1967 AMC 12 真题（P版）

1 答案 C

能被 \(9\) 整除要求 \(5+b+9\) 是 \(9\) 的倍数。因为 \(b\) 是一位数字，所以 \(b=4\)。此时加法为 \(2a3+326=549\)，所以 \(a=2\)。因此 \(a+b=6\)。

因此，正确答案是 C。

English

Divisibility by \(9\) requires \(5+b+9\) to be a multiple of \(9.\) Since \(b\) is a digit, this gives \(b=4.\) The addition is then \(2a3+326=549,\) so \(a=2.\) Hence \(a+b=6.\)

Therefore, the correct answer is C.

2 答案 D

第一个乘积为 \[ xy+\frac{x}{y}+\frac{y}{x}+\frac1{xy}\text{，}\]第二个乘积为 \[ xy-\frac{x}{y}-\frac{y}{x}+\frac1{xy}\text{。}\]两者相加得到 \(2xy+\frac{2}{xy}\)。

因此，正确答案是 D。

English

The first product is \[ xy+\frac{x}{y}+\frac{y}{x}+\frac1{xy}, \] while the second is \[ xy-\frac{x}{y}-\frac{y}{x}+\frac1{xy}. \] Adding gives \(2xy+\frac{2}{xy}.\)

Therefore, the correct answer is D.

3 答案 B

内切圆半径为 \(r=\frac{s\sqrt3}{6}\)。内接正方形的对角线长为 \(2r\)，所以其面积为该对角线平方的一半：\[ \begin{aligned} \frac{(2r)^2}{2} &=2r^2\\ &=2\left(\frac{s\sqrt3}{6}\right)^2\\ &=\frac{s^2}{6} \end{aligned}\text{。}\]

因此，正确答案是 B。

English

The inradius is \(r=\frac{s\sqrt3}{6}.\) The inscribed square has diagonal \(2r,\) so its area is half the square of that diagonal: \[ \begin{aligned} \frac{(2r)^2}{2} &=2r^2\\ &=2\left(\frac{s\sqrt3}{6}\right)^2\\ &=\frac{s^2}{6}. \end{aligned} \]

Therefore, the correct answer is B.

4 答案 C

所给等式说明 \(a=x^p\)、\(b=x^q\)、\(c=x^r\)。因此 \[ \frac{b^2}{ac} =x^{2q-p-r}\text{。}\]因为它等于 \(x^y\)，且 \(x\ne1\)，所以 \(y=2q-p-r\)。

因此，正确答案是 C。

English

The given equalities imply \(a=x^p,\) \(b=x^q,\) and \(c=x^r.\) Therefore \[ \frac{b^2}{ac} =x^{2q-p-r}. \] Since this is \(x^y\) and \(x\ne1,\) we have \(y=2q-p-r.\)

Therefore, the correct answer is C.

5 答案 D

连接圆心与三个切点的半径，把三角形分成若干小三角形，它们的总面积为 \[ K=\frac12r(a+b+c)=\frac{Pr}{2}\text{。}\]因此 \(\frac{P}{K}=\frac{2}{r}\)。

因此，正确答案是 D。

English

The three radii to the points of tangency split the triangle into smaller triangles whose total area is \[ K=\frac12r(a+b+c)=\frac{Pr}{2}. \] Hence \(\frac{P}{K}=\frac{2}{r}.\)

Therefore, the correct answer is D.

6 答案 D

我们有 \(f(x+1)=4^{x+1}=4f(x)\)。因此 \(f(x+1)-f(x)=3f(x)\)。

因此，正确答案是 D。

English

We have \(f(x+1)=4^{x+1}=4f(x).\) Therefore \(f(x+1)-f(x)=3f(x).\)

Thus, the correct answer is D.

7 答案 E

取 \(b=d=1\) 且 \(c=-2\)。不等式变为 \(a\lt2\)，正数、零和负数中的一些 \(a\) 值都能满足它。因此，三种可能的符号都不是必然的。

因此，正确答案是 E。

English

Take \(b=d=1\) and \(c=-2.\) The inequality becomes \(a\lt2,\) which is satisfied by positive, zero, and negative values of \(a.\) Thus none of the three possible signs is forced.

Therefore, the correct answer is E.

