1996 AMC 12 真题（P版）

1 答案 D

三个加数之和为 \(641+852+973=2466\)，比正确结果大 \(10\)。把十位数字 \(7\)（在 \(973\) 中）改为 \(6\)，总和就减少 \(10\)，变成 \(2456\)。题中没有更大的数字能通过改动达到这个效果，所以正确答案是 D。

English

The three addends total \(641+852+973=2466,\) which is \(10\) too large. Changing the tens digit \(7\) in \(973\) to \(6\) lowers the sum by \(10\) and gives \(2456.\) No larger listed digit can make that change, so the correct answer is D.

2 答案 A

十个普通日子会得到 \(10\cdot3=30\) 美元。每个表现特别好的日子会多得 \(5-3=2\) 美元，而实际总额多出 \(6\) 美元。因此共有 \(\frac{6}{2}=3\) 个表现特别好的日子，所以正确答案是 A。

English

Ten ordinary days would pay \(10\cdot3=30\) dollars. Each exceptional day adds \(5-3=2\) dollars, and the actual total is \(6\) dollars higher. Thus there were \(\frac{6}{2}=3\) exceptional days, so the correct answer is A.

3 答案 E

因为 \(3!=6\)，原式为 \(\frac{6!}{6}=5!=120\)。因此正确答案是 E。

English

Since \(3!=6,\) the expression is \(\frac{6!}{6}=5!=120.\) Thus the correct answer is E.

4 答案 D

六个已知数按顺序排列为 \(3,5,5,7,8,9\)。即使三个未给出的整数都大于 \(9\)，完整数列中的第五个数仍为 \(8\)。这个值可以达到，所以正确答案是 D。

English

The six specified values in order are \(3,5,5,7,8,9.\) Even if all three unspecified integers exceed \(9,\) the fifth entry of the full sorted list is \(8.\) This is attainable, so the correct answer is D.

5 答案 E

所列分子中，\(c+d\) 最大；所列分母中，\(a+b\) 最小。因此 \(\frac{c+d}{a+b}\) 大于其余每个正分数，所以正确答案是 E。

English

Among the displayed numerators, \(c+d\) is largest; among the denominators, \(a+b\) is smallest. Therefore \(\frac{c+d}{a+b}\) exceeds every other positive fraction listed, so the correct answer is E.

6 答案 E

直接代入得 \[ \begin{aligned} f(0)&=0,\\ f(-1)&=1,\\ f(-2)&=0,\\ f(-3)&=\frac19 \end{aligned}\text{。}\]它们的和是 \(\frac{10}{9}\)，所以正确答案是 E。

English

Direct substitution gives \[ \begin{aligned} f(0)&=0,\\ f(-1)&=1,\\ f(-2)&=0,\\ f(-3)&=\frac19. \end{aligned} \] Their sum is \(\frac{10}{9},\) so the correct answer is E.

7 答案 B

孩子们的费用为 \(9.45-4.95=4.50\) 美元，所以他们的年龄总和为 \(\frac{4.50}{0.45}=10\)。若每个双胞胎都是 \(t\) 岁，较小的孩子是 \(y\) 岁，则 \(2t+y=10\)，且 \(y\lt t\)。在各选项中，\(y=2\) 给出 \(t=4\)，符合要求。因此正确答案是 B。

English

The children account for \(9.45-4.95=4.50\) dollars, so their ages total \(\frac{4.50}{0.45}=10.\) If each twin is \(t\) and the younger child is \(y,\) then \(2t+y=10\) with \(y\lt t.\) Of the choices, \(y=2\) gives \(t=4,\) which works. Thus the correct answer is B.

8 答案 D

两个方程相除得 \(5=(\frac{4}{2})^r=2^r\)。因此 \(r=\log_2 5\)，所以正确答案是 D。

English

Dividing the equations gives \(5=(\frac{4}{2})^r=2^r.\) Hence \(r=\log_2 5,\) so the correct answer is D.

