10^{-50}\\ =8\cdot10^{-50} \end{gathered}\text{。}\]

因此，正确答案是 C。

English

Using a common power of ten, \[ \begin{gathered} 10^{-49}-2\cdot10^{-50}\\ =(10-2)10^{-50}\\ =8\cdot10^{-50}. \end{gathered} \]

Thus, the correct answer is C.

13 答案 C

因为 \[ \begin{gathered} \sqrt2\approx1.414<1.5,\\ 1.5<1.732\approx\sqrt3 \end{gathered}\text{，}\]所以有理数 \(1.5\) 位于这两个根式之间。

因此，正确答案是 C。

English

Since \[ \begin{gathered} \sqrt2\approx1.414<1.5,\\ 1.5<1.732\approx\sqrt3, \end{gathered} \] the rational number \(1.5\) lies between the two radicals.

Therefore, the correct answer is C.

14 答案 D

两个被开方数分别是 \((x-1)^2\) 与 \((x+1)^2\)。因此 \[ \begin{aligned} y&=\sqrt{(x-1)^2}\\ &\quad+\sqrt{(x+1)^2}\\ &=|x-1|+|x+1| \end{aligned}\text{。}\]

因此，正确答案是 D。

English

The two radicands are \((x-1)^2\) and \((x+1)^2.\) Hence \[ \begin{aligned} y&=\sqrt{(x-1)^2}\\ &\quad+\sqrt{(x+1)^2}\\ &=|x-1|+|x+1|. \end{aligned} \]

Thus, the correct answer is D.

15 答案 B

表中数值满足 \(s=10t^2\)。因此，当 \(t=2.5\) 时，\[ s=10(2.5)^2=10(6.25)=62.5\text{。}\]

因此，正确答案是 B。

English

The entries follow \(s=10t^2.\) Thus, at \(t=2.5,\) \[ s=10(2.5)^2=10(6.25)=62.5. \]

Therefore, the correct answer is B.

16 答案 D

关系 \(c=15n\) 是线性的，但金鱼条数 \(n\) 只能是十二个整数 \(1,2,\ldots,12\) 中的一个。因此，图像由十二个互不相同的点组成，而不是整条直线或线段。

因此，正确答案是 D。

English

The rule \(c=15n\) is linear, but the number \(n\) of goldfish can only be one of the twelve integers \(1,2,\ldots,12.\) The graph therefore consists of twelve distinct points, not an entire line or segment.

Thus, the correct answer is D.

17 答案 A

立方体有八个顶点。一条路径可以依次访问全部八个顶点而不重复，共走过七条棱；随后从最后一个顶点沿一条棱回到起点，这次返回才是第一次重复到达顶点。因此，允许的最长路线共经过八条棱，长度为 \(8\cdot3=24\) 英寸。

因此，正确答案是 A。

English

The cube has eight vertices. A path can visit all eight once, using seven edges, and then traverse the edge from its last vertex back to its starting vertex; the return is the first repeated vertex. Thus the greatest permitted walk uses eight edges. Its length is \(8\cdot3=24\) inches.

Therefore, the correct answer is A.

18 答案 B

角 \(APO\) 与 \(ABM\) 都是直角，且两个三角形共有角 \(A\)。因此三角形 \(APO\) 与 \(ABM\) 相似。由对应边可得 \[ \frac{AP}{AB}=\frac{AO}{AM}\text{。}\]所以 \(AP\cdot AM=AO\cdot AB\)。

因此，正确答案是 B。

English

Angles \(APO\) and \(ABM\) are both right angles, and the two triangles share angle \(A.\) Therefore triangles \(APO\) and \(ABM\) are similar. Corresponding sides give \[ \frac{AP}{AB}=\frac{AO}{AM}. \] Hence \(AP\cdot AM=AO\cdot AB.\)

Thus, the correct answer is B.

19 答案 A

按二进制位值展开，得到 \[ \begin{aligned} (10011)_2 &=1\cdot2^4+1\cdot2^1+1\cdot2^0\\ &=16+2+1=19 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Expanding by binary place value, \[ \begin{aligned} (10011)_2 &=1\cdot2^4+1\cdot2^1+1\cdot2^0\\ &=16+2+1=19. \end{aligned} \]

Therefore, the correct answer is A.

