1993 AMC 12 真题（P版）

1 答案 D

根据定义，\[ \begin{aligned} \boxed{1,-1,2} &=1^{-1}-(-1)^2+2^1\\ &=1-1+2\\ &=2 \end{aligned}\text{。}\]因此正确答案是 D。

English

By the definition, \[ \begin{aligned} \boxed{1,-1,2} &=1^{-1}-(-1)^2+2^1\\ &=1-1+2\\ &=2. \end{aligned} \] Thus the correct answer is D.

2 答案 D

有 \(\angle ABC\) 等于 \(180^\circ-55^\circ-75^\circ=50^\circ\)。因为 \(D\) 和 \(E\) 分别位于 \(\overline{BA}\) 和 \(\overline{BC}\) 上，所以 \(\angle DBE=50^\circ\)。又因为 \(DB=BE\)，\(\triangle DBE\) 的两个底角均为 \[ \frac{180^\circ-50^\circ}{2}=65^\circ\text{。}\]因此正确答案是 D。

English

We have \(\angle ABC\) equal to \(180^\circ-55^\circ-75^\circ=50^\circ.\) Because \(D\) and \(E\) lie on \(\overline{BA}\) and \(\overline{BC},\) respectively, \(\angle DBE=50^\circ.\) Since \(DB=BE,\) the two base angles of \(\triangle DBE\) are \[ \frac{180^\circ-50^\circ}{2}=65^\circ. \] Thus the correct answer is D.

3 答案 E

利用相同的指数，\[ \frac{15^{30}}{45^{15}} =\left(\frac{15^2}{45}\right)^{15} =5^{15}\text{。}\]因此正确答案是 E。

English

Using a common exponent, \[ \frac{15^{30}}{45^{15}} =\left(\frac{15^2}{45}\right)^{15} =5^{15}. \] Thus the correct answer is E.

4 答案 E

对任意实数 \(y\)，都有 \[ 3\circ y=4(3)-3y+3y=12\text{。}\]因此每个实数 \(y\) 都满足条件，所以这样的数多于 \(4\) 个。故正确答案是 E。

English

For every real \(y,\) \[ 3\circ y=4(3)-3y+3y=12. \] Hence every real number \(y\) works, so there are more than \(4\) such numbers. Thus the correct answer is E.

5 答案 A

自行车涨价 \(\$8\)，头盔涨价 \(\$4\)，总计涨价 \(\$12\)。去年的总价为 \(\$200\)，所以涨幅为 \[ \frac{12}{200}\cdot100\%=6\%\text{。}\]因此正确答案是 A。

English

The bicycle increases by \(\$8\) and the helmet by \(\$4,\) for a total increase of \(\$12.\) Last year’s combined cost was \(\$200,\) so the percent increase is \[ \frac{12}{200}\cdot100\%=6\%. \] Thus the correct answer is A.

6 答案 B

将各项都写成 \(2\) 的幂，得到 \[ \begin{aligned} \frac{8^{10}+4^{10}}{8^4+4^{11}} &=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\\ &=\frac{2^{20}(2^{10}+1)} {2^{12}(1+2^{10})}\\ &=2^8 \end{aligned}\text{。}\]它的正平方根是 \(2^4=16\)。因此正确答案是 B。

English

Writing all terms as powers of \(2,\) \[ \begin{aligned} \frac{8^{10}+4^{10}}{8^4+4^{11}} &=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\\ &=\frac{2^{20}(2^{10}+1)} {2^{12}(1+2^{10})}\\ &=2^8. \end{aligned} \] Its positive square root is \(2^4=16.\) Thus the correct answer is B.