8 答案 A

由酸的量守恒可得 \[ \frac{m^2}{100} =\frac{m-10}{100}(m+x)\text{。}\]因此 \(m^2=(m-10)(m+x)\)，所以 \(10m=(m-10)x\)，并且 \[ x=\frac{10m}{m-10}\text{。}\]

因此，正确答案是 A。

English

Conservation of acid gives \[ \frac{m^2}{100} =\frac{m-10}{100}(m+x). \] Hence \(m^2=(m-10)(m+x),\) so \(10m=(m-10)x\) and \[ x=\frac{10m}{m-10}. \]

Therefore, the correct answer is A.

9 答案 E

设短底、高、长底分别为 \(a-d\)、\(a\)、\(a+d\)。则 \[ \begin{aligned} K&=\frac12a\bigl((a-d)+(a+d)\bigr)\\ &=a^2 \end{aligned}\text{。}\]随 \(a\) 的取值不同，它可以是整数、非整数有理数或无理数。选项 (A) 至 (D) 中没有一项必然成立。

因此，正确答案是 E。

English

Let the shorter base, altitude, and longer base be \(a-d,\) \(a,\) and \(a+d.\) Then \[ \begin{aligned} K&=\frac12a\bigl((a-d)+(a+d)\bigr)\\ &=a^2. \end{aligned} \] Depending on \(a,\) this can be an integer, a nonintegral rational number, or an irrational number. No one of choices (A) through (D) must hold.

Therefore, the correct answer is E.

10 答案 A

清除分母得到 \[ \begin{aligned} &a(10^x+2)+b(10^x-1)\\ &\qquad=2\cdot10^x+3 \end{aligned}\text{。}\]因此 \(a+b=2\)，且 \(2a-b=3\)。解得 \(a=\frac{5}{3},\ b=\frac{1}{3}\)，所以 \(a-b=\frac{4}{3}\)。

因此，正确答案是 A。

English

Clearing denominators gives \[ \begin{aligned} &a(10^x+2)+b(10^x-1)\\ &\qquad=2\cdot10^x+3. \end{aligned} \] Therefore \(a+b=2\) and \(2a-b=3.\) Solving gives \(a=\frac{5}{3},\ b=\frac{1}{3},\) so \(a-b=\frac{4}{3}.\)

Therefore, the correct answer is A.

11 答案 B

设相邻两边长为 \(x\) 与 \(10-x\)。则 \[ \begin{aligned} AC^2&=x^2+(10-x)^2\\ &=2(x-5)^2+50 \end{aligned}\text{。}\]当 \(x=5\) 时该式最小，从而 \(AC=\sqrt{50}\)。

因此，正确答案是 B。

English

Let adjacent sides be \(x\) and \(10-x.\) Then \[ \begin{aligned} AC^2&=x^2+(10-x)^2\\ &=2(x-5)^2+50. \end{aligned} \] This is minimized at \(x=5,\) giving \(AC=\sqrt{50}.\)

Therefore, the correct answer is B.

12 答案 B

围成的区域是一个梯形，其两条平行边长为 \(m+4\) 与 \(4m+4\)，间距为 \(3\)。因此 \[ \frac32\bigl((m+4)+(4m+4)\bigr)=7\text{。}\]所以 \(15m+24=14\)，从而 \(m=-\frac{2}{3}\)。两端的高度都为正，符合题中凸区域的要求。

因此，正确答案是 B。

English

The bounded region is a trapezoid with parallel sides \(m+4\) and \(4m+4,\) separated by \(3.\) Thus \[ \frac32\bigl((m+4)+(4m+4)\bigr)=7. \] Hence \(15m+24=14,\) so \(m=-\frac{2}{3}.\) The endpoint heights are positive, as required for the stated convex region.

Therefore, the correct answer is B.

13 答案 E

固定 \(BC=a\)，并从 \(B\) 作一条构成已知角 \(B\) 的射线。点 \(C\) 到该射线的距离是固定的。若这个距离不等于 \(h_c\)，则无解；若等于 \(h_c\)，则可在射线上无限多个位置选择 \(A\)，从而得到无限多个互不全等的三角形。因此 \(N\) 必为零或无限。

因此，正确答案是 E。

English

Fix \(BC=a\) and draw the ray from \(B\) that makes the prescribed angle \(B.\) Its distance from \(C\) is fixed. If that distance is not \(h_c,\) there is no solution. If it is \(h_c,\) then \(A\) may be chosen at infinitely many positions on the ray, producing infinitely many noncongruent triangles. Thus \(N\) must be zero or infinite.

