In the interior of a regular triangle with unit side length, we marked three points such that connecting any of these points with any vertex results in lines that divide the angle at that vertex into four equal parts. Two students came up with different results for the distances between the three points, and the teacher accepted both as correct. How is this possible, and what were the results?
1. We will show that in the interior of the given equilateral triangle $ABC = H$, the three points according to the first sentence of the problem can be marked in two different ways. This immediately explains the question, that the two students gave the distances corresponding to the two different sets of points; we only need to verify that the distances determined by the two sets of points differ from each other.

Draw the angle bisectors of the angles $BAC$, $CBA$, and $ACB$ and mark the intersection points with the opposite sides $BC$, $CA$, and $AB$ as $A_i$, $B_i$, and $C_i$ (i=1,2,3) in the direction of the rotational sense of $A$, $B$, and $C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_1fec7641796de605af68g-1.jpg?height=721&width=805&top_left_y=431&top_left_x=649)

The lines connecting the points of the sought triplet to the vertices of $H$ can only be among these 9 lines, so only points where three such angle bisectors intersect can be considered. There are such points, since $AA_2$, $BB_2$, and $CC_2$ are the symmetry axes of $H$ and each axis reflects the other bisectors into pairs, intersecting on the axis. The intersection points of the pairs $BB_1$, $CC_3$ on the axis $AA_2$, the pairs $BB_2$, $CC_2$, and the pairs $BB_3$, $CC_1$ are denoted by $Q_1$, $P$, and $R_1$ respectively. Similarly, we get the triple intersection points $Q_2$, $P$, $R_2$ on the axis $BB_2$ and $Q_3$, $P$, $R_3$ on the axis $CC_2$.

With these 7 points, we have listed all the triple intersection points of the angle bisectors. It is sufficient to show, based on these symmetries, that there is no further triple intersection point on any of the outer bisectors, for example, on $AA_1$. The line connecting the points $R_2$ and $Q_3$ already established on it with $B$ is $BB_2$ and $BB_3$, so only the intersection point $T$ with $BB_1$ can be considered. However, this point is an internal point of the segment $BQ_1$ between $AB$ and $AA_2$, so $\angle BCT < \angle BCQ_1 = 15^\circ$, hence $T$ does not lie on the bisector of $\angle ACB$; this is what we wanted to show.

Each of the 7 triple intersection points lies on one of the axes of $H$, and $P$ lies on all three axes. Thus, $P$ cannot be in the same triplet as any $Q_i$ or $R_i$ (i=1,2,3), otherwise, connecting the triplet with $A$, $B$, and $C$ would not yield all 9 angle bisectors of $H$, for example, if $P$ were in the same triplet as $Q_1$, then either $AA_1$ or $AA_3$ would not be obtained.

Similarly, it follows that $Q_1$ cannot be in the same triplet as $R_1$ (due to $AA_2$), $R_2$, or $R_3$ (due to $CC_3$ and $BB_1$). The triplet $Q_1, Q_2, Q_3 = H_q$ does, however, produce all 9 angle bisectors of $H$. Similarly, we obtain the triplet $R_1, R_2, R_3 = H_r$ and that no more suitable triplets can be formed from the 7 triple intersection points. This proves the statement made at the beginning of the solution. From the $120^\circ$ rotational symmetry of $H$ around $P$, it also follows that both the $H_q$ and $H_r$ triplets are the vertices of equilateral triangles.

2. Moving on to the anticipated calculation, we only need to calculate one side of the equilateral triangles $H_q$ and $H_r$. These are equal to the $\sqrt{3}$ multiple of the radius of the circumcircle of the triangles $PQ_1$ and $PR_1$.

$PQ_1$ is a part of the side $PA_2$ cut by the angle bisector at $B$ in the right triangle $PBA_2$, and the sides of this triangle are $BA_2 = 1/2$, $A_2P = AA_2 / 3 = \sqrt{3/6}$, and $PB = 2AA_2 / 3 = \sqrt{3}/3$.

Therefore,

\[
\begin{gathered}
PQ_1 : Q_1A_2 = PB : A_2B \\
PQ_1 : (PQ_1 + Q_1A_2) = PQ_1 : PA_2 = PB : (PB + A_2B) \\
PQ_1 = PA_2 \frac{PB}{PB + A_2B} = \frac{2}{\sqrt{3}} - 1
\end{gathered}
\]

$BCR_1$ is an isosceles right triangle by construction, so

\[
PR_1 = A_2R_1 - A_2P = BA_2 - A_2P = \frac{1}{2} - \frac{\sqrt{3}}{6}
\]

From these, based on the previous statements,

\[
\begin{aligned}
Q_1Q_2 & = \sqrt{3} \cdot PQ_1 = 2 - \sqrt{3} (=0.268) \\
R_1R_2 & = \sqrt{3} \cdot PR_1 = \frac{\sqrt{3}}{2} - \frac{1}{2} (=0.366)
\end{aligned}
\]

and our results confirm the thought we mentioned at the beginning of the solution.

Note. If the problem had not asked for the possible numerical results, then instead of point 2, it would have been enough to say: the ratio of the sides of the triangles $H_q$ and $H_r$ from the obviously similar triangles $AQ_3Q_2$ and $AR_2R_3$ is

\[
\frac{Q_3Q_2}{R_2R_3} = \frac{AQ_3}{AR_2} < 1
\]

because $Q_3$ is inside the triangle $ABP$, $AQ_3$ is a part of $AR_2$, so $Q_3Q_2 \neq R_2R_3$.