Shortlisted Problems for the 64th NMO

Let $a$, $b$, $c$ be the complex coordinates of the vertices of the triangle. The given condition is:
$$|a + 2b - 3c| = |a - 4b + 3c|$$
Let $z_1 = a + 2b - 3c$ and $z_2 = a - 4b + 3c$. Then $|z_1| = |z_2|$.

Consider the difference:
$$|z_1|^2 = |z_2|^2$$
Expanding both sides:
$$(a + 2b - 3c)(\overline{a} + 2\overline{b} - 3\overline{c}) = (a - 4b + 3c)(\overline{a} - 4\overline{b} + 3\overline{c})$$
Let us subtract one side from the other:
$$(a + 2b - 3c)(\overline{a} + 2\overline{b} - 3\overline{c}) - (a - 4b + 3c)(\overline{a} - 4\overline{b} + 3\overline{c}) = 0$$
Let us denote $a$, $b$, $c$ as $A$, $B$, $C$ for clarity. Let us expand both products:

First term:
$$(a + 2b - 3c)(\overline{a} + 2\overline{b} - 3\overline{c}) = a\overline{a} + 2a\overline{b} - 3a\overline{c} + 2b\overline{a} + 4b\overline{b} - 6b\overline{c} - 3c\overline{a} - 6c\overline{b} + 9c\overline{c}$$
Second term:
$$(a - 4b + 3c)(\overline{a} - 4\overline{b} + 3\overline{c}) = a\overline{a} - 4a\overline{b} + 3a\overline{c} - 4b\overline{a} + 16b\overline{b} - 12b\overline{c} + 3c\overline{a} - 12c\overline{b} + 9c\overline{c}$$
Subtracting:
$$[a\overline{a} + 2a\overline{b} - 3a\overline{c} + 2b\overline{a} + 4b\overline{b} - 6b\overline{c} - 3c\overline{a} - 6c\overline{b} + 9c\overline{c}] \\ - [a\overline{a} - 4a\overline{b} + 3a\overline{c} - 4b\overline{a} + 16b\overline{b} - 12b\overline{c} + 3c\overline{a} - 12c\overline{b} + 9c\overline{c}] = 0$$
Now, combine like terms:
- $a\overline{a}$ cancels.
- $9c\overline{c} - 9c\overline{c}$ cancels.

$a\overline{b}$: $2a\overline{b} - (-4a\overline{b}) = 6a\overline{b}$
$a\overline{c}$: $-3a\overline{c} - 3a\overline{c} = -6a\overline{c}$
$b\overline{a}$: $2b\overline{a} - (-4b\overline{a}) = 6b\overline{a}$
$b\overline{b}$: $4b\overline{b} - 16b\overline{b} = -12b\overline{b}$
$b\overline{c}$: $-6b\overline{c} - (-12b\overline{c}) = 6b\overline{c}$
$c\overline{a}$: $-3c\overline{a} - 3c\overline{a} = -6c\overline{a}$
$c\overline{b}$: $-6c\overline{b} - (-12c\overline{b}) = 6c\overline{b}$

So the sum is:
$$6a\overline{b} - 6a\overline{c} + 6b\overline{a} - 12b\overline{b} + 6b\overline{c} - 6c\overline{a} + 6c\overline{b} = 0$$
Group terms:
$$6(a\overline{b} + b\overline{a}) - 6(a\overline{c} + c\overline{a}) + 6(b\overline{c} + c\overline{b}) - 12b\overline{b} = 0$$
Divide both sides by $6$:
$$(a\overline{b} + b\overline{a}) - (a\overline{c} + c\overline{a}) + (b\overline{c} + c\overline{b}) - 2b\overline{b} = 0$$
Now, recall that $a\overline{b} + b\overline{a} = 2\operatorname{Re}(a\overline{b})$, etc.
So:
$$2\operatorname{Re}(a\overline{b}) - 2\operatorname{Re}(a\overline{c}) + 2\operatorname{Re}(b\overline{c}) - 2|b|^2 = 0$$
Divide by $2$:
$$\operatorname{Re}(a\overline{b}) - \operatorname{Re}(a\overline{c}) + \operatorname{Re}(b\overline{c}) - |b|^2 = 0$$
Rearrange:
$$\operatorname{Re}(a\overline{b}) - \operatorname{Re}(a\overline{c}) + \operatorname{Re}(b\overline{c}) = |b|^2$$
But $|b|^2 = b\overline{b}$, and $\operatorname{Re}(a\overline{b}) = \frac{1}{2}(a\overline{b} + b\overline{a})$.

Alternatively, interpret geometrically:
Let $A$, $B$, $C$ be the points with complex coordinates $a$, $b$, $c$.
The given condition is:
$$|a + 2b - 3c| = |a - 4b + 3c|$$
Let us denote $z = a + 2b - 3c$, $w = a - 4b + 3c$.

Note that $|z| = |w|$ means that $z$ and $w$ are equidistant from the origin.
But $z - w = (a + 2b - 3c) - (a - 4b + 3c) = 6b - 6c = 6(b - c)$.
So $z$ and $w$ are collinear with $b$ and $c$.

The midpoint of $z$ and $w$ is:
$$\frac{z + w}{2} = \frac{(a + 2b - 3c) + (a - 4b + 3c)}{2} = \frac{2a - 2b}{2} = a - b$$
So the segment $zw$ is centered at $a - b$ and has direction $b - c$.

Since $|z| = |w|$, the segment $zw$ is perpendicular to the line joining the origin to $a - b$.
That is, the vector $b - c$ is perpendicular to $a - b$.

Therefore, $b - c \perp a - b$.
But $a - b$ is the vector from $B$ to $A$, and $b - c$ is the vector from $C$ to $B$.
So the triangle $ABC$ has $\angle ABC = 90^\circ$.

Therefore, the triangle is right-angled.