20 答案 A

设每一程的距离为 \(d\)。平均速度（单位为英里每小时）为 \[ \begin{aligned} \frac{2d}{\frac{d}{50}+\frac{d}{45}} &=\frac{2}{\frac{1}{50}+\frac{1}{45}}\\ &=\frac{900}{19}\\ &=47\frac7{19} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Let each leg have distance \(d.\) The average speed, in miles per hour, is \[ \begin{aligned} \frac{2d}{\frac{d}{50}+\frac{d}{45}} &=\frac{2}{\frac{1}{50}+\frac{1}{45}}\\ &=\frac{900}{19}\\ &=47\frac7{19}. \end{aligned} \]

Thus, the correct answer is A.

21 答案 E

令 \(P\) 表示两个角相等，\(Q\) 表示三角形为等腰三角形。命题 \(3\) 是逆否命题 \(\neg Q\Rightarrow\neg P\)，所以它与 \(P\Rightarrow Q\) 等价。命题 \(4\) 说 \(Q\) 是 \(P\) 的必要条件，这只是 \(P\Rightarrow Q\) 的另一种表述。命题 \(1\) 与 \(2\) 分别是否命题和逆命题。

因此，只有命题 \(3\) 与 \(4\) 等价，正确答案是 E。

English

Let \(P\) mean that two angles are equal and \(Q\) that the triangle is isosceles. Statement \(3\) is the contrapositive \(\neg Q\Rightarrow\neg P,\) so it is equivalent to \(P\Rightarrow Q.\) Statement \(4\) says that \(Q\) is necessary for \(P,\) which is another wording of \(P\Rightarrow Q.\) Statements \(1\) and \(2\) are the inverse and converse.

Thus only statements \(3\) and \(4\) are equivalent, so the correct answer is E.

22 答案 A

移项并平方：\[ \begin{aligned} \sqrt{x+1}&=\sqrt{x-1}+1,\\ x+1&=x-1+2\sqrt{x-1}+1 \end{aligned}\text{。}\]因此 \(2\sqrt{x-1}=1\)，所以 \(x-1=\frac{1}{4}\)，且 \(x=\frac{5}{4}\)。这个值满足原方程。因此 \(4x=5\)。

因此，正确答案是 A。

English

Rearrange and square: \[ \begin{aligned} \sqrt{x+1}&=\sqrt{x-1}+1,\\ x+1&=x-1+2\sqrt{x-1}+1. \end{aligned} \] Thus \(2\sqrt{x-1}=1,\) so \(x-1=\frac{1}{4}\) and \(x=\frac{5}{4}.\) This value satisfies the original equation. Therefore \(4x=5.\)

Thus, the correct answer is A.

23 答案 C

在交点处，\[ 10-x^2=10-x\text{，}\]所以 \(x=0\) 或 \(x=1\)。两个交点为 \((0,10)\) 与 \((1,9)\)。它们之间的距离为 \[ \sqrt{(1-0)^2+(9-10)^2}=\sqrt2\text{。}\]

因此，正确答案是 C。

English

At an intersection, \[ 10-x^2=10-x, \] so \(x=0\) or \(x=1.\) The points are \((0,10)\) and \((1,9).\) Their distance is \[ \sqrt{(1-0)^2+(9-10)^2}=\sqrt2. \]

Therefore, the correct answer is C.

24 答案 B

若两个数字为 \(a,b\)，则 \[ \begin{aligned} &(10a+b)^2-(10b+a)^2\\ &\quad=[9(a-b)][11(a+b)]\\ &\quad=99(a-b)(a+b) \end{aligned}\text{。}\]这个结果总能被 \(9\)、\(11\)、两个数字的和以及两个数字的差整除，但不一定能被两个数字的积整除。例如，\(21^2-12^2=297\) 不能被 \(2\) 整除。

因此，正确答案是 B。

English

If the digits are \(a,b,\) then \[ \begin{aligned} &(10a+b)^2-(10b+a)^2\\ &\quad=[9(a-b)][11(a+b)]\\ &\quad=99(a-b)(a+b). \end{aligned} \] This is always divisible by \(9,\) \(11,\) the digit sum, and the digit difference. It need not be divisible by the digit product; for example, \(21^2-12^2=297\) is not divisible by \(2.\)

Thus, the correct answer is B.