7 答案 E

将这 \(24\) 个数字每四个分为一组，得到 \[ \begin{aligned} R_{24} &=R_4\left(1+10^4+10^8\right.\\ &\qquad\left.{}+10^{12}+10^{16}\right.\\ &\qquad\left.{}+10^{20}\right) \end{aligned}\text{。}\]因此 \(Q\) 中有六个一，每相邻两个一之间有三个零。所以共有 \(5\cdot3=15\) 个零。故正确答案是 E。

English

Grouping the \(24\) digits into six blocks of four gives \[ \begin{aligned} R_{24} &=R_4\left(1+10^4+10^8\right.\\ &\qquad\left.{}+10^{12}+10^{16}\right.\\ &\qquad\left.{}+10^{20}\right). \end{aligned} \] Thus \(Q\) has six ones, with three zeros between each consecutive pair. It therefore has \(5\cdot3=15\) zeros. Thus the correct answer is E.

8 答案 D

两个相切单位圆的圆心相距 \(2\) 个单位。若新圆与单位圆外切，其圆心到单位圆圆心的距离必须为 \(4\)；若内切，则为 \(2\)。距离对 \((4,4)\) 给出两个圆心，\((2,2)\) 也是如此。混合距离对 \((4,2)\) 和 \((2,4)\) 各给出一个圆心，因为相应的圆心轨迹圆内切。因此共有 \(2+2+1+1=6\) 个圆。故正确答案是 D。

English

Let the centers of the tangent unit circles be \(2\) units apart. The new center must be \(4\) units from a unit-circle center for external tangency and \(2\) units away for internal tangency. The distance pair \((4,4)\) gives two centers, as does \((2,2).\) Each mixed pair \((4,2)\) and \((2,4)\) gives one center because the corresponding center-circles are internally tangent. Hence there are \(2+2+1+1=6\) circles. Thus the correct answer is D.

9 答案 D

国家 \(A\) 的人均财富与 \(\frac{d}{c}\) 成正比，而国家 \(B\) 的人均财富与 \(\frac{f}{e}\) 成正比。二者之比为 \[ \frac{\frac{d}{c}}{\frac{f}{e}}=\frac{de}{cf}\text{。}\]因此正确答案是 D。

English

Country \(A\)’s per-capita share is proportional to \(\frac{d}{c},\) while country \(B\)’s is proportional to \(\frac{f}{e}.\) Their ratio is \[ \frac{\frac{d}{c}}{\frac{f}{e}}=\frac{de}{cf}. \] Thus the correct answer is D.

10 答案 C

有 \[ r=(3a)^{3b}=\bigl(27a^3\bigr)^b\text{。}\]另一方面，\(r=a^bx^b=(ax)^b\)。由各量为正，可令两边底数相等，于是 \(ax=27a^3\)，从而 \(x=27a^2\)。因此正确答案是 C。

English

We have \[ r=(3a)^{3b}=\bigl(27a^3\bigr)^b. \] Also \(r=a^bx^b=(ax)^b.\) Positivity allows us to equate the bases, so \(ax=27a^3\) and \(x=27a^2.\) Thus the correct answer is C.

11 答案 A

逐层撤销对数运算，得到 \[ \begin{aligned} \log_2(\log_2 x)&=4,\\ \log_2x&=16,\\ x&=2^{16}=65536 \end{aligned}\text{。}\]这个数的十进制表示有 \(5\) 位。因此正确答案是 A。

English

Undoing the logarithms gives \[ \begin{aligned} \log_2(\log_2 x)&=4,\\ \log_2x&=16,\\ x&=2^{16}=65536. \end{aligned} \] This number has \(5\) decimal digits. Thus the correct answer is A.

12 答案 E

将已知式中的 \(x\) 替换为 \(\frac{x}{2}\)，得到 \[ f(x)=\frac{2}{2+\frac{x}{2}}=\frac{4}{4+x}\text{。}\]因此 \(2f(x)=\frac{8}{4+x}\)。故正确答案是 E。

English

Replacing the given \(x\) by \(\frac{x}{2}\) yields \[ f(x)=\frac{2}{2+\frac{x}{2}}=\frac{4}{4+x}. \] Therefore \(2f(x)=\frac{8}{4+x}.\) Thus the correct answer is E.

