1986 AMC 12 真题（P版）

1 答案 B

展开两个方括号中的式子，得到 \[ \begin{aligned} &[x-(y-x)]-[(x-y)-x]\\ &\qquad=(2x-y)-(-y)=2x\text{。} \end{aligned} \]

所以正确答案是 B。

English

Expanding the two bracketed expressions gives \[ \begin{aligned} &[x-(y-x)]-[(x-y)-x]\\ &\qquad=(2x-y)-(-y)=2x. \end{aligned} \]

Thus the correct answer is B.

2 答案 A

原直线的斜率为 \(\frac{2}{3}\)，\(y\) 轴截距为 \(4\)。因此，\(L\) 的斜率为 \(\frac{1}{3}\)，\(y\) 轴截距为 \(8\)，所以它的方程是 \(y=\frac13x+8\)。

所以正确答案是 A。

English

The original line has slope \(\frac{2}{3}\) and \(y\)-intercept \(4.\) Therefore \(L\) has slope \(\frac{1}{3}\) and \(y\)-intercept \(8,\) so its equation is \(y=\frac13x+8.\)

Thus the correct answer is A.

3 答案 D

由于 \(\angle C=90^\circ\)，且 \(\angle A=20^\circ\)，所以 \(\angle ABC=70^\circ\)。角平分线给出 \(\angle DBC=35^\circ\)。因此，在直角三角形 \(BDC\) 中， \[ \angle BDC=90^\circ-35^\circ=55^\circ\text{。} \]

所以正确答案是 D。

English

Since \(\angle C=90^\circ\) and \(\angle A=20^\circ,\) we have \(\angle ABC=70^\circ.\) The bisector gives \(\angle DBC=35^\circ.\) Therefore, in right triangle \(BDC,\) \[ \angle BDC=90^\circ-35^\circ=55^\circ. \]

Thus the correct answer is D.

4 答案 B

数字 \(33\) 的各位数字之和为 \(6\)，能被 \(6\) 整除。但是 \(33\) 是奇数，所以不能被 \(6\) 整除。因此，\(33\) 使该命题为假。

所以正确答案是 B。

English

The digit sum of \(33\) is \(6,\) which is divisible by \(6.\) However, \(33\) is odd, so it is not divisible by \(6.\) Thus \(33\) makes the implication false.

Therefore the correct answer is B.

5 答案 A

有 \(\sqrt[6]{27}=(3^3)^{\frac{1}{6}}=\sqrt3\)，且 \[ \sqrt{6\frac34}=\sqrt{\frac{27}{4}}=\frac{3\sqrt3}{2}\text{。} \] 因此，原式为 \[ \begin{aligned} \left(\sqrt3-\frac{3\sqrt3}{2}\right)^2 &=\left(-\frac{\sqrt3}{2}\right)^2\\ &=\frac34\text{。} \end{aligned} \]

所以正确答案是 A。

English

We have \(\sqrt[6]{27}=(3^3)^{\frac{1}{6}}=\sqrt3\) and \[ \sqrt{6\frac34}=\sqrt{\frac{27}{4}}=\frac{3\sqrt3}{2}. \] Hence the expression is \[ \begin{aligned} \left(\sqrt3-\frac{3\sqrt3}{2}\right)^2 &=\left(-\frac{\sqrt3}{2}\right)^2\\ &=\frac34. \end{aligned} \]

Thus the correct answer is A.

6 答案 C

设桌高为 \(h\)，木块的长、短边尺寸分别为 \(\ell,w\)。图 \(1\) 给出 \(h+\ell-w=32\)，图 \(2\) 给出 \(h+w-\ell=28\)。两式相加后，木块的尺寸相消： \[ 2h=60\text{，} \] 所以 \(h=30\) 英寸。

所以正确答案是 C。

English

Let \(h\) be the table height and let \(\ell,w\) be the long and short block dimensions. Figure \(1\) gives \(h+\ell-w=32,\) while Figure \(2\) gives \(h+w-\ell=28.\) Adding cancels the block dimensions: \[ 2h=60, \] so \(h=30\) inches.

