2011B AMC 10 真题（P版）

1 答案 C

直接计算：\[\begin{aligned} &\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} \\ &= \dfrac{12}9 - \dfrac 9{12} \\ &= \dfrac {16}{12} - \dfrac 9{12} \\ &=\dfrac 7{12} \end{aligned}\text{。}\]

所以正确答案是 C。

English

Simply solving directly: \[\begin{aligned} &\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} \\ &= \dfrac{12}9 - \dfrac 9{12} \\ &= \dfrac {16}{12} - \dfrac 9{12} \\ &=\dfrac 7{12}. \end{aligned}\]

Thus, the correct answer is C .

2 答案 E

她当前的平均分为 \[\dfrac{90+80+70+60+85}5 = 77\text{，}\]而总分为 \(385\)。目标平均分为 \(80\)，所以六次考试所需的总分为 \(80\cdot 6\)。因此下一次考试至少需要得 \[80\cdot 6-385 = 95\text{。}\]

所以正确答案是 E。

English

Her current average is \[\dfrac{90+80+70+60+85}5 = 77,\] and the sum of her scores is \(385.\) The desired average is then \(80,\) so the sum of scores required is \(80\cdot 6.\) Therefore, the answer is \[80\cdot 6-385 = 95.\]

Thus, the correct answer is E .

3 答案 A

最小可能尺寸为 \(1.5\times 2.5\)，所以面积为 \[1.5\cdot 2.5=3.75\text{。}\]

所以正确答案是 A。

English

The smallest possible dimensions are \(1.5\times 2.5,\) so the area is \[1.5\cdot 2.5=3.75.\]

Thus, the correct answer is A .

4 答案 C

两人每人应承担 \(\dfrac{A+B}2\)，而 LeRoy 已支付 \(A\)。因此他还需向 Bernardo 支付 \[\dfrac{A+B}2-A = \dfrac{B-A}2\] 美元。

所以正确答案是 C。

English

The amount they each would have to pay is \(\dfrac{A+B}2,\) and LeRoy paid \(A.\) Thus, he has to pay \[\dfrac{A+B}2-A = \dfrac{B-A}2\] more.

Thus, the correct answer is C .

5 答案 E

数 \(161\) 等于 \(7\cdot 23\)。\(161\) 除了 \(1\cdot 161\) 外没有其他因数对，所以 \(23\) 是唯一的两位因数。因此 \(23\) 是颠倒后的数，原来的乘法应为 \(32\cdot 7\)，结果是 \(224\)。

所以正确答案是 E。

English

The number \(161\) is equal to \(7\cdot 23.\) There are no other pairs of numbers that multiply to \(161\) besides \(1\cdot 161,\) so \(23\) is the only two digit factor. Thus, \(23\) is the number reversed, so he meant to get \(32\cdot 7\) which is \(224.\)

Thus, the correct answer is E .

6 答案 A

倒推。给妹妹 \(4\) 颗之前，Casper 有 \(12\) 颗糖果，因为第二天结束时剩 \(8\) 颗。

这 \(12\) 颗是第一天结束后糖果数的 \(\dfrac23\)，所以第一天结束后有 \(18\) 颗。

给弟弟 \(2\) 颗之前有 \(20\) 颗。这是原来糖果数的 \(\dfrac23\)，所以一开始有 \(30\) 颗。

所以正确答案是 A。

English

Work backward. Before giving \(4\) candies to his sister, Casper had \(12\) candies, because he ended the second day with \(8\).

Those \(12\) candies were \(\dfrac23\) of what he had after the first day, so after the first day he had \(18\) candies.

Before giving \(2\) candies to his brother, he had \(20\) candies. This was \(\dfrac23\) of his original amount, so he began with \(30\) candies.

Thus, A is the correct answer.

7 答案 B

这两个角之和为 \(\frac 65 \cdot 90 = 108\)。所以第三个角为 \(180-108=72\)。若这两个角中较大的为 \(x\)，较小的为 \(x-30\)，则 \(2x-30=108\)，所以 \(x=69\)。

没有角大于 \(72\)，因此最大角是 \(72\)。

所以正确答案是 B。

English

The two angles add to \(\frac 65 \cdot 90 = 108.\) This makes the other angle \(180-108=72.\) Then, if the larger of the two angles is \(x\) then the smaller of them is \(x-30\) so their sum is \(2x-30=108,\) making \(x=69.\)

This means no angle is larger than \(72,\) making the largest equal to \(72.\)

Thus, the correct answer is B .

