2006 AIME I Problem 14

Attempt Problem 14 of the 2006 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME I solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

A tripod has three legs each of length 55 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 44 feet from the ground. In setting up the tripod, the lower 11 foot of one leg breaks off. Let hh be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then hh can be written in the form mn,\frac{m}{\sqrt{n}}, where mm and nn are positive integers and nn is not divisible by the square of any prime. Find m+n.\lfloor m + \sqrt{n} \rfloor. (The notation x\lfloor x \rfloor denotes the greatest integer that is less than or equal to x.x.)

Answer: 183
Concepts:3D geometrycoordinate geometrydistance formulasymmetry
Difficulty rating: 3270
Solution:

Place the top at T=(0,0,4).T = (0, 0, 4). Each leg has length 5,5, so each foot is 33 from the origin: A=(3,0,0),A = (3, 0, 0), B=(32,332,0),B = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}, 0\right), C=(32,332,0).C = \left(-\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 0\right). The broken leg has length 4,4, so its tip is A=45(3,0,0)A' = \frac{4}{5}(3, 0, 0) +15(0,0,4)+ \frac{1}{5}(0, 0, 4) =(125,0,45),= \left(\frac{12}{5}, 0, \frac{4}{5}\right), and the tripod now stands on the plane ABC,A'BC, with hh equal to the distance from TT to that plane.

The plane ABCA'BC contains BC,\overline{BC}, which is parallel to the yy-axis and passes through the midpoint M=(32,0,0),M = \left(-\frac{3}{2}, 0, 0\right), so its distance from TT can be measured in the xzxz-plane: it is the distance from (0,4)(0, 4) to the line through (32,0)\left(-\frac{3}{2}, 0\right) and (125,45),\left(\frac{12}{5}, \frac{4}{5}\right), whose equation is 8x39z+12=0.8x - 39z + 12 = 0. Therefore h=80394+1282+392=1441585. \begin{aligned} h &= \frac{|8 \cdot 0 - 39 \cdot 4 + 12|}{\sqrt{8^2 + 39^2}} \\ &= \frac{144}{\sqrt{1585}}. \end{aligned}

Here m=144m = 144 and n=1585=5317n = 1585 = 5 \cdot 317 is squarefree. Since 392=1521<1585<1600,39^2 = 1521 \lt 1585 \lt 1600, we get 144+1585\lfloor 144 + \sqrt{1585} \rfloor =144+39=183.= 144 + 39 = 183.

← Problem 13#13
Full Exam

Problem 14 in Other Years