13 答案 D

外、内正方形的边长分别为 \(7\) 和 \(5\)。在任意一个内正方形顶点处，外正方形一条边被分成长度为 \(x\) 和 \(y\) 的两段，且 \[ x+y=7,\qquad x^2+y^2=25\text{。}\]因此 \(2xy=49-25=24\)，所以 \(\{x,y\}=\{3,4\}\)。从位于一条边上的内正方形顶点到较远的对面外正方形顶点，水平位移为 \(4\)，竖直位移为 \(7\)。所以最大距离为 \(\sqrt{4^2+7^2}=\sqrt{65}\)。故正确答案是 D。

English

The outer and inner side lengths are \(7\) and \(5.\) At any inner vertex, the two pieces of the outer side have lengths \(x\) and \(y\) with \[ x+y=7,\qquad x^2+y^2=25. \] Hence \(2xy=49-25=24,\) so \(\{x,y\}=\{3,4\}.\) From an inner vertex on one side, the farther opposite outer vertex has horizontal displacement \(4\) and vertical displacement \(7.\) The greatest distance is therefore \(\sqrt{4^2+7^2}=\sqrt{65}.\) Thus the correct answer is D.

14 答案 B

取 \(A=(0,0)\)、\(B=(2,0)\)。由两个 \(120^\circ\) 角以及边长 \(EA=BC=2\)，得到 \[ E=(-1,\sqrt3),\qquad C=(3,\sqrt3)\text{。}\]因此 \(ABCE\) 是高为 \(\sqrt3\)、两底长为 \(2\) 和 \(4\) 的梯形，其面积为 \(3\sqrt3\)。又因为 \(CE=4=CD=DE\)，所以 \(\triangle CDE\) 是面积为 \(4\sqrt3\) 的等边三角形。总面积为 \(7\sqrt3\)。因此正确答案是 B。

English

Place \(A=(0,0)\) and \(B=(2,0).\) The \(120^\circ\) angles and side lengths \(EA=BC=2\) give \[ E=(-1,\sqrt3),\qquad C=(3,\sqrt3). \] Thus \(ABCE\) is a trapezoid of height \(\sqrt3\) and bases \(2\) and \(4,\) so its area is \(3\sqrt3.\) Also \(CE=4=CD=DE,\) making \(\triangle CDE\) equilateral with area \(4\sqrt3.\) The total area is \(7\sqrt3.\) Thus the correct answer is B.

15 答案 D

内角为 \(180^\circ-\frac{360^\circ}{n}\)，其度数为整数当且仅当 \(360\) 能被 \(n\) 整除。因为 \(360=2^3\cdot3^2\cdot5\)，它有 \[ (3+1)(2+1)(1+1)=24 \] 个正因数。排除 \(n=1\) 和 \(n=2\) 后，剩下 \(22\) 个可行值。因此正确答案是 D。

English

The interior angle is \(180^\circ-\frac{360^\circ}{n},\) which is integral exactly when \(360\) is divisible by \(n.\) Since \(360=2^3\cdot3^2\cdot5,\) it has \[ (3+1)(2+1)(1+1)=24 \] positive divisors. Excluding \(n=1\) and \(n=2\) leaves \(22\) allowable values. Thus the correct answer is D.

16 答案 D

最后一个 \(62\) 出现在位置 \[ \frac{62\cdot63}{2}=1953\text{，}\]而最后一个 \(63\) 出现在位置 \(\frac{63\cdot64}{2}=2016\)。因此第 \(1993\) 项是 \(63\)，它除以 \(5\) 的余数为 \(3\)。故正确答案是 D。

English

The last \(62\) occurs in position \[ \frac{62\cdot63}{2}=1953, \] while the last \(63\) occurs in position \(\frac{63\cdot64}{2}=2016.\) Thus the \(1993\)rd term is \(63,\) whose remainder modulo \(5\) is \(3.\) Thus the correct answer is D.