8 答案 B

原命题是：若气温至少 \(80^\circ F\) 且晴朗，则海滩拥挤。

因为海滩不拥挤，这两个条件不可能同时成立。因此气温低于 \(80^\circ F\)，或者天气不晴朗，或者两者都成立。

所以正确答案是 B。

English

The statement says that if the weather was at least \(80^\circ F\) and sunny, then the beach was crowded.

Because the beach was not crowded, the two conditions could not both have been true. Therefore the temperature was cooler than \(80^\circ F\), or it was not sunny, or both.

Thus, B is the correct answer.

9 答案 D

由角角相似，\(BDE \sim BCA \)。

面积比为 \(\frac 13\)，所以对应边长比为 \(\frac{1}{\sqrt 3}\)。

因此 \[\dfrac{BD}{BC} = \dfrac{BD}4 = \dfrac{1}{\sqrt 3}\text{，}\]从而 \[BD = \dfrac 4{ \sqrt 3} = \dfrac{4\sqrt{3}}{3}\text{。}\]

所以正确答案是 D。

English

By angle angle similarity, we have \(BDE \sim BCA .\)

Then, since the ratio of the areas is \(\frac 13,\) the ratio of the sidelengths is \(\frac{1}{\sqrt 3}.\)

As such, \[\dfrac{BD}{BC} = \dfrac{BD}4 = \dfrac{1}{\sqrt 3}, \] making \[BD = \dfrac 4{ \sqrt 3} = \dfrac{4\sqrt{3}}{3} .\]

Thus, the correct answer is D .

10 答案 B

最大元素为 \(10^{10}\)。其余十个元素之和为 \[S=\sum_{i=0}^9 10^i=\dfrac{10^{10}-1}{9}\text{。}\]因此所求比值为 \[\dfrac{10^{10}}{S}=9\cdot\dfrac{10^{10}}{10^{10}-1}\text{，}\]它略大于 \(9\)，所以最接近 \(9\)。

所以正确答案是 B。

English

The largest number is \(10^{10}.\) The other ten numbers have sum \[S=\sum_{i=0}^9 10^i=\dfrac{10^{10}-1}{9}.\] Therefore their ratio is \[\dfrac{10^{10}}{S}=9\cdot\dfrac{10^{10}}{10^{10}-1},\] which is just slightly greater than \(9\) and hence closest to \(9.\)

Thus, the correct answer is B .

11 答案 D

\(n\geq 6\) 不一定成立，因为可以让 \(5\) 人的生日落在前 \(4\) 个月中的每个月，而其余八个月各有 \(4\) 人生日。

但一定有某个月至少 \(5\) 人生日，因为平均每月人数为 \(\frac{52}{12}\)，大于 \(4\)。

所以正确答案是 D。

English

It isn’t necessarily true for \(n\geq 6\) as we could have \(5\) people born in the first \(4\) months and \(4\) people born in the subsequent months.

However, one month must be greater than or equal to \(5\) as the average of the number of people born in each month is \(\frac{52}{12}\) which is greater than \(4,\) and some month must be above average.

Thus, the correct answer is D .

12 答案 A

设内侧半圆半径为 \(r\)。

内外路径的直线部分总长相同，所以距离差只来自两个半圆端。两个内侧半圆总长为 \(2\pi r\)，两个外侧半圆总长为 \(2\pi(r+6)\)，差为 \(12\pi\)。

Keiko 多用 \(36\) 秒走了 \(12\pi\) 米，所以速度为 \(\frac{12\pi}{36}=\dfrac\pi3\) 米每秒。

所以正确答案是 A。

English

Let the inner semicircle radius be \(r\). The straight parts of the inside and outside paths have the same total length, so only the semicircular ends change the distance.

The two inner semicircles have total length \(2\pi r\), while the two outer semicircles have total length \(2\pi(r+6)\). The outside path is therefore \(12\pi\) meters longer.

Keiko takes \(36\) more seconds to walk \(12\pi\) more meters, so her speed is \(\frac{12\pi}{36}=\dfrac\pi3\) meters per second.

Thus, A is the correct answer.

