I have this problem about a regular hexagon being folded so that three vertices meet at the center O. I need to find out what shape is formed after folding. The hexagon has some vertices marked, and when folded, those three points come together at O.

First, I should recall that a regular hexagon has all sides equal and all internal angles 120 degrees. It can be divided into six equilateral triangles. The center O is the common vertex of all these triangles.

Now, when we fold the paper so that three vertices meet at O, it means we're bringing those three corner points to the center. Since it's a fold, it's like reflecting part of the hexagon over a line to bring those points together.

The problem says "folded," which probably means a crease is made, and one part is folded over that crease so that the vertices land on O. But it says three vertices meet at O, so it might be that we're making folds such that after folding, those three points coincide at O.

Perhaps it's a single fold that brings all three vertices to the center simultaneously? But with a flat paper, folding it so three points meet at one point might require multiple folds or a specific type of fold.

Maybe it's a fold along a line, and when you fold, one vertex goes to O, but the other two might not, unless the fold line is chosen such that all three are mapped to O.

I think I need to think differently. Perhaps it's about folding the hexagon in such a way that the three vertices are brought to the center, and we need to see what the resulting shape is.

Another idea: in origami or paper folding, when you fold a point to another point, the fold line is the perpendicular bisector of the segment joining those two points. But here, we have three points all going to the same point O.

So, for each vertex, if we want to fold it to O, the fold line would be the perpendicular bisector of the line from that vertex to O.

But since all three vertices are being folded to O, it might be that we have three separate fold lines, but that could be messy.

Perhaps for the three vertices to meet at O after folding, the fold must be such that O is the image, and the vertices are symmetric in some way.

Let me label the vertices. Let's say the regular hexagon has vertices A, B, C, D, E, F, with O the center. The problem shows three vertices marked; from the diagram, it looks like one at the top, one at the bottom-left, and one on the right, but I can't see the diagram, so I need to assume which vertices are marked.

The problem says "the three marked vertices," but since there's no diagram, I should assume a standard position. Typically, in a regular hexagon, we might mark every second vertex or something.

Perhaps in the diagram, the three marked vertices are such that they form an equilateral triangle or something.

Since it's a regular hexagon, and we're folding so that three vertices meet at the center, it might be that we're folding along lines that are axes of symmetry.

A regular hexagon has rotational symmetry and reflection symmetry. It has six reflection axes: three through opposite vertices and three through the midpoints of opposite sides.

Now, if I want to fold so that a vertex meets the center, but the center is not a vertex, so folding a vertex to the center means we're creating a new point or something.

I think I need to consider the fold line. When you fold a point P to a point Q, the fold line is the perpendicular bisector of PQ. So for each vertex, if we fold it to O, the fold line is the perpendicular bisector of the segment from that vertex to O.

In a regular hexagon, the distance from a vertex to the center is equal to the side length, since it's made of equilateral triangles.

Let the side length be S. Then, distance from vertex to center is also S.

So, for a vertex V, the segment VO has length S. The perpendicular bisector of VO will be the set of points equidistant to V and O.

When we fold along that perpendicular bisector, V maps to O.

But for three vertices, if we have three such fold lines, folding along them might not bring all three to O at the same time unless the fold lines are related.

Perhaps for the three vertices to be folded to O simultaneously, the three perpendicular bisectors must be the same line or something, but that might not be possible.

Another thought: if the three vertices are symmetric with respect to the center, then their perpendicular bisectors to O might coincide or be concurrent.

Let's assume the three marked vertices are, say, A, C, E, which are every second vertex, so they form an equilateral triangle.

In a regular hexagon, vertices A, C, E are separated by 120 degrees, so they form an equilateral triangle with side length S * √3, I think? Let's calculate.

In a regular hexagon with side S, the distance from center to vertex is S. Distance between two adjacent vertices is S. Distance between vertices two apart, like A to C, with one in between, so angle at center is 120 degrees, so distance A to C is 2 * S * sin(60°) = 2 * S * √3/2 = S√3.

Similarly, for vertices three apart, opposite, it's 2S.

But for three vertices every second, like A, C, E, they are 120 degrees apart, so the triangle ACE is equilateral with side S√3.

Now, the center O.

For each of these vertices, say A, the segment AO has length S. The perpendicular bisector of AO.

Since O is the center, and A is a vertex, let's set coordinates to make it easier.

