What happened to that nice thread?

[krtxmrtz]

I sort of liked that very nice geometry problem (thread entitled "gayest geometry problem ever" if my memory is not failing me) -the harder the problem, the more enjoyable- I think it was far from being satisfactorily resolved and I'm wondering whatever happened to it, I can't spot it anywhere... Have the administrators transferred it to another forum or what?

[krtxmrtz]

What context did you grab this problem from? Is it related to some other (major?) kind of problem?

[bugzpodder]

its a contest practice question. at least i am not aware of its relativity to anything major, and i doubt that. i'll try to remember to post its solutions when it comes out

[opus]

I'm still trying to solve the thread!

[krtxmrtz]

Originally posted by opus
I'm still trying to solve the thread!

Well, I'm too busy lately... I keep trying... but only on weekends

[krtxmrtz]

Originally posted by bugzpodder
its a contest practice question. at least i am not aware of its relativity to anything major, and i doubt that. i'll try to remember to post its solutions when it comes out

I think I have found the solution but, you know, it was so easy that I'm not sure if I still remember exactly what the problem was.

"Given a triangle with all sides equal to one (equilateral), prove that for any point inside it (or at one of the sides), the product of its distances to the three vertices is less than or equal to one".

Is this right?

(I suppose it's not for, the solution is so obvious that I can't believe it kept me baffled so long)...

Maybe it's not the product <=1 but the sum<=3...?

[opus]

yea, I think that was the original problem (hard to remember, if the thread has vanished)

[krtxmrtz]

Originally posted by opus
yea, I think that was the original problem (hard to remember, if the thread has vanished)

You mean, with the sum?

[opus]

No , the product.

[krtxmrtz]

If it's the product, then it's very easy.

Probably the most obvious way to see it is by considering a circle of radius 1 with center at any one of the vertices. The circle will pass through the other 2 vertices. The distance of any point to the center of that circle will be > 1 only for points outside the circle so, for all the points in the triangle it will be <1 (and =1 in the other 2 vertices). We'll end up with a product of 3 numbers <=1, so the result will be <=1. (I suppose the last statement needs no proof).

[opus]

Stupid me!
So that "proofs" it even for point that are somewhat out of the triangle, but inside the area that is part of all three cricles.

[krtxmrtz]

Originally posted by opus
Stupid me!

Stupid all of us/ass(es) !

It had taken me 2 full weeks of trigonometry, calculus, algebra and whatnot last year before I gave up Yesterday all of a sudden I remembered about it and the solution came to me like a flash of lightning in a few seconds.

Must have been like this when Archimedes was in the bathtub and cried "Eureka!"

[bugzpodder]

the question is actually: given an unit equilateral triangle ABC, and some point P, prove PA.PB+PB.PC+PC.PA>=1

i wish it is that easy. trust me, its not easy.

[bugzpodder]

for the "new" question you proposed here is my solution, any point P inside the triangle, PA<=1, PB<=1, PC<=1. therefore the product is less than 1. done.

[Guv]

Am I misunderstanding something? This seems very simple.

The furthest you can get from any vertex is another vertex, which is one unit. The product of three distances equal to or less than one must be equal to or less than one.

To analyze a bit more: There are only three lines with length equal to one (the sides), and they do not meet. Hence, at least one line must be shorter than one, and the other two cannot be longer than one. Hence the product must be less than one.

[krtxmrtz]

Originally posted by bugzpodder
the question is actually: given an unit equilateral triangle ABC, and some point P, prove PA.PB+PB.PC+PC.PA>=1

i wish it is that easy. trust me, its not easy.

Thank God the problem is still standing!

As I didn't exactly remember it I felt kind of disappointed that it could be that easy and, at the same time I was feeling ashamed that I could have spent so long to solve it.

Fortunately the actual problem is far from easy as you rightly point out.

[bugzpodder]

well i know a solution to it, so if you wanna know, let me know...

[krtxmrtz]

Originally posted by bugzpodder
well i know a solution to it, so if you wanna know, let me know...

Oh no! At least not yet, This is like having a brand new toy and I want to rack my brain for awhile, just trying to solve it is a lot of fun, even if the solution is hard to arrive at.

[krtxmrtz]

I arrived at a solution, yet a "brute force" one.

The way I've done it:

1. Take point [0, 1/2sqrt(3)] located at the geometrical center of the triangle.
2. Calculate the value of PA.PB+PB.PC+PC.PA at this point. It turns out to be 1.
3. Prove that it is a minimum, therefore at any other point PA.PB+PB.PC+PC.PA>=1

Now, to prove it's a minimum you must write some algebra dealing with first and second derivative calculations. I've done it on paper and it's a very straightforward calculation yet cumbersome. If there is interest in it I can type it in a separate post.

Now, I admit it's not a very "smart" solution so, Bugz, I'd finally like to see your own solution.