Therefore, the correct answer is E.

14 答案 C

由 \(y=\frac{x}{1-x}\) 可得 \(y-yx=x\)，所以 \[ x=\frac{y}{1+y}\text{。}\]因为 \(f(-y)=-\frac{y}{1+y}\)，这就是 \(-f(-y)\)。

因此，正确答案是 C。

English

From \(y=\frac{x}{1-x}\) we obtain \(y-yx=x,\) so \[ x=\frac{y}{1+y}. \] Since \(f(-y)=-\frac{y}{1+y},\) this is \(-f(-y).\)

Therefore, the correct answer is C.

15 答案 D

设边长比例因子为整数 \(k\gt1\)，较小面积为整数 \(T\)。则 \[ T(k^2-1)=18\text{。}\]因此 \(k^2-1\) 是 \(18\) 的正因数。所得 \(k^2\) 中唯一的完全平方数是 \(4\)，所以 \(k=2\)。对应边长为 \(3k=6\)。

因此，正确答案是 D。

English

Let the side-scale factor be the integer \(k\gt1\) and the smaller area be the integer \(T.\) Then \[ T(k^2-1)=18. \] Thus \(k^2-1\) is a positive divisor of \(18.\) The only resulting square \(k^2\) is \(4,\) so \(k=2.\) The corresponding side is \(3k=6.\)

Therefore, the correct answer is D.

16 答案 B

乘积方程为 \[ \begin{aligned} &(b+2)(b+5)(b+6)\\ &\qquad=3b^3+b^2+4b+6 \end{aligned}\text{。}\]化简为 \[ \begin{aligned} b^3-6b^2-24b-27&=0,\\ (b-9)(b^2+3b+3)&=0 \end{aligned}\text{。}\]因此，有效的进制底数为 \(b=9\)。所求的和为 \(3b+13=4b+4\)，也就是 \(44_b\)。

因此，正确答案是 B。

English

The product equation is \[ \begin{aligned} &(b+2)(b+5)(b+6)\\ &\qquad=3b^3+b^2+4b+6. \end{aligned} \] It simplifies to \[ \begin{aligned} b^3-6b^2-24b-27&=0,\\ (b-9)(b^2+3b+3)&=0. \end{aligned} \] Thus the valid base is \(b=9.\) The required sum is \(3b+13=4b+4,\) which is \(44_b.\)

Therefore, the correct answer is B.

17 答案 A

判别式条件为 \(p^2-32\gt0\)，所以 \(\lvert p\rvert\gt4\sqrt2\)。由韦达定理，\(r_1+r_2=-p\)，因此 \[ \lvert r_1+r_2\rvert=\lvert p\rvert\gt4\sqrt2\text{。}\]

因此，正确答案是 A。

English

The discriminant condition is \(p^2-32\gt0,\) so \(\lvert p\rvert\gt4\sqrt2.\) By Vieta’s formulas, \(r_1+r_2=-p,\) and therefore \[ \lvert r_1+r_2\rvert=\lvert p\rvert\gt4\sqrt2. \]

Therefore, the correct answer is A.

18 答案 B

不等式 \((x-2)(x-3)\lt0\) 给出 \(2\lt x\lt3\)。在该区间上，\(P=x^2+5x+6\) 递增，两个端点值为 \(P(2)=20\) 与 \(P(3)=30\)。由于端点不包含在内，所以 \(20\lt P\lt30\)。

因此，正确答案是 B。

English

The inequality \((x-2)(x-3)\lt0\) gives \(2\lt x\lt3.\) On this interval \(P=x^2+5x+6\) is increasing, with endpoint values \(P(2)=20\) and \(P(3)=30.\) The endpoints are excluded, so \(20\lt P\lt30.\)

Therefore, the correct answer is B.

19 答案 E

两个条件为 \[ \begin{aligned} lw&=\left(l+\frac52\right)\left(w-\frac23\right),\\ lw&=\left(l-\frac52\right)\left(w+\frac43\right) \end{aligned}\text{。}\]展开后，得到 \(-4l+15w=10\) 与 \(8l-15w=20\)。因此 \(l=\frac{15}{2}\)、\(w=\frac{8}{3}\)，且 \(lw=20\)。

因此，正确答案是 E。

English

The two conditions are \[ \begin{aligned} lw&=\left(l+\frac52\right)\left(w-\frac23\right),\\ lw&=\left(l-\frac52\right)\left(w+\frac43\right). \end{aligned} \] After expanding, these give \(-4l+15w=10\) and \(8l-15w=20.\) Hence \(l=\frac{15}{2},\) \(w=\frac{8}{3},\) and \(lw=20.\)

Therefore, the correct answer is E.

