1971 AMC 12 真题（P版）

1 答案 B

有 \(2^{12}5^8=2^4(2^85^8)\)，且 \(16\cdot10^8=1{,}600{,}000{,}000\)，该数共有 \(10\) 位。

因此，正确答案为 B。

English

We have \(2^{12}5^8=2^4(2^85^8)\) and \(16\cdot10^8=1{,}600{,}000{,}000,\) which has \(10\) digits.

Therefore, the correct answer is B.

2 答案 D

每个工日的效率是 \(\frac{f}{bc}\) 块砖。因此，\(c\) 名工人每天砌 \(\frac{f}{b}\) 块砖，所以砌好 \(b\) 块砖需要 \[ \frac{b}{\frac{f}{b}}=\frac{b^2}{f} \] 天。

因此，正确答案为 D。

English

The rate is \(\frac{f}{bc}\) bricks per man-day. Thus \(c\) men lay \(\frac{f}{b}\) bricks per day, so laying \(b\) bricks takes \[ \frac{b}{\frac{f}{b}}=\frac{b^2}{f} \] days.

Therefore, the correct answer is D.

3 答案 E

过 \((0,8)\) 与 \((-4,0)\) 的直线斜率为 \(2\)，所以直线方程为 \(y=2x+8\)。代入 \(y=-4\)，得 \(-4=2x+8\)，因此 \(x=-6\)。

因此，正确答案为 E。

English

The slope through \((0,8)\) and \((-4,0)\) is \(2,\) so the line is \(y=2x+8.\) Substituting \(y=-4\) gives \(-4=2x+8,\) hence \(x=-6.\)

Therefore, the correct answer is E.

4 答案 A

若 \(P\) 是本金，则 \[ \begin{aligned} 255.31&=P\left(1+\frac{0.05}{6}\right)\\ &=P\frac{121}{120}\text{。} \end{aligned} \] 因此 \(P=253.20\)，利息为 \(\$2.11\)，其中的美分数为 \(11\)。

因此，正确答案为 A。

English

If \(P\) is the principal, then \[ \begin{aligned} 255.31&=P\left(1+\frac{0.05}{6}\right)\\ &=P\frac{121}{120}. \end{aligned} \] Hence \(P=253.20,\) so the interest is \(\$2.11\) and its cents part is \(11.\)

Therefore, the correct answer is A.

5 答案 C

令 \(m\overset{\frown}{AC}=u\)。由圆外两割线所成角公式， \[ m\angle P=\frac12\left(m\overset{\frown}{BD}-u\right)\text{，} \] 而圆周角定理给出 \(m\angle Q=\frac{u}{2}\)。因此 \[ \begin{aligned} m\angle P+m\angle Q &=\frac12m\overset{\frown}{BD}\\ &=\frac12(42^\circ+38^\circ)\\ &=40^\circ\text{。} \end{aligned} \]

因此，正确答案为 C。

English

Let \(m\overset{\frown}{AC}=u.\) The external-secant formula gives \[ m\angle P=\frac12\left(m\overset{\frown}{BD}-u\right), \] while the inscribed-angle theorem gives \(m\angle Q=\frac{u}{2}.\) Therefore \[ \begin{aligned} m\angle P+m\angle Q &=\frac12m\overset{\frown}{BD}\\ &=\frac12(42^\circ+38^\circ)\\ &=40^\circ. \end{aligned} \]

Therefore, the correct answer is C.

6 答案 E

该运算满足交换律，并且 \[ (a*b)*c=4abc=a*(b*c)\text{，} \] 所以它也满足结合律。其单位元为 \(\frac{1}{2}\)。\(a\) 的逆元必须满足 \(2ax=\frac{1}{2}\)，故逆元为 \(x=\frac{1}{4a}\)，而不是 \(\frac{1}{2a}\)。

因此，正确答案为 E。

English

The operation is commutative, and \[ (a*b)*c=4abc=a*(b*c), \] so it is associative. Its identity is \(\frac{1}{2}.\) The inverse of \(a\) must satisfy \(2ax=\frac{1}{2},\) so it is \(x=\frac{1}{4a},\) not \(\frac{1}{2a}.\)

Therefore, the correct answer is E.

