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        riddles >> easy >> Interior Point 
        
(Message started by: THUDandBLUNDER on Jan 14^th, 2005, 12:32pm)

Title: Interior Point
Post by THUDandBLUNDER on Jan 14^th, 2005, 12:32pm

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.)

Title: Re: Interior Point
Post by TenaliRaman on Jan 15^th, 2005, 1:17am

~~Why cant they both be very very close to zero?~~
OOPS!! ;D

-- AI

Title: Re: Interior Point
Post by THUDandBLUNDER on Jan 15^th, 2005, 1:31am

on 01/15/05 at 01:17:42, TenaliRaman wrote:

Why cant they both be very very close to zero?

Define 'very very close'.
Anyway, they are both integers.

And if you mean
a = 0
b = 1
c = 1
k = 1
is P 'inside'?

Title: Re: Interior Point
Post by TenaliRaman on Jan 15^th, 2005, 3:59am

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 ;D

-- AI

Title: Re: Interior Point
Post by Sir Col on Jan 15^th, 2005, 7:18am

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...

::[hide]
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.
[/hide]::

Title: Re: Interior Point
Post by Grimbal on Jan 15^th, 2005, 7:25am

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]

Title: Re: Interior Point
Post by Sir Col on Jan 15^th, 2005, 8:41am

It may not be entirely irrelevant. I was playing with that idea before when I was trying to connect it with this (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1104021436) 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.

Title: Re: Interior Point
Post by Barukh on Jan 15^th, 2005, 9:26am

Right now, I can tell 2 things:

1. There are infinitely many solutions of this problem.
2. It's not Easy.

Title: Re: Interior Point
Post by TenaliRaman on Jan 15^th, 2005, 11:42am

on 01/15/05 at 07:18:28, Sir Col wrote:

How did you get that, TenaliRaman?

Consider the diagram below,(hope u dont mind using ur diagram ;D)
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

Title: Re: Interior Point
Post by THUDandBLUNDER on Jan 16^th, 2005, 7:21pm

on 01/15/05 at 07:18:28, 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.

This does not answer the question, but:
:[hide]
a = 57
b = 65
c = 73
k = 112

OR

a = 73
b = 88
c = 95
k = 147
[/hide]