Place the hexagon with center O at (0,0). Let vertex A be at (S, 0), B at (S/2, (S√3)/2), C at (-S/2, (S√3)/2), D at (-S, 0), E at (-S/2, -(S√3)/2), F at (S/2, -(S√3)/2).

Standard position: let’s say A at (1,0), B at (0.5, √3/2), C at (-0.5, √3/2), D at (-1,0), E at (-0.5, -√3/2), F at (0.5, -√3/2), with S=1 for simplicity. Distance from O to A is 1.

So O at (0,0).

Now, the three marked vertices, from the diagram, it might be, say, the top, bottom-left, and right or something. But since no diagram, let's assume it's A, C, E for simplicity, as they are symmetric.

A (1,0), C (-0.5, √3/2), E (-0.5, -√3/2)? Let's list:

A: (1, 0)

B: (0.5, √3/2)

C: (-0.5, √3/2)

D: (-1, 0)

E: (-0.5, -√3/2)

F: (0.5, -√3/2)

If we take every second vertex, say A, C, E: A (1,0), C (-0.5, √3/2), E (-0.5, -√3/2)

But these are not all at the same height; A is on the x-axis, C and E are symmetric.

The problem says "the three marked vertices meet at the centre O", so let's assume it's A, C, E for now.

For vertex A (1,0), to fold to O (0,0), the segment AO from (1,0) to (0,0), midpoint is (0.5, 0), and the perpendicular bisector is the vertical line x = 0.5.

Because AO is along the x-axis from (1,0) to (0,0), so perpendicular is vertical, bisector at (0.5,0), so line x=0.5.

Similarly, for vertex C (-0.5, √3/2), segment CO from (-0.5, √3/2) to (0,0), midpoint is (-0.25, √3/4), and the direction of CO is from (0,0) to (-0.5, √3/2), which is angle 120 degrees, so the segment has slope (√3/2 - 0)/(-0.5 - 0) = (√3/2)/(-0.5) = -√3, so slope of CO is -√3.

Thus, perpendicular slope is 1/(√3) = √3/3.

So perpendicular bisector: passes through midpoint (-0.25, √3/4), slope √3/3.

Similarly for E (-0.5, -√3/2), segment EO to (0,0), midpoint (-0.25, -√3/4), slope of EO is (-√3/2 - 0)/(-0.5 - 0) = (-√3/2)/(-0.5) = √3, so perpendicular slope is -1/√3 = -√3/3.

But these perpendicular bisectors are different lines, so folding along each would bring that vertex to O, but not all at once.

The problem says "folded so that the three marked vertices meet at the centre O", which might imply a single fold that brings all three to O simultaneously, but that seems impossible unless they are collinear or something, which they are not.

Perhaps it's a series of folds, but the problem says "folded", singular, so probably one fold.

Another possibility: "folded" might mean that we create a crease and fold, but the crease is such that when we fold, the three vertices are brought to O.

For that to happen with one fold, the fold line must be the perpendicular bisector for each vertex to O, meaning that all three vertices must be on the same perpendicular bisector, which is not the case.

Unless the three vertices are symmetric with respect to the fold line.

Suppose we have a fold line L. When we fold over L, each point is reflected over L. For the three vertices to be mapped to O after reflection, that means that the reflection of each vertex over L is O.

But reflection is a function, so for each vertex V, ref_L(V) = O.

But O is a fixed point, so ref_L(V) = O for each V.

That means that V and O are symmetric with respect to L, so L is the perpendicular bisector of VO for each vertex.

But for different vertices, VO are different segments, so unless all VO have the same perpendicular bisector, which only if all V are the same point, which is not.

So it's impossible for a single reflection to map multiple points to the same point unless those points are the same.

For example, if I have two points, I can reflect one to the other, but not both to a third point unless they are symmetric.

Specifically, ref_L(V) = O for one V means L is perp bisector of VO.

For another W, ref_L(W) = O means L is perp bisector of WO.

But perp bisector of VO and WO are different unless V and W are symmetric w.r.t. O or something.

Suppose ref_L(V) = O and ref_L(W) = O, then ref_L(V) = ref_L(W), so V = W, contradiction.

Exactly, reflection is invertible, so if ref_L(V) = O and ref_L(W) = O, then V = W.

So it's impossible for a single reflection to map two different points to the same point.

Therefore, with one fold, you can't bring three vertices to the center at the same time.

But the problem must be possible, so I think I misinterpreted.