7 答案 C

提取公因式可得 \[ 2^{-(2k+1)}(1-4+2)=-2^{-(2k+1)}\text{。} \]

因此，正确答案为 C。

English

Factoring gives \[ 2^{-(2k+1)}(1-4+2)=-2^{-(2k+1)}. \]

Therefore, the correct answer is C.

8 答案 B

该不等式等价于 \[ \begin{aligned} 6x^2+5x-4&\lt0,\\ (3x+4)(2x-1)&\lt0\text{。} \end{aligned} \] 两个根是 \(-\frac{4}{3}\) 和 \(\frac{1}{2}\)，开口向上的二次函数在两根之间为负。

因此，正确答案为 B。

English

The inequality is \[ \begin{aligned} 6x^2+5x-4&\lt0,\\ (3x+4)(2x-1)&\lt0. \end{aligned} \] The roots are \(-\frac{4}{3}\) and \(\frac{1}{2},\) and the upward-opening quadratic is negative between them.

Therefore, the correct answer is B.

9 答案 D

公外切线与连接圆心和切点的两条半径构成一个直角三角形，其两条直角边长为 \(24\) 和 \(14-4=10\)。因此圆心距为 \[ \sqrt{24^2+10^2}=\sqrt{676}=26\text{。} \]

因此，正确答案为 D。

English

The common external tangent and the radii to its contact points form a right triangle whose legs are \(24\) and \(14-4=10.\) Thus the center distance is \[ \sqrt{24^2+10^2}=\sqrt{676}=26. \]

Therefore, the correct answer is D.

10 答案 E

共有 \(50-31=19\) 名金发女孩，其中 \(19-14=5\) 名是棕眼睛。因此，棕眼睛且深色头发的女孩有 \(18-5=13\) 名。

因此，正确答案为 E。

English

There are \(50-31=19\) blondes, of whom \(19-14=5\) are brown-eyed. Hence the number of brown-eyed brunettes is \(18-5=13.\)

Therefore, the correct answer is E.

11 答案 D

等式为 \(4a+7=7b+4\)，即 \(4a-7b=-3\)。由于两个数码都必须有效，所以 \(a,b\gt7\)。模 \(7\) 化简得 \(a\equiv1\pmod7\)。第一个有效值是 \(a=15\)，对应 \(b=9\)。因此 \(a+b=24\)，写成罗马数字为 \(\mathrm{XXIV}\)。

因此，正确答案为 D。

English

The equality is \(4a+7=7b+4,\) or \(4a-7b=-3.\) Since both digits must be valid, \(a,b\gt7.\) Reducing modulo \(7\) gives \(a\equiv1\pmod7.\) The first valid value is \(a=15,\) which gives \(b=9.\) Thus \(a+b=24,\) written \(\mathrm{XXIV}.\)

Therefore, the correct answer is D.

12 答案 B

模数 \(N\) 同时整除 \(90-69=21\) 与 \(125-90=35\)。由于 \(N\gt1\)，只能有 \(N=7\)。于是 \(81\equiv4\pmod7\)。

因此，正确答案为 B。

English

The modulus \(N\) divides both \(90-69=21\) and \(125-90=35.\) Since \(N\gt1,\) this forces \(N=7.\) Then \(81\equiv4\pmod7.\)

Therefore, the correct answer is B.