Thus the correct answer is C.

7 答案 E

若 \(x\) 是整数，则 \(\lfloor x\rfloor+\lceil x\rceil=2x\)，不可能等于 \(5\)。若 \(x\) 不是整数，则 \[ \lfloor x\rfloor+\lceil x\rceil=2\lfloor x\rfloor+1\text{。} \] 它等于 \(5\) 的充要条件是 \(\lfloor x\rfloor=2\)，所以 \(2\lt x\lt3\)。

所以正确答案是 E。

English

If \(x\) is an integer, then \(\lfloor x\rfloor+\lceil x\rceil=2x,\) which cannot equal \(5.\) If \(x\) is not an integer, then \[ \lfloor x\rfloor+\lceil x\rceil=2\lfloor x\rfloor+1. \] This equals \(5\) exactly when \(\lfloor x\rfloor=2,\) so \(2\lt x\lt3.\)

Thus the correct answer is E.

8 答案 E

平均值约为 \[ \begin{aligned} &\frac{(3.615\times10^6)(5280)^2} {2.265\times10^8}\\ &\qquad\approx 4.45\times10^5 \end{aligned} \] 平方英尺/人。在各选项中，它最接近 \(500{,}000\)。

所以正确答案是 E。

English

The average is approximately \[ \begin{aligned} &\frac{(3.615\times10^6)(5280)^2} {2.265\times10^8}\\ &\qquad\approx 4.45\times10^5 \end{aligned} \] square feet per person. Of the choices, this is closest to \(500{,}000.\)

Thus the correct answer is E.

9 答案 C

对每个 \(n\)， \[ 1-\frac1{n^2}=\frac{n-1}{n}\cdot\frac{n+1}{n}\text{。} \] 因此乘积发生连锁消去： \[ \begin{aligned} \prod_{n=2}^{10}\left(1-\frac1{n^2}\right) &=\left(\frac12\right) \left(\frac{11}{10}\right)\\ &=\frac{11}{20}\text{。} \end{aligned} \]

所以正确答案是 C。

English

For each \(n,\) \[ 1-\frac1{n^2}=\frac{n-1}{n}\cdot\frac{n+1}{n}. \] Therefore the product telescopes: \[ \begin{aligned} \prod_{n=2}^{10}\left(1-\frac1{n^2}\right) &=\left(\frac12\right) \left(\frac{11}{10}\right)\\ &=\frac{11}{20}. \end{aligned} \]

Thus the correct answer is C.

10 答案 E

每个首字母对应一组 \(4!=24\) 个单词。第 \(73\) 至第 \(96\) 个单词以 \(M\) 开头，所以第 \(86\) 个单词是这一组中的第 \(14\) 个。前 \(6\) 个以 \(MA\) 开头，接下来 \(6\) 个以 \(ME\) 开头，第 \(13\) 至第 \(18\) 个以 \(MH\) 开头。其中前两个是 \(MHAES\) 和 \(MHASE\)。因此，第 \(86\) 个单词以 E 结尾。

所以正确答案是 E。

English

Each first letter occupies a block of \(4!=24\) words. Positions \(73\) through \(96\) begin with \(M,\) so the \(86\)th word is the \(14\)th word in that block. The first \(6\) begin with \(MA,\) the next \(6\) with \(ME,\) and the \(13\)th through \(18\)th begin with \(MH.\) The first two of those are \(MHAES\) and \(MHASE.\) Thus the \(86\)th word ends in E.

Therefore the correct answer is E.

11 答案 B

三角形 \(AHB\) 在 \(H\) 处为直角，且 \(M\) 是其斜边 \(AB\) 的中点。直角三角形的斜边中点到三个顶点的距离相等，所以 \[ HM=\frac{AB}{2}=\frac{13}{2}=6.5\text{。} \]

所以正确答案是 B。

English

Triangle \(AHB\) is right at \(H,\) and \(M\) is the midpoint of its hypotenuse \(AB.\) The midpoint of a right triangle’s hypotenuse is equidistant from all three vertices, so \[ HM=\frac{AB}{2}=\frac{13}{2}=6.5. \]

Thus the correct answer is B.

