2011 AMC 12B 第 22 题

先试着解答 2011 AMC 12B 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2011 AMC 12B 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

22.

T1T_1 是边长为 2011,20122011, 201220132013 的三角形。对 n1n\ge1Tn=ABCT_n=\triangle ABC,且 DDEEFF 分别是 ABC\triangle ABC 的内切圆与边 ABABBCBCACAC 的切点,那么若三角形存在,Tn+1T_{n+1} 的边长为 ADADBEBECFCF。序列 (Tn)(T_n) 中最后一个三角形的周长是多少?

Let T1T_1 be a triangle with sides 2011,2012,2011, 2012, and 2013.2013. For n1,n\ge1, if Tn=ABCT_n=\triangle ABC and D,D, E,E, and FF are the points of tangency of the incircle of ABC\triangle ABC to the sides AB,AB, BC,BC, and AC,AC, respectively, then Tn+1T_{n+1} is a triangle with side lengths AD,AD, BE,BE, and CF,CF, if it exists. What is the perimeter of the last triangle in the sequence (Tn)?(T_n)?

15098\dfrac{1509}{8}

150932\dfrac{1509}{32}

150964\dfrac{1509}{64}

1509128\dfrac{1509}{128}

1509256\dfrac{1509}{256}

答案:D
知识点:内切圆、内心与内切圆半径递推三角不等式
难度评级:2350
解答:

对边长为 a,b,ca,b,c 的三角形,切线段长度为 AD=12(b+ca)AD=\tfrac12(b+c-a)BE=12(a+cb)BE=\tfrac12(a+c-b)CF=12(a+bc)CF=\tfrac12(a+b-c)。若 TnT_n 的边长为 (y1,y,y+1)(y-1,y,y+1),则 Tn+1T_{n+1} 的边长为 (y21,y2,y2+1)\left(\tfrac{y}{2}-1,\tfrac{y}{2},\tfrac{y}{2}+1\right)

从中间边为 20122012T1T_1 开始,中间边每步减半,且 Tn+1T_{n+1} 的周长是 TnT_n 周长的 12\tfrac12。这种形式的三角形存在当且仅当中间边大于 22

TnT_n 的中间边为 20122n1\dfrac{2012}{2^{n-1}}。它在 n=11n=11 时第一次不超过 22,所以最后一个有效三角形是 T10T_{10},其中间边为 201229\dfrac{2012}{2^9},周长为 3201229=6036512=1509128. 3\cdot\dfrac{2012}{2^9}=\dfrac{6036}{512}=\dfrac{1509}{128}.

所以正确答案是 D

For a triangle with sides a,b,c,a,b,c, the tangent lengths are AD=12(b+ca),AD=\tfrac12(b+c-a), BE=12(a+cb),BE=\tfrac12(a+c-b), and CF=12(a+bc).CF=\tfrac12(a+b-c). If TnT_n has sides (y1,y,y+1),(y-1,y,y+1), then Tn+1T_{n+1} has sides (y21,y2,y2+1).\left(\tfrac{y}{2}-1,\tfrac{y}{2},\tfrac{y}{2}+1\right).

Starting from T1T_1 with middle side 2012,2012, the middle side halves each step and the perimeter of Tn+1T_{n+1} is 12\tfrac12 the perimeter of Tn.T_n. A triangle of this form exists only while its middle side exceeds 2.2.

The middle side of TnT_n is 20122n1.\dfrac{2012}{2^{n-1}}. This first drops to 22 or below at n=11,n=11, so the last valid triangle is T10,T_{10}, whose middle side is 201229\dfrac{2012}{2^9} and whose perimeter is 3201229=6036512=1509128. 3\cdot\dfrac{2012}{2^9}=\dfrac{6036}{512}=\dfrac{1509}{128}.

Thus, the correct answer is D.

← 第 21 题#21
完整试卷

其他年份的第 22 题