[bugzpodder]

i missed this post last time i was here, sorry. anywayz, i'd like to see your solution if you dont mind. and the following isnt my solution, so i dont take credit for it (well i guess its sorta mine now that i know how to do it).

the the length in the middle be a,b,c and angles be A,B,C
therefore A+B+C=360
we have:

a^2+b^2-2abcosA=1
a^2+c^2-2accosB=1
b^2+c^2-2bccosC=1

so

cosA+cosB+cosC=(a^2+b^2-1)/2ab+(a^2+c^2-1)/2ac+(b^2+c^2-1)/2bc

the minimum of cosA+cosB+cosC occurs at A=B=C=120, using jensen's inequality.

therefore:

-3/2<=(a^2+b^2-1)/2ab+(a^2+c^2-1)/2ac+(b^2+c^2-1)/2bc

-3(abc)<=a^2c+b^2c-c+a^2b+bc^2-b+ab^2+ac^2-a
and behold, this factors:

a^2c+b^2c-c+a^2b+bc^2-b+ab^2+ac^2-a+3abc>=0
(ab+bc+ca-1)(a+b+c)>=0
since (a+b+c)>0
so ab+bc+ca>=1

cute little 5 lines, wouldnt you say?

[krtxmrtz]

It's a very nice little proof because it's simple. I wasn't aware of Jensen's inequality. Proving this inequality would require perhaps a good deal of equations and that's why I had to write so much to solve your problem.

I'll try to type my own equations as soon as I have a break but I'm afraid it won't be till next week for my boss is kind of bugging me these days.

[bugzpodder]

actually, proving jensen's inequality isnt more than just a few lines. just draw a diagram, any convex function, and connect any two points on the curve. the midpoint of the secant is always greater than the curve. this can be easily derived from the definition of a convex function also but really all you gotta state is the theorem

[krtxmrtz]

At last I was able to transcribe my calculations. I couldn't swear there aren't any typos so, please, bear with me and... brace yourselves!

I have used this notation:

PA = p
PB = q
PC = r

so that what must be proved is:

F = pq + qr + rp >= 1

Let the triangle be located such that the coordinates of its vertices are:

(-1/2,0), (1/2,0) and (0,sqrt(3)/2)

Hypothesis: F has its minimum at the center of mass of the triangle (call
this point "P") with coordinates x = 0, y = 1/(2sqrt(3)).
Because F = 1 at P, the consequence will be that for any other point F >= 1.

The following must be proved:

1) dF/dx = 0 and dF/dy = 0 at point P.
2) J > 0 and (d²F/dx² > 0 or d²F/dy² > 0 ) at P, where

J = (d²F/dx²)*(d²F/dy²) - (d²F/dxdy)²

Let's write the explicit formulas for p, q and r:

p = sqrt[(1/2 + x)² + y²]
q = sqrt[(1/2 - x)² + y²]
r = sqrt[x² + (sqrt(3)/2 - y)²]

Now the first derivatives:

dp/dx = (x + 1/2) / p
dq/dx = (x - 1/2) / q
dr/dx = x / r

dp/dy = y / p
dq/dy = y / q
dr/dy = (y - sqrt(3)/2) / r

It is easily verified that, at P:

p = q = r = 1/srqt(3)
F = 1

dp/dx = sqrt(3)/2
dq/dx = -sqrt(3)/2
dr/dx = 0

and

dp/dy = 1/2
dq/dy = 1/2
dr/dy = -1

The derivatives of F are:

dF/dx = p dq/dx + q dp/dx + q dr/dx + r dq/dx + r dp/dx + p dr/dx =
p/q (x - 1/2) + q/p (x + 1/2) + q/r x + r/q (x - 1/2) + r/p (x + 1/2) + p/r x

dF/dy = p dq/dy + q dp/dy + q dr/dy + r dq/dy + r dp/dy + p dr/dy =
p/q y + q/p y + q/r (y - sqrt(3)/2) + r/q y + r/p y + p/r (y - sqrt(3)/2)

Using the above values one gets dF/dx = 0 and dF/dy = 0 at P.

Now, the second derivatives.

d²F/dx² = p/q + q/p + q/r + r/q + r/p + p/r +
(x - 1/2)*(q dp/dx - p dq/dx)/q² + (x + 1/2)*(p dq/dx - q dp/dx)/p² +
x*(r dq/dx - q dr/dx)/r² + (x - 1/2)*(q dr/dx - r dq/dx)/q² +
(x + 1/2)*(p dr/dx - r dp/dx)/p² + x*(r dp/dx - p dr/dx)/r²

d²F/dy² = p/q + q/p + q/r + r/q + r/p + p/r +
y*(q dp/dy - p dq/dy)/q² + y*(p dq/dy - q dp/dy)/p² +
(y - sqrt(3)/2)*(r dq/dy - q dr/dy)/r² + y*(q dr/dy - r dq/dy)/q² +
y*(p dr/dy - r dp/dy)/p² + (y - sqrt(3)/2)*(r dp/dy - p dr/dy)/r²

d²F/dxdy =
(x - 1/2)*(q dp/dy - p dq/dy)/q² + (x + 1/2)*(p dq/dy - q dp/dy)/p² +
x*(r dq/dy - q dr/dy)/r² + (x - 1/2)*(q dr/dy - r dq/dy)/q² +
(x + 1/2)*(p dr/dy - r dp/dy)/p² + x*(r dp/dy - p dr/dy)/r²

All these terms have already been evaluated for point P so that, finally, at P
(take your time to carefully do the substitutions):

d²F/dx² = 3/2 > 0
d²F/dy² = 3/2 > 0
d²F/dxdy = 0

J = (d²F/dx²)(d²F/dy²) - (d²F/dxdy)² = (3/2)2 - (0)2 = 9/4 > 0

...and that's it...