Perhaps "folded" means that we fold the paper so that the vertices are brought to the center, but it might be that we fold the paper in a way that the vertices meet at O when we bring them together, but it could be a fold along a line that is not necessarily the perp bisector for each.

Another idea: in some contexts, "folding to a point" might mean creasing so that the point is fixed, but I think I need to think about the resulting shape.

Perhaps the fold is such that the three vertices are folded to O, meaning that we make three separate folds, each bringing one vertex to O, and then the paper is folded, and we have a new shape with those points at O.

But the problem says "the three marked vertices meet at the centre O", and "the figure that is formed", so after folding, the shape has those three points at O.

But with multiple folds, it might be layered.

The problem says "a piece of paper" and "folded", so it might be that after folding, the paper has a new shape with the vertices at O.

But let's read the problem: "a piece of paper in the shape of a regular hexagon, as shown, is folded so that the three marked vertices meet at the centre O of the hexagon. What is the shape of the figure that is formed?"

And "the figure that is formed" might mean the shape after folding, which could be a polygon with those points identified or something.

Perhaps it's like in origami, when you fold, the visible shape changes.

But I think for this problem, it's about the geometric shape formed by the crease or the boundary after folding.

Another thought: when you fold a vertex to the center, you might be creating a new edge or something.

Perhaps for a regular hexagon, folding three vertices to the center can be done by folding along the lines from the center to the midpoints or something.

Let's think about the axes of symmetry.

Suppose the three marked vertices are not all the same type. For example, in the diagram, it might be that the three vertices are every other one but not including the center.

I recall that in a regular hexagon, if you connect the midpoints of every other side or something.

Perhaps the three marked vertices are the ones that are not adjacent to a particular vertex or something.

Let's assume that the three marked vertices are A, C, and E, as I had.

But with A, C, E, they are not symmetric in a way that allows a single fold.

Unless we fold along a line that is the perpendicular bisector for each, but it's impossible as I thought.

Perhaps "meet at the centre" means that after folding, the vertices are brought to O, but it could be that the fold is not a reflection to O, but rather the paper is folded such that the vertices are mapped to O, but with different fold lines for each, but that would require multiple folds.

The problem says "folded", singular, so likely one fold.

Another idea: perhaps the fold is along a line that passes through O, and when you fold, the vertices are reflected, but not necessarily to O.

For example, if I fold along a line through O, say the x-axis, then points are reflected over x-axis. Vertex A at (1,0) is on the axis, so stays. C at (-0.5, √3/2) goes to (-0.5, -√3/2) which is E, not O. So not to O.

Similarly, E goes to C, not O. So only if a vertex is on the fold line, it stays, but O is not a vertex.

So not.

Perhaps the fold is not through O.

Let's think about the condition for three points to be folded to a single point.

I think it's impossible with one fold for three points to map to one point.

Unless the three points are on the fold line, but then they stay, not move to O.

So that doesn't work.

Perhaps "folded so that the three marked vertices meet at the centre" means that we fold the paper in a way that the three vertices are brought together to the center point, but it might be that we are creating a new figure where those points are at O.

I think I need to consider that after folding, the paper has a crease, and the shape is the union of the unfolded and folded parts, but with the vertices at O.

But that might be messy.

Perhaps for a regular hexagon, there is a way to fold it so that three vertices meet at the center by folding along a line that is not a straight line, but the problem likely assumes a straight fold.

The problem says "folded", so probably a straight crease.

Let's look for similar problems or think about the geometry.

Another thought: in a regular hexagon, the lines from the center to the vertices are radii.

The perpendicular bisector from a vertex to O is the line perpendicular to the radius at the midpoint.

For vertex A (1,0), perp bisector is x=0.5.

For C (-0.5, √3/2), let's calculate the perp bisector.

Midpoint of C and O: C (-0.5, √3/2), O (0,0), midpoint M_c = (-0.25, √3/4)

Slope of CO: from (0,0) to (-0.5, √3/2), slope = (√3/2 - 0)/( -0.5 - 0) = (√3/2)/(-0.5) = -√3

So perp slope is 1/√3 = √3/3

So perp bisector: line through (-0.25, √3/4) with slope √3/3

Similarly for E (-0.5, -√3/2), midpoint M_e = (-0.25, -√3/4), slope of EO is ( -√3/2 -0)/( -0.5 -0) = (-√3/2)/(-0.5) = √3, perp slope -1/√3 = -√3/3

Now, these three lines are different, so no single line.