20 答案 A

第一个圆的面积为 \(\pi(\frac{m}{2})^2=\frac{\pi m^2}{4}\)。每个内接正方形使下一个圆的半径平方，从而也使其面积缩小为原来的 \(\frac{1}{2}\)。因此极限和为 \[ \begin{aligned} &\frac{\pi m^2}{4} \left(1+\frac12+\frac14+\cdots\right)\\ &\qquad=\frac{\pi m^2}{2} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

The first circle has area \(\pi(\frac{m}{2})^2=\frac{\pi m^2}{4}.\) Each inscribed square reduces the next circle’s squared radius, and hence its area, by a factor of \(\frac{1}{2}.\) Thus the limiting sum is \[ \begin{aligned} &\frac{\pi m^2}{4} \left(1+\frac12+\frac14+\cdots\right)\\ &\qquad=\frac{\pi m^2}{2}. \end{aligned} \]

Therefore, the correct answer is A.

21 答案 B

因为 \(BC=4\)，由角平分线定理可得 \(A_1B:A_1C=5:3\)。因此 \(A_1B=\frac{5}{2}\)，且 \(A_1C=\frac{3}{2}\)，所以三角形 \(PQR\) 是三角形 \(ABC\) 缩小一半的副本。

对于原三角形中从 \(A\) 引出的角平分线，\[ \begin{aligned} AA_1^2 &=5\cdot3\left(1-\frac{4^2}{8^2}\right)\\ &=\frac{45}{4} \end{aligned}\text{，}\]所以 \(AA_1=\frac{3\sqrt5}{2}\)。因此 \(PP_1\) 是它的一半，即 \(\frac{3\sqrt5}{4}\)。

因此，正确答案是 B。

English

Since \(BC=4,\) the angle-bisector theorem gives \(A_1B:A_1C=5:3.\) Hence \(A_1B=\frac{5}{2}\) and \(A_1C=\frac{3}{2},\) so triangle \(PQR\) is a half-scale copy of \(ABC.\)

For the angle bisector from \(A\) in the original triangle, \[ \begin{aligned} AA_1^2 &=5\cdot3\left(1-\frac{4^2}{8^2}\right)\\ &=\frac{45}{4}, \end{aligned} \] so \(AA_1=\frac{3\sqrt5}{2}.\) Therefore \(PP_1\) is half of this, or \(\frac{3\sqrt5}{4}.\)

Thus, the correct answer is B.

22 答案 A

代入得到 \[ \begin{aligned} P&=(Q'D'+R')D+R\\ &=Q'(DD')+(R'D+R) \end{aligned}\text{。}\]余数的范围说明 \(R'D+R\lt DD'\)，所以这确实是余数。

因此，正确答案是 A。

English

Substitution gives \[ \begin{aligned} P&=(Q'D'+R')D+R\\ &=Q'(DD')+(R'D+R). \end{aligned} \] The remainder bounds imply \(R'D+R\lt DD',\) so this is indeed the remainder.

Therefore, the correct answer is A.

23 答案 B

该差为 \[ \log_3\left(\frac{6x-5}{2x+1}\right)\text{。}\]分数趋近于 \(3\)，所以原式趋近于 \(\log_3 3=1\)。

因此，正确答案是 B。

English

The difference is \[ \log_3\left(\frac{6x-5}{2x+1}\right). \] The fraction approaches \(3,\) so the expression approaches \(\log_3 3=1.\)

Therefore, the correct answer is B.

24 答案 A

对方程模 \(5\)，可得 \(3x\equiv1\)，所以 \(x\equiv2\pmod5\)。写成 \(x=2+5k\)。则 \[ y=\frac{501-3x}{5}=99-3k\text{。}\]正性要求 \(k=0,1,\ldots,32\)，因此共有 \(33\) 对。

因此，正确答案是 A。

English

Modulo \(5,\) the equation gives \(3x\equiv1,\) so \(x\equiv2\pmod5.\) Write \(x=2+5k.\) Then \[ y=\frac{501-3x}{5}=99-3k. \] Positivity requires \(k=0,1,\ldots,32,\) giving \(33\) pairs.

