1954 AMC 12 真题（P版）

1 答案 E

展开可得 \[ \begin{aligned} &\left(5-\sqrt{y^2-25}\right)^2\\ &\quad=25-10\sqrt{y^2-25}\\ &\qquad{}+y^2-25\\ &\quad=y^2-10\sqrt{y^2-25} \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Expanding gives \[ \begin{aligned} &\left(5-\sqrt{y^2-25}\right)^2\\ &\quad=25-10\sqrt{y^2-25}\\ &\qquad{}+y^2-25\\ &\quad=y^2-10\sqrt{y^2-25}. \end{aligned} \]

Thus, the correct answer is E.

2 答案 C

变换后的二次方程可因式分解为 \((x-1)(x-4)=0\)。但原方程的分母含有 \(x-1\)，所以在 \(x=1\) 时无定义。代入 \(x=4\) 合法，并且满足原方程。

因此，唯一的根是 \(4\)，正确答案是 C。

English

The transformed quadratic factors as \((x-1)(x-4)=0.\) However, the original equation has denominator \(x-1,\) so it is undefined at \(x=1.\) Substitution of \(x=4\) is valid and satisfies the original equation.

Thus, the only root is \(4,\) and the correct answer is C.

3 答案 C

由于 \(y\) 与 \(z^{\frac{1}{5}}\) 成正比，它的立方与 \(z^{\frac{3}{5}}\) 成正比。又因为 \(x\) 与 \(y^3\) 成正比，所以 \[ x\propto z^{\frac{3}{5}}\text{。}\]因此 \(n=\frac{3}{5}\)。

因此，正确答案是 C。

English

Since \(y\) is proportional to \(z^{\frac{1}{5}},\) its cube is proportional to \(z^{\frac{3}{5}}.\) Because \(x\) is proportional to \(y^3,\) \[ x\propto z^{\frac{3}{5}}. \] Hence \(n=\frac{3}{5}.\)

Thus, the correct answer is C.

4 答案 E

由欧几里得算法可得 \[ \gcd(6432,132)=12\text{。}\]再减去 \(8\)，得到 \(12-8=4\)。

因此，正确答案是 E。

English

The Euclidean algorithm gives \[ \gcd(6432,132)=12. \] Diminishing this by \(8\) gives \(12-8=4.\)

Thus, the correct answer is E.

5 答案 A

这个正六边形由六个边长为 \(10\) 的等边三角形组成。它的面积为 \[ 6\left(\frac{\sqrt3}{4}\cdot10^2\right) =150\sqrt3 \] 平方英寸。

因此，正确答案是 A。

English

The hexagon consists of six equilateral triangles of side \(10.\) Its area is \[ 6\left(\frac{\sqrt3}{4}\cdot10^2\right) =150\sqrt3 \] square inches.

Thus, the correct answer is A.

6 答案 D

对于符合定义条件的 \(a\)，四项依次为 \[ \frac1{16},\qquad 1,\qquad \frac18,\qquad \frac1{16}\text{。}\]因此原表达式为 \[ \frac1{16}+1-\frac18-\frac1{16}=\frac78\text{。}\]

因此，正确答案是 D。

English

For permissible \(a,\) the four terms are \[ \frac1{16},\qquad 1,\qquad \frac18,\qquad \frac1{16}. \] Therefore the expression is \[ \frac1{16}+1-\frac18-\frac1{16}=\frac78. \]

Thus, the correct answer is D.

7 答案 B

原价为 \(\$27.50\)。节省的比例为 \[ \frac{2.50}{27.50}=\frac1{11}\approx0.0909\text{，}\]约为 \(9\%\)。

因此，正确答案是 B。

English

The original price was \(\$27.50.\) The fraction saved was \[ \frac{2.50}{27.50}=\frac1{11}\approx0.0909, \] or about \(9\%.\)

Thus, the correct answer is B.

8 答案 C

若正方形边长为 \(s\)，则三角形底边长为 \(2s\)。由面积相等可得 \[ s^2=\frac12(2s)h=sh\text{。}\]因此 \(h=s\)，所求比值为 \(1\)。

正确答案是 C。

English

If the square side is \(s,\) the triangle base is \(2s.\) Equal areas give \[ s^2=\frac12(2s)h=sh. \] Thus \(h=s,\) so the requested ratio is \(1.\)

The correct answer is C.