25 答案 B

带符号的面积二倍为 \[ \begin{vmatrix} 0&a&1\\ b&0&1\\ c&d&1 \end{vmatrix} =ab-ac-bd\text{。}\]直线 \(PQ\) 的方程为 \(ax+by=ab\)。由于原点代入左边所得值小于 \(ab\)，而 \(R\) 位于另一侧，所以 \(ac+bd>ab\)。因此面积为 \[ \frac{ac+bd-ab}{2}\text{。}\]

因此，正确答案是 B。

English

The signed doubled area is \[ \begin{vmatrix} 0&a&1\\ b&0&1\\ c&d&1 \end{vmatrix} =ab-ac-bd. \] The line \(PQ\) has equation \(ax+by=ab.\) Since the origin gives a value below \(ab\) and \(R\) is on the opposite side, \(ac+bd>ab.\) Thus the area is \[ \frac{ac+bd-ab}{2}. \]

Therefore, the correct answer is B.

26 答案 E

三条中线交于重心，并把三角形分成六个等面积的小三角形。以重心和原三角形的一条边为顶点的三个三角形，各由其中两个小三角形组成，所以面积相等。反过来，三个面积相等会给出相等的重心坐标，从而唯一确定重心。

因此，所求的充要点是三条中线的交点，正确答案是 E。

English

The three medians meet at the centroid and divide the triangle into six small triangles of equal area. Each of the three triangles having the centroid and one side of the original triangle consists of two of those small triangles, so their areas are equal. Conversely, equal areas give equal barycentric coordinates, which uniquely locate the centroid.

Thus, the necessary and sufficient point is the intersection of the medians, and the correct answer is E.

27 答案 A

若两根为 \(r,s\)，则由韦达定理可得 \(r+s=-p\) 与 \(rs=q\)。因此 \[ \frac1r+\frac1s=\frac{r+s}{rs}=-\frac pq\text{。}\]

因此，正确答案是 A。

English

If the roots are \(r,s,\) then Vieta’s formulas give \(r+s=-p\) and \(rs=q.\) Therefore \[ \frac1r+\frac1s=\frac{r+s}{rs}=-\frac pq. \]

Thus, the correct answer is A.

28 答案 B

根据对数的定义，\[ b^{\log_b a}=a\text{。}\]因此，这个值取决于 \(a\)，而不取决于所选的合适底数 \(b\)。

因此，正确答案是 B。

English

By the definition of a logarithm, \[ b^{\log_b a}=a. \] Thus the value depends on \(a,\) not on the choice of admissible base \(b.\)

Therefore, the correct answer is B.

29 答案 D

当 \(x=0\) 时，乘积为零。对于 \(x\ne0\)，因子 \(x^2\) 为正，所以乘积非负当且仅当 \[ x^2-1\ge0\text{，}\]即 \(x\le-1\) 或 \(x\ge1\)。再加入孤立值 \(x=0\)，就得到选项 D 中的集合。

因此，正确答案是 D。

English

At \(x=0,\) the product is zero. For \(x\ne0,\) the factor \(x^2\) is positive, so the product is nonnegative exactly when \[ x^2-1\ge0, \] or \(x\le-1\) or \(x\ge1.\) Including the isolated value \(x=0\) gives the set in choice D.

Thus, the correct answer is D.

30 答案 D

标准公式为 \[ \begin{aligned} 1^2+2^2+\cdots+n^2 &=\frac{n(n+1)}6\\ &\quad\cdot(2n+1) \end{aligned}\text{。}\]因此 \(c=1\)，且 \(k=1\)。

因此，正确答案是 D。

English

The standard formula is \[ \begin{aligned} 1^2+2^2+\cdots+n^2 &=\frac{n(n+1)}6\\ &\quad\cdot(2n+1). \end{aligned} \] Therefore \(c=1\) and \(k=1.\)

Thus, the correct answer is D.

31 答案 B

设每个角上的小三角形直角边长为 \(x\)。八边形的水平边和竖直边长为 \(1-2x\)，而四条斜边长为 \(x\sqrt2\)。因此 \[ 1-2x=x\sqrt2\text{，}\]所以 \[ x=\frac1{2+\sqrt2}=\frac{2-\sqrt2}{2}\text{。}\]

因此，正确答案是 B。

English

Let \(x\) be a leg of each corner triangle. The horizontal and vertical octagon sides have length \(1-2x,\) while the four slanted sides have length \(x\sqrt2.\) Thus \[ 1-2x=x\sqrt2, \] so \[ x=\frac1{2+\sqrt2}=\frac{2-\sqrt2}{2}. \]

Therefore, the correct answer is B.