Therefore, the correct answer is A.

25 答案 A

因为 \(p\gt1\) 为奇数，所以 \(n=\frac{p-1}{2}\) 是正整数。模 \(p-2\) 时，有 \(p-1\equiv1\)。因此 \[ \begin{aligned} (p-1)^n-1 &\equiv1^n-1\\ &\equiv0\pmod{p-2} \end{aligned}\text{。}\]所以选项 (A) 总是成立。

因此，正确答案是 A。

English

Because \(p\gt1\) is odd, \(n=\frac{p-1}{2}\) is a positive integer. Modulo \(p-2,\) we have \(p-1\equiv1.\) Therefore \[ \begin{aligned} (p-1)^n-1 &\equiv1^n-1\\ &\equiv0\pmod{p-2}. \end{aligned} \] Thus choice (A) always holds.

Therefore, the correct answer is A.

26 答案 C

由 \(10^3\lt2^{10}\) 可得 \(3\lt10\log_{10}2\)，所以 \(\log_{10}2\gt\frac{3}{10}\)。由 \(2^{13}\lt10^4\) 可得 \(13\log_{10}2\lt4\)，所以 \(\log_{10}2\lt\frac{4}{13}\)。这是表中信息能够推出的选项中最窄的区间。

因此，正确答案是 C。

English

From \(10^3\lt2^{10}\) we get \(3\lt10\log_{10}2,\) so \(\log_{10}2\gt\frac{3}{10}.\) From \(2^{13}\lt10^4\) we get \(13\log_{10}2\lt4,\) so \(\log_{10}2\lt\frac{4}{13}.\) This is the narrowest listed interval justified by the table.

Therefore, the correct answer is C.

27 答案 C

经过 \(t\) 小时后，燃烧较快和较慢的蜡烛剩余比例分别为 \(1-\frac{t}{3}\) 与 \(1-\frac{t}{4}\)。较慢蜡烛的剩余长度必须是较快蜡烛的两倍：\[ 1-\frac t4=2\left(1-\frac t3\right)\text{。}\]因此 \(t=\frac{12}{5}=2\) 小时 \(24\) 分钟。从下午 \(4\) 点向前推，得到下午 \(1{:}36\)。

因此，正确答案是 C。

English

After \(t\) hours, the faster and slower candles have fractions \(1-\frac{t}{3}\) and \(1-\frac{t}{4}\) remaining. The slower stub must be twice the faster: \[ 1-\frac t4=2\left(1-\frac t3\right). \] Thus \(t=\frac{12}{5}=2\) hours \(24\) minutes. Counting back from \(4\) P.M. gives \(1{:}36\) P.M.

Therefore, the correct answer is C.

28 答案 E

这些前提只说明 \(M\) 中有某个元素不在 \(N\) 中，并且 \(N\cap V=\varnothing\)。令 \(M=V\) 的模型在满足前提的同时使 (A)、(B)、(C) 都为假；令 \(M\cap V=\varnothing\) 的模型则使 (D) 为假。因此，选项 (A) 至 (D) 都不是必然结论。

因此，正确答案是 E。

English

The hypotheses say only that some element of \(M\) lies outside \(N,\) and that \(N\cap V=\varnothing.\) A model with \(M=V\) makes (A), (B), and (C) false while satisfying the hypotheses. A model with \(M\cap V=\varnothing\) makes (D) false. Hence none of choices (A) through (D) is forced.

Therefore, the correct answer is E.

29 答案 C

设直径 \(d=AB\)。因为 \(AC\) 与 \(BD\) 的交点在圆上，所以由泰勒斯定理，这两条直线互相垂直。此外，切线 \(AD\) 与 \(BC\) 平行。由此得到的直角三角形 \(ADB\) 与 \(BCA\) 相似，所以 \[ \frac da=\frac bd\text{。}\]因此 \(d^2=ab\)，且 \(d=\sqrt{ab}\)。

因此，正确答案是 C。

English

Let \(d=AB\) be the diameter. Since the intersection of \(AC\) and \(BD\) lies on the circle, those two lines are perpendicular by Thales’ theorem. Also the tangents \(AD\) and \(BC\) are parallel. The resulting right triangles \(ADB\) and \(BCA\) are similar, so \[ \frac da=\frac bd. \] Hence \(d^2=ab\) and \(d=\sqrt{ab}.\)

Therefore, the correct answer is C.