9 答案 C

点 \(P\) 的幂为 \[ PQ\cdot PR=9(16)=144\text{。}\]它也等于 \(13^2-r^2\)，所以 \(169-r^2=144\)。因此 \(r^2=25\)，且 \(r=5\) 英寸。

因此，正确答案是 C。

English

The power of \(P\) is \[ PQ\cdot PR=9(16)=144. \] It is also \(13^2-r^2,\) so \(169-r^2=144.\) Hence \(r^2=25\) and \(r=5\) inches.

Thus, the correct answer is C.

10 答案 C

令 \(a=b=1\)，则每个单项式都等于 \(1\)，因此展开式的值等于各系数之和。所以这个和为 \[ (1+1)^6=64\text{。}\]

正确答案是 C。

English

Setting \(a=b=1\) makes every monomial equal to \(1,\) so the value of the expansion equals the sum of its coefficients. Thus the sum is \[ (1+1)^6=64. \]

The correct answer is C.

11 答案 A

若标价为 \(M\)，则售价为 \(\frac{2M}{3}\)。因此成本为 \[ \frac34\cdot\frac23M=\frac12M\text{。}\]成本与标价之比为 \(\frac{1}{2}\)。

因此，正确答案是 A。

English

If the marked price is \(M,\) then the selling price is \(\frac{2M}{3}.\) The cost is therefore \[ \frac34\cdot\frac23M=\frac12M. \] The cost-to-marked-price ratio is \(\frac{1}{2}.\)

Thus, the correct answer is A.

12 答案 C

第二个方程的左边恰好是第一个方程左边的两倍，但右边不是 \(7\) 的两倍。这个方程组要求同一个表达式同时等于 \(14\) 和 \(20\)，这是不可能的。

因此方程组无解，正确答案是 C。

English

The left side of the second equation is exactly twice the left side of the first, but its right side is not twice \(7.\) The equations would require the same expression to equal both \(14\) and \(20,\) which is impossible.

Thus, there is no solution, and the correct answer is C.

13 答案 B

设四段弧的度数为 \(a,b,c,d\)，其和为 \(360^\circ\)。位于度数为 \(a\) 的圆弧上的角截得另外三段弧，所以其度数为 \(\frac{360^\circ-a}{2}\)，其余三个角同理。四个角之和为 \[ \begin{aligned} &\frac{4(360^\circ)-(a+b+c+d)}2\\ &\qquad=\frac{1080^\circ}{2}\\ &\qquad=540^\circ \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Let the four arcs have measures \(a,b,c,d,\) whose sum is \(360^\circ.\) The angle placed in the arc of measure \(a\) intercepts the other three arcs, so it measures \(\frac{360^\circ-a}{2},\) and similarly for the others. Their sum is \[ \begin{aligned} &\frac{4(360^\circ)-(a+b+c+d)}2\\ &\qquad=\frac{1080^\circ}{2}\\ &\qquad=540^\circ. \end{aligned} \]

Thus, the correct answer is B.

14 答案 E

当 \(x\ne0\) 时， \[ \begin{aligned} &1+\left(\frac{x^4-1}{2x^2}\right)^2\\ &\quad=\frac{4x^4+(x^4-1)^2}{4x^4}\\ &\quad=\frac{(x^4+1)^2}{4x^4} \end{aligned}\text{。}\]因为 \(x^2\gt0\)，所以它的平方根为 \[ \frac{x^4+1}{2x^2} =\frac{x^2}{2}+\frac1{2x^2}\text{。}\]

因此，正确答案是 E。

English

For \(x\ne0,\) \[ \begin{aligned} &1+\left(\frac{x^4-1}{2x^2}\right)^2\\ &\quad=\frac{4x^4+(x^4-1)^2}{4x^4}\\ &\quad=\frac{(x^4+1)^2}{4x^4}. \end{aligned} \] Because \(x^2\gt0,\) its square root is \[ \frac{x^4+1}{2x^2} =\frac{x^2}{2}+\frac1{2x^2}. \]

Thus, the correct answer is E.

15 答案 D

利用对数运算律， \[ \begin{aligned} \log125 &=\log\left(\frac{1000}{8}\right)\\ &=3-\log(2^3)\\ &=3-3\log2 \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Using logarithm laws, \[ \begin{aligned} \log125 &=\log\left(\frac{1000}{8}\right)\\ &=3-\log(2^3)\\ &=3-3\log2. \end{aligned} \]

Thus, the correct answer is D.