12 答案 B

设答对、答错和未作答的题数分别为 \(c,w,u\)。旧计分制下的分数和总题数给出 \[ \begin{aligned} 30+4c-w&=84,\\ c+w+u&=30\text{。} \end{aligned} \] 消去 \(w\)，得到 \(5c+u=84\)。新计分制下的分数为 \(5c+2u=93\)。两式相减，得到 \(u=9\)。

所以正确答案是 B。

English

Let \(c,w,u\) be the numbers correct, wrong and unanswered. The old score and total number of questions give \[ \begin{aligned} 30+4c-w&=84,\\ c+w+u&=30. \end{aligned} \] Eliminating \(w\) yields \(5c+u=84.\) The new score is \(5c+2u=93.\) Subtracting these equations gives \(u=9.\)

Thus the correct answer is B.

13 答案 E

将方程写成 \(y=a(x-4)^2+2\)。代入 \((2,0)\)，得到 \(0=4a+2\)，所以 \(a=-\frac{1}{2}\)。展开得 \[ y=-\frac12x^2+4x-6\text{。} \] 因此 \(b=4\)、\(c=-6\)，且 \(abc=(-\frac{1}{2})(4)(-6)=12\)。

所以正确答案是 E。

English

Write the equation as \(y=a(x-4)^2+2.\) Substituting \((2,0)\) gives \(0=4a+2,\) so \(a=-\frac{1}{2}.\) Expanding, \[ y=-\frac12x^2+4x-6. \] Thus \(b=4,\) \(c=-6,\) and \(abc=(-\frac{1}{2})(4)(-6)=12.\)

Therefore the correct answer is E.

14 答案 D

由三个关系可得 \[ \begin{aligned} 1\text{ 米}&=\frac fg\text{ 个跳步},\\ 1\text{ 个跳步}&=\frac ed\text{ 个跃步},\\ 1\text{ 个跃步}&=\frac cb\text{ 个跨步}\text{。} \end{aligned} \] 将换算因子相乘，得到 \(\frac fg\cdot\frac ed\cdot\frac cb=\frac{cef}{bdg}\) 个跨步。

所以正确答案是 D。

English

From the three relations, \[ \begin{aligned} 1\text{ meter}&=\frac fg\text{ jumps},\\ 1\text{ jump}&=\frac ed\text{ hops},\\ 1\text{ hop}&=\frac cb\text{ skips}. \end{aligned} \] Multiplying the conversion factors gives \(\frac fg\cdot\frac ed\cdot\frac cb=\frac{cef}{bdg}\) skips.

Thus the correct answer is D.

15 答案 C

该学生得到的结果为 \(\frac{x+y+2z}{4}\)，正确的平均数为 \(\frac{x+y+z}{3}\)。两者之差为 \[ \begin{aligned} &\frac{x+y+2z}{4}-\frac{x+y+z}{3}\\ &\qquad=\frac{2z-x-y}{12}\text{。} \end{aligned} \] 由于 \(z\gt x\)，且 \(z\gt y\)，分子为正。因此，该学生的结果总是大于 \(A\)。

所以正确答案是 C。

English

The student’s result is \(\frac{x+y+2z}{4},\) while the true average is \(\frac{x+y+z}{3}.\) Their difference is \[ \begin{aligned} &\frac{x+y+2z}{4}-\frac{x+y+z}{3}\\ &\qquad=\frac{2z-x-y}{12}. \end{aligned} \] Since \(z\gt x\) and \(z\gt y,\) the numerator is positive. The student’s result is therefore always greater than \(A.\)

Thus the correct answer is C.