9 答案 B

因为 \(PA^2+PB^2\) \(=3^2+4^2=5^2=AB^2\)，所以三角形 \(PAB\) 在 \(P\) 处为直角。在正方形 \(ABCD\) 中，\(AD=5\)，且 \(AD\perp AB\)。由于两个平面沿 \(AB\) 互相垂直，\(AD\) 垂直于 \(PAB\) 所在平面，因而也垂直于 \(AP\)。所以 \(PD^2=PA^2+AD^2\) \(=3^2+5^2=34\)。正确答案是 B。

English

Since \(PA^2+PB^2\) \(=3^2+4^2=5^2=AB^2,\) triangle \(PAB\) is right at \(P.\) In square \(ABCD,\) \(AD=5\) and \(AD\perp AB.\) Because the two planes are perpendicular along \(AB,\) \(AD\) is perpendicular to the plane of \(PAB,\) and hence to \(AP.\) Thus \(PD^2=PA^2+AD^2\) \(=3^2+5^2=34.\) The correct answer is B.

10 答案 D

立方体的 \(8\) 个顶点中，每个无序点对都确定一条线段，其中包括棱和对角线。共有 \(\binom82=28\) 条，所以正确答案是 D。

English

Every unordered pair of the cube’s \(8\) vertices determines one segment, including edges and diagonals. There are \(\binom82=28,\) so the correct answer is D.

11 答案 D

每条线段的中心是距圆心 \(2\) 个单位的切点，并沿切线向两个方向各延伸 \(1\) 个单位。当切点绕圆转动时，这些线段填满一个内半径为 \(2\)、外半径为 \(\sqrt{2^2+1^2}=\sqrt5\) 的圆环。其面积为 \(\pi(5-4)=\pi\)，所以正确答案是 D。

English

Each segment is centered at a tangency point \(2\) units from the circle’s center and extends \(1\) unit along the tangent in both directions. As the tangency point rotates, the segments fill the annulus with inner radius \(2\) and outer radius \(\sqrt{2^2+1^2}=\sqrt5.\) Its area is \(\pi(5-4)=\pi,\) so the correct answer is D.

12 答案 B

因为 \(k\) 是奇数，所以 \(f(k)=k+3\) 是偶数，于是 \(f(f(k))=\frac{k+3}{2}\)。若这个值是奇数，加上 \(3\) 不可能得到 \(27\)，因为这会要求该值为 \(24\)。因此它是偶数，除以二后得到 \(27\)，所以 \(\frac{k+3}{2}=54\)。由此 \(k=105\)，其各位数字之和为 \(6\)。正确答案是 B。

English

Since \(k\) is odd, \(f(k)=k+3\) is even, so \(f(f(k))=\frac{k+3}{2}.\) If this value were odd, adding \(3\) could not produce \(27\) because that would require the value \(24.\) Therefore it is even and is halved to \(27,\) so \(\frac{k+3}{2}=54.\) Hence \(k=105,\) whose digits sum to \(6.\) The correct answer is B.

13 答案 D

假设月光跑了 \(x\) 米。经过的时间为 \(\frac{x}{mv}\)，所以从月光出发起，桑妮又跑了 \(\frac{v\cdot x}{mv}=\frac{x}{m}\) 米。追上时，\(x=h+\frac{x}{m}\)。解得 \(x=\frac{hm}{m-1}\)，所以正确答案是 D。

English

Suppose Moonbeam runs \(x\) meters. The elapsed time is \(\frac{x}{mv},\) so Sunny runs \(\frac{v\cdot x}{mv}=\frac{x}{m}\) meters after the start. At the catch, \(x=h+\frac{x}{m}.\) Solving gives \(x=\frac{hm}{m-1},\) so the correct answer is D.

14 答案 C

从 \(00\) 到 \(99\)，每个数字在两个数位上都各出现 \(10\) 次。正偶数数字的和为 \(2+4+6+8=20\)，所以总贡献为 \(2\cdot10\cdot20=400\)。数 \(100\) 不贡献任何正偶数数字，所以正确答案是 C。

English

From \(00\) through \(99,\) each digit occurs \(10\) times in each of the two positions. The positive even digits sum to \(2+4+6+8=20,\) so the total contribution is \(2\cdot10\cdot20=400.\) The number \(100\) contributes no even positive digit, so the correct answer is C.

15 答案 B

三角形 \(A\) 的底为 \(\frac{w}{n}\)，高为 \(\frac{h}{2}\)，所以面积为 \(\frac{wh}{4n}\)。三角形 \(B\) 的底为 \(\frac{h}{m}\)，高为 \(\frac{w}{2}\)，所以面积为 \(\frac{wh}{4m}\)。二者之比为 \(\frac{m}{n}\)，正确答案是 B。

English

Triangle \(A\) has base \(\frac{w}{n}\) and altitude \(\frac{h}{2},\) so its area is \(\frac{wh}{4n}.\) Triangle \(B\) has base \(\frac{h}{m}\) and altitude \(\frac{w}{2},\) so its area is \(\frac{wh}{4m}.\) Their ratio is \(\frac{m}{n},\) and the correct answer is B.