13 答案 E

二项式展开给出 \[ \begin{gathered} (1+0.0025)^{10}\\ =1+10(0.0025)\\ \quad+45(0.0025)^2\\ \quad+120(0.0025)^3+\cdots\\ =1.025283125\ldots \end{gathered} \] 四舍五入后为 \(1.02528\)。小数点后第五位数字是 \(8\)。

因此，正确答案为 E。

English

The binomial expansion gives \[ \begin{gathered} (1+0.0025)^{10}\\ =1+10(0.0025)\\ \quad+45(0.0025)^2\\ \quad+120(0.0025)^3+\cdots\\ =1.025283125\ldots \end{gathered} \] which rounds to \(1.02528.\) The fifth decimal digit is \(8.\)

Therefore, the correct answer is E.

14 答案 C

因为 \(48\) 是 \(6\) 的倍数，所以 \[ 2^{48}-1 \] 是 \(63=2^6-1\) 的倍数。又有 \(2^{12}-1=4095=63\cdot65\)，且 \(48\) 是 \(12\) 的倍数，所以 \(2^{48}-1\) 是 \(65\) 的倍数。因此这两个数是 \(63\) 和 \(65\)。

因此，正确答案为 C。

English

Because \(48\) is divisible by \(6,\) the number \[ 2^{48}-1 \] is divisible by \(63=2^6-1.\) Also \(2^{12}-1=4095=63\cdot65,\) and \(48\) is divisible by \(12,\) so \(2^{48}-1\) is divisible by \(65.\) Thus the two numbers are \(63\) and \(65.\)

Therefore, the correct answer is C.

15 答案 B

设水族箱长为 \(L\)，水平放置时的水深为 \(h\)。倾斜时，水的纵向截面是一个底为 \(\frac{3L}{4}\)、高为 \(8\) 的三角形。因此 \[ 10\left(\frac12\cdot\frac{3L}{4}\cdot8\right)=10Lh\text{，} \] 所以 \(h=3\) 英寸。

因此，正确答案为 B。

English

Let the aquarium length be \(L\) and the level-water depth be \(h.\) In the tilted position the longitudinal cross-section of the water is a triangle with base \(\frac{3L}{4}\) and height \(8.\) Therefore \[ 10\left(\frac12\cdot\frac{3L}{4}\cdot8\right)=10Lh, \] so \(h=3\) inches.

Therefore, the correct answer is B.

16 答案 A

若真实平均数为 \(x\)，则原来的总和为 \(35x\)。再加入 \(x\) 本身后，\(36\) 个数的总和为 \(36x\)，所以新平均数仍为 \(x\)。所求比为 \(1:1\)。

因此，正确答案为 A。

English

If the true average is \(x,\) then the original sum is \(35x.\) Including \(x\) itself gives a sum of \(36x\) over \(36\) numbers, so the new average is still \(x.\) The ratio is \(1:1.\)

Therefore, the correct answer is A.

17 答案 E

这些半径先形成 \(2n\) 个扇形。一条不过圆心的割线在 \(n\) 对反向半径中的每一对里至多与一条相交，因此至多与 \(n\) 条半径相交。这些交点把圆内的割线弦段分成 \(n+1\) 段，每段都会增加一个区域。因此最大区域数为 \(2n+(n+1)=3n+1\)。

因此，正确答案为 E。

English

The radii first make \(2n\) sectors. A secant not through the center can meet at most one radius in each of \(n\) opposite pairs, hence at most \(n\) radii. Those intersections divide the secant chord into \(n+1\) pieces, each of which adds one region. The maximum is therefore \(2n+(n+1)=3n+1.\)

Therefore, the correct answer is E.

18 答案 D

若 \(v\) 是船在静水中的速度，则 \[ \frac4{v+3}+\frac4{v-3}=1\text{。} \] 化简得 \(v^2-8v-9=0\)，所以满足条件的正根是 \(v=9\)。顺流与逆流速度分别为 \(12\) 和 \(6\)，其比为 \(2:1\)。

因此，正确答案为 D。

English

If \(v\) is the still-water speed, then \[ \frac4{v+3}+\frac4{v-3}=1. \] This simplifies to \(v^2-8v-9=0,\) so the positive admissible solution is \(v=9.\) The downstream and upstream rates are \(12\) and \(6,\) whose ratio is \(2:1.\)

Therefore, the correct answer is D.