16 答案 C

题目给出的相似顺序说明 \[ \frac{PA}{PC}=\frac{PB}{PA}=\frac{AB}{CA}=\frac43\text{。} \] 令 \(PC=t\)。则 \(PA=\frac{4t}{3}\)，且 \(PB=\frac{16t}{9}\)。由于 \(PB=PC+CB=t+7\)， \[ \frac{16t}{9}=t+7\text{，} \] 解得 \(t=9\)。

所以正确答案是 C。

English

The stated order of similarity gives \[ \frac{PA}{PC}=\frac{PB}{PA}=\frac{AB}{CA}=\frac43. \] Let \(PC=t.\) Then \(PA=\frac{4t}{3}\) and \(PB=\frac{16t}{9}.\) Since \(PB=PC+CB=t+7,\) \[ \frac{16t}{9}=t+7, \] which gives \(t=9.\)

Thus the correct answer is C.

17 答案 B

取出 \(23\) 只袜子时，数量为奇数的颜色种数也必须为奇数，所以至多为 \(3\)。因此，至多有 \(3\) 只袜子未配对，余下至少 \(20\) 只袜子可组成 \(10\) 双。但是，取出 \(22\) 只还不够：若四种颜色的数量分别为 \(7,5,5,5\)，则只能组成 \(3+2+2+2=9\) 双。因此，最小数量为 \(23\)。

所以正确答案是 B。

English

With \(23\) selected socks, the number of colors having an odd count must itself be odd, so it is at most \(3.\) Thus at most \(3\) socks are unpaired, leaving at least \(20\) socks in \(10\) pairs. But \(22\) socks do not suffice: color counts \(7,5,5,5\) produce only \(3+2+2+2=9\) pairs. Therefore the minimum is \(23.\)

Thus the correct answer is B.

18 答案 E

直圆柱的椭圆平面截面的短轴是一条圆柱直径，因此其长度为 \(2\)。长轴比它长 \(50\%\)，所以长轴的长度为 \[ 2+0.50(2)=3\text{。} \]

所以正确答案是 E。

English

The minor axis of an elliptical plane section of a right circular cylinder is a diameter of the cylinder. Its length is therefore \(2.\) The major axis is \(50\%\) longer, so its length is \[ 2+0.50(2)=3. \]

Thus the correct answer is E.

19 答案 A

令第一条边沿水平方向。三段有向路程对应的向量为 \[ \begin{gathered} (2,0),\qquad (1,\sqrt3),\\ \left(-\frac12,\frac{\sqrt3}{2}\right)\text{。} \end{gathered} \] 它们的和为 \((\frac{5}{2},\frac{3\sqrt3}{2})\)。其长度的平方为 \[ \left(\frac52\right)^2+\left(\frac{3\sqrt3}{2}\right)^2=13\text{。} \] 因此距离为 \(\sqrt{13}\) 千米。

所以正确答案是 A。

English

Choose the first side in the horizontal direction. The three directed portions of the walk have vectors \[ \begin{gathered} (2,0),\qquad (1,\sqrt3),\\ \left(-\frac12,\frac{\sqrt3}{2}\right). \end{gathered} \] Their sum is \((\frac{5}{2},\frac{3\sqrt3}{2}).\) Its squared length is \[ \left(\frac52\right)^2+\left(\frac{3\sqrt3}{2}\right)^2=13. \] The distance is therefore \(\sqrt{13}\) km.

Thus the correct answer is A.

20 答案 E

\(x\) 的新值为 \(\frac{x(100+p)}{100}\)。因此，\(y\) 的新值为 \[ y\cdot\frac{100}{100+p}\text{。} \] 减少的比例为 \(1-\frac{100}{100+p}=\frac{p}{100+p}\)。化成百分数，就是 \(\frac{100p}{100+p}\%\)。

所以正确答案是 E。

English

The new value of \(x\) is \(\frac{x(100+p)}{100}.\) Hence the new value of \(y\) is \[ y\cdot\frac{100}{100+p}. \] The fractional decrease is \(1-\frac{100}{100+p}=\frac{p}{100+p}.\) Expressed as a percentage, this is \(\frac{100p}{100+p}\%.\)

Thus the correct answer is E.

