1964 AMC 12 真题（P版）

1 答案 E

由于 \(\log_{10}100=2\)，方括号内的表达式为 \(\log_{10}(5\cdot2)=\log_{10}10=1\)。它的平方是 \(1\)。

因此，正确答案是 E。

English

Since \(\log_{10}100=2,\) the expression inside the square brackets is \(\log_{10}(5\cdot2)=\log_{10}10=1.\) Its square is \(1.\)

Therefore, the correct answer is E.

2 答案 C

分解因式得 \[x^2-4y^2=(x-2y)(x+2y)\text{。}\]因此每个点都位于 \(x=2y\) 或 \(x=-2y\) 上，图像是两条直线。

因此，正确答案是 C。

English

Factoring gives \[x^2-4y^2=(x-2y)(x+2y).\] Thus every point lies on \(x=2y\) or \(x=-2y,\) a pair of straight lines.

Therefore, the correct answer is C.

3 答案 D

根据带余除法，\(x=uy+v\)。因此 \[x+2uy=3uy+v\text{。}\]第一项能被 \(y\) 整除，所以余数是 \(v\)。

因此，正确答案是 D。

English

The division algorithm gives \(x=uy+v.\) Hence \[x+2uy=3uy+v.\] The first term is divisible by \(y,\) so the remainder is \(v.\)

Therefore, the correct answer is D.

4 答案 A

由于 \(P+Q=2x\) 且 \(P-Q=2y\)，原式化为 \[\frac{x}{y}-\frac{y}{x}=\frac{x^2-y^2}{xy}\text{。}\]

因此，正确答案是 A。

English

Since \(P+Q=2x\) and \(P-Q=2y,\) the expression becomes \[\frac{x}{y}-\frac{y}{x}=\frac{x^2-y^2}{xy}.\]

Therefore, the correct answer is A.

5 答案 A

正比例关系给出 \(y=kx\)。由 \(8=4k\) 得 \(k=2\)。因此当 \(x=-8\) 时，\(y=2(-8)=-16\)。

因此，正确答案是 A。

English

Direct variation gives \(y=kx.\) From \(8=4k,\) \(k=2.\) Therefore, when \(x=-8,\) \(y=2(-8)=-16.\)

Thus, the correct answer is A.

6 答案 B

等比中项关系给出 \[(2x+2)^2=x(3x+3)\text{。}\]按本试题对等比数列的公比约定，相邻项之比必须有定义，所以 \(x\ne-1\)。两边除以 \(x+1\)，得 \(4(x+1)=3x\)，从而 \(x=-4\)。前三项为 \(-4,-6,-9\)，公比为 \(\frac{3}{2}\)。第四项是 \(-9(\frac{3}{2})=-\frac{27}{2}\)。

因此，正确答案是 B。

English

The geometric-mean relation gives \[(2x+2)^2=x(3x+3).\] Under the examination’s ratio convention for a geometric progression, the consecutive ratios must be defined, so \(x\ne-1.\) Dividing by \(x+1\) yields \(4(x+1)=3x\) and \(x=-4.\) The first three terms are \(-4,-6,-9,\) with ratio \(\frac{3}{2}.\) The fourth is \(-9(\frac{3}{2})=-\frac{27}{2}.\)

Therefore, the correct answer is B.

7 答案 C

两根相等要求 \[(-p)^2-4p=p(p-4)=0\text{。}\]因此 \(p=0\) 或 \(p=4\)，共有两个实数值。

因此，正确答案是 C。

English

Equal roots require \[(-p)^2-4p=p(p-4)=0.\] Thus \(p=0\) or \(p=4,\) giving two real values.

Therefore, the correct answer is C.

8 答案 C

合并两项并分解因式得 \[\left(x-\frac34\right)\left(2x-\frac54\right)=0\text{。}\]因此两根为 \(\frac{3}{4}\) 和 \(\frac{5}{8}\)，其中 \(\frac{5}{8}\) 较小。

因此，正确答案是 C。

English

Combining the two terms and factoring gives \[\left(x-\frac34\right)\left(2x-\frac54\right)=0.\] Hence the roots are \(\frac{3}{4}\) and \(\frac{5}{8},\) of which \(\frac{5}{8}\) is smaller.