19 答案 C

代入可得 \[ (1+4m^2)x^2+8mx+3=0\text{。} \] 相切要求判别式为零： \[ 64m^2-12(1+4m^2)=0\text{。} \] 因此 \(16m^2=12\)，所以 \(m^2=\frac{3}{4}\)。

因此，正确答案为 C。

English

Substitution gives \[ (1+4m^2)x^2+8mx+3=0. \] Tangency requires its discriminant to vanish: \[ 64m^2-12(1+4m^2)=0. \] Thus \(16m^2=12,\) so \(m^2=\frac{3}{4}.\)

Therefore, the correct answer is C.

20 答案 E

设两根为 \(r,s\)，由韦达定理，\(r+s=-2h\) 且 \(rs=-3\)。于是 \[ \begin{aligned} 4h^2=(r+s)^2 &=r^2+s^2+2rs\\ &=10-6=4\text{。} \end{aligned} \] 因此 \(|h|=1\)。选项 A 为 \(-1\)，所以选项 A 至 D 中均没有该值。

因此，正确答案为 E。

English

For roots \(r,s,\) Vieta’s formulas give \(r+s=-2h\) and \(rs=-3.\) Hence \[ \begin{aligned} 4h^2=(r+s)^2 &=r^2+s^2+2rs\\ &=10-6=4. \end{aligned} \] Therefore \(|h|=1,\) which is not among choices A-D because choice A is \(-1.\)

Therefore, the correct answer is E.

21 答案 C

从最外层向内逐层撤销对数，可得 \[ \begin{gathered} x=4^3=64,\\ y=2^4=16,\\ z=3^2=9\text{。} \end{gathered} \] 因此 \(x+y+z=64+16+9=89\)。

因此，正确答案为 C。

English

Undoing the logarithms from the outside inward gives \[ \begin{gathered} x=4^3=64,\\ y=2^4=16,\\ z=3^2=9. \end{gathered} \] Therefore \(x+y+z=64+16+9=89.\)

Therefore, the correct answer is C.

22 答案 A

由于 \(w\ne1\) 且 \(w^3=1\)，有 \(1+w+w^2=0\)。因此 \[ \begin{gathered} 1-w+w^2=-2w,\\ 1+w-w^2=-2w^2\text{。} \end{gathered} \] 两式之积为 \(4w^3=4\)。

因此，正确答案为 A。

English

Because \(w\ne1\) and \(w^3=1,\) we have \(1+w+w^2=0.\) Thus \[ \begin{gathered} 1-w+w^2=-2w,\\ 1+w-w^2=-2w^2. \end{gathered} \] Their product is \(4w^3=4.\)

Therefore, the correct answer is A.

23 答案 A

队 \(A\) 在序列 BBB、ABBB、BABB 和 BBAB 中输掉系列赛。这些序列的总概率为 \[ \frac18+3\left(\frac1{16}\right)=\frac5{16}\text{。} \] 因此队 \(A\) 获胜的概率为 \(\frac{11}{16}\)，其有利赔率为 \(11:5\)。

因此，正确答案为 A。

English

Team \(A\) loses in the sequences BBB, ABBB, BABB, and BBAB. Their total probability is \[ \frac18+3\left(\frac1{16}\right)=\frac5{16}. \] Thus Team \(A\) wins with probability \(\frac{11}{16},\) so the odds in its favor are \(11:5.\)

Therefore, the correct answer is A.