21 答案 B

令圆的半径为 \(r=AC\)，其上方阴影扇形的面积为 \(\frac{\theta r^2}{2}\)。另一个阴影部分是从三角形 \(ABC\) 中去掉一个面积相同的扇形。因此，两个阴影部分面积相等的条件可写为： \[ 2\left(\frac{\theta r^2}{2}\right) =\frac12r\cdot AB \] 该式等价于 \(\frac{AB}{r}=2\theta\)。又因为 \(AB\) 在 \(A\) 点与圆相切，三角形 \(ABC\) 在 \(A\) 点为直角，并且 \(\frac{AB}{r}=\tan\theta\)，所以所需条件为 \(\tan\theta=2\theta\)。

因此正确答案是 B。

English

Let \(r=AC.\) The upper shaded sector has area \(\frac{\theta r^2}{2}.\) The lower shaded region is triangle \(ABC\) with an equal sector removed. Thus the two shaded regions are equal exactly when \[ 2\left(\frac{\theta r^2}{2}\right) =\frac12r\cdot AB. \] This is equivalent to \(\frac{AB}{r}=2\theta.\) Since \(AB\) is tangent at \(A,\) triangle \(ABC\) is right at \(A,\) and \(\frac{AB}{r}=\tan\theta.\) Therefore the condition is \(\tan\theta=2\theta.\)

Thus the correct answer is B.

22 答案 C

共有 \(\binom{10}{6}=210\) 个可能的集合。若第二小的元素是 \(3\)，则必须选取 \(3\)，再从 \(\{1,2\}\) 中选一个，并从 \(\{4,5,\ldots,10\}\) 中选四个。因此满足条件的集合有 \[ 2\binom74=70 \] 个。所求概率为 \(\frac{70}{210}=\frac{1}{3}\)。

因此正确答案是 C。

English

There are \(\binom{10}{6}=210\) possible sets. If the second-smallest element is \(3,\) then \(3\) is selected, one element is chosen from \(\{1,2\},\) and four are chosen from \(\{4,5,\ldots,10\}.\) This gives \[ 2\binom74=70 \] favorable sets. The probability is \(\frac{70}{210}=\frac{1}{3}.\)

Thus the correct answer is C.

23 答案 E

由二项式定理， \[ \begin{aligned} N&=(69+1)^5=70^5\\ &=2^5\cdot5^5\cdot7^5 \end{aligned}\text{。}\]构造一个因数时，可以分别为这三个质因数选择从 \(0\) 到 \(5\) 的指数。因此，\(N\) 有 \(6^3=216\) 个正因数。

因此正确答案是 E。

English

By the binomial theorem, \[ \begin{aligned} N&=(69+1)^5=70^5\\ &=2^5\cdot5^5\cdot7^5. \end{aligned} \] A divisor independently chooses an exponent from \(0\) through \(5\) for each of the three primes. Thus \(N\) has \(6^3=216\) positive divisors.

Therefore the correct answer is E.

24 答案 D

公因式 \(p(x)\) 整除 \[ \begin{aligned} &3(x^4+6x^2+25)\\ &\quad{}-(3x^4+4x^2+28x+5)\\ &\qquad=14(x^2-2x+5) \end{aligned}\text{。}\]由于 \(p(x)\) 是整系数首一多项式，由高斯引理可知它整除 \(x^2-2x+5\)。这两个多项式都是同次的首一多项式，所以 \(p(x)=x^2-2x+5\)。因此 \(p(1)=1-2+5=4\)。

因此正确答案是 D。

English

The common factor \(p(x)\) divides \[ \begin{aligned} &3(x^4+6x^2+25)\\ &\quad{}-(3x^4+4x^2+28x+5)\\ &\qquad=14(x^2-2x+5). \end{aligned} \] Because \(p(x)\) is monic with integer coefficients, Gauss’s lemma implies that it divides \(x^2-2x+5.\) The two polynomials are monic and have the same degree, so \(p(x)=x^2-2x+5.\) Hence \(p(1)=1-2+5=4.\)

Thus the correct answer is D.