16 答案 D

当第三次点数依次为 \(2,3,4,5,6\) 时，前两次点数的有序数对数量依次为 \(1,2,3,4,5\)，所以共有 \(15\) 个等可能的条件结果。第三次掷出 \(2\) 对应 \((1,1,2)\)。第一次掷出 \(2\) 有 \(4\) 个结果，第二次掷出 \(2\) 也有 \(4\) 个结果，而 \((2,2,4)\) 被重复计算。因此有利结果共有 \(1+4+4-1=8\) 个，正确答案是 D。

English

For third tosses \(2,3,4,5,6,\) the numbers of ordered first-two-toss pairs are \(1,2,3,4,5,\) for \(15\) equally likely conditional outcomes. A third toss of \(2\) contributes \((1,1,2).\) A \(2\) in the first position gives \(4\) outcomes and a \(2\) in the second gives \(4,\) with \((2,2,4)\) counted twice. Thus there are \(1+4+4-1=8\) favorable outcomes, and the correct answer is D.

17 答案 E

设长方形的宽为 \(w\)，高为 \(h\)。在直角三角形 \(CBE\) 中，\(\tan30^\circ=\frac{BE}{CB}=\frac{6}{h}\)，所以 \(h=6\sqrt3\)。在由 \(C,F,D\) 构成的三角形中，水平距离为 \(w\)，竖直下降距离为 \(h-2\)，所以 \(\tan60^\circ=\frac{w}{h-2}\)。因此 \[ w=\sqrt3(6\sqrt3-2)=18-2\sqrt3\text{。}\]面积为 \(6\sqrt3(18-2\sqrt3)\) \(=108\sqrt3-36\approx151.1\)，最接近 \(150\)。所以正确答案是 E。

English

Let the rectangle have width \(w\) and height \(h.\) In right triangle \(CBE,\) \(\tan30^\circ=\frac{BE}{CB}=\frac{6}{h},\) so \(h=6\sqrt3.\) In the triangle using \(C,F,D,\) the horizontal run is \(w\) and the vertical drop is \(h-2,\) so \(\tan60^\circ=\frac{w}{h-2}.\) Hence \[ w=\sqrt3(6\sqrt3-2)=18-2\sqrt3. \] The area is \(6\sqrt3(18-2\sqrt3)\) \(=108\sqrt3-36\approx151.1,\) closest to \(150.\) Thus the correct answer is E.

18 答案 D

把上方的公切线写成 \(mx-y+b=0\)，并令 \(s=\sqrt{m^2+1}\)。两个圆心到直线的距离给出 \[ 2m+b=2s,\qquad 5m+b=s\text{。}\]因此 \(3m=-s\)，所以 \(9m^2=m^2+1\)，且 \(m=-\frac{1}{\sqrt8}\)。由第二个方程，\[ \begin{aligned} b&=s-5m\\ &=\frac3{\sqrt8}+\frac5{\sqrt8}=2\sqrt2 \end{aligned}\text{。}\]所以正确答案是 D。

English

Write the upper common tangent as \(mx-y+b=0\) and let \(s=\sqrt{m^2+1}.\) The two center-to-line distances give \[ 2m+b=2s,\qquad 5m+b=s. \] Thus \(3m=-s,\) so \(9m^2=m^2+1\) and \(m=-\frac{1}{\sqrt8}.\) From the second equation, \[ \begin{aligned} b&=s-5m\\ &=\frac3{\sqrt8}+\frac5{\sqrt8}=2\sqrt2. \end{aligned} \] Therefore the correct answer is D.

19 答案 D

两个相邻中点与它们共有的顶点组成一个三角形，其中两边均为 \(\frac{s}{2}\)，夹角为 \(120^\circ\)。其对边长度的平方为 \[ \begin{aligned} \frac{s^2}{4}+\frac{s^2}{4} &-2\left(\frac{s}{2}\right)^2\cos120^\circ\\ &=\frac{3s^2}{4} \end{aligned}\text{。}\]因此较小六边形的缩放比例为 \(\frac{\sqrt3}{2}\)，面积之比为 \(\frac{3}{4}\)。正确答案是 D。

English

Two adjacent midpoints and their shared vertex form a triangle with two sides \(\frac{s}{2}\) and included angle \(120^\circ.\) Its opposite side has squared length \[ \begin{aligned} \frac{s^2}{4}+\frac{s^2}{4} &-2\left(\frac{s}{2}\right)^2\cos120^\circ\\ &=\frac{3s^2}{4}. \end{aligned} \] Thus the smaller hexagon has scale factor \(\frac{\sqrt3}{2},\) and its area ratio is \(\frac{3}{4}.\) The correct answer is D.