Therefore, the correct answer is C.

9 答案 E

成本为 \(24(1-\frac{1}{8})=21\) 美元。增加三分之一的利润后，目标售价为 \(21(\frac{4}{3})=28\) 美元。设标价为 \(M\)，则 \(0.8M=28\)，所以 \(M=35\)。选项中没有这个数值。

因此，正确答案是 E。

English

The cost is \(24(1-\frac{1}{8})=21\) dollars. A gain of one third makes the desired selling price \(21(\frac{4}{3})=28\) dollars. If \(M\) is the marked price, then \(0.8M=28,\) so \(M=35.\) This is not listed.

Thus, the correct answer is E.

10 答案 A

三角形的底是正方形的对角线，长为 \(s\sqrt2\)。设高为 \(h\)，由面积相等得 \[\frac12(s\sqrt2)h=s^2\text{，}\]所以 \(h=s\sqrt2\)。

因此，正确答案是 A。

English

The triangle’s base is the square’s diagonal, \(s\sqrt2.\) If its altitude is \(h,\) equality of areas gives \[\frac12(s\sqrt2)h=s^2,\] so \(h=s\sqrt2.\)

Therefore, the correct answer is A.

11 答案 D

将两个方程写成底数相同的形式，得 \[ \begin{aligned} x&=3y+3,\\ 2y&=x-9\text{。} \end{aligned} \] 代入可得 \(y=6\) 且 \(x=21\)，所以 \(x+y=27\)。

因此，正确答案是 D。

English

Writing both equations with common bases gives \[ \begin{aligned} x&=3y+3,\\ 2y&=x-9. \end{aligned} \] Substitution yields \(y=6\) and \(x=21,\) so \(x+y=27.\)

Therefore, the correct answer is D.

12 答案 E

全称命题的否定是存在命题，而 \(x^2\gt0\) 的否定是 \(x^2\leq0\)。因此否定命题是：存在某个 \(x\)，使 \(x^2\leq0\)。

因此，正确答案是 E。

English

The negation of a universal statement is an existential statement, and the negation of \(x^2\gt0\) is \(x^2\leq0.\) Thus the negation is: for some \(x,\) \(x^2\leq0.\)

Therefore, the correct answer is E.

13 答案 A

半周长为 \(\frac{8+13+17}{2}=19\)。长度为 \(8\) 的边两端对应的切线段长分别为 \(19-13=6\) 和 \(19-17=2\)。因此 \(r:s=2:6=1:3\)。

因此，正确答案是 A。

English

The semiperimeter is \(\frac{8+13+17}{2}=19.\) At the endpoints of the side of length \(8,\) the tangent lengths are \(19-13=6\) and \(19-17=2.\) Thus \(r:s=2:6=1:3.\)

Therefore, the correct answer is A.

14 答案 C

设总成本为 \(C\)。由于前 \(700\) 只羊共卖得 \(C\)，每只售价为 \(\frac{C}{700}\)。因此总收入为 \(\frac{749C}{700}=1.07C\)，利润为 \(7\%\)。

因此，正确答案是 C。

English

Let the total cost be \(C.\) Since the first \(700\) sheep sell for \(C,\) each sheep sells for \(\frac{C}{700}.\) Total revenue is therefore \(\frac{749C}{700}=1.07C,\) a \(7\%\) gain.

Thus, the correct answer is C.

15 答案 B

设 \(y\) 轴截距为 \(b\)。第二象限内三角形的面积为 \(\frac{ab}{2}=T\)，所以 \(b=\frac{2T}{a}\)。直线的截距式为 \[\frac{x}{-a}+\frac{y}{\frac{2T}{a}}=1\text{。}\]消去分母得 \(2Tx-a^2y+2aT=0\)。

因此，正确答案是 B。

English

Let the \(y\)-intercept be \(b.\) The second-quadrant triangle has area \(\frac{ab}{2}=T,\) so \(b=\frac{2T}{a}.\) The intercept form is \[\frac{x}{-a}+\frac{y}{\frac{2T}{a}}=1.\] Clearing denominators gives \(2Tx-a^2y+2aT=0.\)

Therefore, the correct answer is B.

