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Topic: Tripod Word Problem (Read 447 times) |
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CornMuffin
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Tripod Word Problem
« on: May 11th, 2006, 2:37pm » |
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I have no idea on how to even begin this problem
*note* i put (n)^(1/2) because i did not no how to make the square root symbol so i did n raised to the 1 half power
Also the original problem did not have the ¬ symbol but since i did not no how to make that particular symbol i just put ¬ instead, it still does the same thing as in the original problem
"A tripod has three legs 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 or the top of the tripod from the ground when the broken tripod is set up. Then h can be writen in the form m/(n)^(1/2), where m and n are positive integers and n is not divisible by the square of any prime. Find ¬m + (n)^(1/2)¬. (the notation ¬x¬ denotes the greatest integer that is less then or equal to x)"
*problem from The Mathematical Association of America*
Tell me how you would do this problem and don't just the answer, I already know the answer just not how to do it
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Icarus
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Re: Tripod Word Problem
« Reply #1 on: May 11th, 2006, 5:29pm » |
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One straightforward approach (though rather nasty in the math):
First consider the situation when the tripod is unbroken. Call the points at the bottoms of the tripod legs A, B, C and the apex P. Let Q be the intersection of the vertical line through P and the ground. So PQA, PQB, and PQC are all right triangles, with the right angle at Q. Further, PQ = 4, and PA = PB = PC = 5, so QA = QB = QC = 3.
Assume A is the leg that gets broken, and define D to be the midpoint of BC. Let E be the point of the break along PA, so PE = 4 and EA = 1. When the leg breaks, the tripod rotates about the line through B, D, and C. So the new height is the height of the triangle EPD. To find it, we first find the lengths PD and ED.
PD is the same for the broken and unbroken tripods, so we can look at the unbroken version. First, look at ground plane and consider the angles <AQB, <BQC, and <CQA. By symmetry, the three angles are the same, and they add up to a full 360o, so each must be 120o. The segment QD bisects the angle <BQC, so <DQC is 60o, and triangle CQD is a 30-60-90 right triangle with right angle at D and hypothenuse QC = 3. It follows that QD = (QC)cos 60o = 3/2. Now PQD is also a right triangle with legs QD = 3/2 and PQ = 4, so PD = sqrt(73)/2. Note also that CD = (QC)sin 60o = 3sqrt(3)/2. so BC = 2(DC) = 3 sqrt(3). By symmetry, AB = AC = 3 sqrt(3) as well.
To find DE, I found it easiest to first find CE. Consider the triangle CPA. We need to find the cosine of the angle at P. Let p, a, c be the lengths of the sides opposite the corresponding angles. So a = PC = 5, c = PA = 5, and p = AC = 3 sqrt(3). The law of cosines says that
p2 = a2 + c2 - 2ac cos(P)
So 27 = 25 + 25 - 2*25*cos(P), and cos(P) = 23/50.
Now consider the triangle CPE. The angle at P is the same, so cos(P) = 23/50. CP = 5, and PE = 4. We apply the law of cosines again to find CE:
(CE)2 = 52 + 42 - 2*5*4*cos(P) = 41 + 40(23/50) = 297/5. So CE = sqrt(297/5).
Now look at the right triangle CDE. CD = 3sqrt(3)/2 = sqrt(27/4), CE = sqrt(297/5). It follows that DE = sqrt(297/5 - 27/4) = sqrt(1053/20) = 9sqrt(13/5)/2.
Finally, we can look at triangle DPE. The plane of this triangle is perpendicular to the ground, so its height is the height of the broken tripod. First, find the cosine of one of the base angles. The one at E gives slightly simpler math. Apply the law of cosines again:
(PD)2 = (DE)2 + (PE)2 - 2(DE)(PE)cos E.
73/4 = 1053/20 + 16 - 36sqrt(13/5)cos E
Which reduces to cos E = 7/sqrt(65). So, sin E = sqrt(1 - 49/65) = 4/sqrt(65).
Last, the height of the triangle is (PE)sin E = 16/sqrt(65).
Thus m = 16, n = 65, m + sqrt(n) = 24.062..., and [ m + sqrt(n) ] = 24.
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