20 答案 C

每个端点到圆心的距离都是 \(10\)，所以每条切线段的长度为 \(\sqrt{10^2-5^2}=5\sqrt3\)。在切点处的直角三角形中，圆心处的角为 \(60^\circ\)。由于从圆心到两个端点的射线方向相反，两个切点之间的小弧所对圆心角为 \(180^\circ-2(60^\circ)=60^\circ\)，所以弧长为 \(\frac{5\pi}{3}\)。总长度为 \(10\sqrt3+\frac{5\pi}{3}\)，正确答案是 C。

English

Each endpoint is \(10\) units from the circle’s center, so each tangent segment has length \(\sqrt{10^2-5^2}=5\sqrt3.\) In the right triangle at a tangency point, the angle at the center is \(60^\circ.\) Since the endpoint rays are opposite, the intervening minor arc subtends \(180^\circ-2(60^\circ)=60^\circ,\) so its length is \(\frac{5\pi}{3}.\) The total is \(10\sqrt3+\frac{5\pi}{3},\) and the correct answer is C.

21 答案 D

设 \(\angle BAC=\alpha\)。由于 \(AB=AC\)，\[ \angle C=\frac{180^\circ-\alpha}{2}=90^\circ-\frac{\alpha}{2}\text{。}\]由于 \(BD\perp AC\)，\(BA\) 与 \(BD\) 的夹角为 \(90^\circ-\alpha\)。在等腰三角形 \(ABD\) 中，\(AB=BD\)，所以两个底角为 \[ \begin{aligned} \angle D &=\frac{180^\circ-(90^\circ-\alpha)}2\\ &=45^\circ+\frac{\alpha}{2} \end{aligned}\text{。}\]二者之和为 \(135^\circ\)，所以正确答案是 D。

English

Let \(\angle BAC=\alpha.\) Since \(AB=AC,\) \[ \angle C=\frac{180^\circ-\alpha}{2}=90^\circ-\frac{\alpha}{2}. \] Because \(BD\perp AC,\) the angle between \(BA\) and \(BD\) is \(90^\circ-\alpha.\) In isosceles triangle \(ABD,\) \(AB=BD,\) so its two base angles are \[ \begin{aligned} \angle D &=\frac{180^\circ-(90^\circ-\alpha)}2\\ &=45^\circ+\frac{\alpha}{2}. \end{aligned} \] Their sum is \(135^\circ,\) so the correct answer is D.

22 答案 B

对于任意固定的四个点，把它们的标号分成两组无序弦对共有三种方式。恰有一种配对连接圆周上交错排列的点，因此两弦相交。指定的弦对 \(\overline{AB},\overline{CD}\) 等可能地对应这三种配对中的任意一种，所以概率为 \(\frac{1}{3}\)。正确答案是 B。

English

For any fixed four points, there are three ways to partition their labels into two unordered chord pairs. Exactly one pairing joins alternating points around the circle and therefore crosses. The named pair \(\overline{AB},\overline{CD}\) is equally likely to be any of these three pairings, so the probability is \(\frac{1}{3}.\) The correct answer is B.

23 答案 B

棱长之和给出 \(4(a+b+c)=140\)，所以 \(a+b+c=35\)。空间对角线给出 \(a^2+b^2+c^2=21^2=441\)。因此总表面积为 \[ \begin{gathered} 2(ab+bc+ca)\\ {}=(a+b+c)^2\\ {}-(a^2+b^2+c^2)\\ {}=35^2-441=784 \end{gathered}\text{。}\]所以正确答案是 B。

English

The edge sum gives \(4(a+b+c)=140,\) so \(a+b+c=35.\) The space diagonal gives \(a^2+b^2+c^2=21^2=441.\) Therefore the surface area is \[ \begin{gathered} 2(ab+bc+ca)\\ {}=(a+b+c)^2\\ {}-(a^2+b^2+c^2)\\ {}=35^2-441=784. \end{gathered} \] Thus the correct answer is B.