13 答案 D

选中的数为负数的概率是 \(\frac{20}{30}=\frac{2}{3}\)，为正数的概率是 \(\frac{10}{30}=\frac{1}{3}\)。（恰好选中 \(0\) 的概率为 \(0\)。）乘积为正当且仅当两个数同号，所以所求概率为 \[\left(\dfrac23\right)^2+\left(\dfrac13\right)^2=\dfrac49+\dfrac19=\dfrac59\text{。}\]

所以正确答案是 D。

English

A selected number is negative with probability \(\frac{20}{30}=\frac{2}{3}\) and positive with probability \(\frac{10}{30}=\frac{1}{3}.\) (Selecting exactly \(0\) has probability \(0.\)) The product is positive exactly when both numbers have the same sign, so the probability is \[\left(\dfrac23\right)^2+\left(\dfrac13\right)^2=\dfrac49+\dfrac19=\dfrac59.\]

Thus, the correct answer is D .

14 答案 C

设长方形边长为 \(l,w\)。要求的是 \(2(l+w)\)。由勾股定理，\[25 = \sqrt{l^2+w^2}\] \[l^2+w^2 = 625\] 并且 \(lw = 168\)。

因此 \[\begin{aligned}l^2+2lw+ w^2 &= (l+w)^2 \\&= 961 \\&= 31^2\end{aligned}\text{。}\]所以 \(l+w = 31\)，周长为 \(31\cdot 2=62\)。

所以正确答案是 C。

English

Let the side lengths be \(l,w.\) We wish to find \(2(l+w).\) From the Pythagorean Theorem, we get \[25 = \sqrt{l^2+w^2}\]\[l^2+w^2 = 625\] We also know \(lw = 168.\)

As such \[\begin{aligned}l^2+2lw+ w^2 &= (l+w)^2 \\&= 961 \\&= 31^2.\end{aligned}\] This makes \(l+w = 31,\) and as such, our answer is \(31\cdot 2=62.\)

Thus, the correct answer is C .

15 答案 E

I 的左边为 \(\dfrac{x+y+z}2\)，右边为 \(x+\dfrac{y+z}2\)，不恒相等。

II 的左右两边都等于 \(x+\dfrac{y+z}2\)，所以恒成立。

III 的左右两边都等于 \(\dfrac{2x+y+z}4\)，所以也恒成立。

所以正确答案是 E。

English

In statement I, the left-hand side equals \(\dfrac{x+y+z}2\) and the right-hand side equals \(x+\dfrac{y+z}2,\) so they are not always equal.

In statement II, both sides equal \(x+\dfrac{y+z}2,\) so it holds.

In statement III, both sides equal \(\dfrac{2x+y+z}4,\) so it also holds.

Thus, the correct answer is E .

16 答案 A

设中心正方形边长为 \(1\)。则中心面积为 \(1\)。

八边形可看作一个大正方形去掉 \(4\) 个等腰直角三角形。大正方形边长为 \[1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2\text{，}\]所以面积为 \[(1+\sqrt 2)^2 = 3 + 2 \sqrt 2\text{。}\]

这些直角三角形的直角边长为 \(\dfrac {1}{\sqrt 2} \)，每个面积为 \[\dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14\text{。}\]四个总面积为 \(1\)，所以八边形面积为 \(2+ 2 \sqrt 2\)。

因此比值为 \[\begin{aligned}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2} \end{aligned}\text{。}\]

所以正确答案是 A。

English

Let the side length be \(1.\) Then, the area of the center is \(1.\)

Then, we must find the area of the octagon. It can be found as a square with \(4\) isosceles right triangles taken out. The side length of this square is \[1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2 .\] It has an area of \[(1+\sqrt 2)^2 = 3 + 2 \sqrt 2.\]

Then, the side length of the right triangles is \(\dfrac {1}{\sqrt 2} ,\) making the area of one equal to \[ \dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14 .\] This makes them have a total combined area of \(1,\) so the area of the octagon is \(2+ 2 \sqrt 2.\)

Thus, the ratio is \[\begin{aligned}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{aligned}\]

Thus, the correct answer is A .

