Question 3

We will start with an equilateral triangle and carefully cut its corners to find the resulting shape.

Step 1 — Setting up the triangle and marks

Let us consider an equilateral triangle, ABC. Let the length of each side of triangle ABC be $\mathbf{3s}$ units. We divide each side into three equal parts. Each part will have a length of $s$ units.

On side AB, let the marks be $D_1$ and $D_2$. $AD_1 = D_1D_2 = D_2B = s$ On side BC, let the marks be $E_1$ and $E_2$. $BE_1 = E_1E_2 = E_2C = s$ On side CA, let the marks be $F_1$ and $F_2$. $CF_1 = F_1F_2 = F_2A = s$

Step 2 — Identifying the cut-off corners

The problem asks us to cut off each corner "as far as the marks". This means we cut along the lines connecting the marks near each corner. For corner A, we cut along $D_1F_2$. This removes triangle $AD_1F_2$. For corner B, we cut along $E_1D_2$. This removes triangle $BE_1D_2$. For corner C, we cut along $F_1E_2$. This removes triangle $CF_1E_2$.

Let us examine triangle $AD_1F_2$: Side $AD_1 = s$ (from Step 1). Side $AF_2 = s$ (from Step 1). The angle at vertex A is $\mathbf{60^\circ}$, because ABC is an equilateral triangle. Since two sides are equal ($AD_1 = AF_2 = s$) and the included angle is $60^\circ$, triangle $AD_1F_2$ is an equilateral triangle. So, the third side $D_1F_2$ also has a length of $\mathbf{s}$ units.

Similarly, for triangle $BE_1D_2$: Side $BD_2 = s$ (from Step 1). Side $BE_1 = s$ (from Step 1). The angle at vertex B is $\mathbf{60^\circ}$. So, triangle $BE_1D_2$ is an equilateral triangle, and its third side $E_1D_2$ has a length of $\mathbf{s}$ units.

And for triangle $CF_1E_2$: Side $CE_2 = s$ (from Step 1). Side $CF_1 = s$ (from Step 1). The angle at vertex C is $\mathbf{60^\circ}$. So, triangle $CF_1E_2$ is an equilateral triangle, and its third side $F_1E_2$ has a length of $\mathbf{s}$ units.

Step 3 — Describing the resulting shape

After cutting off the three corner triangles ($AD_1F_2$, $BE_1D_2$, $CF_1E_2$), the remaining shape is a polygon. The vertices of this new shape are $D_1, F_2, F_1, E_2, E_1, D_2$. This shape has six sides, so it is a hexagon.

Let us find the length of each side of this hexagon: The side $D_1F_2$ is the base of the cut-off triangle $AD_1F_2$. $D_1F_2 = s$ The side $F_2F_1$ is part of the original side AC. $F_2F_1 = AC - AF_2 - CF_1$ $= 3s - s - s$ $= \mathbf{s}$ The side $F_1E_2$ is the base of the cut-off triangle $CF_1E_2$. $F_1E_2 = s$ The side $E_2E_1$ is part of the original side BC. $E_2E_1 = BC - CE_2 - BE_1$ $= 3s - s - s$ $= \mathbf{s}$ The side $E_1D_2$ is the base of the cut-off triangle $BE_1D_2$. $E_1D_2 = s$ The side $D_2D_1$ is part of the original side AB. $D_2D_1 = AB - AD_1 - BD_2$ $= 3s - s - s$ $= \mathbf{s}$ All six sides of the hexagon are equal in length, each being $\mathbf{s}$ units.

Now, let us find the interior angles of the hexagon. Consider the angle at vertex $D_1$. The original side AB was a straight line. The angle $\angle AD_1F_2$ is an angle of the equilateral triangle $AD_1F_2$, so it is $60^\circ$. The angle $\angle AD_1D_2$ is a straight angle, which is $180^\circ$. The interior angle of the hexagon at $D_1$ is $\angle F_2D_1D_2$. $\angle F_2D_1D_2 = \angle AD_1D_2 - \angle AD_1F_2$ $= 180^\circ - 60^\circ$

$\boxed{\mathbf{120^\circ}}$ By symmetry, all six interior angles of the hexagon will be $\mathbf{120^\circ}$. Since all sides are equal and all interior angles are equal, the resulting shape is a regular hexagon.

Answer

The shape you get is a regular hexagon.