17 答案 A

令正方形为 \([-1,1]^2\)，其中心为原点。\(1\) 点方向的射线与上边相交于 \((\frac{1}{\sqrt3},1)\)，所以 \[ t=\frac12\left(\frac1{\sqrt3}\right)(1)=\frac1{2\sqrt3}\text{。}\]角部四边形的顶点为 \((0,0)\)、\((\frac{1}{\sqrt3},1)\)、\((1,1)\)、\((1,\frac{1}{\sqrt3})\)。其面积为 \(q=1-\frac1{\sqrt3}\)。因此 \[ \frac qt=\frac{1-\frac{1}{\sqrt3}}{\frac{1}{2\sqrt3}}=2\sqrt3-2\text{。}\]故正确答案是 A。

English

Let the square be \([-1,1]^2\) and its center be the origin. The \(1\) o’clock ray meets the top side at \((\frac{1}{\sqrt3},1),\) so \[ t=\frac12\left(\frac1{\sqrt3}\right)(1)=\frac1{2\sqrt3}. \] The corner quadrilateral has vertices \((0,0),\) \((\frac{1}{\sqrt3},1),\) \((1,1),\) \((1,\frac{1}{\sqrt3}).\) Its area is \(q=1-\frac1{\sqrt3}.\) Hence \[ \frac qt=\frac{1-\frac{1}{\sqrt3}}{\frac{1}{2\sqrt3}}=2\sqrt3-2. \] Thus the correct answer is A.

18 答案 E

合并后的规律每 \(20\) 天重复一次。阿尔在第 \(4\)、\(8\)、\(12\)、\(16\)、\(20\) 天休息，而芭布在第 \(8\)、\(9\)、\(10\)、\(18\)、\(19\)、\(20\) 天休息。两人共同的休息日是第 \(8\) 天和第 \(20\) 天，每个周期有两天。共有 \(\frac{1000}{20}=50\) 个周期，因此共同休息日有 \(2\cdot50=100\) 天。故正确答案是 E。

English

The combined pattern repeats every \(20\) days. Al rests on days \(4,\) \(8,\) \(12,\) \(16,\) \(20,\) while Barb rests on days \(8,\) \(9,\) \(10,\) \(18,\) \(19,\) \(20.\) Their common rest days are \(8\) and \(20,\) two per cycle. There are \(\frac{1000}{20}=50\) cycles, giving \(2\cdot50=100\) common rest days. Thus the correct answer is E.

19 答案 D

消去分母并整理，得到 \[ \begin{aligned} mn-4n-2m=0 &\quad\Longleftrightarrow\quad\\ (m-4)(n-2)=8 \end{aligned}\text{。}\]两个因数都必须为正。\(8\) 的四个有序正因数对给出 \[ \begin{aligned} (m,n)={}&(5,10),(6,6),\\ &(8,4),(12,3) \end{aligned}\text{。}\]因此正确答案是 D。

English

Clearing denominators and rearranging gives \[ \begin{aligned} mn-4n-2m=0 &\quad\Longleftrightarrow\quad\\ (m-4)(n-2)=8. \end{aligned} \] Both factors must be positive. The four ordered positive factor pairs of \(8\) give \[ \begin{aligned} (m,n)={}&(5,10),(6,6),\\ &(8,4),(12,3). \end{aligned} \] Thus the correct answer is D.

20 答案 B

两个根为 \[ z=\frac{3i\pm\sqrt{-9+40k}}{20}\text{。}\]若 \(k\) 是负实数，则 \(-9+40k\) 是负实数，所以它的平方根是纯虚数。因此所得两个 \(z\) 值都是纯虚数。其余全称命题均不成立，例如可视情况分别取 \(k=1\)、\(i\) 或 \(0\)。故正确答案是 B。

English

The roots are \[ z=\frac{3i\pm\sqrt{-9+40k}}{20}. \] If \(k\) is negative and real, then \(-9+40k\) is negative real, so its square roots are pure imaginary. Both resulting values of \(z\) are therefore pure imaginary. The other universal claims fail, for example by taking \(k=1,\) \(i,\) or \(0\) as appropriate. Thus the correct answer is B.