24 答案 B

到第 \(k\) 个数块为止，有 \(k\) 个一和 \(\frac{k(k+1)}{2}\) 个二，因此共有 \(\frac{k(k+3)}{2}\) 项。当 \(k=48\) 时，项数为 \(1224\)。这些项的和为 \[ 48+2\left(\frac{48\cdot49}{2}\right)=2400\text{。}\]接下来的 \(10\) 项是一个 \(1\) 和九个 \(2\)，和为 \(19\)。所求总和为 \(2419\)，所以正确答案是 B。

English

Through block \(k\) there are \(k\) ones and \(\frac{k(k+1)}{2}\) twos, hence \(\frac{k(k+3)}{2}\) terms. For \(k=48\) this is \(1224.\) Their sum is \[ 48+2\left(\frac{48\cdot49}{2}\right)=2400. \] The next \(10\) terms are one \(1\) and nine \(2\)’s, with sum \(19.\) The requested sum is \(2419,\) so the correct answer is B.

25 答案 B

配方得 \[ (x-7)^2+(y-3)^2=64\text{。}\]在圆心处，\(3x+4y=3(7)+4(3)=33\)。移动 \(8\) 个单位，方向取为向量 \((3,4)\) 的方向，表达式会增加 \(8\sqrt{3^2+4^2}=40\)。最大值为 \(73\)，所以正确答案是 B。

English

Completing the square gives \[ (x-7)^2+(y-3)^2=64. \] At the center, \(3x+4y=3(7)+4(3)=33.\) Moving \(8\) units in the direction of the vector \((3,4)\) increases the expression by \(8\sqrt{3^2+4^2}=40.\) The maximum is \(73,\) so the correct answer is B.

26 答案 B

相等的概率具有相同的总样本数，因此有利选法数满足 \(\binom r4=w\binom r3\) \(=wb\binom r2=wbgr\)。逐次取比值得 \(w=\frac{r-3}{4}\)、\(b=\frac{r-2}{3}\) 和 \(g=\frac{r-1}{2}\)。使三者均为正整数的最小 \(r\ge4\) 是 \(r=11\)。此时 \((w,b,g)=(2,3,5)\)，弹珠总数为 \(11+2+3+5=21\)。所以正确答案是 B。

English

Equal probabilities have the same common denominator, so their favorable selection counts satisfy \(\binom r4=w\binom r3\) \(=wb\binom r2=wbgr.\) Successive ratios give \(w=\frac{r-3}{4},\) \(b=\frac{r-2}{3},\) and \(g=\frac{r-1}{2}.\) The least \(r\ge4\) making all three positive integers is \(r=11.\) Then \((w,b,g)=(2,3,5),\) for \(11+2+3+5=21\) marbles. Thus the correct answer is B.

27 答案 D

第一个球体允许的整数高度为 \(5\) 到 \(16\)，第二个允许的整数高度为 \(-3\) 到 \(5\)，所以交集中的格点必须满足 \(z=5\)。在这个高度，两个限制为 \[ \begin{aligned} x^2+y^2 &\le36-\left(5-\frac{21}{2}\right)^2\\ &=\frac{23}{4},\\ x^2+y^2 &\le\frac{81}{4}-(5-1)^2\\ &=\frac{17}{4} \end{aligned}\text{。}\]因此 \(x^2+y^2\) 可以是 \(0,1,2\) 或 \(4\)。它们分别给出 \(1,4,4\) 和 \(4\) 个有序整数对，共有 \(13\) 个点。正确答案是 D。

English

The first ball permits integer heights \(5\) through \(16,\) while the second permits \(-3\) through \(5,\) so an intersection lattice point must have \(z=5.\) At that height the two bounds are \[ \begin{aligned} x^2+y^2 &\le36-\left(5-\frac{21}{2}\right)^2\\ &=\frac{23}{4},\\ x^2+y^2 &\le\frac{81}{4}-(5-1)^2\\ &=\frac{17}{4}. \end{aligned} \] Thus \(x^2+y^2\) can be \(0,1,2,\) or \(4.\) These give \(1,4,4,\) and \(4\) ordered integer pairs, respectively, for \(13\) points. The correct answer is D.