25 答案 B

对于 \(0\le k\le9\)，恰有 \(2^k\) 个整数 \(N\) 满足 \(2^k\le N\lt2^{k+1}\)，每个都贡献 \(k\)。最后一个整数 \(1024=2^{10}\) 贡献 \(10\)。因此 \[ \sum_{N=1}^{1024}\lfloor\log_2N\rfloor =\sum_{k=0}^{9}k2^k+10\text{。}\]由有限几何级数的恒等式，\(\sum_{k=0}^{9}k2^k=8194\)，所以所求的和为 \(8204\)。

因此正确答案是 B。

English

For \(0\le k\le9,\) exactly \(2^k\) integers \(N\) satisfy \(2^k\le N\lt2^{k+1},\) and each contributes \(k.\) The final integer \(1024=2^{10}\) contributes \(10.\) Hence \[ \sum_{N=1}^{1024}\lfloor\log_2N\rfloor =\sum_{k=0}^{9}k2^k+10. \] The finite geometric-sum identity gives \(\sum_{k=0}^{9}k2^k=8194,\) so the requested sum is \(8204.\)

Thus the correct answer is B.

26 答案 C

将直角顶点置于 \((0,0)\)，另两个顶点分别置于 \((a,0)\) 和 \((0,b)\)。通向两条直角边中点的中线斜率分别为 \(-\frac{2b}{a}\) 和 \(-\frac{b}{2a}\)，二者之比为 \(4\)。因此，若一条中线的斜率为 \(3\)，另一条的斜率可以为 \(12\) 或 \(\frac{3}{4}\)。两种情况都能实现：先选取具有所需两条斜率的三角形，再将它的重心平移到两条给定直线的交点即可。因此 \(m\) 有两个可能值。

因此正确答案是 C。

English

Place the right-angle vertex at \((0,0)\) and the other vertices at \((a,0)\) and \((0,b).\) The medians to the legs have slopes \(-\frac{2b}{a}\) and \(-\frac{b}{2a},\) whose ratio is \(4.\) Therefore, if one median has slope \(3,\) the other can have slope \(12\) or \(\frac{3}{4}.\) Both occur: choose a triangle with the required pair of slopes and translate its centroid to the intersection of the two specified lines. Thus there are two possible values of \(m.\)

Therefore the correct answer is C.

27 答案 C

两个三角形在 \(E\) 点所用的角相同，所以 \[ \frac{[CDE]}{[ABE]} =\frac{CE\cdot DE}{AE\cdot BE}\text{。}\]由相交弦定理，\(AE\cdot CE=BE\cdot DE\)，所以这个比值化为 \((\frac{DE}{AE})^2\)。连接 \(AD\)。由于 \(AB\) 是直径，\(\angle ADB=90^\circ\)，且 \(D,E,B\) 三点共线。因此三角形 \(ADE\) 在 \(D\) 点为直角，并且 \(\frac{DE}{AE}=\cos\alpha\)。所求比值为 \(\cos^2\alpha\)。

因此正确答案是 C。

English

The triangles use the same angle at \(E,\) so \[ \frac{[CDE]}{[ABE]} =\frac{CE\cdot DE}{AE\cdot BE}. \] Intersecting chords give \(AE\cdot CE=BE\cdot DE,\) so this ratio becomes \((\frac{DE}{AE})^2.\) Draw \(AD.\) Since \(AB\) is a diameter, \(\angle ADB=90^\circ,\) and \(D,E,B\) are collinear. Thus triangle \(ADE\) is right at \(D,\) and \(\frac{DE}{AE}=\cos\alpha.\) The required ratio is \(\cos^2\alpha.\)

Thus the correct answer is C.