17 答案 C

因为 \(EB\) 是直径，\(\angle EAB=90^\circ\)。由 \(\angle AEB:\angle ABE=4:5\)，得 \(\angle AEB=40^\circ\)，\(\angle ABE=50^\circ\)。

因为 \(AB\parallel ED\)，所以 \(\angle DEB=50^\circ\)。又 \(EB\parallel DC\)，四边形 \(EBCD\) 是等腰梯形，所以 \(\angle BCD=\angle CDE\)。

角 \(DEB\) 与 \(CDE\) 互补，所以 \(\angle CDE=180^\circ-50^\circ=130^\circ\)，因此 \(\angle BCD=130^\circ\)。

所以正确答案是 C。

English

Since \(EB\) is a diameter, \(\angle EAB=90^\circ\). The ratio \(\angle AEB:\angle ABE=4:5\) then gives \(\angle AEB=40^\circ\) and \(\angle ABE=50^\circ\).

Because \(AB\parallel ED\), \(\angle DEB=50^\circ\). Since \(EB\parallel DC\), quadrilateral \(EBCD\) is an isosceles trapezoid, so \(\angle BCD=\angle CDE\).

Angles \(DEB\) and \(CDE\) are supplementary, so \(\angle CDE=180^\circ-50^\circ=130^\circ\). Hence \(\angle BCD=130^\circ\).

Thus, C is the correct answer.

18 答案 E

角 \(\angle AMD\) 与 \(\angle MDC\) 相等，因为 \(AB \parallel DC\)。

因此 \(\angle MDC = \angle DMC \)，所以三角形 \(MDC\) 为等腰三角形，\(MC = DC = 6\)。

在直角三角形中，由边长比例可得 \(\sin (\angle CMB) = \frac 12\)，因此 \(\angle CMB = 30^\circ \)。

因此 \(\angle AMC = 150^\circ \)。由于 \(\angle AMD\) 是它的一半，\(\angle AMD = 75^\circ \)。

所以正确答案是 E。

English

The angles \(\angle AMD \) and \(\angle MDC \) are equal since \(AB \parallel DC.\)

As such, \(\angle MDC = \angle DMC ,\) making \(MDC \) isosceles and \(MC = DC = 6.\)

As we can see, \(\sin (\angle CMB) = \frac 12,\) making \(\angle CMB = 30^\circ .\)

Therefore, \(\angle AMC = 150^\circ .\) Since \(\angle AMD \) is half of that, \(\angle AMD = 75^\circ .\)

Thus, the correct answer is E .

19 答案 A

因为 \(x^2=|x|^2\)，将方程两边平方可得 \[\begin{aligned} 5|x|+8 &= |x|^2-16,\\ |x|^2-5|x|-24&=0,\\ (|x|-8)(|x|+3)&=0 \end{aligned}\text{。}\]\(|x|\) 在代数上可能为 \(8\) 或 \(-3\)，但绝对值不能为负。因此 \(|x|=8\)，两个根为 \(8\) 和 \(-8\)，乘积为 \(-64\)。

所以正确答案是 A。

English

Because \(x^2=|x|^2,\) squaring the equation gives \[\begin{aligned} 5|x|+8 &= |x|^2-16,\\ |x|^2-5|x|-24&=0,\\ (|x|-8)(|x|+3)&=0. \end{aligned}\] The algebraic possibilities for \(|x|\) are \(8\) and \(-3,\) but an absolute value cannot be negative. Hence \(|x|=8,\) so the roots are \(8\) and \(-8\), with product \(-64.\)

Thus, the correct answer is A .

20 答案 C

比到其他顶点更接近 \(B\) 的点由 \(BA\)、\(BC\)、\(BD\) 的垂直平分线围定。\(BD\) 的垂直平分线是对角线 \(AC\)，所以所求区域位于 \(\triangle ABC\) 中。

三角形 \(ABC\) 面积为 \(\dfrac12\cdot2\cdot2\sin120^\circ=\sqrt3\)。\(BA\) 和 \(BC\) 的垂直平分线切去两个全等的 \(30\)-\(60\)-\(90\) 三角形，每个面积为 \(\dfrac{\sqrt3}{6}\)。

所求面积为 \(\sqrt3-2\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}\)。

所以正确答案是 C。

English

The points closer to \(B\) than to another vertex are bounded by the perpendicular bisectors of \(BA\), \(BC\), and \(BD\). The bisector of \(BD\) is diagonal \(AC\), so the desired region lies in \(\triangle ABC\).