32 答案 E

因式分解得 \[ \begin{aligned} n^5-n &=n(n-1)(n+1)\\ &\quad\cdot(n^2+1) \end{aligned}\text{。}\]三个连续整数中有一个能被 \(3\) 整除，且至少有一个是偶数，所以该式能被 \(6\) 整除。又因为对每个整数 \(n\) 都有 \(n^5\equiv n\pmod5\)，所以它也能被 \(5\) 整除。因此每一项都能被 \(30\) 整除。取 \(n=2\) 时，\(2^5-2=30\)，所以选项中没有更大的整数能整除每一项。

因此，正确答案是 E。

English

Factor \[ \begin{aligned} n^5-n &=n(n-1)(n+1)\\ &\quad\cdot(n^2+1). \end{aligned} \] Among three consecutive integers, one is divisible by \(3\) and at least one is even, so the expression is divisible by \(6.\) Also \(n^5\equiv n\pmod5\) for every integer \(n,\) so it is divisible by \(5.\) Hence every term is divisible by \(30.\) Taking \(n=2\) gives \(2^5-2=30,\) so no larger listed integer can divide every term.

Thus, the correct answer is E.

33 答案 E

提取公因式 \(9^x\)，得到 \[ 81\cdot9^x-9^x=240\text{，}\]所以 \(80\cdot9^x=240\)，且 \(9^x=3\)。因为 \(9^{\frac{1}{2}}=3\)，所以 \(x=\frac{1}{2}=0.5\)。

因此，正确答案是 E。

English

Factoring \(9^x,\) \[ 81\cdot9^x-9^x=240, \] so \(80\cdot9^x=240\) and \(9^x=3.\) Since \(9^{\frac{1}{2}}=3,\) we have \(x=\frac{1}{2}=0.5.\)

Therefore, the correct answer is E.

34 答案 C

该不等式描述以原点为圆心、半径为 \(5\) 的圆内部的开圆盘。直线 \(x+y=1\) 与圆相交于两点。直线在圆盘内的部分是两交点之间的线段，但严格不等式排除了两个端点。

因此，正确答案是 C。

English

The inequality describes the open disk inside the circle of radius \(5\) centered at the origin. The line \(x+y=1\) crosses the circle in two points. Its portion inside the disk is the segment between those points, but the strict inequality excludes the endpoints.

Thus, the correct answer is C.

35 答案 D

由相似性可知，当 \(k=1,\ldots,7\) 时，第 \(k\) 个分点处的线段长为 \(\frac{10k}{8}\)。它们的和为 \[ \begin{aligned} \frac{10}{8}(1+2+\cdots+7) &=\frac{10}{8}\cdot28\\ &=35 \end{aligned}\text{。}\]

因此，正确答案是 D。

English

By similarity, the segment at the \(k\)th division point has length \(\frac{10k}{8}\) for \(k=1,\ldots,7.\) Their sum is \[ \begin{aligned} \frac{10}{8}(1+2+\cdots+7) &=\frac{10}{8}\cdot28\\ &=35. \end{aligned} \]

Therefore, the correct answer is D.

36 答案 D

我们有 \[ \begin{aligned} xy=x(1-x) &=\frac14\\ &\quad-\left(x-\frac12\right)^2 \end{aligned}\text{。}\]平方项非负，所以乘积的最大可能值为 \(\frac{1}{4}=0.25\)。

因此，正确答案是 D。

English

We have \[ \begin{aligned} xy=x(1-x) &=\frac14\\ &\quad-\left(x-\frac12\right)^2. \end{aligned} \] The squared term is nonnegative, so the largest possible product is \(\frac{1}{4}=0.25.\)

Thus, the correct answer is D.

37 答案 C

由相似性，\[ MN=\frac5{12}AM=\frac{5x}{12}\text{。}\]另外，\(NP=MC=AC-AM\) \(=12-x\)。因此 \[ \begin{aligned} y=MN+NP &=\frac{5x}{12}+12-x\\ &=\frac{144-7x}{12} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

By similarity, \[ MN=\frac5{12}AM=\frac{5x}{12}. \] Also \(NP=MC=AC-AM\) \(=12-x.\) Hence \[ \begin{aligned} y=MN+NP &=\frac{5x}{12}+12-x\\ &=\frac{144-7x}{12}. \end{aligned} \]

Therefore, the correct answer is C.