16 答案 E

有 \(f(s)=(s+1)(s+2)\)，这是两个连续整数的乘积，所以总是偶数。除非 \(s\equiv0\pmod3\)，否则它能被 \(3\) 整除。在 \(0,1,\ldots,25\) 中，有九个数是 \(3\) 的倍数。因此有 \(26-9=17\) 个数符合条件。

因此，正确答案是 E。

English

We have \(f(s)=(s+1)(s+2),\) a product of consecutive integers, so it is always even. It is divisible by \(3\) unless \(s\equiv0\pmod3.\) Among \(0,1,\ldots,25,\) nine values are multiples of \(3.\) Thus \(26-9=17\) values work.

Therefore, the correct answer is E.

17 答案 D

当 \(P\) 和 \(Q\) 不共线时，关系 \(\overrightarrow{OR}=\overrightarrow{OP}+\overrightarrow{OQ}\) 使 \(OPRQ\) 的两组对边分别平行且相等，所以图形是平行四边形。若两个向量线性相关，则四点都在一条直线上。它不可能是严格意义上的梯形。

因此，正确答案是 D。

English

The relation \(\overrightarrow{OR}=\overrightarrow{OP}+\overrightarrow{OQ}\) makes opposite sides of \(OPRQ\) parallel and equal whenever \(P\) and \(Q\) are not collinear, so the figure is a parallelogram. If the two vectors are dependent, all four points lie on a straight line. It cannot be a genuine trapezoid.

Thus, the correct answer is D.

18 答案 C

要使图像相同，必须有 \((c,-2,12)=k(3,b,c)\)。因此 \(c=3k\) 且 \(12=kc=3k^2\)，所以 \(k=\pm2\)。每个值都通过 \(b=-\frac{2}{k}\) 和 \(c=3k\) 确定一组数值。因此共有两对。

因此，正确答案是 C。

English

For the same graph, \((c,-2,12)=k(3,b,c).\) Thus \(c=3k\) and \(12=kc=3k^2,\) so \(k=\pm2.\) Each value determines one pair through \(b=-\frac{2}{k}\) and \(c=3k.\) Hence there are two pairs.

Therefore, the correct answer is C.

19 答案 A

解方程得 \(y=3z\) 和 \(x=5z\)。因此 \[\frac{x^2+3xy}{y^2+z^2}=\frac{25z^2+45z^2}{9z^2+z^2}=7\text{。}\]

因此，正确答案是 A。

English

Solving the equations gives \(y=3z\) and \(x=5z.\) Therefore \[\frac{x^2+3xy}{y^2+z^2}=\frac{25z^2+45z^2}{9z^2+z^2}=7.\]

Thus, the correct answer is A.

20 答案 B

令 \(x=y=1\)。所有数值系数之和就是 \[(1-2)^{18}=(-1)^{18}=1\text{。}\]

因此，正确答案是 B。

English

Set \(x=y=1.\) The sum of all numerical coefficients is then \[(1-2)^{18}=(-1)^{18}=1.\]

Therefore, the correct answer is B.

21 答案 D

令 \(t=\log_bx\)。则 \(\log_{b^2}x=\frac{t}{2}\) 且 \(\log_{x^2}b=\frac{1}{2t}\)。因此 \[t+\frac1t=2\text{，}\]所以 \((t-1)^2=0\) 且 \(t=1\)。故 \(x=b\)。

因此，正确答案是 D。

English

Let \(t=\log_bx.\) Then \(\log_{b^2}x=\frac{t}{2}\) and \(\log_{x^2}b=\frac{1}{2t}.\) Thus \[t+\frac1t=2,\] so \((t-1)^2=0\) and \(t=1.\) Hence \(x=b.\)

Therefore, the correct answer is D.

22 答案 C

设平行四边形面积为 \(S=|u\times v|\)，并取 \(A=0\)、\(B=u\)、\(D=v\)。则 \(E=\frac{u+v}{2}\)，\(F=\frac{2v}{3}\)。由行列式得 \([DFE]=\frac{S}{12}\)。对于四边形 \(ABEF\)，两部分面积给出 \[ [ABEF]=\frac S4+\frac S6=\frac{5S}{12}\text{。}\]所求比为 \(1:5\)。

因此，正确答案是 C。

English

Let the parallelogram’s area be \(S=|u\times v|,\) with \(A=0,\) \(B=u,\) and \(D=v.\) Then \(E=\frac{u+v}{2}\) and \(F=\frac{2v}{3}.\) Determinants give \([DFE]=\frac{S}{12}.\) For quadrilateral \(ABEF,\) the two determinant contributions give \[ [ABEF]=\frac S4+\frac S6=\frac{5S}{12}. \] The desired ratio is \(1:5.\)

Therefore, the correct answer is C.

