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I was told this question by a friend, who said that their friend had thought about it on and off for six months without any luck. I have then had it for a while without any luck either. It is in the style of a contest question, but it is unlike any other, as really nothing seems to gain any ground.

Here's the question:

Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuous function satisfying the following property: If $ABC$ is the equilateral triangle with side lengths 1, we have for any point $P$ inside $ABC$, $f(\overline{AP})+f(\overline{BP})+f(\overline{CP})=0$ where $\overline{AP}$ is the distance from point P to vertex $A$. (Example: by taking $P=A$ we see $f(0)=-2f(1)$.)

Prove or disprove: $f$ must be identically zero on its domain

Notice that continuity is critical since otherwise I could just take the function $f(x)=0$ whenever $x\in (0,1)$ and $f(0)=1$, $f(1)=\frac{-1}{2}$.

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3 Answers

up vote 8 down vote accepted
+50

Chandru has provided the link to a complete solution, but as Eric notes, it is somewhat involved. Some months ago, being unaware of that solution, I had found a less involved proof if you assume that $f$ is not only continuous but also differentiable. I reproduce that proof below.

Let $\triangle$ be the set of points inside or on the triangle, and $a, b, c : \triangle \to [0,1]$ be scalar fields on $\triangle$ mapping any point $P$ to the distances $\overline{AP}$, $\overline{BP}$, $\overline{CP}$ respectively. Define the scalar field $$g := f \circ a + f \circ b + f \circ c = 0.$$ It is easily shown that $\nabla a$ is a unit vector pointing away from $A$, and similarly for $b$ and $c$. Then $$\nabla g = f'(a) \nabla a + f'(b) \nabla b + f'(c) \nabla c,$$ where $f'(x) = df/dx$. But $g$ is constant, so $\nabla g = 0$.

Take $P$ to be a point on the side $AB$. Then $\nabla a$ and $\nabla b$ are antiparallel, while $\nabla c$ is linearly independent. For their linear combination to be $0$, the coefficient of $\nabla c$, which is $f'(c)$, must be $0$. Since $P$ is an arbitrary point on $AB$, $c$ is any value between $\sqrt{3}/2$ and $1$, so over that range $f'$ is zero, and $f$ is constant.

Once we've shown that $f'$ is zero between $\sqrt{3}/2$ and $1$, for any other value $a$ it's fairly easy to pick a point inside the triangle that's $a$ units from $A$ and over $\sqrt{3}/2$ units from $C$. Now $\nabla g = f'(a) \nabla a + f'(b) \nabla b + f'(c) \nabla c = 0$. The last term drops out because $f'(c) = 0$. $\nabla a$ and $\nabla b$ are linearly independent in the interior of the triangle, so their coefficients $f'(a)$ and $f'(b)$ must be zero. Thus $f' = 0$ everywhere, so $f$ is constant. In particular, $f(1/\sqrt{3}) = 0$, so $f = 0$ everywhere.

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Excellent another solution! I like it. Out of curiosity where did you first hear of the question? –  Eric Naslund Jan 24 '11 at 1:24
    
@Eric: Someone posted the problem on Reddit, at the link I gave in the first paragraph. They didn't know the original source. –  Rahul Jan 24 '11 at 1:35
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@Rahul: +1 Really Nice proof! –  user17762 Jan 24 '11 at 1:52
    
I don't like this proof ; you supposed $f$ was differentiable. –  Patrick Da Silva Jun 9 '11 at 3:53
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Can you make a random edit so that I can switch my vote from down to up? =S –  Patrick Da Silva Jun 13 '11 at 20:04
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This is a problem asked in the American Mathematical Monthly. May be this link could be of some help:

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The link is in the solution for advanced problems. –  anonymous Jan 23 '11 at 18:27
    
Pentacyclic sets of points. I guess it is comforting that the solution is this involved. –  Eric Naslund Jan 23 '11 at 18:29
    
@Eric Naslund: If you just copy and paste your question into google, the jstor link comes up. –  PEV Jan 23 '11 at 18:36
    
@PEV: Well, said PEV! –  anonymous Jan 23 '11 at 18:38
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Though Chandru has provided the link to the solution, I was wondering if the following approach could be of any help.

Instead of choosing $P$ to coincide with a vertex, let $P$ to fall on one of the sides say $AB$.

Let $AP = x$. Since $ABC$ is an equilateral triangle $BP = 1-x$ and $CP = \sqrt{1+x^2 - x}$.

So $\forall x \in [0,1]$, the function has to satisfy $$f(x) + f(1-x) + f(\sqrt{1+x^2 - x}) = 0$$

Can we squeeze a proof that $f(x)$ has to be zero out of this?

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I don't believe so, but I could be wrong. There's a few other symmetries we can use too. First by considering the center of the triangle we find $f(\frac{1}{\sqrt{3}})=0$. Also along one of the perpendicular bisectors we have a similar equation: $2f(x)+f(\frac{\sqrt{3}}{2}-\sqrt{x^2-0.25})=0$. Both of these equations let me to consider circles centered at the vertices of the triangle, but stuff was just too messy to use. (Square roots ruined everything). Having said that I did find a relatively simple argument which showed that $f$ was a monotonic function. –  Eric Naslund Jan 23 '11 at 19:26
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