24 答案 D

前 \(n\) 行共有 \(1+2+\cdots+n=\frac{n(n+1)}{2}\) 项。边界上有 \(2n-1\) 个 \(1\)，所以其余项的个数为 \[ \begin{gathered} \frac{n(n+1)}2-(2n-1)\\ =\frac{n^2+n-4n+2}{2}\\ =\frac{n^2-3n+2}{2}\text{。} \end{gathered} \] 除以 \(2n-1\)，得到 \(\frac{n^2-3n+2}{4n-2}\)。

因此，正确答案为 D。

English

The first \(n\) rows contain \(1+2+\cdots+n=\frac{n(n+1)}{2}\) entries. There are \(2n-1\) boundary \(1\)’s, so the number of other entries is \[ \begin{gathered} \frac{n(n+1)}2-(2n-1)\\ =\frac{n^2+n-4n+2}{2}\\ =\frac{n^2-3n+2}{2}. \end{gathered} \] Dividing by \(2n-1\) gives \(\frac{n^2-3n+2}{4n-2}.\)

Therefore, the correct answer is D.

25 答案 D

设父亲和少年的年龄分别为 \(f\) 与 \(b\)。由于 \(f\gt b\)， \[ \begin{aligned} 100f+b-(f-b)&=4289,\\ 99f+2b&=4289\text{。} \end{aligned} \] 少年处于十几岁，所以 \(13\le b\le19\)。模 \(9\) 化简得 \(2b\equiv5\pmod9\)，因此 \(b=16\)。于是 \(f=\frac{4289-32}{99}=43\)，且 \(f+b=59\)。

因此，正确答案为 D。

English

Let the father and boy be \(f\) and \(b\) years old. Since \(f\gt b,\) \[ \begin{aligned} 100f+b-(f-b)&=4289,\\ 99f+2b&=4289. \end{aligned} \] The boy is a teenager, so \(13\le b\le19.\) Reducing modulo \(9\) gives \(2b\equiv5\pmod9,\) hence \(b=16.\) Then \(f=\frac{4289-32}{99}=43,\) and \(f+b=59.\)

Therefore, the correct answer is D.

26 答案 B

使用质量点法。由于 \(AF:FC=1:2\)，给 \(A\) 与 \(C\) 分别赋质量 \(2\) 与 \(1\)，于是 \(F\) 处的质量为 \(3\)。因为 \(G\) 是 \(BF\) 的中点，所以 \(B\) 处的质量也为 \(3\)。因此 \[ BE:EC=m_C:m_B=1:3\text{。} \]

因此，正确答案为 B。

English

Use mass points. Since \(AF:FC=1:2,\) assign masses \(2\) and \(1\) to \(A\) and \(C,\) so the mass at \(F\) is \(3.\) Because \(G\) is the midpoint of \(BF,\) the mass at \(B\) is also \(3.\) Therefore \[ BE:EC=m_C:m_B=1:3. \]

Therefore, the correct answer is B.

27 答案 E

设三种筹码的数量分别为 \(r,w,b\)。条件给出 \[ \begin{gathered} w\le2b,\\ r\ge3b,\\ w+b\ge55\text{。} \end{gathered} \] 因此 \(3b\ge w+b\ge55\)，所以 \(b\ge19\)，且 \(r\ge57\)。取 \((w,b,r)=(36,19,57)\) 时可以达到等号，因此最小值为 \(57\)。

因此，正确答案为 E。

English

Let the counts be \(r,w,b.\) The conditions give \[ \begin{gathered} w\le2b,\\ r\ge3b,\\ w+b\ge55. \end{gathered} \] Hence \(3b\ge w+b\ge55,\) so \(b\ge19\) and \(r\ge57.\) Equality is possible with \((w,b,r)=(36,19,57),\) so the minimum is \(57.\)

Therefore, the correct answer is E.