28 答案 C

设边长为 \(s\)。由于边心距 \(OP=1\)，由以中心为顶点的五个三角形可得五边形的面积为 \(\frac{5s}{2}\)。同一个五边形也是三角形 \(ABC\)、\(ACD\) 和 \(ADE\) 的并，这三个三角形以五边形边长为底时的高分别为 \(AQ,AP,AR\)。因此 \[ \frac{s}{2}(AQ+AP+AR)=\frac{5s}{2}\text{，}\]所以 \(AQ+AP+AR=5\)。此外，\(AP=AO+OP=AO+1\)。因此 \(AO+AQ+AR=4\)。

因此正确答案是 C。

English

Let the side length be \(s.\) Since the apothem \(OP=1,\) the five central triangles give pentagon area \(\frac{5s}{2}.\) The same pentagon is the union of triangles \(ABC,\) \(ACD\) and \(ADE,\) whose respective altitudes to side-length bases are \(AQ,AP,AR.\) Hence \[ \frac{s}{2}(AQ+AP+AR)=\frac{5s}{2}, \] so \(AQ+AP+AR=5.\) Also \(AP=AO+OP=AO+1.\) Therefore \(AO+AQ+AR=4.\)

Thus the correct answer is C.

29 答案 B

设第三条高为 \(h\)。由于每条边的长度为公共面积的两倍除以对应的高，三条边的长度成比例于 \[ \frac14,\qquad\frac1{12},\qquad\frac1h\text{。}\]两个非平凡的三角形不等式给出 \[ \begin{aligned} \frac1h&\lt\frac14+\frac1{12}=\frac13,\\ \frac1h&\gt\frac14-\frac1{12}=\frac16 \end{aligned}\text{。}\]因此 \(3\lt h\lt6\)。其中最大的整数为 \(5\)，而三条高互不相同，符合不等边三角形的要求。

因此正确答案是 B。

English

Let the third altitude be \(h.\) Since each side equals twice the common area divided by its altitude, the side lengths are proportional to \[ \frac14,\qquad\frac1{12},\qquad\frac1h. \] The two nontrivial triangle inequalities give \[ \begin{aligned} \frac1h&\lt\frac14+\frac1{12}=\frac13,\\ \frac1h&\gt\frac14-\frac1{12}=\frac16. \end{aligned} \] Thus \(3\lt h\lt6.\) The largest integral possibility is \(5,\) and its three altitudes are distinct as required for a scalene triangle.

Therefore the correct answer is B.

30 答案 B

每个表达式 \(t+\frac{17}{t}\) 都与 \(t\) 同号，所以四个变量必定同号。先假设它们全为正数。由算术平均值与几何平均值不等式，每个变量都至少为 \(\sqrt{17}\)。当 \(t\gt\sqrt{17}\) 时， \[ \frac{t+\frac{17}{t}}{2}\lt t\text{。}\]如果任一变量大于 \(\sqrt{17}\)，这些方程就会导出不可能成立的严格循环不等式 \(x\gt y\gt z\gt w\gt x\)。所以唯一的正数解是 \[ x=y=z=w=\sqrt{17}\text{。}\]将四个变量同时取相反数后方程组不变，因此恰好还有一个负数解，其中所有变量都等于 \(-\sqrt{17}\)。所以共有两个实数解。

因此正确答案是 B。

English

Each expression \(t+\frac{17}{t}\) has the same sign as \(t,\) so all four variables have the same sign. Suppose first that they are positive. By AM-GM, every variable is at least \(\sqrt{17}.\) For \(t\gt\sqrt{17},\) \[ \frac{t+\frac{17}{t}}{2}\lt t. \] If any variable exceeded \(\sqrt{17},\) the equations would give the impossible strict cycle \(x\gt y\gt z\gt w\gt x.\) Hence the only positive solution is \[ x=y=z=w=\sqrt{17}. \] Negating all four variables preserves the system, giving exactly one negative solution, with all variables equal to \(-\sqrt{17}.\) Thus there are two real solutions.

Therefore the correct answer is B.