Triangle \(ABC\) has area \(\dfrac12\cdot2\cdot2\sin120^\circ=\sqrt3\). The perpendicular bisectors of \(BA\) and \(BC\) cut off two congruent \(30\)-\(60\)-\(90\) triangles, each with area \(\dfrac{\sqrt3}{6}\).

Therefore the desired area is \(\sqrt3-2\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}\).

Thus, C is the correct answer.

21 答案 B

最大差为 \(9\)，所以 \(w-z=9\)。对任一中间数 \(n\)，差 \(w-n\) 与 \(n-z\) 的和必须为 \(9\)。

可用的和为 \(9\) 的配对为 \(3+6\) 和 \(4+5\)，剩下两个中间数之间的差为 \(1\)。

一种集合为 \(\{w,w-5,w-6,w-9\}\)，给出 \(4w-20=44\)，所以 \(w=16\)。另一种为 \(\{w,w-3,w-4,w-9\}\)，给出 \(4w-16=44\)，所以 \(w=15\)。

可能的 \(w\) 之和为 \(16+15=31\)。

所以正确答案是 B。

English

The largest difference is \(9\), so \(w-z=9\). For either middle number \(n\), the two differences \(w-n\) and \(n-z\) must add to \(9\).

The available pairs of differences that add to \(9\) are \(3+6\) and \(4+5\), and the remaining difference between the two middle numbers is \(1\).

One possible set is \(\{w,w-5,w-6,w-9\}\), giving \(4w-20=44\) and \(w=16\). The other is \(\{w,w-3,w-4,w-9\}\), giving \(4w-16=44\) and \(w=15\).

The sum of the possible values of \(w\) is \(16+15=31\).

Thus, B is the correct answer.

22 答案 A

设立方体边长为 \(x\)。作金字塔的一个竖直对角截面，它经过正方形底面的两个相对顶点。

这个截面是斜边为 \(\sqrt2\) 的等腰直角三角形。立方体在其中表现为高 \(x\)、宽 \(\sqrt2x\) 的长方形，两侧留下两个全等的、直角边为 \(x\) 的等腰直角三角形。

因此 \(\sqrt2=\sqrt2x+2x\)，所以 \(x=\sqrt2-1\)，立方体体积为 \(x^3=(\sqrt2-1)^3=5\sqrt2-7\)。

所以正确答案是 A。

English

Let the cube have side length \(x\). Take a vertical diagonal cross-section of the pyramid through opposite vertices of the square base.

This cross-section is an isosceles right triangle with hypotenuse \(\sqrt2\). The cube appears as a rectangle of height \(x\) and width \(\sqrt2x\), leaving two congruent right isosceles triangles of leg \(x\).

Thus \(\sqrt2=\sqrt2x+2x\), so \(x=\sqrt2-1\). The cube volume is \(x^3=(\sqrt2-1)^3=5\sqrt2-7\).

Thus, A is the correct answer.

23 答案 D

因为 \(2011\equiv11\pmod{1000}\)，只需计算 \(11^{2011}=(10+1)^{2011}\) 在模 \(1000\) 下的值。

含 \(10^3\) 或更高幂的项都能被 \(1000\) 整除，所以只需前三项：\[1+2011\cdot10+\binom{2011}{2}10^2\text{。}\]

模 \(1000\) 下，这等于 \[ \begin{aligned} &1+11\cdot10+\dfrac{2011\cdot2010}{2}\cdot100 \\ &\quad \equiv1+110+500=611 \end{aligned}\text{。}\]

因此百位数字为 \(6\)。

所以正确答案是 D。

English

Because \(2011\equiv11\pmod{1000}\), it is enough to find \(11^{2011}=(10+1)^{2011}\) modulo \(1000\).

All terms with \(10^3\) or higher are divisible by \(1000\), so only the first three terms matter: \[1+2011\cdot10+\binom{2011}{2}10^2.\]

Modulo \(1000\), this is \[ \begin{aligned} &1+11\cdot10+\dfrac{2011\cdot2010}{2}\cdot100 \\ &\quad \equiv1+110+500=611 \end{aligned} \].

The hundreds digit is therefore \(6\).

Thus, D is the correct answer.