28 答案 C

设总面积为 \(K\)。最下面条带上方的三角形与原三角形相似，线性比例为 \(\frac{9}{10}\)，所以其面积为 \(\frac{81K}{100}\)。因此最大条带的面积为 \[ K-\frac{81K}{100}=\frac{19K}{100}=38\text{，} \] 从而 \(K=200\)。

因此，正确答案为 C。

English

If the whole area is \(K,\) the triangle above the bottom strip is similar to the original with scale \(\frac{9}{10},\) so its area is \(\frac{81K}{100}.\) Thus the largest strip has area \[ K-\frac{81K}{100}=\frac{19K}{100}=38, \] giving \(K=200.\)

Therefore, the correct answer is C.

29 答案 E

该乘积为 \[ 10^{\frac{1+2+\cdots+n}{11}}=10^{\frac{n(n+1)}{22}}\text{。} \] 它大于 \(100{,}000=10^5\) 当且仅当 \(n(n+1)\gt110\)。当 \(n=10\) 时恰好相等，而 \(n=11\) 时满足条件。

因此，正确答案为 E。

English

The product is \[ 10^{\frac{1+2+\cdots+n}{11}}=10^{\frac{n(n+1)}{22}}. \] It exceeds \(100{,}000=10^5\) exactly when \(n(n+1)\gt110.\) For \(n=10\) there is equality, while \(n=11\) works.

Therefore, the correct answer is E.

30 答案 D

该变换可逆。由 \(f_{35}=f_5\)，与 \(f_5^{-1}\) 复合可知 \(f_{30}\) 是恒等变换。由 \(y=\frac{2x-1}{x+1}\) 解出 \(x\)，得 \[ g(y)=f_1^{-1}(y)=\frac{y+1}{2-y}\text{。} \] 由于迭代的周期为 \(30\)，\(f_{28}=f_{-2}=g^2\)。直接复合得到 \[ g(g(x))=\frac{1}{1-x}\text{。} \]

因此，正确答案为 D。

English

The transformation is invertible. From \(f_{35}=f_5,\) composing with \(f_5^{-1}\) shows that \(f_{30}\) is the identity. Solving \(y=\frac{2x-1}{x+1}\) for \(x\) gives \[ g(y)=f_1^{-1}(y)=\frac{y+1}{2-y}. \] Since iterates have period \(30,\) \(f_{28}=f_{-2}=g^2.\) Direct composition gives \[ g(g(x))=\frac{1}{1-x}. \]

Therefore, the correct answer is D.

31 答案 A

该圆的半径为 \(2\)。设等弦 \(AB\) 与 \(BC\) 所对的圆心角都为 \(2t\)。则 \[ 1=4\sin t\text{，} \] 所以 \(\sin t=\frac{1}{4}\)。沿半圆从 \(C\) 到 \(D\) 的剩余圆心角为 \(\pi-4t\)，故 \[ \begin{aligned} CD&=4\sin\left(\frac{\pi-4t}{2}\right)\\ &=4\cos(2t)\\ &=4\left(1-2\cdot\frac1{16}\right)\\ &=\frac72\text{。} \end{aligned} \]

因此，正确答案为 A。

English

The circle has radius \(2.\) Let the central angles subtending the equal chords \(AB\) and \(BC\) each be \(2t.\) Then \[ 1=4\sin t, \] so \(\sin t=\frac{1}{4}.\) The remaining central angle from \(C\) to \(D\) along the semicircle is \(\pi-4t,\) hence \[ \begin{aligned} CD&=4\sin\left(\frac{\pi-4t}{2}\right)\\ &=4\cos(2t)\\ &=4\left(1-2\cdot\frac1{16}\right)\\ &=\frac72. \end{aligned} \]

Therefore, the correct answer is A.

32 答案 A

令 \(x=2^{-\frac{1}{32}}\)。则 \[ \begin{aligned} (1-x)s&=1-x^{32}\\ &=1-\frac12=\frac12\text{。} \end{aligned} \] 因此 \[ s=\frac12(1-2^{-\frac{1}{32}})^{-1}\text{。} \]

因此，正确答案为 A。

English

Let \(x=2^{-\frac{1}{32}}.\) Then \[ \begin{aligned} (1-x)s&=1-x^{32}\\ &=1-\frac12=\frac12. \end{aligned} \] Therefore \[ s=\frac12(1-2^{-\frac{1}{32}})^{-1}. \]

Therefore, the correct answer is A.

