Problem:
A tripod has three legs each of length 5 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 4 feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let h be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then h can be written in the form nm, where m and n are positive integers and n is not divisible by the square of any prime. Find ⌊m+n⌋. (The notation ⌊x⌋ denotes the greatest integer that is less than or equal to x.)
Solution:
The feet of the unbroken tripod are the vertices of an equilateral triangle ABC, and the foot of the perpendicular from the top to the plane of this triangle is at the center of the triangle. By the Pythagorean Theorem, the distance from the center to each vertex of the triangle is 3 . Place a coordinate system so that the coordinates of the top T are (0,0,4) and the coordinates of A,B, and C are (3,0,0),(−3/2,33/2,0), and (−3/2,−33/2,0), respectively. Let the break point A′ be on TA. Then TA′:A′A=4:1. Thus the coordinates of A′ are
54(3,0,0)+51(0,0,4)=(12/5,0,4/5)
Note that the coordinates of M, the midpoint of BC, are (−3/2,0,0). The perpendicular from T to the plane of △A′BC will intersect this plane at a point on MA′. This segment lies in the xz-plane and has equation 8x−39z+12=0 in this plane. Then h is the distance from T to line MA′, and is equal to
82+(−39)2∣8⋅0−39⋅4+12∣=1585144
so ⌊m+n⌋=144+39=183.
OR
Place a coordinate system as in the first solution. Note that △A′MT is in the xz-plane. In this plane, circumscribe a rectangle around △A′MT with its sides parallel to the axes. Then
Thus h=A′M2[A′MT]=15.8514.4=1585144, so ⌊m+n⌋=144+39=183.
OR
The feet of the unbroken tripod are the vertices of an equilateral triangle ABC, and the foot of the perpendicular from the vertex to the plane of this triangle is at the center of the triangle. Let T be the top of the tripod, let O be the center of △ABC, let A′ be the break point on TA, and let M be the midpoint of BC. Apply the Pythagorean Theorem to △TOA to conclude that OA=3. Therefore △ABC has sides of length 33. Notice that A′,M, and T are all equidistant from B and C, so the plane determined by △TA′M is perpendicular to BC, and so h is the length of the altitude from T in △TA′M. Because
21A′M⋅h=[TA′M]=21A′T⋅TMsin∠A′TM
it follows that
h=A′MA′T⋅TMsin∠A′TM.
The length of A′T is 4, and TM=TB2−BM2=25−(27/4)=73/2. To find A′M, note that A′M2=A′C2−CM2, and that A′C2=A′T2+TC2−2A′T⋅TCcos∠A′TC.
But cos∠A′TC=cos∠ATC=2⋅5⋅552+52−(33)2=5023,
so A′C2=42+52−2⋅4⋅5⋅(23/50)=113/5, and A′M=5113−427=20317.
Now cos∠A′TM=2⋅4⋅73/216+73/4−317/20=57323,
so sin2∠A′TM=1−25⋅73232, and sin∠A′TM=57336. Thus