21 答案 B

因为等差数列中关于某项对称的各项平均值等于该中间项，所以 \[ 3a_7=17,\qquad 11a_9=77\text{。}\]因此 \(a_7=\frac{17}{3}\)、\(a_9=7\)，公差为 \(d=\frac{7-\frac{17}{3}}{2}=\frac{2}{3}\)。由于 \(13-7=6=9d\)，有 \(a_k=a_{9+9}=a_{18}\)。故正确答案是 B。

English

Because symmetric terms of an arithmetic sequence average to the middle term, \[ 3a_7=17,\qquad 11a_9=77. \] Thus \(a_7=\frac{17}{3}\) and \(a_9=7,\) so the common difference is \(d=\frac{7-\frac{17}{3}}{2}=\frac{2}{3}.\) Since \(13-7=6=9d,\) we have \(a_k=a_{9+9}=a_{18}.\) Thus the correct answer is B.

22 答案 C

逐层展开各个和，顶层数值为 \(6c+3(e_1+e_2+e_3\) \({}+e_4+e_5+e_6)\) \({}+(v_1+v_2+v_3)\)，其中 \(c\) 是底层中心积木，\(e_i\) 是六个非角落的边界积木，\(v_i\) 是三个角落积木。为使这个加权和最小，将 \(1\) 放在系数为 \(6\) 的位置，将 \(2\)、\(\ldots\)、\(7\) 放在系数为 \(3\) 的位置，并将 \(8\)、\(9\)、\(10\) 放在系数为 \(1\) 的位置。最小值为 \(6+3(2+3+4\) \({}+5+6+7)\) \({}+(8+9+10)=114\)。因此正确答案是 C。

English

Expanding the sums layer by layer, the top value is \(6c+3(e_1+e_2+e_3\)\({}+e_4+e_5+e_6)\)\({}+(v_1+v_2+v_3),\) where \(c\) is the center bottom block, the \(e_i\) are the six non-corner boundary blocks, and the \(v_i\) are the three corner blocks. To minimize this weighted sum, assign \(1\) to the coefficient-\(6\) position, \(2,\) \(\ldots,\) \(7\) to the coefficient-\(3\) positions, and \(8,\) \(9,\) \(10\) to the coefficient-\(1\) positions. The minimum is \(6+3(2+3+4\)\({}+5+6+7)\)\({}+(8+9+10)=114.\) Thus the correct answer is C.

23 答案 B

因为 \(BX=CX\)，直线 \(AD\) 平分 \(\angle BXC\)，由对称性还可得 \(\angle BAD=6^\circ\)。因为直径 \(AD=1\)，所以 \(\angle ABD=90^\circ\)，从而 \(AB=\cos6^\circ\)。此外，\[ \begin{gathered} \angle AXB=180^\circ-18^\circ=162^\circ, \\ \angle ABX=180^\circ-162^\circ\\ {}-6^\circ=12^\circ \end{gathered}\text{。}\]在 \(\triangle ABX\) 中应用正弦定理，得到 \[ \begin{aligned} AX&=\frac{AB\sin12^\circ}{\sin162^\circ}\\ &=\cos6^\circ\sin12^\circ\\ &\qquad{}\csc18^\circ \end{aligned}\text{。}\]因此正确答案是 B。

English

Since \(BX=CX,\) the line \(AD\) bisects \(\angle BXC,\) and symmetry also gives \(\angle BAD=6^\circ.\) Because \(AD=1\) is a diameter, \(\angle ABD=90^\circ,\) so \(AB=\cos6^\circ.\) Also \[ \begin{gathered} \angle AXB=180^\circ-18^\circ=162^\circ, \\ \angle ABX=180^\circ-162^\circ\\ {}-6^\circ=12^\circ. \end{gathered} \] The Law of Sines in \(\triangle ABX\) gives \[ \begin{aligned} AX&=\frac{AB\sin12^\circ}{\sin162^\circ}\\ &=\cos6^\circ\sin12^\circ\\ &\qquad{}\csc18^\circ. \end{aligned} \] Thus the correct answer is B.