16 答案 D

展开并消去同类项可得 \[ \begin{gathered} f(x+h)-f(x)\\ =5(x+h)^2-2(x+h)-1\\ {}-(5x^2-2x-1)\\ =10xh+5h^2-2h\\ =h(10x+5h-2) \end{gathered}\text{。}\]

因此，正确答案是 D。

English

Expanding and canceling gives \[ \begin{gathered} f(x+h)-f(x)\\ =5(x+h)^2-2(x+h)-1\\ {}-(5x^2-2x-1)\\ =10xh+5h^2-2h\\ =h(10x+5h-2). \end{gathered} \]

Thus, the correct answer is D.

17 答案 A

首项 \(2x^3\) 决定图像两端的走向。当 \(x\) 趋于 \(+\infty\) 时，\(f(x)\) 趋于 \(+\infty\)；当 \(x\) 趋于 \(-\infty\) 时，\(f(x)\) 趋于 \(-\infty\)。因此图像向右上方、向左下方延伸。

正确答案是 A。

English

The leading term \(2x^3\) controls the end behavior. As \(x\) tends to \(+\infty,\) \(f(x)\) tends to \(+\infty;\) as \(x\) tends to \(-\infty,\) \(f(x)\) tends to \(-\infty.\) Thus the graph goes up to the right and down to the left.

The correct answer is A.

18 答案 D

两边加上 \(x+3\)，得到 \[ 3x\gt10\text{，}\]所以 \(x\gt\frac{10}{3}\)。

因此，正确答案是 D。

English

Adding \(x+3\) to both sides gives \[ 3x\gt10, \] so \(x\gt\frac{10}{3}.\)

Thus, the correct answer is D.

19 答案 D

通向两个相邻切点的半径分别垂直于对应边。这两条半径所夹的圆心角为 \(180^\circ-A\)。接触三角形中相应的角等于所截圆弧度数的一半，即 \[ 90^\circ-\frac A2\text{。}\]另外两个角分别为 \(90^\circ-\frac{B}{2}\) 和 \(90^\circ-\frac{C}{2}\)。因为三角形的每个角都严格介于 \(0^\circ\) 与 \(180^\circ\) 之间，所以这三个角都是锐角。

因此，正确答案是 D。

English

The radii to two adjacent tangency points are perpendicular to the corresponding sides. The central angle between those radii is \(180^\circ-A.\) The relevant angle of the contact triangle is half the intercepted arc, giving \[ 90^\circ-\frac A2. \] The other two angles are \(90^\circ-\frac{B}{2}\) and \(90^\circ-\frac{C}{2}.\) Since every triangle angle lies strictly between \(0^\circ\) and \(180^\circ,\) all three are acute.

Thus, the correct answer is D.

20 答案 B

该多项式可因式分解为 \[ \begin{gathered} x^3+6x^2+11x+6\\ =(x+1)(x+2)(x+3) \end{gathered}\text{。}\]它的根为 \(-1,-2,-3\)，所以没有正实根。

因此，正确答案是 B。

English

The polynomial factors as \[ \begin{gathered} x^3+6x^2+11x+6\\ =(x+1)(x+2)(x+3). \end{gathered} \] Its roots are \(-1,-2,-3,\) so it has no positive real roots.

Thus, the correct answer is B.

21 答案 C

乘以 \(\sqrt{x}\)，得到 \[ 2x+2=5\sqrt{x}\text{。}\]两边平方并化简， \[ 4x^2+8x+4=25x\text{，}\]所以 \[ 4x^2-17x+4=0\text{。}\]

因此，正确答案是 C。

English

Multiplying by \(\sqrt{x}\) gives \[ 2x+2=5\sqrt{x}. \] Squaring and simplifying, \[ 4x^2+8x+4=25x, \] so \[ 4x^2-17x+4=0. \]

Thus, the correct answer is C.

22 答案 B

合并分子，得到 \[ \begin{gathered} 2x^2-x-(4+x)\\ =2x^2-2x-4\\ =2(x+1)(x-2) \end{gathered}\text{。}\]对于允许的值 \(x\ne-1,2\)，约去因式后，表达式等于 \(2\)。

因此，正确答案是 B。

English

Combining the numerators gives \[ \begin{gathered} 2x^2-x-(4+x)\\ =2x^2-2x-4\\ =2(x+1)(x-2). \end{gathered} \] For the allowed values \(x\ne-1,2,\) the factors cancel and the expression equals \(2.\)

Thus, the correct answer is B.