23 答案 D

设差为 \(k\)，和为 \(7k\)。两个数为 \(4k\) 和 \(3k\)，所以乘积为 \(12k^2\)。而比值还说明乘积为 \(24k\)。由于两数不同，\(k\ne0\)，故 \(k=2\)，乘积为 \(48\)。

因此，正确答案是 D。

English

Let the difference be \(k\) and the sum be \(7k.\) The numbers are \(4k\) and \(3k,\) so their product is \(12k^2.\) But the ratio also says the product is \(24k.\) For distinct numbers \(k\ne0,\) so \(k=2\) and the product is \(48.\)

Therefore, the correct answer is D.

24 答案 A

展开并配方得 \[y=2\left(x-\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{2}\text{。}\]当 \(x=\frac{a+b}{2}\) 时，平方项最小。

因此，正确答案是 A。

English

Expanding and completing the square gives \[y=2\left(x-\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{2}.\] The squared term is minimized at \(x=\frac{a+b}{2}.\)

Therefore, the correct answer is A.

25 答案 B

将分解式写成 \((x+ay+b)(x+cy+d)\)。比较 \(y^2\) 和 \(xy\) 的系数，得 \(ac=0\) 且 \(a+c=3\)，可取 \(a=0\)、\(c=3\)。其余系数给出 \[ \begin{aligned} b+d&=1,\\ bd&=-m,\\ 3b&=m\text{。} \end{aligned} \] 因而 \(b(d+3)=0\)。若 \(b=0\)，则 \(m=0\)；否则 \(d=-3\)、\(b=4\)，且 \(m=12\)。这两个值都能得到有效的整系数因式分解。

因此，正确答案是 B。

English

Write the factorization as \((x+ay+b)(x+cy+d).\) Comparing the \(y^2\) and \(xy\) coefficients gives \(ac=0\) and \(a+c=3,\) so take \(a=0,\) \(c=3.\) The other coefficients give \[ \begin{aligned} b+d&=1,\\ bd&=-m,\\ 3b&=m. \end{aligned} \] Hence \(b(d+3)=0.\) If \(b=0,\) then \(m=0;\) otherwise \(d=-3,\) \(b=4,\) and \(m=12.\) Both values produce valid integer factorizations.

Therefore, the correct answer is B.

26 答案 C

相对于第一名的速度，第二名的速度为 \(\frac{8}{10}=\frac{4}{5}\)，第三名的速度为 \(\frac{6}{10}=\frac{3}{5}\)。因此第三名的速度是第二名的 \(\frac{3}{4}\)。当第二名跑完 \(10\) 英里时，第三名跑了 \(10(\frac{3}{4})=7\dfrac12\) 英里，所以第二名领先 \(2\dfrac12\) 英里。

因此，正确答案是 C。

English

Relative to First’s speed, Second’s speed is \(\frac{8}{10}=\frac{4}{5}\) and Third’s is \(\frac{6}{10}=\frac{3}{5}.\) Thus Third runs \(\frac{3}{4}\) as fast as Second. When Second finishes \(10\) miles, Third has run \(10(\frac{3}{4})=7\dfrac12\) miles, so Second wins by \(2\dfrac12\) miles.

Therefore, the correct answer is C.

27 答案 E

由三角不等式，\[|x-4|+|x-3|\geq|4-3|=1\text{。}\]对 \(3\) 与 \(4\) 之间的每个 \(x\)，等号都成立。因此严格不等式恰在 \(a\gt1\) 时有实数解。

因此，正确答案是 E。

English

By the triangle inequality, \[|x-4|+|x-3|\geq|4-3|=1.\] Equality holds for every \(x\) between \(3\) and \(4.\) Therefore the strict inequality has a real solution exactly when \(a\gt1.\)

Thus, the correct answer is E.