38 答案 D

写成 \(N=10a+b\)，正的差为 \[ \begin{aligned} &(10a+b)-(10b+a)\\ &\quad=9(a-b) \end{aligned}\text{。}\]它至多为 \(81\)。在 \(1\)、\(8\)、\(27\) 和 \(64\) 中，只有 \(27\) 能被 \(9\) 整除，所以 \(a-b=3\)。数字对为 \[ \begin{gathered} (3,0),(4,1),(5,2),(6,3),\\ (7,4),(8,5),(9,6) \end{gathered}\text{，}\]因而恰有七个 \(N\) 值。

因此，正确答案是 D。

English

Writing \(N=10a+b,\) the positive difference is \[ \begin{aligned} &(10a+b)-(10b+a)\\ &\quad=9(a-b). \end{aligned} \] It is at most \(81.\) Among \(1,\) \(8,\) \(27,\) and \(64,\) only \(27\) is divisible by \(9,\) so \(a-b=3.\) The digit pairs are \[ \begin{gathered} (3,0),(4,1),(5,2),(6,3),\\ (7,4),(8,5),(9,6), \end{gathered} \] giving exactly seven values of \(N.\)

Thus, the correct answer is D.

39 答案 E

在 \(t\) 小时内，第二人每小时走过的路程组成首项为 \(2\)、末项为 \(2+\frac{t-1}{2}\) 的等差数列。他走过的路程为 \[ \frac t2\left(4+\frac{t-1}{2}\right) =\frac{t(t+7)}4\text{。}\]两人合计走过 \[ 4t+\frac{t(t+7)}4=72\text{，}\]即 \(t^2+23t-288=0\)。正根为 \(t=9\)。此时第一人走了 \(4(9)=36\) 英里，恰为原距离的一半。

因此，正确答案是 E。

English

In \(t\) hours, the second man’s hourly distances form an arithmetic progression with first term \(2\) and last term \(2+\frac{t-1}{2}.\) His distance is \[ \frac t2\left(4+\frac{t-1}{2}\right) =\frac{t(t+7)}4. \] Together the men cover \[ 4t+\frac{t(t+7)}4=72, \] or \(t^2+23t-288=0.\) The positive root is \(t=9.\) The first man then walks \(4(9)=36\) miles, exactly half the original distance.

Therefore, the correct answer is E.

40 答案 D

顶点在 \(x\) 轴上，当且仅当该二次方程有一个重根。其判别式必须满足 \[ b^2-4(-1)(-8)=b^2-32=0\text{。}\]因此 \(b=\pm\sqrt{32}=\pm4\sqrt2\)，是正无理数或负无理数。

因此，正确答案是 D。

English

The vertex lies on the \(x\)-axis exactly when the quadratic has one repeated root. Its discriminant must satisfy \[ b^2-4(-1)(-8)=b^2-32=0. \] Hence \(b=\pm\sqrt{32}=\pm4\sqrt2,\) a positive or negative irrational number.

Thus, the correct answer is D.

41 答案 D

系数行列式为 \[ \begin{aligned} &a(-a)-(a-1)(a+1)\\ &\quad=1-2a^2 \end{aligned}\text{。}\]当 \(a=\pm\frac{\sqrt2}{2}\) 时，它等于零。对这两个值中的任一个，两个系数行成比例，但右端的两个数不满足相同比例，所以方程组不相容。

因此，正确答案是 D。

English

The coefficient determinant is \[ \begin{aligned} &a(-a)-(a-1)(a+1)\\ &\quad=1-2a^2. \end{aligned} \] It vanishes when \(a=\pm\frac{\sqrt2}{2}.\) For either value, the two coefficient rows are proportional but the two right sides are not in the same ratio, so the system is inconsistent.

Thus, the correct answer is D.

42 答案 C

若 \(n\) 为偶数，则 \(i^n\) 为 \(1\) 或 \(-1\)，且等于其倒数，所以 \(S=2\) 或 \(-2\)。若 \(n\) 为奇数，则两项按某种顺序分别为 \(i\) 与 \(-i\)，所以 \(S=0\)。因此，不同的值为 \[ -2,\ 0,\ 2\text{，}\]共三个。

因此，正确答案是 C。

English

If \(n\) is even, \(i^n\) is \(1\) or \(-1\) and equals its reciprocal, so \(S=2\) or \(-2.\) If \(n\) is odd, the two terms are \(i\) and \(-i,\) in some order, so \(S=0.\) Thus the distinct values are \[ -2,\ 0,\ 2, \] three values.