23 答案 D

因为 \(S=C+M=C+\frac{C}{n}\)， \[ \begin{aligned} S&=\frac{n+1}{n}C,\\ C&=\frac n{n+1}S \end{aligned}\text{。}\]所以 \[ M=\frac Cn=\frac1{n+1}S\text{。}\]

因此，正确答案是 D。

English

Since \(S=C+M=C+\frac{C}{n},\) \[ \begin{aligned} S&=\frac{n+1}{n}C,\\ C&=\frac n{n+1}S. \end{aligned} \] Therefore \[ M=\frac Cn=\frac1{n+1}S. \]

Thus, the correct answer is D.

24 答案 A

要有相等实根，必须满足 \[ (1-k)^2-4(2)(8)=0\text{。}\]因此 \((1-k)^2=64\)，所以 \(1-k=\pm8\)。两个值为 \(k=9\) 和 \(k=-7\)。

因此，正确答案是 A。

English

A repeated real root requires \[ (1-k)^2-4(2)(8)=0. \] Thus \((1-k)^2=64,\) so \(1-k=\pm8.\) The two values are \(k=9\) and \(k=-7.\)

Thus, the correct answer is A.

25 答案 D

由韦达定理，两根之积为 \[ \frac{c(a-b)}{a(b-c)}\text{。}\]因为其中一个根是 \(1\)，所以另一个根等于这个乘积。

因此，正确答案是 D。

English

By Vieta’s formulas, the product of the two roots is \[ \frac{c(a-b)}{a(b-c)}. \] Since one root is \(1,\) the other root equals this product.

Thus, the correct answer is D.

26 答案 B

设 \(\overline{CB}=u\)，则 \(\overline{AC}=3u\)，且 \(\overline{AB}=4u\)。从 \(A\) 起量，两圆心的位置分别为 \(\frac{3u}{2}\) 和 \(\frac{7u}{2}\)，两半径之比为 \(3:1\)。若 \(AD=z\)，公切线使 \(D\) 成为两圆的外位似中心，所以 \[ \frac{z-\frac32u}{z-\frac72u}=3\text{。}\]因此 \(z=\frac{9u}{2}\)，且 \[ BD=z-4u=\frac u2\text{，}\]这正是较小圆的半径。

因此，正确答案是 B。

English

Let \(\overline{CB}=u,\) so \(\overline{AC}=3u\) and \(\overline{AB}=4u.\) Measured from \(A,\) the circle centers are at \(\frac{3u}{2}\) and \(\frac{7u}{2},\) and their radii are in the ratio \(3:1.\) If \(AD=z,\) the common external tangent makes \(D\) the external center of similitude, so \[ \frac{z-\frac32u}{z-\frac72u}=3. \] Hence \(z=\frac{9u}{2},\) and \[ BD=z-4u=\frac u2, \] which is the radius of the smaller circle.

Thus, the correct answer is B.

27 答案 D

体积条件为 \[ \frac13\pi r^2h=\frac12\left(\frac43\pi r^3\right)\text{。}\]约去公因式可得 \(h=2r\)，所以所求比为 \(2:1\)。

因此，正确答案是 D。

English

The volume condition is \[ \frac13\pi r^2h=\frac12\left(\frac43\pi r^3\right). \] Cancelling the common factors gives \(h=2r,\) so the requested ratio is \(2:1.\)

Thus, the correct answer is D.

28 答案 B

将两个已知比值相乘， \[ \frac{mr}{nt}=\frac43\cdot\frac9{14}=\frac67\text{。}\]令 \(mr=6k\) 且 \(nt=7k\)。则 \[ \begin{aligned} \frac{3mr-nt}{4nt-7mr} &=\frac{18k-7k}{28k-42k}\\ &=-\frac{11}{14} \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Multiplying the two given ratios, \[ \frac{mr}{nt}=\frac43\cdot\frac9{14}=\frac67. \] Put \(mr=6k\) and \(nt=7k.\) Then \[ \begin{aligned} \frac{3mr-nt}{4nt-7mr} &=\frac{18k-7k}{28k-42k}\\ &=-\frac{11}{14}. \end{aligned} \]

Thus, the correct answer is B.