28 答案 D

等差数列求和公式化为 \[153=n(a+n-1)\text{。}\]因此 \(n\) 必须整除 \(153\)，而每个这样的 \(n\) 都给出整数 \(a\)。大于 \(1\) 的约数为 \(3\)、\(9\)、\(17\)、\(51\)、\(153\)，所以共有五种可能。

因此，正确答案是 D。

English

The arithmetic-series formula simplifies to \[153=n(a+n-1).\] Thus \(n\) must divide \(153,\) and every such \(n\) gives an integer \(a.\) The divisors greater than \(1\) are \(3,\) \(9,\) \(17,\) \(51,\) and \(153,\) so there are five possibilities.

Therefore, the correct answer is D.

29 答案 E

在相等的夹角两侧，\[ \begin{gathered} \frac{FR}{FD}=\frac54,\\ \frac{FS}{DR}=\frac{15}{12}=\frac54 \end{gathered}\text{。}\]因此由边角边，\(\triangle RFS\sim\triangle FDR\)。边 \(RS\) 对应 \(FR=5\)，所以 \(RS=(\frac{5}{4})\cdot5=\frac{25}{4}=6\frac14\)。

因此，正确答案是 E。

English

At the equal included angles, \[ \begin{gathered} \frac{FR}{FD}=\frac54,\\ \frac{FS}{DR}=\frac{15}{12}=\frac54. \end{gathered} \] Thus \(\triangle RFS\sim\triangle FDR\) by SAS. Side \(RS\) corresponds to \(FR=5,\) so \(RS=(\frac{5}{4})\cdot5=\frac{25}{4}=6\frac14.\)

Therefore, the correct answer is E.

30 答案 D

令 \(A=7+4\sqrt3=(2+\sqrt3)^2\)，且 \(B=2+\sqrt3\)。判别式为 \[B^2-4A(-2)=A+8A=9A\text{。}\]因而两根之差为 \[\frac{\sqrt{9A}}A=\frac3{2+\sqrt3}=6-3\sqrt3\text{。}\]

因此，正确答案是 D。

English

Let \(A=7+4\sqrt3=(2+\sqrt3)^2\) and \(B=2+\sqrt3.\) The discriminant is \[B^2-4A(-2)=A+8A=9A.\] Hence the root difference is \[\frac{\sqrt{9A}}A=\frac3{2+\sqrt3}=6-3\sqrt3.\]

Therefore, the correct answer is D.

31 答案 B

每个底数 \(r=\frac{1\pm\sqrt5}{2}\) 都满足 \(r^2=r+1\)，所以 \(r^2-1=r\)。因此 \[r^{n+1}-r^{n-1}=r^{n-1}(r^2-1)=r^n\text{。}\]将此恒等式逐项用于定义中的线性组合，得 \(f(n+1)-f(n-1)=f(n)\)。

因此，正确答案是 B。

English

Each base \(r=\frac{1\pm\sqrt5}{2}\) satisfies \(r^2=r+1,\) so \(r^2-1=r.\) Therefore \[r^{n+1}-r^{n-1}=r^{n-1}(r^2-1)=r^n.\] Applying this identity term by term to the defining linear combination gives \(f(n+1)-f(n-1)=f(n).\)

Thus, the correct answer is B.

32 答案 C

交叉相乘得 \((a+b)(a+d)=(b+c)(c+d)\)。移去右边并分解因式，得 \[(a-c)(a+b+c+d)=0\text{。}\]因此要么 \(a=c\)，要么总和为零，也可能两者都成立。

因此，正确答案是 C。

English

Cross-multiplication gives \((a+b)(a+d)=(b+c)(c+d).\) Subtracting the right side and factoring yields \[(a-c)(a+b+c+d)=0.\] Hence either \(a=c\) or the sum is zero, and both may occur.

Therefore, the correct answer is C.

33 答案 B

对矩形内任一点，英国国旗定理给出 \[PA^2+PC^2=PB^2+PD^2\text{。}\]因此 \(9+25=PB^2+16\)，所以 \(PB^2=18\) 且 \(PB=3\sqrt2\)。

因此，正确答案是 B。

English

For any point in a rectangle, the British flag theorem gives \[PA^2+PC^2=PB^2+PD^2.\] Thus \(9+25=PB^2+16,\) so \(PB^2=18\) and \(PB=3\sqrt2.\)

Therefore, the correct answer is B.

