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Topic: Interior Point (Read 622 times) |
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ThudnBlunder
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Interior Point
« on: Jan 14th, 2005, 12:32pm » |
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Let P be a point inside an equilateral triangle whose sides have integer length k.
If the distance of P from the 3 vertices are integers a,b,c what is the smallest possible value of:
1) a
2) k
(I haven't tried this, so I don't know how Easy it is.)
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| « Last Edit: Jan 14th, 2005, 11:44pm by ThudnBlunder » |
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TenaliRaman
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Re: Interior Point
« Reply #1 on: Jan 15th, 2005, 1:17am » |
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Why cant they both be very very close to zero?
OOPS!! 
-- AI
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| « Last Edit: Jan 15th, 2005, 2:02am by TenaliRaman » |
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ThudnBlunder
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Re: Interior Point
« Reply #2 on: Jan 15th, 2005, 1:31am » |
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on Jan 15th, 2005, 1:17am, TenaliRaman wrote:| Why cant they both be very very close to zero? |
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Define 'very very close'.
Anyway, they are both integers.
And if you mean
a = 0
b = 1
c = 1
k = 1
is P 'inside'?
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| « Last Edit: Jan 15th, 2005, 1:35am by ThudnBlunder » |
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TenaliRaman
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Re: Interior Point
« Reply #3 on: Jan 15th, 2005, 3:59am » |
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I think i have managed to show that,
3(a4 + b4 + c4 + k4)=(a2 + b2 + c2 + k2)2
So what we are looking for is smallest integer solution to this equation.
Hmm, i think i will leave it for others to find the solution
-- AI
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Sir Col
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How did you get that, TenaliRaman? I've set my computer searching for a solution and so far it hasn't found any that work for the triangle. There are lots of solutions to the equation, but they all have a+b=k, which means that P lies on the edge.
I could well be missing something obvious, but I don't think that this is Easy, T&B.
I've attached a diagram for people to use for future reference.
We shall let the angles around point P, be A, B, and C.
I've made little progress, but so far...
::
Using the cosine rule,
k2 = a2+b2-2ab cosC
k2 = a2+c2-2ac cosB
k2 = b2+c2-2bc cosA
As a,b,c are integer, we need cosA, cosB, cosC all to be rational if k is to be integer.
Interestingly, if we allow P to lie on the perpendicular bisector of the base, we get b=c and B=C, and we can show that cosA will be rational if cosB = cosC is rational.
Proof:
Let cosB = m/n, as A = 360-(B+C), A = 360-2B, cosA = cos(360-2B) = cos(2B) = 2cos2B-1 = 2(m2-n2)/n2, which is rational.
I don't know if this restriction is fruitful, but it may simplify the search slightly. It at least ensures that k2 will always be rational, but it may exclude the minimal solution.
I've tried equating the formulae from above (using b=c and B=C):
k2 = a2+b2-2ab cosB = 2b2+b2 cosA.
Therefore, a2-2ab cosB = b2+b2 cosA.
But even getting integer solutions in this does not guarantee that k will be integer.
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Grimbal
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Re: Interior Point
« Reply #5 on: Jan 15th, 2005, 7:25am » |
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It seems to me that the sum of the distances from the sides of an equilateral triangle is a contant. It is sqrt(3/4)*side. So, if k is an integer, a+b+c cannot be integer, and a, b, c can not be all integers. Unless they are all zero.
[e] oops, this is totally irrelevant to the problem... [/e]
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| « Last Edit: Jan 15th, 2005, 7:32am by Grimbal » |
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Sir Col
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Re: Interior Point
« Reply #6 on: Jan 15th, 2005, 8:41am » |
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It may not be entirely irrelevant. I was playing with that idea before when I was trying to connect it with this problem.
In that problem, perpendiculars were dropped from P to each side. Let the points be X (in triangle with angle A), Y (in triangle with angle B), and Z. Let the segments on each side be x1, x2, y1, y2, z1, and z2, respectively.
We showed a couple of results:
2(x1+y1+z1) = 2(x2+y2+z2) = 3k
x12+y12+z12 = x22+y22+z22
As 3k is fixed x1+y1+z1 must be fixed, as is the sum of perpendicular segments. In other words, for any interior point, P, the sum of the lengths of the legs of the right angle triangles formed by dropping perpendiculars is constant.
I couldn't really do much else with it, as it doesn't tell us a great deal about the hypotenuse lengths (a, b, and c), but maybe someone else can take this and run with it.
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Barukh
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Re: Interior Point
« Reply #7 on: Jan 15th, 2005, 9:26am » |
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Right now, I can tell 2 things:
1. There are infinitely many solutions of this problem.
2. It's not Easy.
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TenaliRaman
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on Jan 15th, 2005, 7:18am, Sir Col wrote:| How did you get that, TenaliRaman? |
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Consider the diagram below,(hope u dont mind using ur diagram )
find cos(x) and cos(y) using cosine law interms of a,b,c,k.
cos(x+y) = cos(x)cos(y)-sin(x)sin(y)
sin(x)sin(y) = cos(x)cos(y) - cos(x+y) ... (1)
Note that x+y = 60
write sin(x) as sqrt(1-cos(x)^2) and same with sin(y)
Square both sides of (1) and sub in cos(x) and cos(y) we found out interms of a,b,c,k. A bit of algebraic simplification gives the equation i gave.
(I am just worried whether i oversimplified it)
-- AI
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ThudnBlunder
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Re: Interior Point
« Reply #9 on: Jan 16th, 2005, 7:21pm » |
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on Jan 15th, 2005, 7:18am, Sir Col wrote:How did you get that, TenaliRaman? I've set my computer searching for a solution and so far it hasn't found any that work for the triangle.
I could well be missing something obvious, but I don't think that this is Easy, T&B. |
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This does not answer the question, but:
:
a = 57
b = 65
c = 73
k = 112
OR
a = 73
b = 88
c = 95
k = 147
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| « Last Edit: Jan 16th, 2005, 11:57pm by ThudnBlunder » |
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