29 答案 A

若两条直角边长为 \(a\) 和 \(b\)，则它们在斜边上的投影长为 \(\frac{a^2}{c}\) 和 \(\frac{b^2}{c}\)。因此两段之比为 \(a^2:b^2\)。当 \(a:b=1:2\) 时，该比为 \(1:4\)。

因此，正确答案是 A。

English

If the legs have lengths \(a\) and \(b,\) their projections on the hypotenuse have lengths \(\frac{a^2}{c}\) and \(\frac{b^2}{c}.\) Their ratio is therefore \(a^2:b^2.\) With \(a:b=1:2,\) this is \(1:4.\)

Thus, the correct answer is A.

30 答案 B

设 \(a,b,c\) 为三人各自每天的工作效率。则 \[ \begin{aligned} a+b&=\frac12,\\ b+c&=\frac14,\\ a+c&=\frac5{12} \end{aligned}\text{。}\]因此 \[ \begin{aligned} 2a&=(a+b)+(a+c)\\ &\quad{}-(b+c)\\ &=\frac12+\frac5{12}-\frac14\\ &=\frac23 \end{aligned}\text{，}\]所以 \(a=\frac{1}{3}\)。因此 \(A\) 需要 \(3\) 天。

因此，正确答案是 B。

English

Let \(a,b,c\) be the individual daily rates. Then \[ \begin{aligned} a+b&=\frac12,\\ b+c&=\frac14,\\ a+c&=\frac5{12}. \end{aligned} \] Thus \[ \begin{aligned} 2a&=(a+b)+(a+c)\\ &\quad{}-(b+c)\\ &=\frac12+\frac5{12}-\frac14\\ &=\frac23, \end{aligned} \] so \(a=\frac{1}{3}.\) Therefore \(A\) needs \(3\) days.

Thus, the correct answer is B.

31 答案 A

等腰三角形的两个底角为 \[ \angle ABC=\angle BCA=70^\circ\text{。}\]令 \(\angle OBC=\angle OCA=x\)。则 \(\angle OCB=70^\circ-x\)。在三角形 \(BOC\) 中， \[ \begin{aligned} \angle BOC &=180^\circ-x\\ &\quad{}-(70^\circ-x)\\ &=110^\circ \end{aligned}\text{。}\]

因此，正确答案是 A。

English

The base angles of the isosceles triangle are \[ \angle ABC=\angle BCA=70^\circ. \] Put \(\angle OBC=\angle OCA=x.\) Then \(\angle OCB=70^\circ-x.\) In triangle \(BOC,\) \[ \begin{aligned} \angle BOC &=180^\circ-x\\ &\quad{}-(70^\circ-x)\\ &=110^\circ. \end{aligned} \]

Thus, the correct answer is A.

32 答案 E

利用平方差公式， \[ \begin{aligned} x^4+64 &=(x^2+8)^2-16x^2\\ &=(x^2-4x+8)\\ &\quad{}\cdot(x^2+4x+8) \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Using a difference of squares, \[ \begin{aligned} x^4+64 &=(x^2+8)^2-16x^2\\ &=(x^2-4x+8)\\ &\quad{}\cdot(x^2+4x+8). \end{aligned} \]

Thus, the correct answer is E.

33 答案 D

未偿余额从 \(\$120\) 到 \(\$10\)，每次等额减少 \(\$10\)，所以平均未偿余额约为 \[ \frac{120+10}{2}=65\text{。}\]因此年费用率约为 \[ \frac6{65}\cdot100\%\approx9.2\%\text{。}\]最接近的选项是 \(9\%\)。

因此，正确答案是 D。

English

The balance decreases in equal \(\$10\) steps from \(\$120\) to \(\$10,\) so its average outstanding value is approximately \[ \frac{120+10}{2}=65. \] The annual charge is therefore about \[ \frac6{65}\cdot100\%\approx9.2\%. \] The nearest choice is \(9\%.\)

Thus, the correct answer is D.

34 答案 D

准确地说， \[ \begin{gathered} \frac13-\frac{33333333}{10^8}\\ =\frac{10^8-3(33333333)} {3\cdot10^8}\\ =\frac1{3\cdot10^8} \end{gathered}\text{。}\]因此 \(\frac{1}{3}\) 恰好多出上述数值。

因此，正确答案是 D。

English

Exactly, \[ \begin{gathered} \frac13-\frac{33333333}{10^8}\\ =\frac{10^8-3(33333333)} {3\cdot10^8}\\ =\frac1{3\cdot10^8}. \end{gathered} \] Thus \(\frac{1}{3}\) is greater by the stated amount.

Thus, the correct answer is D.

