Convex sum order

Question

If I have a strictly convex function $f(x)$ with $f''(x)>0$

and if I know that for some $a\le b \le c$ and $x \le y \le z$ I have

$$a+b+c = x+y+z$$ $$f(a)+f(b)+f(c)=f(x)+f(y)+f(z)$$

can I conclude that at least one of the following must be true about the order?

$$ a \le x \le y \le b \le c \le z$$ $$ x \le a \le b \le y \le z \le c$$

Perhaps an equivalent question is to consider coincident centroids of two triangles with vertices on the convex curve, looking something like this diagram with pairs of vertices on the inside of the order

though I realise that I cannot push this analogy too far as it would not be necessarily true if say the curve was a circle and the triangles equilateral

Answer 1

For simplicity, let us assume that all the points are different, i.e., $a < b < c$, $x < y < z$ and the all the points $a,b,c$ are different from $x,y,z$. (Maybe some arguments can be transferred to the general case).

Now, the order of the points can be described by a permutation of $abcxyz$ which preserves the order of $abc$ and $xyz$.

By symmetry, we can assume w.l.o.g. that the permutation starts with $a$.

Let us consider permutations starting with $ab$.

• $abc$ can only be completed to $abcxyz$ and this cannot satisfy $a + b + c = x + y + z$.
• $abx$ can be completed to
• • $abxycz$: this cannot satisfy $a + b + c = x + y + z$.
• • $abxcyz$: this cannot satisfy $a + b + c = x + y + z$.
• • $abxyzc$: by your triangly analogy, the entire triangle $xyz$ lies below the segment $bc$, hence the centroids of $abc$ and $xyz$ cannot coincide.

These arguments show that the permutation has to start with $ax$. By similar arguments we can conclude that the permutation has to end with $zc$ or $cz$.

Hence, four possibilities remain:

• $ax by cz$
• $ax yb cz$
• $ax by zc$
• $ax yb zc$

The first case cannot satisfy $a + b + c = x + y + z$. The second case is what you would like to have. Therefore, it remains to rule out the last two cases. By symmetry, it is enough to rule out $axbyzc$.

The strict convexity gives \begin{align} f(x) &< f(a) \frac{b-x}{b-a} + f(b) \frac{x-a}{b-a}, \\ f(y) &< f(b) \frac{c-y}{c-b} + f(c) \frac{y-b}{c-b}, \\ f(z) &< f(b) \frac{c-z}{c-b} + f(c) \frac{z-b}{c-b}. \end{align} Adding these inequalities and using $f(a) + f(b) + f(c) = f(x) + f(y) + f(z)$ gives \begin{equation*} f(a) + f(b) + f(c) < f(a) \frac{b-x}{b-a} + f(b) \Big[ \frac{x-a}{b-a} + \frac{2c-z-y}{c-b}\Big] + f(c) \frac{z+y-2b}{c-b}. \end{equation*} Rearranging terms and using $a + b + c = x + y + z$ gives \begin{equation*} 0 < (x-a) \Bigg[ \frac{f(b) - f(a)}{b-a} - \frac{f(c) - f(b)}{c-b} \Bigg]. \end{equation*} This, however, is a contradiction to the strict convexity which yields \begin{equation*} \frac{f(b) - f(a)}{b-a} < \frac{f(c) - f(b)}{c-b}. \end{equation*} This rules out the case $axbyzc$.

Hence, if $a$ is the smallest point, the permutation has to be $axybcz$, i.e., $a < x < y < b < c < z$.

Answer 2

If $A=(a,f(a)),\ B,\ C$ are vertexes in triangle on $f$ and if so are $X,\ Y,\ Z$, then assume that we have order : $A,\ X,\ Y,\ B,\ Z,\ C$.

In further, assume that they share center of mass $G$.

If $A',\ X'$ is a mid point of a segment $[BC],\ [YZ]$, then $[AA'],\ [XX']$ intersect at a point $G$.

If $f(a)>f(x)$, then first coordinate of $A'$ is smaller than that of $X'$. But by assumption, $$\frac{b+c}{2} > \frac{y+z}{2}$$

Answer 3

In geometry a triangle $(ABC)$ on the plane $xOy$ with coordinates $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$ has its gravity center (baricenter) the point $$ G=\left(\frac{a_1+b_1+c_1}{3},\frac{a_2+b_2+c_2}{3}\right). $$ Hence what you ask is:

Given a curve (function) $c_f:y=f(x)$ or $M=(x,f(x))$, with positive curvature, find all triangles $(ABC)$ inscribed in $c_f$ such that they have the same gravity center.

In geometry, the gravity center of (ABC) is defined as the point of intersection of all the dimedians i.e. In $A$ edge the dimedian is the sigment $AM_1$ with $M_1$ being the median of $BC$ side. In $B$ edge the coresponding dimedian is $BM_2$, where $M_2$ is the median of $AC$ and in $C$ edge the sigment $CM_3$.

For a given baricenter there are finite number of triangles inscribed in $c_f$ with the same baricenter $G=\left(\frac{a+b+c}{3},\frac{f(a)+f(b)+f(c)}{3}\right)$.

A theorem of Leibniz says

Theorem.

If $M$ is an arbitrary point of plane of the triangle $(ABC)$ and $G$ is the baricenter, then $$ MA^2+MB^2+MC^2=3MG^2+\frac{1}{3}(AB^2+BC^2+CD^2) $$

Hence your broblem reduces to find all points $A=(a,f(a))$, $B=(b,f(b))$, $C=(c,f(c))$ of curve $y=f(x)$ such $$ GA^2+GB^2+GC^2=\frac{1}{3}(AB^2+BC^2+CA^2) $$ or equivalently for given $f$ find all $a,b,c$ such $$ (g_1-a)^2+(g_2-f(a))^2+(g_1-b)^2+(g_2-f(b))^2+(g_1-c)^2+(g_2-f(c))^2=\frac{1}{3}\left((a-b)^2+(f(a)-f(b))^2+(b-c)^2+(f(b)-f(c))^2+(c-a)^2+(f(c)-f(a))^2\right),\tag 1 $$ where $$ g_1=\frac{a+b+c}{3}\textrm{ and }g_2=\frac{f(a)+f(b)+f(c)}{3}.\tag 2 $$ Hence we have three equations with three unknowns $a,b,c$. Hence the number of points will be finite.

For example assume the parabola $y=x^2$. Then equations $(2)$ become $$ a=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right) $$ $$ b=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right) $$ or $$ a=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right) $$ $$ b=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right) $$ The value of $c$ is given from $$ \frac{4 a^3}{3}-\frac{2 a^2 b^2}{3}-\frac{2 a^2 c^2}{3}-\frac{2 a b}{3}-\frac{2 a c}{3}+\frac{4 b^3}{3}-\frac{2 b^2 c^2}{3}-\frac{2 b c}{3}+\frac{4 c^3}{3}=0 $$