But perhaps for the three vertices to be folded to O, we need to fold along a line such that the reflection of each is O, but impossible.

Unless the three vertices are on the perpendicular bisector of O and another point, but I'm confused.

Perhaps "meet at the centre" means that after folding, the three vertices are at the same location as O, so we are collapsing them to O.

But with one fold, only one point can be moved to O.

Unless the fold line is chosen so that the three vertices are mapped to O, but that requires them to be the same point.

I think there's a mistake in interpretation.

Let's read the problem again: "folded so that the three marked vertices meet at the centre O"

Perhaps it means that when we fold the paper, we bring the three vertices to the center point O, so we are creating a new figure where those three points are identified at O.

In that case, after folding, the shape has the three vertices at O, so the boundary is folded.

But with one fold, it's like reflecting part of the paper over a line.

So, suppose we have a fold line L. When we fold over L, one part is reflected, and the three vertices are reflected to some points. For them to meet at O, their reflections must be at O.

But as before, only if they are the same point.

Unless for some specific L, the reflection of each vertex is O, but impossible for multiple vertices.

Perhaps for some L, the reflection of one vertex is O, and the others are also mapped to O by coincidence, but in a hexagon, that doesn't happen.

For example, if I take L to be the perp bisector of AO, then ref_L(A) = O, but ref_L(C) is some other point, not O.

Similarly for others.

So not.

Another idea: perhaps "the three marked vertices meet at the centre" means that after folding, the three vertices are brought to the same point, which is O, but not necessarily that each is mapped to O, but that their positions coincide at O.

But with one fold, reflection, points are mapped to different points, so they can't coincide unless they were the same.

So impossible.

I think I need to consider that the fold is not a reflection to the plane, but a physical folding that brings the points together, so the paper is creased, and the vertices are at O.

Perhaps it's like in paper folding, when you fold a point to another, you create a crease, but for multiple points, you might have multiple creases.

But the problem says "folded", singular.

Let's look at the diagram description. The user said "as shown", but in text, it's not shown, so I need to assume.

Perhaps the three marked vertices are not all vertices; but the problem says "vertices".

Another thought: in some contexts, for a hexagon, folding vertices to the center might involve folding along the lines of symmetry.

Suppose the three marked vertices are A, D, and say, the top and bottom, but A and D are opposite.

Let's assume the three marked vertices are, for example, the vertices at 0°, 120°, and 240° positions, so A, C, E as before.

But same issue.

Perhaps the fold is not a straight line, but a curve, but that seems unlikely for this problem.

I recall that in a regular hexagon, if you fold every other vertex to the center, you might get a smaller hexagon or something.

But how to do that with one fold.

Perhaps it's a single fold along a line that is the perpendicular bisector for one vertex, and for the others, it's different.

I think I need to search for a different approach.

Let's think about the resulting shape.

Suppose we have the hexagon, and we fold it so that three vertices meet at O. After folding, the paper has a crease, and the shape is a polygon with the three vertices at O, so the area is divided, and the folded part is on top or something.

But the "figure that is formed" might be the boundary of the paper after folding.

For example, when you fold, the visible shape might have a different outline.

But it could be that after folding, the three vertices are at O, so the points are identified, and the new shape has fewer vertices.

For instance, with three vertices at O, the new boundary consists of the other three vertices and the crease lines.

But let's assume one fold.

Suppose we fold along a line L. After folding, the paper has two parts: the unfolded part and the reflected part over L.

The three marked vertices: if they are on one side, they are reflected to new positions.

For them to be at O, their reflected positions must be O.

But only if they are on the perp bisector, but not all can be.

Unless for some L, the reflection of the three points are all at O, which is impossible.

Perhaps "meet at the centre" means that the vertices are brought to the center area, not necessarily to the point O, but the problem says "at the centre O", so to the point.

I think I have to consider that the fold is such that the three vertices are folded to O, but with multiple folds, and the figure formed is the crease pattern or something.

But the problem says "the figure that is formed", and "a piece of paper", so likely the physical shape after folding.

Perhaps after folding, the paper is a polygon with the three vertices at O, so it's like a triangle or something with the points pinched at O.

But with one fold, it's not pinched; it's folded flat.

Let's assume that the three marked vertices are the ones that are not adjacent to the top or something.

Perhaps in the diagram, the three marked vertices are the bottom-left, bottom-right, and top, for example.