35 答案 A

由题设条件和勾股定理可得 \[ x+\sqrt{d^2+(h+x)^2}=h+d\text{。}\]将根式单独放在一边并平方后， \[ d^2+(h+x)^2=(h+d-x)^2\text{。}\]消去相同项，得到 \[ 4hx+2dx=2hd\text{，}\]所以 \(x=\frac{hd}{2h+d}\)。

因此，正确答案是 A。

English

The condition and the Pythagorean theorem give \[ x+\sqrt{d^2+(h+x)^2}=h+d. \] After isolating the radical and squaring, \[ d^2+(h+x)^2=(h+d-x)^2. \] Cancelling common terms leaves \[ 4hx+2dx=2hd, \] so \(x=\frac{hd}{2h+d}.\)

Thus, the correct answer is A.

36 答案 C

若单程距离为 \(d\)，则总时间为 \[ \frac d{20}+\frac d{10}=\frac{3d}{20}\text{。}\]因此往返平均速度为 \[ \frac{2d}{\frac{3d}{20}}=\frac{40}{3}\text{。}\]它与 \(15\) 之比为 \[ \frac{\frac{40}{3}}{15}=\frac89\text{。}\]

因此，正确答案是 C。

English

For a one-way distance \(d,\) the total time is \[ \frac d{20}+\frac d{10}=\frac{3d}{20}. \] Thus the round-trip average speed is \[ \frac{2d}{\frac{3d}{20}}=\frac{40}{3}. \] Its ratio to \(15\) is \[ \frac{\frac{40}{3}}{15}=\frac89. \]

Thus, the correct answer is C.

37 答案 B

\(R\) 处的角为 \[ 180^\circ-\angle p-\angle q\text{。}\]因为 \(\overline{RS}\) 平分这个角，所以 \(\overline{PR}\) 与 \(\overline{RS}\) 的夹角为 \[ 90^\circ-\frac12(\angle p+\angle q)\text{。}\]构成 \(\angle m\) 的直线垂直于 \(\overline{RS}\)，所以 \[ \angle m=\frac12(\angle p+\angle q)\text{。}\]

因此，正确答案是 B。

English

The angle at \(R\) is \[ 180^\circ-\angle p-\angle q. \] Because \(\overline{RS}\) bisects this angle, the angle between \(\overline{PR}\) and \(\overline{RS}\) is \[ 90^\circ-\frac12(\angle p+\angle q). \] The line forming \(\angle m\) is perpendicular to \(\overline{RS},\) so \[ \angle m=\frac12(\angle p+\angle q). \]

Thus, the correct answer is B.

38 答案 B

方程化为 \(3^x=5\)，所以 \[ \begin{aligned} x&=\frac{\log5}{\log3}\\ &=\frac{1-\log2}{\log3}\\ &=\frac{0.6990}{0.4771}\\ &\approx1.47 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

The equation reduces to \(3^x=5,\) so \[ \begin{aligned} x&=\frac{\log5}{\log3}\\ &=\frac{1-\log2}{\log3}\\ &=\frac{0.6990}{0.4771}\\ &\approx1.47. \end{aligned} \]

Thus, the correct answer is B.

39 答案 E

当端点 \(X\) 在以 \(O\) 为圆心的圆上移动时，它与定点 \(P\) 所成线段的中点 \(M\) 是 \(X\) 在以 \(P\) 为中心、比例因子为 \(\frac{1}{2}\) 的位似下的像。因此该轨迹是一个圆，其圆心为 \(\overline{PO}\) 的中点，半径为 \(\frac{r}{2}\)。

因此，正确答案是 E。

English

As the endpoint \(X\) moves on the circle centered at \(O,\) its midpoint \(M\) with the fixed point \(P\) is the image of \(X\) under a dilation of scale \(\frac{1}{2}\) centered at \(P.\) Therefore the locus is a circle whose center is the midpoint of \(\overline{PO}\) and whose radius is \(\frac{r}{2}.\)

Thus, the correct answer is E.

40 答案 C

令 \(t=a+\frac{1}{a}\)。由标准立方恒等式可得 \[ a^3+\frac1{a^3}=t^3-3t=t(t^2-3)\text{。}\]因为 \(t^2=3\)，所以该表达式等于 \(0\)。

因此，正确答案是 C。

English

Let \(t=a+\frac{1}{a}.\) The standard cubic identity gives \[ a^3+\frac1{a^3}=t^3-3t=t(t^2-3). \] Since \(t^2=3,\) this expression is \(0.\)

Thus, the correct answer is C.