30 答案 D

\(n-2\) 台正常销售带来 \((n-2)(\frac{d}{n}+8)\) 的收入，两台义卖收音机合计带来 \(\frac{d}{n}\) 的收入。由于总收入为 \(d+72\)，\[ \begin{aligned} &(n-2)\left(\frac dn+8\right)+\frac dn\\ &\qquad=d+72 \end{aligned}\text{。}\]化简得 \(d=8n(n-11)\)。正性要求 \(n\gt11\)，而 \(n=12\) 给出正整数 \(d=96\)。

因此，正确答案是 D。

English

The \(n-2\) regular sales bring \((n-2)(\frac{d}{n}+8),\) while the two bazaar radios bring \(\frac{d}{n}\) together. Since the total intake is \(d+72,\) \[ \begin{aligned} &(n-2)\left(\frac dn+8\right)+\frac dn\\ &\qquad=d+72. \end{aligned} \] This reduces to \(d=8n(n-11).\) Positivity requires \(n\gt11,\) and \(n=12\) gives the positive integer \(d=96.\)

Therefore, the correct answer is D.

31 答案 C

取 \(b=a+1\) 且 \(c=a(a+1)\)，则 \[ \begin{aligned} D&=a^2+(a+1)^2+a^2(a+1)^2\\ &=(a^2+a+1)^2 \end{aligned}\text{。}\]因此 \(\sqrt D=a^2+a+1\)。因为 \(a(a+1)\) 为偶数，所以这个整数总是奇数。

因此，正确答案是 C。

English

With \(b=a+1\) and \(c=a(a+1),\) \[ \begin{aligned} D&=a^2+(a+1)^2+a^2(a+1)^2\\ &=(a^2+a+1)^2. \end{aligned} \] Thus \(\sqrt D=a^2+a+1.\) Since \(a(a+1)\) is even, this integer is always odd.

Therefore, the correct answer is C.

32 答案 E

在三角形 \(AOB\) 中，\[ \begin{aligned} \cos\angle AOB &=\frac{8^2+4^2-6^2}{2\cdot8\cdot4}\\ &=\frac{11}{16} \end{aligned}\text{。}\]因此 \(\cos\angle AOD=-\frac{11}{16}\)。在三角形 \(AOD\) 中应用余弦定理可得 \[ \begin{aligned} AD^2 &=8^2+6^2\\ &\quad+2(8)(6)\frac{11}{16}\\ &=166 \end{aligned}\text{。}\]所以 \(AD=\sqrt{166}\)。

因此，正确答案是 E。

English

In triangle \(AOB,\) \[ \begin{aligned} \cos\angle AOB &=\frac{8^2+4^2-6^2}{2\cdot8\cdot4}\\ &=\frac{11}{16}. \end{aligned} \] Therefore \(\cos\angle AOD=-\frac{11}{16}.\) Applying the law of cosines in triangle \(AOD\) gives \[ \begin{aligned} AD^2 &=8^2+6^2\\ &\quad+2(8)(6)\frac{11}{16}\\ &=166. \end{aligned} \] Hence \(AD=\sqrt{166}.\)

Therefore, the correct answer is E.

33 答案 D

设 \(AC=u\)、\(CB=v\)。阴影面积等于大半圆的面积减去两个小半圆的面积：\[ \frac{\pi}{8}\bigl((u+v)^2-u^2-v^2\bigr) =\frac{\pi uv}{4}\text{。}\]因为 \(D\) 位于以 \(AB\) 为直径的半圆上，所以三角形 \(ADB\) 是直角三角形，且其高满足 \(CD^2=uv\)。因此，以 \(CD\) 为半径的圆面积为 \(\pi uv\)。所求比为 \(1:4\)。

因此，正确答案是 D。

English

Let \(AC=u\) and \(CB=v.\) The shaded area is the large semicircle minus the two smaller ones: \[ \frac{\pi}{8}\bigl((u+v)^2-u^2-v^2\bigr) =\frac{\pi uv}{4}. \] Since \(D\) lies on the semicircle with diameter \(AB,\) triangle \(ADB\) is right, and its altitude satisfies \(CD^2=uv.\) A circle of radius \(CD\) therefore has area \(\pi uv.\) The required ratio is \(1:4.\)

Therefore, the correct answer is D.