24 答案 B

将图像下移 \(2\)。问题等价于找最小的 \(m\gt\dfrac12\)，使直线 \(y=mx\) 经过某个满足 \(0\lt x\le100\) 的格点。

对固定整数 \(x\)，最小整数 \(y\) 必须满足 \(\frac{y}{x}\gt\frac{1}{2}\)。它为 \(\frac{x}{2}+1\)（若 \(x\) 为偶数），以及 \(\frac{x+1}{2}\)（若 \(x\) 为奇数）。

候选斜率为 \(\dfrac12+\dfrac1x\)（当 \(x\) 为偶数），在 \(x=100\) 时最小为 \(\dfrac{51}{100}\)。候选斜率为 \(\dfrac12+\dfrac1{2x}\)（若 \(x\) 为奇数），在 \(x=99\) 时最小为 \(\dfrac{50}{99}\)。

两类中较小的端点是 \(\dfrac{50}{99}\)。因此当 \(m\) 满足 \(\dfrac12\lt m\lt\dfrac{50}{99}\) 时不会经过这些格点，而这个上端点已经是可能的最大值。

所以正确答案是 B。

English

Shift the graph down by \(2\). The problem is equivalent to finding the smallest slope \(m\gt\dfrac12\) for which \(y=mx\) passes through a lattice point with \(0\lt x\le100\).

For a fixed integer \(x\), the smallest integer \(y\) with \(\frac{y}{x}\gt\frac{1}{2}\) is \(\frac{x}{2}+1\) when \(x\) is even, and \(\frac{x+1}{2}\) when \(x\) is odd.

Thus the candidate slopes are \(\dfrac12+\dfrac1x\) for even \(x\), minimized at \(x=100\) as \(\dfrac{51}{100}\), and \(\dfrac12+\dfrac1{2x}\) for odd \(x\), minimized at \(x=99\) as \(\dfrac{50}{99}\).

The smaller of these is \(\dfrac{50}{99}\), so every \(m\) with \(\dfrac12\lt m\lt\dfrac{50}{99}\) avoids such lattice points, and this upper endpoint is best possible.

Thus, B is the correct answer.

25 答案 D

对边长 \(a=BC\)、\(b=CA\)、\(c=AB\) 的三角形，同一顶点到内切圆的切线段相等，所以下一个三角形边长为 \[ \begin{gathered} \dfrac{b+c-a}{2}, \\ \quad \dfrac{a+c-b}{2}, \\ \quad \dfrac{a+b-c}{2} \end{gathered}\text{。}\]

若当前边长为 \(s-1,s,s+1\)，则下一组边长为 \(\dfrac{s}{2}-1,\dfrac{s}{2},\dfrac{s}{2}+1\)。因此这种形式会一直保持，而中间边每次减半。

对 \(T_n\)，中间边为 \(\frac{2012}{2^{n-1}}\)。形如 \(s-1,s,s+1\) 的三角形存在，当且仅当 \(s>2\)。

最后一个有效三角形满足 \(\frac{2012}{2^{n-1}}>2\)，但下一次不满足。此时 \(n=10\)，中间边为 \(\frac{2012}{2^9}=\frac{503}{128}\)。

周长为 \(3\cdot\dfrac{503}{128}=\dfrac{1509}{128}\)。

所以正确答案是 D。

English

For a triangle with side lengths \(a=BC\), \(b=CA\), and \(c=AB\), equal tangents from the same vertex give the next side lengths \[ \begin{gathered} \dfrac{b+c-a}{2}, \\ \quad \dfrac{a+c-b}{2}, \\ \quad \dfrac{a+b-c}{2}. \end{gathered} \]

If the current side lengths are \(s-1,s,s+1\), then the next side lengths are \(\dfrac{s}{2}-1,\dfrac{s}{2},\dfrac{s}{2}+1\). Thus the same form persists while the middle side halves each time.

For \(T_n\), the middle side is \(\frac{2012}{2^{n-1}}\). A triangle of the form \(s-1,s,s+1\) exists exactly when \(s>2\).

The last valid triangle has \(\frac{2012}{2^{n-1}}>2\), but the next one does not. This gives \(n=10\), with middle side \(\frac{2012}{2^9}=\frac{503}{128}\).

The perimeter is \(3\cdot\dfrac{503}{128}=\dfrac{1509}{128}\).

Thus, D is the correct answer.