41 答案 B

对于 \(ax^3+bx^2+\cdots=0\)，根之和为 \(-\frac{b}{a}\)。本题中为 \[ -\frac{-8}{4}=2\text{。}\]

因此，正确答案是 B。

English

For \(ax^3+bx^2+\cdots=0,\) the sum of the roots is \(-\frac{b}{a}.\) Here it is \[ -\frac{-8}{4}=2. \]

Thus, the correct answer is B.

42 答案 D

\(y=x^2+bx+2\) 的顶点的 \(x\) 坐标为 \(-\frac{b}{2}\)。因此图像 \((1)\) 的顶点满足 \(x=\frac{1}{4}\)，而图像 \((2)\) 的顶点满足 \(x=-\frac{1}{4}\)。两个顶点的纵坐标相等，所以图像 \((1)\) 是同一条抛物线向右平移所得。

因此，正确答案是 D。

English

The vertex of \(y=x^2+bx+2\) has \(x\)-coordinate \(-\frac{b}{2}.\) Thus graph \((1)\) has its vertex at \(x=\frac{1}{4},\) while graph \((2)\) has its vertex at \(x=-\frac{1}{4}.\) Their vertex heights are equal, so graph \((1)\) is the same parabola shifted to the right.

Thus, the correct answer is D.

43 答案 B

对于直角三角形， \[ r=\frac{a+b-c}{2}\text{。}\]代入 \(r=1\) 和 \(c=10\)，得到 \(a+b=12\)。因此周长为 \[ a+b+c=12+10=22\text{。}\]

因此，正确答案是 B。

English

For a right triangle, \[ r=\frac{a+b-c}{2}. \] With \(r=1\) and \(c=10,\) this gives \(a+b=12.\) Hence the perimeter is \[ a+b+c=12+10=22. \]

Thus, the correct answer is B.

44 答案 E

年份 \(x^2\) 必须位于十九世纪，因此 \(x=43\)，因为 \(43^2=1849\)，而相邻整数的平方不在相关范围内。所以他的出生年份为 \[ x^2-x=1849-43=1806\text{，}\]确实位于该世纪上半叶。

因此，正确答案是 E。

English

The year \(x^2\) must lie in the nineteenth century, so \(x=43\) because \(43^2=1849\) while the neighboring squares fall outside the relevant range. His birth year is therefore \[ x^2-x=1849-43=1806, \] which is in the first half of the century.

Thus, the correct answer is E.

45 答案 D

从 \(A\) 出发时，平行截线的长度为 \(0\)，然后线性增加，直至达到对角线 \(\overline{BD}\) 的长度。继续向对面顶点移动时，线段长度线性减小回 \(0\)。因此图像由两条组成倒 \(V\) 形的线段构成。

因此，正确答案是 D。

English

Starting at \(A,\) the parallel cross section has length \(0\) and grows linearly until it reaches the diagonal \(\overline{BD}.\) Continuing toward the opposite vertex, its length decreases linearly back to \(0.\) The graph is therefore made of two line segments forming an inverted \(V.\)

Thus, the correct answer is D.

46 答案 E

半径为 \(\frac{3}{16}\) 英寸。圆心位于角平分线上。到任一斜边的垂直半径与从顶点到圆心的线段构成一个含 \(30^\circ\) 角的直角三角形，所以圆心位于顶点上方 \[ \frac{\frac{3}{16}}{\sin30^\circ}=\frac38 \] 英寸处。因此上方切线位于顶点上方 \(\frac{3}{8}+\frac{3}{16}=\frac{9}{16}\) 英寸处。由于下方水平线位于顶点上方 \(\frac{1}{2}\) 英寸处， \[ x=\frac9{16}-\frac12=\frac1{16}\text{ 英寸} \]

因此，正确答案是 E。

English

The radius is \(\frac{3}{16}\) inch. The center lies on the angle bisector. The perpendicular radius to either sloping side and the segment from the vertex to the center form a right triangle with a \(30^\circ\) angle, so the center is \[ \frac{\frac{3}{16}}{\sin30^\circ}=\frac38 \] inch above the vertex. Hence the top tangent is \(\frac{3}{8}+\frac{3}{16}=\frac{9}{16}\) inch above the vertex. Since the ledge is \(\frac{1}{2}\) inch above the vertex, \[ x=\frac9{16}-\frac12=\frac1{16}\text{ in.} \]

Thus, the correct answer is E.

