2004 AIME I 第 14 题

先试着解答 2004 AIME I 第 14 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 2004 AIME I 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。

14.

一只独角兽被一根 2020 英尺长的银绳拴在一座魔法师圆柱形塔的底部,塔的半径为 88 英尺。 绳子一端固定在塔的地面高度处,另一端系在独角兽身上,离地高度为 44 英尺。独角兽把绳子拉紧, 绳子的末端距塔上最近点 44 英尺,且绳子接触塔的长度为 abc\frac{a - \sqrt{b}}{c} 英尺,其中 aabbcc 是正整数,且 cc 是质数。 求 a+b+ca + b + c

A unicorn is tethered by a 2020-foot silver rope to the base of a magician's cylindrical tower whose radius is 88 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 44 feet. The unicorn has pulled the rope taut, the end of the rope is 44 feet from the nearest point on the tower, and the length of the rope that is touching the tower is abc\frac{a - \sqrt{b}}{c} feet, where a,a, b,b, and cc are positive integers, and cc is prime. Find a+b+c.a + b + c.

答案:813
知识点:圆柱展开图(立体几何)切线勾股定理
难度评级:3270
解答:

绳子从塔底固定点 AA 出发,沿墙面贴到一点 PP,再直线连到末端 QQ,其高度为 44,到塔轴的距离为 8+4=128 + 4 = 12。将圆柱侧面展开成平面:绷紧的绳子成为一条长度为 2020、上升 44 英尺的直线段,所以其水平投影长度为 20242=86\sqrt{20^2 - 4^2} = 8\sqrt{6},而绳子的每一段都有相同的长度与水平投影之比 2086=526\frac{20}{8\sqrt{6}} = \frac{5}{2\sqrt{6}}

从上方看,自由段 PQPQ 与半径为 88 的圆相切,切点为 PP,其另一端距圆心 1212,所以其水平投影长度为 12282=45\sqrt{12^2 - 8^2} = 4\sqrt{5}。因此 PQ=52645=1056=5303. \begin{aligned} PQ &= \frac{5}{2\sqrt{6}} \cdot 4\sqrt{5} \\ &= \frac{10\sqrt{5}}{\sqrt{6}} \\ &= \frac{5\sqrt{30}}{3}. \end{aligned}

接触塔的绳长为 205303=60750320 - \frac{5\sqrt{30}}{3} = \frac{60 - \sqrt{750}}{3}, 且 c=3c = 3 是质数,所以 a+b+c=60+750+3=813a + b + c = 60 + 750 + 3 = 813

The rope runs from its anchor AA at the base of the tower, hugs the wall up to a point P,P, then goes straight to its end Q,Q, which is at height 44 and at distance 8+4=128 + 4 = 12 from the tower's axis. Unroll the cylinder's wall into a plane: a taut rope becomes a single straight segment of length 2020 rising 44 feet, so its horizontal projection has length 20242=86,\sqrt{20^2 - 4^2} = 8\sqrt{6}, and every piece of the rope has the same ratio 2086=526\frac{20}{8\sqrt{6}} = \frac{5}{2\sqrt{6}} of length to horizontal projection.

Viewed from above, the free portion PQPQ is tangent to the circle of radius 88 at PP from a point at distance 12,12, so its horizontal projection has length 12282=45.\sqrt{12^2 - 8^2} = 4\sqrt{5}. Therefore PQ=52645=1056=5303. \begin{aligned} PQ &= \frac{5}{2\sqrt{6}} \cdot 4\sqrt{5} \\ &= \frac{10\sqrt{5}}{\sqrt{6}} \\ &= \frac{5\sqrt{30}}{3}. \end{aligned}

The rope touching the tower has length 205303=607503,20 - \frac{5\sqrt{30}}{3} = \frac{60 - \sqrt{750}}{3}, and c=3c = 3 is prime, so a+b+c=60+750+3=813.a + b + c = 60 + 750 + 3 = 813.

← 第 13 题#13
完整试卷

其他年份的第 14 题