33 答案 B

把这些项写成 \(a,ar,\ldots,ar^{n-1}\)。将倒数之和反向排列可得 \[ S'=\frac{S}{a^2r^{n-1}}\text{，} \] 所以 \(\frac{S}{S'}=a^2r^{n-1}\)。另一方面， \[ \begin{aligned} P&=a^nr^{\frac{n(n-1)}{2}}\\ &=\left(a^2r^{n-1}\right)^{\frac{n}{2}}\\ &=\left(\frac{S}{S'}\right)^{\frac{n}{2}}\text{。} \end{aligned} \]

因此，正确答案为 B。

English

Write the terms as \(a,ar,\ldots,ar^{n-1}.\) Reversing the reciprocal sum gives \[ S'=\frac{S}{a^2r^{n-1}}, \] so \(\frac{S}{S'}=a^2r^{n-1}.\) Meanwhile \[ \begin{aligned} P&=a^nr^{\frac{n(n-1)}{2}}\\ &=\left(a^2r^{n-1}\right)^{\frac{n}{2}}\\ &=\left(\frac{S}{S'}\right)^{\frac{n}{2}}. \end{aligned} \]

Therefore, the correct answer is B.

34 答案 B

正常情况下，相邻两次指针重合的实际间隔为 \(\frac{720}{11}\) 分钟，而这只慢钟在两次重合之间显示经过 \(69\) 分钟。因此，慢钟显示 \(12\) 小时时，实际经过 \(759\) 分钟，因为 \[ 11\cdot69=759=12\cdot60+39\text{。}\] 所以慢钟显示的八小时实际耗时 \(8\) 小时 \(26\) 分钟，加班时间为 \(26\) 分钟。按每小时 \(\$6\) 计算，加班工资为 \(\$2.60\)。

因此，正确答案为 B。

English

Successive hand-overlaps are \(\frac{720}{11}\) real minutes apart, while this slow clock displays \(69\) minutes between them. Thus \(12\) displayed hours correspond to \(759\) real minutes, since \[ 11\cdot69=759=12\cdot60+39. \] Eight displayed hours therefore take \(8\) hours \(26\) minutes, so the overtime is \(26\) minutes. At \(\$6\) per hour, that pays \(\$2.60.\)

Therefore, the correct answer is B.

35 答案 C

设相邻两圆半径为 \(r\gt r'\)，则它们的圆心位于角平分线上，距顶点分别为 \(r\sqrt2\) 和 \(r'\sqrt2\)。外切条件给出 \[ \sqrt2(r-r')=r+r'\text{，} \] 所以 \(\frac{r'}{r}=3-2\sqrt2\)。相邻两圆的面积比为 \[ q=(3-2\sqrt2)^2\text{。} \] 因此，第一个圆的面积除以后续所有圆的面积之和为 \[ \begin{aligned} \frac{1-q}{q} &=\frac1q-1\\ &=(3+2\sqrt2)^2-1\\ &=16+12\sqrt2\text{。} \end{aligned} \]

因此，正确答案为 C。

English

If consecutive radii are \(r\gt r',\) their centers lie on the angle bisector at distances \(r\sqrt2\) and \(r'\sqrt2\) from the vertex. External tangency gives \[ \sqrt2(r-r')=r+r', \] so \(\frac{r'}{r}=3-2\sqrt2.\) The ratio of successive areas is \[ q=(3-2\sqrt2)^2. \] Therefore the first area divided by the sum of all later areas is \[ \begin{aligned} \frac{1-q}{q} &=\frac1q-1\\ &=(3+2\sqrt2)^2-1\\ &=16+12\sqrt2. \end{aligned} \]

Therefore, the correct answer is C.