Let's assume the hexagon has vertices: let's say F (0.5, -√3/2), A (1,0), B (0.5, √3/2), C (-0.5, √3/2), D (-1,0), E (-0.5, -√3/2), with O (0,0).

Suppose the three marked vertices are, say, F, B, and D, for example, which are every other vertex starting from the bottom-right.

F (0.5, -√3/2), B (0.5, √3/2), D (-1,0).

B and F have the same x-coordinate, D is on the negative x-axis.

Not symmetric.

Perhaps it's the vertices at the corners of one triangle.

Another idea: in a regular hexagon, you can fold it along a line that is the line from a vertex to the opposite side midpoint or something.

Let's think about the crease.

Suppose we want to fold vertex A to O. The fold line is the perpendicular bisector of AO.

With S=1, A (1,0), O (0,0), AO from (1,0) to (0,0), midpoint (0.5,0), perp bisector x=0.5.

If we fold along x=0.5, then A is mapped to O, and the right part is reflected over x=0.5.

For example, point B (0.5, √3/2) is on the line, so it stays, but B is at (0.5, √3/2), on x=0.5, so when we fold, points on the fold line stay, points on the right are reflected left.

Vertex A (1,0) is mapped to (0,0) O.

Vertex F (0.5, -√3/2) is on the line, so stays.

But the three marked vertices are not all on the right; for example, if we have three vertices to fold, say A, C, E, but C and E are on the left, so when we fold along x=0.5, C and E are not moved; they are on the left side, so they stay in place, not moved to O.

Only A is moved to O.

C at (-0.5, √3/2) is not on the right, so it is not reflected; it stays.

Similarly for E.

So only one vertex can be moved to O with one fold along x=0.5.

If we want to move another vertex, we need a different fold.

But for three, we need three separate folds.

Perhaps for the three vertices A, C, E, we can fold each to O with their own perp bisector, but that would create three creases, and the paper is folded with multiple layers.

Then after folding, the shape has the three vertices at O, and the crease lines form a smaller shape.

Let's try that.

Suppose we have the hexagon, and we fold along the perp bisector for A to O, which is x=0.5.

Then A is mapped to O.

Similarly, for C to O, the perp bisector is the line through (-0.5, √3/4) with slope √3/3.

Similarly for E.

But these lines are different.

For example, perp bisector for C: through (-0.25, √3/4), slope √3/3.

Equation: y - √3/4 = (√3/3)(x + 0.25) since x - (-0.25) = x+0.25

Similarly, for A, x=0.5.

For E, through (-0.25, -√3/4), slope -√3/3, y + √3/4 = (-√3/3)(x + 0.25)

Now, if we fold along all three lines, but that might not be a single fold.

Perhaps the three perp bisectors intersect or something.

Let's find the intersection.

First, perp bisector for A: x=0.5

For C: y - √3/4 = (√3/3)(x + 0.25)

At x=0.5, y = √3/4 + (√3/3)(0.5 + 0.25) = √3/4 + (√3/3)(0.75) = √3/4 + (√3/3)(3/4) = √3/4 + √3/4 = 2√3/4 = √3/2

So point (0.5, √3/2), which is exactly vertex B.

Similarly, for E: y + √3/4 = (-√3/3)(x + 0.25)

At x=0.5, y = -√3/4 + (-√3/3)(0.5 + 0.25) = -√3/4 - (√3/3)(0.75) = -√3/4 - √3/4 = -2√3/4 = -√3/2, which is vertex F.

So the perp bisector for C passes through B, and for E passes through F.

But B and F are vertices.

Not helpful.

Perhaps for the three vertices, the perp bisectors form a triangle.

I recall that in a regular hexagon, the three perp bisectors for the vertices to the center form a larger equilateral triangle or something.

Let's find the three perp bisectors.

For A (1,0), perp bisector: x=0.5

For C (-0.5, √3/2), perp bisector: y - √3/4 = (√3/3)(x + 0.25)

For E (-0.5, -√3/2), perp bisector: y + √3/4 = (-√3/3)(x + 0.25)

Now, these three lines form a triangle themselves.

For example, intersection of x=0.5 and the perp bisector of C: as above, at (0.5, √3/2) B

x=0.5 and perp bisector of C: at (0.5, √3/2)

Similarly, x=0.5 and perp bisector of E: at (0.5, -√3/2) F

Now, intersection of perp bisector of C and perp bisector of E.