24 答案 E

三枚有光泽的便士所占的 \(3\) 元子集，是从 \(7\) 个抽取位置中等可能选出的，共有 \(\binom73=35\) 种可能。第三枚有光泽的便士不迟于第 \(4\) 次抽取出现，当且仅当三个有光泽便士的位置都在前四个位置中，这有 \(\binom43=4\) 种情形。所求概率为 \[ 1-\frac4{35}=\frac{31}{35}\text{。}\]因此 \(a+b=31+35=66\)。故正确答案是 E。

English

The shiny pennies occupy a uniformly chosen \(3\)-element subset of the \(7\) draw positions. There are \(\binom73=35\) possibilities. The third shiny penny appears by draw \(4\) exactly when all three shiny positions lie among the first four, which occurs in \(\binom43=4\) cases. The requested probability is \[ 1-\frac4{35}=\frac{31}{35}. \] Therefore \(a+b=31+35=66.\) Thus the correct answer is E.

25 答案 E

设 \(O\) 为角的顶点。在一条射线上取点 \(A\)，使 \(OA=OP\)；因为 \(OP\) 平分 \(120^\circ\) 角，所以 \(\triangle AOP\) 是等边三角形。对 \(\overline{AO}\) 上任意一点 \(Q\)，在另一条射线上取点 \(R\)，使 \(OR=AQ\)。由边角边可得 \(\triangle POR\cong\triangle PAQ\)。因此 \(PQ=PR\)，且 \(\angle QPR=\angle APO=60^\circ\)，所以 \(\triangle PQR\) 是等边三角形。连续变化的 \(Q\) 会给出不同的三角形，因此这样的三角形当然多于 \(15\) 个。故正确答案是 E。

English

Let \(O\) be the vertex. Choose \(A\) on one ray so that \(OA=OP;\) because \(OP\) bisects the \(120^\circ\) angle, \(\triangle AOP\) is equilateral. For any \(Q\) on \(\overline{AO},\) choose \(R\) on the other ray with \(OR=AQ.\) Then \(\triangle POR\cong\triangle PAQ\) by SAS. Consequently \(PQ=PR\) and \(\angle QPR=\angle APO=60^\circ,\) so \(\triangle PQR\) is equilateral. Continuously many choices of \(Q\) give distinct triangles, hence certainly more than \(15.\) Thus the correct answer is E.

26 答案 C

两个根式同时为实数当且仅当 \(6\le x\le8\)。分解并有理化，得到 \[ \begin{aligned} f(x) &=\sqrt{8-x}\bigl(\sqrt x-\sqrt{x-6}\bigr)\\ &=\frac{6\sqrt{8-x}}{\sqrt x+\sqrt{x-6}} \end{aligned}\text{。}\]在此区间内，分子递减而分母递增，所以最大值在 \(x=6\) 时取得，其值为 \[ f(6)=\sqrt{12}=2\sqrt3\text{。}\]因此正确答案是 C。

English

Both radicals are real exactly when \(6\le x\le8.\) Factoring and rationalizing, \[ \begin{aligned} f(x) &=\sqrt{8-x}\bigl(\sqrt x-\sqrt{x-6}\bigr)\\ &=\frac{6\sqrt{8-x}}{\sqrt x+\sqrt{x-6}}. \end{aligned} \] On this interval the numerator decreases while the denominator increases, so the maximum occurs at \(x=6.\) Its value is \[ f(6)=\sqrt{12}=2\sqrt3. \] Thus the correct answer is C.

