Problem A. 933. (April 2026)

A. 933. Let $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ be real numbers in the interval $\displaystyle (1,4)$. Consider the following three inequalities: $\displaystyle {bx(x+y-z)}+{cx(x+z-y)}\geq {a(2yz+x)}$, $\displaystyle {cy(y+z-x)}+{ay(y+x-z)} \geq {b(2zx+y)}$, $\displaystyle {az(z+x-y)}+{bz(z+y-x)} \geq {c(2xy+z)}$.

a) Determine, in terms of $\displaystyle a$, $\displaystyle b$, $\displaystyle c$, the maximum number of these inequalities that can hold simultaneously for positive real numbers $\displaystyle x$, $\displaystyle y$, $\displaystyle z$ satisfying $\displaystyle x+y+z=1$.

b) Find an explicit formula for $\displaystyle x$, $\displaystyle y$, $\displaystyle z$ in terms of $\displaystyle a$, $\displaystyle b$, $\displaystyle c$, without using case distinctions, such that $\displaystyle x$, $\displaystyle y$, $\displaystyle z>0$, $\displaystyle x+y+z=1$, and none of the three inequalities holds.

Proposed by Áron Bán-Szabó, Palaiseau

(7 pont)

Deadline expired on May 11, 2026.

Solution. We shall use barycentric coordinates. Recall the well-known distance formula stating that if a triangle has side lengths $\displaystyle a,b,c$, and $\displaystyle X=(x_1,x_2,x_3)$ and $\displaystyle Y=(y_1,y_2,y_3)$ are barycentric coordinates satisfying $\displaystyle x_1+x_2+x_3=y_1+y_2+y_3=1$, then

$\displaystyle |XY|^2=-a^2z_2z_3-b^2z_3z_1-c^2z_1z_2, $

where

$\displaystyle \overrightarrow{XY}=(z_1,z_2,z_3)=(x_1-y_1,x_2-y_2,x_3-y_3) $

is the displacement vector.

Observe that since $\displaystyle a,b,c\in(1,4)$, we have $\displaystyle \sqrt a,\sqrt b,\sqrt c\in(1,2)$, hence there exists a triangle $\displaystyle ABC$ with side lengths

$\displaystyle |AB|=\sqrt c,\qquad |AC|=\sqrt b,\qquad |BC|=\sqrt a. $

Then

$\displaystyle A=(1,0,0),\qquad B=(0,1,0),\qquad C=(0,0,1). $

Let $\displaystyle P=(x,y,z)$; we may assume this since $\displaystyle x+y+z=1$. Since $\displaystyle x,y,z$ are positive, the point $\displaystyle P$ lies strictly inside the triangle.

Then

$\displaystyle \overrightarrow{PA}=(x-1,y,z),\qquad \overrightarrow{PB}=(x,y-1,z),\qquad \overrightarrow{PC}=(x,y,z-1), $

hence

$$\begin{align*} |PA|^2 &= -ayz-b(x-1)z-c(x-1)y \\ &= -ayz-bxz+bz-cxy+cy, \\ |PB|^2 &= -a(y-1)z-bxz-cx(y-1) \\ &= -ayz+az-bxz-cxy+cx, \\ |PC|^2 &= -ay(z-1)-bx(z-1)-cxy \\ &= -ayz+ay-bxz+bx-cxy. \end{align*}$$

It follows that

$$\begin{align*} \angle BPC\le90^\circ &\iff |PB|^2+|PC|^2\ge|BC|^2 \\ &\iff -2ayz+az+ay-2bxz+bx+cx-2cxy\ge a \\ &\iff bx(1-2z)+cx(1-2y)\ge a(1+2yz-y-z) \\ &\iff bx(x+y-z)+cx(x+z-y)\ge a(2yz+x), \end{align*}$$

where in the last step we used that $\displaystyle x+y+z=1$. Similarly, one can show that the conditions

$\displaystyle \angle CPA\le90^\circ \qquad\text{and}\qquad \angle APB\le90^\circ $

are equivalent to the other two inequalities appearing in the problem statement.

Thus the question in part (a) is precisely how many of the angles

$\displaystyle \angle BPC,\qquad \angle CPA,\qquad \angle APB $

can be acute. Clearly all three cannot be acute, since their sum is $\displaystyle 360^\circ$, so one of them must be at least $\displaystyle 120^\circ$. Nor can exactly two of them be acute, because then their sum would be at most $\displaystyle 180^\circ$, forcing the third angle to be at least $\displaystyle 180^\circ$, which is impossible since $\displaystyle P$ lies strictly inside the triangle (recall that $\displaystyle x,y,z$ are positive).

Of course, exactly one acute angle is possible. Indeed, choose a vertex of the triangle whose interior angle is small, at most $\displaystyle 60^\circ$ (such a vertex exists since the three interior angles sum to $\displaystyle 180^\circ$). If $\displaystyle P$ is sufficiently close to this vertex, then the opposite side will subtend an acute angle from $\displaystyle P$.

Finally, we show that it is also possible that none of the angles is acute, and in doing so we also answer part (b) by giving an explicit formula. Observe that the incenter $\displaystyle I$ of the triangle is seen under an obtuse angle from every side, since

$\displaystyle \angle BIC = 90^\circ+\frac{\angle BAC}{2}. $

Fortunately, the barycentric coordinates of $\displaystyle I$ can be expressed explicitly in terms of the side lengths:

$\displaystyle I= \left( \dfrac{\sqrt a}{\sqrt a+\sqrt b+\sqrt c}, \dfrac{\sqrt b}{\sqrt a+\sqrt b+\sqrt c}, \dfrac{\sqrt c}{\sqrt a+\sqrt b+\sqrt c} \right). $

Statistics:

9 students sent a solution.

7 points: Aravin Peter, Bodor Mátyás, Diaconescu Tashi, Kis Ágoston, Sami El-Hajjar.

6 points: Áron Bence.

5 points: 1 student.

0 point: 1 student.

Not shown because of missing birth date or parental permission: 1 solutions.

Problems in Mathematics of KöMaL, April 2026

9 students sent a solution.
7 points:	Aravin Peter, Bodor Mátyás, Diaconescu Tashi, Kis Ágoston, Sami El-Hajjar.
6 points:	Áron Bence.
5 points:	1 student.
0 point:	1 student.
Not shown because of missing birth date or parental permission:	1 solutions.

Already signed up?
New to KöMaL?