28 答案 C

令 \(D=(0,0,0)\)，并把三个相邻顶点取为 \((4,0,0),(0,4,0),(0,0,3)\)。它们所在的平面为 \[ \frac{x}{4}+\frac{y}{4}+\frac{z}{3}=1\text{。}\]该平面到原点的距离为 \[ \begin{aligned} \frac1{\sqrt{\begin{gathered} (\frac{1}{4})^2+(\frac{1}{4})^2\\ {}+(\frac{1}{3})^2 \end{gathered}}} &=\frac{12}{\sqrt{34}}\\ &\approx2.06 \end{aligned}\text{，}\]最接近 \(2.1\)。所以正确答案是 C。

English

Put \(D=(0,0,0)\) and take the adjacent vertices as \((4,0,0),(0,4,0),(0,0,3).\) Their plane is \[ \frac{x}{4}+\frac{y}{4}+\frac{z}{3}=1. \] Its distance from the origin is \[ \begin{aligned} \frac1{\sqrt{\begin{gathered} (\frac{1}{4})^2+(\frac{1}{4})^2\\ {}+(\frac{1}{3})^2 \end{gathered}}} &=\frac{12}{\sqrt{34}}\\ &\approx2.06, \end{aligned} \] closest to \(2.1.\) Thus the correct answer is C.

29 答案 C

写成 \(n=2^a3^bq\)，其中 \(\gcd(q,6)=1\)，并令 \(t=\tau(q)\)。于是 \[ \begin{aligned} (a+2)(b+1)t&=28,\\ (a+1)(b+2)t&=30 \end{aligned}\text{。}\]因此 \(t\) 是 \(1\) 或 \(2\)。检验 \(14,15\) 的因数对，发现当 \(t=2\) 时无解；检验 \(28,30\) 的因数对，发现当 \(t=1\) 时得到唯一解 \(a=5,b=3\)。所以 \[ \begin{aligned} \tau(6n)&=(a+2)(b+2)t\\ &=7\cdot5=35 \end{aligned}\text{，}\]正确答案是 C。

English

Write \(n=2^a3^bq\) with \(\gcd(q,6)=1,\) and let \(t=\tau(q).\) Then \[ \begin{aligned} (a+2)(b+1)t&=28,\\ (a+1)(b+2)t&=30. \end{aligned} \] Hence \(t\) is \(1\) or \(2.\) Checking the factor pairs of \(14,15\) gives no solution when \(t=2.\) Checking those of \(28,30\) when \(t=1\) gives the unique solution \(a=5,b=3.\) Therefore \[ \begin{aligned} \tau(6n)&=(a+2)(b+2)t\\ &=7\cdot5=35, \end{aligned} \] so the correct answer is C.

30 答案 E

设圆的半径为 \(R\)，并设 \(\alpha,\beta\) 分别为长度 \(3,5\) 的边所对半圆心角。则 \(3=2R\sin\alpha\)、\(5=2R\sin\beta\)，且 \(\alpha+\beta=60^\circ\)。因此 \(\frac53=\frac{\sin(60^\circ-\alpha)}{\sin\alpha}\) \(=\frac{\sqrt3}{2}\cot\alpha-\frac12\)，所以 \(\tan\alpha=\frac{3\sqrt3}{13}\)，且 \(\sin^2\alpha=\frac{27}{196}\)。分割弦跨过连续三条长度为 \(3\) 的边，所以 \(\frac L3=\frac{\sin3\alpha}{\sin\alpha}\) \(=3-4\sin^2\alpha=3-\frac{27}{49}=\frac{120}{49}\)。因此 \(L=\frac{360}{49}\)，所以 \(m+n=409\)。正确答案是 E。

English

Let the circle have radius \(R,\) and let \(\alpha,\beta\) be the half-central angles for the sides \(3,5.\) Then \(3=2R\sin\alpha,\) \(5=2R\sin\beta,\) and \(\alpha+\beta=60^\circ.\) Thus \(\frac53=\frac{\sin(60^\circ-\alpha)}{\sin\alpha}\) \(=\frac{\sqrt3}{2}\cot\alpha-\frac12,\) so \(\tan\alpha=\frac{3\sqrt3}{13}\) and \(\sin^2\alpha=\frac{27}{196}.\) The dividing chord spans the three consecutive sides of length \(3,\) so \(\frac L3=\frac{\sin3\alpha}{\sin\alpha}\) \(=3-4\sin^2\alpha=3-\frac{27}{49}=\frac{120}{49}.\) Therefore \(L=\frac{360}{49},\) so \(m+n=409.\) The correct answer is E.