47 答案 B

令 \(MT=t\)。因为 \(AM=BM=RT=\frac{p}{2}\)，由勾股定理可得 \[ t^2+q^2=\frac{p^2}{4}\text{。}\]此外， \[ AT=\frac p2+t,\qquad TB=\frac p2-t\text{。}\]两者之和为 \(p\)，乘积为 \[ \frac{p^2}{4}-t^2=q^2\text{。}\]因此它们是 \(x^2-px+q^2=0\) 的两个根。

因此，正确答案是 B。

English

Let \(MT=t.\) Since \(AM=BM=RT=\frac{p}{2},\) the Pythagorean theorem gives \[ t^2+q^2=\frac{p^2}{4}. \] Also, \[ AT=\frac p2+t,\qquad TB=\frac p2-t. \] Their sum is \(p,\) and their product is \[ \frac{p^2}{4}-t^2=q^2. \] Therefore they are the roots of \(x^2-px+q^2=0.\)

Thus, the correct answer is B.

48 答案 C

设正常速度为 \(v\)，全程长度为 \(L\)。一段路程以 \(\frac{3v}{4}\) 而非 \(v\) 的速度行驶，会增加该路段正常行驶时间的三分之一。第一种情况下， \[ \frac12+\frac13\left(\frac Lv-1\right)=\frac72\text{，}\]所以 \(\frac{L}{v}=10\)。若事故在前方 \(90\) 英里处发生，则 \[ \frac12+\frac13\left(\frac Lv-1-\frac{90}{v}\right)=3\text{。}\]代入 \(\frac{L}{v}=10\)，得到 \(\frac{90}{v}=\frac{3}{2}\)，因此 \(v=60\)，且 \[ L=10v=600\text{。}\]

因此，正确答案是 C。

English

Let the normal speed be \(v\) and the trip length be \(L.\) Traveling a distance at \(\frac{3v}{4}\) rather than \(v\) adds one-third of its normal travel time. In the first case, \[ \frac12+\frac13\left(\frac Lv-1\right)=\frac72, \] so \(\frac{L}{v}=10.\) If the accident happens \(90\) miles farther along, \[ \frac12+\frac13\left(\frac Lv-1-\frac{90}{v}\right)=3. \] Substituting \(\frac{L}{v}=10\) gives \(\frac{90}{v}=\frac{3}{2},\) hence \(v=60\) and \[ L=10v=600. \]

Thus, the correct answer is C.

49 答案 C

展开并重新分组， \[ \begin{aligned} &(2a+1)^2-(2b+1)^2\\ &\qquad=4[a(a+1)-b(b+1)] \end{aligned}\text{。}\]\(a(a+1)\) 和 \(b(b+1)\) 都是偶数，所以两者之差也是偶数。因此所示表达式能被 \(4\cdot2=8\) 整除。

因此，正确答案是 C。

English

Expanding and regrouping, \[ \begin{aligned} &(2a+1)^2-(2b+1)^2\\ &\qquad=4[a(a+1)-b(b+1)]. \end{aligned} \] Each of \(a(a+1)\) and \(b(b+1)\) is even, so their difference is even. The displayed expression is consequently divisible by \(4\cdot2=8.\)

Thus, the correct answer is C.

50 答案 A

\(7{:}00\) 过 \(t\) 分钟后，时针位于 \(210^\circ+0.5t^\circ\)，分针位于 \(6t^\circ\)。因此 \[ |210-5.5t|=84\text{。}\]两个解为 \[ \begin{aligned} t&=\frac{126}{5.5}\approx22.91,\\ t&=\frac{294}{5.5}\approx53.45 \end{aligned}\text{。}\]精确到最近一分钟，两个时刻为 \(7{:}23\) 和 \(7{:}53\)。

因此，正确答案是 A。

English

At \(t\) minutes after \(7{:}00,\) the hour hand is at \(210^\circ+0.5t^\circ\) and the minute hand is at \(6t^\circ.\) Thus \[ |210-5.5t|=84. \] The two solutions are \[ \begin{aligned} t&=\frac{126}{5.5}\approx22.91,\\ t&=\frac{294}{5.5}\approx53.45. \end{aligned} \] To the nearest minute, the times are \(7{:}23\) and \(7{:}53.\)

Thus, the correct answer is A.