34 答案 C

写成 \(n=4m\)。对 \(j=0,\ldots,m-1\)，指数从 \(4j\) 到 \(4j+3\) 的四项之和为 \[ \begin{aligned} &(4j+1)+(4j+2)i\\ &\quad-(4j+3)-(4j+4)i\\ &=-2-2i \end{aligned}\text{。}\]最后一项为 \(4m+1\)。因此 \[ \begin{aligned} s&=m(-2-2i)+(4m+1)\\ &=\frac{n+2-ni}{2} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Write \(n=4m.\) For \(j=0,\ldots,m-1,\) the four terms with exponents \(4j\) through \(4j+3\) sum to \[ \begin{aligned} &(4j+1)+(4j+2)i\\ &\quad-(4j+3)-(4j+4)i\\ &=-2-2i. \end{aligned} \] The final term is \(4m+1.\) Therefore \[ \begin{aligned} s&=m(-2-2i)+(4m+1)\\ &=\frac{n+2-ni}{2}. \end{aligned} \]

Thus, the correct answer is C.

35 答案 B

由海伦公式可得面积为 \(84\)，所以 \(AD=\frac{2(84)}{14}=12\)。邻接的边长为 \(13\) 的边在底边上的投影长为 \(\sqrt{13^2-12^2}=5\)，底边剩余部分长为 \(9\)。设 \(D=(0,0),\ A=(0,12)\)、\(B=(-5,0),\ C=(9,0)\)。直线 \(AC\) 的斜率为 \(-\frac{4}{3}\)，所以经过 \(B\) 的高的斜率为 \(\frac{3}{4}\)，并在高度 \(\frac{15}{4}\) 处与 \(AD\) 相交。因此 \[ \begin{gathered} HD=\frac{15}{4},\\ HA=12-\frac{15}{4}=\frac{33}{4} \end{gathered}\text{，}\]从而 \(HD:HA=5:11\)。

因此，正确答案是 B。

English

Heron’s formula gives area \(84,\) so \(AD=\frac{2(84)}{14}=12.\) The adjacent \(13\)-side has projection \(\sqrt{13^2-12^2}=5,\) leaving \(9\) on the base. Put \(D=(0,0),\ A=(0,12),\) \(B=(-5,0),\ C=(9,0).\) Line \(AC\) has slope \(-\frac{4}{3},\) so the altitude through \(B\) has slope \(\frac{3}{4}\) and meets \(AD\) at height \(\frac{15}{4}.\) Thus \[ \begin{gathered} HD=\frac{15}{4},\\ HA=12-\frac{15}{4}=\frac{33}{4}, \end{gathered} \] giving \(HD:HA=5:11.\)

Therefore, the correct answer is B.

36 答案 E

设 \(O\) 为圆心。\(O\) 和 \(C\) 到 \(AB\) 的距离都等于三角形的高，所以 \(CO\parallel AB\)。延长 \(NC\) 过 \(C\)，使其再次与圆交于 \(D\)。由于 \(CD\) 与 \(CN\) 方向相反，\[ \begin{aligned} \angle MCD &=180^\circ-\angle MCN\\ &=120^\circ \end{aligned}\text{。}\]平行于 \(AB\) 的直线 \(CO\) 平分这个角。关于 \(CO\) 的反射使圆保持不变，并交换射线 \(CM\) 与 \(CD\)，所以也交换点 \(M\) 与 \(D\)。因此 \(CM=CD\)，等腰三角形 \(MCD\) 的两个底角均为 \(30^\circ\)。

由于 \(D,C,N\) 共线，\(\angle MDN=30^\circ\)。这个圆周角所对的是弧 \(MTN\)，因此在圆的每个允许位置，弧的度数都为 \(60^\circ\)。