Therefore, the correct answer is C.

43 答案 B

对于 \(0\)、\(1\)、\(2\)、\(3\) 和 \(4\) 中的每个 \(x\)，从 \(y=0\) 到 \(y=x^2\) 共有 \(x^2+1\) 个整数值。因此格点数为 \[ \begin{aligned} &(0^2+1)+(1^2+1)+(2^2+1)\\ &\quad +(3^2+1)+(4^2+1)\\ &=1+2+5+10+17\\ &=35 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

For each \(x\) among \(0,\) \(1,\) \(2,\) \(3,\) and \(4,\) there are \(x^2+1\) integer values from \(y=0\) through \(y=x^2.\) Therefore the number of lattice points is \[ \begin{aligned} &(0^2+1)+(1^2+1)+(2^2+1)\\ &\quad +(3^2+1)+(4^2+1)\\ &=1+2+5+10+17\\ &=35. \end{aligned} \]

Thus, the correct answer is B.

44 答案 E

令 \(\alpha=\angle CAB\)、\(\beta=\angle ABC\)，且 \(x=\angle BAD\)。因为 \(D\) 在 \(BC\) 上，所以作为三角形 \(ABD\) 的一个外角，\[ \angle ADC=x+\beta\text{。}\]由于 \(AC=CD\)，三角形 \(ACD\) 为等腰三角形，所以 \(\angle CAD=\angle ADC\)。而 \(\angle CAD=\alpha-x\)。因此 \[ \alpha-x=x+\beta\text{，}\]所以 \(2x=\alpha-\beta=30^\circ\)，且 \(x=15^\circ\)。

因此，正确答案是 E。

English

Let \(\alpha=\angle CAB,\) \(\beta=\angle ABC,\) and \(x=\angle BAD.\) Since \(D\) lies on \(BC,\) \[ \angle ADC=x+\beta \] as an exterior angle of triangle \(ABD.\) Because \(AC=CD,\) triangle \(ACD\) is isosceles, so \(\angle CAD=\angle ADC.\) But \(\angle CAD=\alpha-x.\) Hence \[ \alpha-x=x+\beta, \] so \(2x=\alpha-\beta=30^\circ\) and \(x=15^\circ.\)

Thus, the correct answer is E.

45 答案 A

因为 \(y\ne0\)，两边乘以 \(y\)，得到 \[ x=xy-y^2\text{，}\]即 \(y^2-xy+x=0\)。要使 \(y\) 为实数，判别式必须满足 \[ x^2-4x=x(x-4)\ge0\text{。}\]因此 \(x\le0\) 或 \(x\ge4\)。

因此，正确答案是 A。

English

Since \(y\ne0,\) multiplying by \(y\) gives \[ x=xy-y^2, \] or \(y^2-xy+x=0.\) For a real value of \(y,\) its discriminant must satisfy \[ x^2-4x=x(x-4)\ge0. \] Thus \(x\le0\) or \(x\ge4.\)

Therefore, the correct answer is A.

46 答案 E

将交点置于 \((0,0)\)，四个端点为 \((-3,0)\)、\((4,0)\)、\((0,-2)\) 和 \((0,6)\)。两条弦的垂直平分线分别为 \(x=\frac{1}{2}\) 与 \(y=2\)，所以圆心为 \((\frac{1}{2},2)\)。使用端点 \((4,0)\)，得到 \[ \begin{aligned} r^2 &=\left(4-\frac12\right)^2+(0-2)^2\\ &=\frac{65}{4} \end{aligned}\text{。}\]因此直径为 \(2r=\sqrt{65}\)。

因此，正确答案是 E。

English

Place the intersection at \((0,0)\) with endpoints \((-3,0),\) \((4,0),\) \((0,-2),\) and \((0,6).\) The perpendicular bisectors of the chords are \(x=\frac{1}{2}\) and \(y=2,\) so the center is \((\frac{1}{2},2).\) Using the endpoint \((4,0),\) \[ \begin{aligned} r^2 &=\left(4-\frac12\right)^2+(0-2)^2\\ &=\frac{65}{4}. \end{aligned} \] Therefore the diameter is \(2r=\sqrt{65}.\)

Thus, the correct answer is E.