34 答案 A

在每个顶点处，小三角形所占相邻两边的比例分别为 \(\frac{1}{n+1}\) 和 \(\frac{n}{n+1}\)。因此，每个顶角处的小三角形面积都是 \([ABC]\) 的 \(\frac{n}{(n+1)^2}\) 倍。所以 \[ \begin{aligned} \frac{[DEF]}{[ABC]} &=1-\frac{3n}{(n+1)^2}\\ &=\frac{n^2-n+1}{(n+1)^2} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

At each vertex, the two adjacent side fractions used by the corner triangle are \(\frac{1}{n+1}\) and \(\frac{n}{n+1}.\) Thus each corner triangle has area \(\frac{n}{(n+1)^2}\) times \([ABC].\) Therefore \[ \begin{aligned} \frac{[DEF]}{[ABC]} &=1-\frac{3n}{(n+1)^2}\\ &=\frac{n^2-n+1}{(n+1)^2}. \end{aligned} \]

Therefore, the correct answer is A.

35 答案 B

设三个根为 \(t-d\)、\(t\)、\(t+d\)。它们的和为 \(3t=\frac{144}{64}=\frac{9}{4}\)，所以 \(t=\frac{3}{4}\)。它们的积为 \[ t(t^2-d^2)=\frac{15}{64}\text{。}\]代入 \(t=\frac{3}{4}\) 得 \(d^2=\frac{1}{4}\)，所以最大根与最小根之差为 \(2\lvert d\rvert=1\)。

因此，正确答案是 B。

English

Let the roots be \(t-d,\) \(t,\) and \(t+d.\) Their sum is \(3t=\frac{144}{64}=\frac{9}{4},\) so \(t=\frac{3}{4}.\) Their product is \[ t(t^2-d^2)=\frac{15}{64}. \] Substituting \(t=\frac{3}{4}\) gives \(d^2=\frac{1}{4},\) so the difference between the extreme roots is \(2\lvert d\rvert=1.\)

Therefore, the correct answer is B.

36 答案 C

设最简形式的公比为 \(\frac{c}{d}\)，中间项为 \(a\)。因为五项都是整数，所以 \(a\) 是 \(c^2d^2\) 的倍数，可写成 \(a=kc^2d^2\)。此时总和 \(211\) 能被 \(k\) 整除，所以 \(k\) 为 \(1\) 或 \(211\)。但中间项 \(a\) 小于 \(100\)，故 \(k=1\)。

因此 \[ \begin{aligned} d^4+d^3c+d^2c^2&\\ \quad+dc^3+c^4&=211 \end{aligned}\text{。}\]由范围限制可知 \(c,d\lt4\)。若其中一个为 \(1\)，可能的总和为 \(5\)、\(31\) 或 \(121\)，都不是 \(211\)。再结合互质条件，只剩 \(c\) 与 \(d\) 分别为 \(2\) 和 \(3\)，次序可以互换。这个数列为 \(16\)、\(24\)、\(36\)、\(54\)、\(81\)。其中完全平方数项之和为 \(16+36+81=133\)。

因此，正确答案是 C。

English

Let the common ratio be \(\frac{c}{d}\) in lowest terms and the middle term be \(a.\) Since all five terms are integers, \(a\) is divisible by \(c^2d^2,\) so write \(a=kc^2d^2.\) The sum \(211\) is then divisible by \(k.\) Thus \(k\) is \(1\) or \(211.\) But the middle term \(a\) is less than \(100,\) so \(k=1.\)

Thus \[ \begin{aligned} d^4+d^3c+d^2c^2&\\ \quad+dc^3+c^4&=211. \end{aligned} \] The bound gives \(c,d\lt4.\) If either is \(1,\) the possible sums are \(5,\) \(31,\) or \(121,\) not \(211.\) Coprimality therefore leaves \(c\) and \(d\) equal to \(2\) and \(3\) in either order. The progression is \(16,\) \(24,\) \(36,\) \(54,\) \(81.\) Its square terms sum to \(16+36+81=133.\)

Therefore, the correct answer is C.