27 答案 B

圆心 \(P\) 始终与所接触的边相距一个单位，因此它的轨迹是三条向内平移 \(1\) 个单位的平行线所围成的三角形。这个三角形与 \(\triangle ABC\) 相似。原直角三角形的面积为 \(24\)，半周长为 \(12\)，所以内切圆半径为 \(\frac{24}{12}=2\)。轨迹三角形的内切圆半径为 \(2-1=1\)，因此线性缩放比例为 \(\frac{1}{2}\)。它的周长，也就是所走距离，为 \[ \frac12(6+8+10)=12\text{。}\]故正确答案是 B。

English

The center \(P\) remains one unit from the side it touches, so its path is the triangle formed by the three inward parallel lines at distance \(1.\) This triangle is similar to \(\triangle ABC.\) The original right triangle has area \(24\) and semiperimeter \(12,\) hence inradius \(\frac{24}{12}=2.\) The path triangle has inradius \(2-1=1,\) so its linear scale is \(\frac{1}{2}.\) Its perimeter, and therefore the distance traveled, is \[ \frac12(6+8+10)=12. \] Thus the correct answer is B.

28 答案 D

共有 \(\binom{16}{3}=560\) 个格点三元组。四行和四列贡献了 \[ 8\binom43=32 \] 个共线三元组。对斜率 \(1\)，能容纳三个点的对角线长度依次为 \(3\)、\(4\)、\(3\)，贡献 \(\binom33+\binom43+\binom33=6\) 个；斜率 \(-1\) 再贡献 \(6\) 个。在一个 \(4\) 乘 \(4\) 的点阵中，其他斜率都不能容纳三个格点。因此面积为正的三角形数为 \(560-32-6-6=516\)。故正确答案是 D。

English

There are \(\binom{16}{3}=560\) triples of grid points. The four rows and four columns contribute \[ 8\binom43=32 \] collinear triples. For slope \(1,\) the diagonal lengths capable of containing three points are \(3,\) \(4,\) \(3,\) contributing \(\binom33+\binom43+\binom33=6;\) slope \(-1\) contributes another \(6.\) No other slope fits three lattice points inside a \(4\)-by-\(4\) point array. Thus the number of positive-area triangles is \(560-32-6-6=516.\) Thus the correct answer is D.

29 答案 B

若棱长为 \(a\)、\(b\)、\(c\)，则三条表面对角线长度的平方为 \(a^2+b^2\)、\(a^2+c^2\)、\(b^2+c^2\)。对按大小排列的对角线 \(p\le q\le r\)，有 \[ a^2=\frac{p^2+q^2-r^2}{2}\text{，}\]因此必须满足 \(r^2\lt p^2+q^2\)；类似公式也说明这个条件充分。只有 \(\{4,5,7\}\) 不满足，因为 \(7^2=49\gt16+25=41\)。所以正确答案是 B。

English

If the edge lengths are \(a,\) \(b,\) \(c,\) then the three face-diagonal squares are \(a^2+b^2,\) \(a^2+c^2,\) \(b^2+c^2.\) For sorted diagonals \(p\le q\le r,\) \[ a^2=\frac{p^2+q^2-r^2}{2}, \] so necessarily \(r^2\lt p^2+q^2;\) the analogous formulas show this is also sufficient. Only \(\{4,5,7\}\) fails, since \(7^2=49\gt16+25=41.\) Thus the correct answer is B.

30 答案 D

递推关系为 \(x_n=\{2x_{n-1}\}\)，所以 \(x_5=\{32x_0\}\)。等式 \(x_5=x_0\) 要求 \[ 32x_0-x_0=\lfloor32x_0\rfloor\text{，}\]因而 \(31x_0\) 是整数。值 \(x_0=\frac{k}{31}\) 在 \(k=0\)、\(1\)、\(\ldots\)、\(30\) 时都满足递推关系并位于 \([0,1)\) 内。共有 \(31\) 个值。因此正确答案是 D。

English

The recurrence is \(x_n=\{2x_{n-1}\},\) so \(x_5=\{32x_0\}.\) The equality \(x_5=x_0\) requires \[ 32x_0-x_0=\lfloor32x_0\rfloor, \] hence \(31x_0\) is an integer. The values \(x_0=\frac{k}{31}\) for \(k=0,\) \(1,\) \(\ldots,\) \(30\) all satisfy the recurrence and lie in \([0,1).\) There are \(31\) values. Thus the correct answer is D.