47 答案 A

直线 \(XY\) 是 \(AB\) 的垂直平分线，所以 \(MA=MB\)。因为 \(M\) 在以 \(AB\) 为直径的半圆上，所以 \(\angle AMB=90^\circ\)。因此三角形 \(AMB\) 为等腰直角三角形，且 \(\angle ABM=45^\circ\)。因为 \(B,M,D\) 共线，所以圆周角 \(\angle ABD=45^\circ\)，其所对的弧 \(AD\) 为 \(90^\circ\)。因此弦 \(AD\) 是内接正方形的一条边，长度为 \[ AD=r\sqrt2\text{。}\]

因此，正确答案是 A。

English

Line \(XY\) is the perpendicular bisector of \(AB,\) so \(MA=MB.\) Since \(M\) lies on the semicircle with diameter \(AB,\) \(\angle AMB=90^\circ.\) Thus triangle \(AMB\) is an isosceles right triangle and \(\angle ABM=45^\circ.\) Because \(B,M,D\) are collinear, the inscribed angle \(\angle ABD=45^\circ,\) so its intercepted arc \(AD\) measures \(90^\circ.\) Therefore chord \(AD\) is a side of an inscribed square and has length \[ AD=r\sqrt2. \]

Thus, the correct answer is A.

48 答案 A

在圆内接四边形 \(ABMC\) 中，托勒密定理给出 \[ \begin{aligned} AM\cdot BC &=AB\cdot CM\\ &\quad+AC\cdot BM \end{aligned}\text{。}\]因为 \(ABC\) 是等边三角形，所以 \(AB=BC=AC\)。两边除以这个公共长度，得到 \[ AM=CM+BM\text{。}\]

因此，正确答案是 A。

English

In cyclic quadrilateral \(ABMC,\) Ptolemy’s theorem gives \[ \begin{aligned} AM\cdot BC &=AB\cdot CM\\ &\quad+AC\cdot BM. \end{aligned} \] Since \(ABC\) is equilateral, \(AB=BC=AC.\) Dividing by this common length yields \[ AM=CM+BM. \]

Therefore, the correct answer is A.

49 答案 C

设长度为 \(4\) 与 \(6\) 的两条腰的上段分别为 \(4t\) 与 \(6t\)。分割线段在两个周长中各出现一次，可以相消。因此，由周长相等可得 \[ \begin{aligned} 3+4t+6t &=9+4(1-t)\\ &\quad+6(1-t) \end{aligned}\text{。}\]所以 \(20t=16\)，从而 \(t=\frac{4}{5}\)。每条腰被分割的两段之比为 \[ t:(1-t)=\frac45:\frac15=4:1\text{。}\]

因此，正确答案是 C。

English

Let the upper pieces of the legs of lengths \(4\) and \(6\) be \(4t\) and \(6t,\) respectively. The dividing segment appears once in each perimeter and cancels. Equal perimeters therefore give \[ \begin{aligned} 3+4t+6t &=9+4(1-t)\\ &\quad+6(1-t). \end{aligned} \] Hence \(20t=16,\) so \(t=\frac{4}{5}.\) Each leg is divided in the ratio \[ t:(1-t)=\frac45:\frac15=4:1. \]

Thus, the correct answer is C.

50 答案 B

令 \(A=(0,0)\)、\(B=(L,0)\)，且 \(G=(g,0)\)。在同一侧作垂线，得到 \[ A'=(0,g),\qquad B'=(L,L-g)\text{。}\]它们的中点为 \[ \begin{aligned} O' &=\left(\frac L2,\frac{g+L-g}{2}\right)\\ &=\left(\frac L2,\frac L2\right) \end{aligned}\text{，}\]与 \(g\) 无关。因此 \(O'\) 保持不动。

因此，正确答案是 B。

English

Let \(A=(0,0),\) \(B=(L,0),\) and \(G=(g,0).\) The perpendicular constructions on the same side give \[ A'=(0,g),\qquad B'=(L,L-g). \] Their midpoint is \[ \begin{aligned} O' &=\left(\frac L2,\frac{g+L-g}{2}\right)\\ &=\left(\frac L2,\frac L2\right), \end{aligned} \] independent of \(g.\) Thus \(O'\) remains stationary.

Therefore, the correct answer is B.