所以，正确答案是 E。

English

Let \(O\) be the circle’s center. Both \(O\) and \(C\) are one triangle altitude above \(AB,\) so \(CO\parallel AB.\) Extend \(NC\) through \(C\) to meet the circle again at \(D.\) Since \(CD\) is opposite to \(CN,\) \[ \begin{aligned} \angle MCD &=180^\circ-\angle MCN\\ &=120^\circ. \end{aligned} \] The line \(CO,\) parallel to \(AB,\) bisects this angle. Reflection across \(CO\) fixes the circle and interchanges rays \(CM\) and \(CD,\) so it interchanges \(M\) and \(D.\) Hence \(CM=CD,\) and isosceles triangle \(MCD\) has base angles \(30^\circ.\)

Because \(D,C,N\) are collinear, \(\angle MDN=30^\circ.\) This inscribed angle subtends arc \(MTN,\) whose measure is therefore \(60^\circ\) for every permissible circle position.

Thus, the correct answer is E.

37 答案 D

我们有 \[ \frac{a+b}{2}-\sqrt{ab} =\frac{(\sqrt b-\sqrt a)^2}{2}\text{。}\]约去正因子 \((\sqrt b-\sqrt a)^2\) 后，与选项 D 的比较化为 \[ \frac12 <\frac{(\sqrt b+\sqrt a)^2}{8a}\text{。}\]由于 \(\sqrt b+\sqrt a\gt2\sqrt a\)，这个不等式成立。因此，A.M. 与 G.M. 之差总小于选项 D 中的式子。

所以，正确答案是 D。

English

We have \[ \frac{a+b}{2}-\sqrt{ab} =\frac{(\sqrt b-\sqrt a)^2}{2}. \] After canceling the positive factor \((\sqrt b-\sqrt a)^2,\) comparison with choice D reduces to \[ \frac12 <\frac{(\sqrt b+\sqrt a)^2}{8a}. \] This is true because \(\sqrt b+\sqrt a\gt2\sqrt a.\) Thus A.M. minus G.M. is always less than the expression in choice D.

Therefore, the correct answer is D.

38 答案 D

由阿波罗尼斯定理，\[ \begin{aligned} QR^2 &=2(4^2+7^2)-4\left(\frac72\right)^2\\ &=81 \end{aligned}\text{。}\]因此 \(QR=9\)。

所以，正确答案是 D。

English

Apollonius’s theorem gives \[ \begin{aligned} QR^2 &=2(4^2+7^2)-4\left(\frac72\right)^2\\ &=81. \end{aligned} \] Therefore \(QR=9.\)

Thus, the correct answer is D.

39 答案 A

对于一个固定顶点，到该顶点的距离平方沿对边是凸函数，所以最大值在端点处取得。由于每条顶点连线与对边的交点都在该边内部，\[ \begin{gathered} AA'\lt\max(AB,AC)=b,\\ BB'\lt\max(BA,BC)=a,\\ CC'\lt\max(CA,CB)=a \end{gathered}\text{。}\]三式相加得 \(s\lt2a+b\)。

因此，正确答案是 A。

English

For a fixed vertex, squared distance is a convex function along the opposite side, so its maximum occurs at an endpoint. Since each cevian endpoint is interior to its side, \[ \begin{gathered} AA'\lt\max(AB,AC)=b,\\ BB'\lt\max(BA,BC)=a,\\ CC'\lt\max(CA,CB)=a. \end{gathered} \] Adding gives \(s\lt2a+b.\)

Therefore, the correct answer is A.

40 答案 A

手表走时速率是正确速率的 \(\frac{575}{576}\)。手表显示的经过时间为 \(5\) 天 \(20\) 小时，即 \(8400\) 分钟。因此，实际经过时间为 \(8400(\frac{576}{575})\)，校正量为 \[ \begin{aligned} n&=8400\left(\frac{576}{575}-1\right)\\ &=\frac{8400}{575} =\frac{336}{23}\\ &=14\frac{14}{23} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

The watch runs at \(\frac{575}{576}\) of the correct rate. The displayed elapsed time is \(5\) days \(20\) hours, or \(8400\) minutes. Thus the real elapsed time is \(8400(\frac{576}{575}),\) and the correction is \[ \begin{aligned} n&=8400\left(\frac{576}{575}-1\right)\\ &=\frac{8400}{575} =\frac{336}{23}\\ &=14\frac{14}{23}. \end{aligned} \]

Therefore, the correct answer is A.