37 答案 A

由于 \(RS\) 不与三角形相交，三个垂直距离的符号相同。点到固定直线的有向距离是仿射函数，而重心是三个顶点的平均。因此，重心到直线的距离等于三个距离的平均值：\[ x=\frac{10+6+24}{3}=\frac{40}{3}\text{。}\]

因此，正确答案是 A。

English

Because \(RS\) does not intersect the triangle, the three perpendicular distances have the same sign. Signed distance to a fixed line is affine, and the centroid is the average of the vertices. Therefore its distance is the average \[ x=\frac{10+6+24}{3}=\frac{40}{3}. \]

Therefore, the correct answer is A.

38 答案 E

将四个 pib 标记为 \(1\)、\(2\)、\(3\)、\(4\)。由 \(\mathrm P_2\) 和 \(\mathrm P_3\) 可知，每个 maa 恰好是一对无序 pib 所共有的 maa。因此共有 \(\binom42=6\) 个 maa，这就证明了 \(\mathrm T_1\)。

固定一个 pib \(i\)，它含有三个形如 \(ij\) 的 maa，其中 \(j\ne i\)，这证明了 \(\mathrm T_2\)。对于 maa \(ij\)，唯一一个与它不共属任一 pib 的 maa，就是属于其余两个 pib 所成之互补对的 maa，这证明了 \(\mathrm T_3\)。因此三个定理都能推出。

因此，正确答案是 E。

English

Label the four pibs \(1,\) \(2,\) \(3,\) and \(4.\) By \(\mathrm P_2\) and \(\mathrm P_3,\) every maa is exactly the common maa of one unordered pair of pibs. Thus there are \(\binom42=6\) maas, proving \(\mathrm T_1.\)

A fixed pib \(i\) contains the three maas \(ij\) with \(j\ne i,\) proving \(\mathrm T_2.\) For maa \(ij,\) the unique maa sharing neither pib is the one belonging to the complementary pair of pibs, proving \(\mathrm T_3.\) All three follow.

Therefore, the correct answer is E.

39 答案 B

第 \(n\) 个集合以 \(\frac{n(n+1)}{2}\) 为末项，并含有 \(n\) 个连续整数。其总和为 \[ \begin{aligned} S_n &=n\left(\frac{n(n+1)}2\right)\\ &\quad-\frac{n(n-1)}2\\ &=\frac{n(n^2+1)}2 \end{aligned}\text{。}\]因此 \(S_{21}=\frac{21(442)}{2}=4641\)。

因此，正确答案是 B。

English

The \(n\)th set ends at \(\frac{n(n+1)}{2}\) and contains \(n\) consecutive integers. Its sum is \[ \begin{aligned} S_n &=n\left(\frac{n(n+1)}2\right)\\ &\quad-\frac{n(n-1)}2\\ &=\frac{n(n^2+1)}2. \end{aligned} \] Thus \(S_{21}=\frac{21(442)}{2}=4641.\)

Therefore, the correct answer is B.

40 答案 D

绕 \(A\) 将 \(P\) 旋转 \(60^\circ\) 到 \(P'\)。由于该旋转把 \(C\) 映到 \(B\)，所以 \(P'B=PC=10\)、\(PP'=PA=8\)，且 \(PB=6\)。因此三角形 \(PP'B\) 是直角三角形。又因为三角形 \(APP'\) 是等边三角形，所以 \(\angle APB=60^\circ+90^\circ=150^\circ\)。

设等边三角形边长为 \(s\)，在三角形 \(APB\) 中应用余弦定理可得 \[ \begin{aligned} s^2 &=6^2+8^2-2(6)(8)\cos150^\circ\\ &=100+48\sqrt3 \end{aligned}\text{。}\]它的面积为 \[ \frac{\sqrt3}{4}s^2=36+25\sqrt3\text{，}\]最接近的整数是 \(79\)。

因此，正确答案是 D。

English

Rotate \(P\) by \(60^\circ\) about \(A\) to \(P'.\) Since this rotation sends \(C\) to \(B,\) we have \(P'B=PC=10,\) \(PP'=PA=8,\) and \(PB=6.\) Thus triangle \(PP'B\) is right. Also triangle \(APP'\) is equilateral, so \(\angle APB=60^\circ+90^\circ=150^\circ.\)

If the equilateral triangle has side \(s,\) the law of cosines in triangle \(APB\) gives \[ \begin{aligned} s^2 &=6^2+8^2-2(6)(8)\cos150^\circ\\ &=100+48\sqrt3. \end{aligned} \] Its area is \[ \frac{\sqrt3}{4}s^2=36+25\sqrt3, \] whose nearest integer is \(79.\)

Therefore, the correct answer is D.