Set the equations:

For C: y = (√3/3)(x + 0.25) + √3/4

For E: y = (-√3/3)(x + 0.25) - √3/4

Set equal: (√3/3)(x + 0.25) + √3/4 = (-√3/3)(x + 0.25) - √3/4

Multiply both sides by 3/√3 = √3 to simplify:

(1)(x + 0.25) + (3/4)(√3)(√3/√3) better to divide by √3 first.

Equation: (√3/3)(x + 0.25) + √3/4 = (-√3/3)(x + 0.25) - √3/4

Bring all terms to one side: (√3/3)(x + 0.25) + √3/4 + (√3/3)(x + 0.25) + √3/4 = 0

From:

Left - right = 0

[ (√3/3)(x + 0.25) + √3/4 ] - [ (-√3/3)(x + 0.25) - √3/4 ] = 0

= (√3/3)(x + 0.25) + √3/4 + (√3/3)(x + 0.25) + √3/4 = (2√3/3)(x + 0.25) + 2√3/4 = (2√3/3)(x + 0.25) + √3/2

Set equal to 0:

(2√3/3)(x + 0.25) + √3/2 = 0

Divide both sides by √3: (2/3)(x + 0.25) + 1/2 = 0

Multiply both sides by 6 to eliminate denominators: 4(x + 0.25) + 3 = 0? 6* (2/3)(x+0.25) = 4(x+0.25), 6*(1/2) = 3, so 4(x+0.25) + 3 = 0

4x + 1 + 3 = 0? 4(x+0.25) = 4x +1, so 4x+1 +3 =0, 4x+4=0, 4x=-4, x=-1

Then from perp bisector of C, y = (√3/3)(-1 + 0.25) + √3/4 = (√3/3)(-0.75) + √3/4 = (√3/3)(-3/4) + √3/4 = -√3/4 + √3/4 = 0

So intersection at (-1,0), which is vertex D.

Similarly, perp bisector of C and perp bisector of E intersect at D (-1,0)

Similarly, perp bisector of C and x=0.5 at B, etc.

But the three perp bisectors form a triangle with vertices at B, F, and D.

B (0.5, √3/2), F (0.5, -√3/2), D (-1,0)

So it's a triangle.

But this is the crease lines.

If we fold along these three lines, we can bring the three vertices to O.

For example, fold along x=0.5, then A is mapped to O.

Fold along the perp bisector of C, which passes through B and D, but it's a line, so when we fold along that line, C is mapped to O.

Similarly for E.

But with three separate folds, the paper is folded, and the vertices are at O.

After folding, the shape might be the three creased lines forming a smaller shape with the points at O.

Perhaps the figure formed is an equilateral triangle.

Let's see the points.

After folding, the three vertices are at O, and the creases are the lines B-D, etc.

The crease lines are the perp bisectors: for A, x=0.5, which is the line from (0.5,0) to (0.5, √3/2) and (0.5, -√3/2), but (0.5,0) is not a vertex, it's the midpoint of A and F or something.

With S=1, A (1,0), F (0.5, -√3/2), O (0,0), the perp bisector of AO is x=0.5, which is a vertical line through the midpoint of A and O.

Similarly, perp bisector of CO is the line from B to D, as we found.

B (0.5, √3/2), D (-1,0), and the line passes through (-0.25, √3/4), etc.

Similarly for E, perp bisector is from F to D, since at x=0.5, y=-√3/2, and at (-1,0).

So the three crease lines are:

1. The line x=0.5, which is the vertical line through the midpoint of A and O, and it passes through B and F.

2. The line from B to D, which is the perp bisector for C.

3. The line from F to D, which is the perp bisector for E.

So these three lines: x=0.5, BD, and FD, but BD and FD are the same as the perp bisectors.

BD is the perp bisector for C, and FD for E.

And x=0.5 for A.

Now, these three lines intersect at B, F, and D.

B, F, D are three vertices.

If we fold along these three lines, we can bring A, C, E to O.

After folding, the paper has creases, and the three points at O.

The resulting figure might be a triangle formed by the creases or the folded paper.

Perhaps the shape is a smaller equilateral triangle.

Let's find the points.

The crease lines form a triangle with vertices at B, D, F.

B, D, F are three vertices of the hexagon.

B (0.5, √3/2), D (-1,0), F (0.5, -√3/2)

Distance between B and F: both at x=0.5, y from -√3/2 to √3/2, distance √3.

B and D: from (0.5, √3/2) to (-1,0), distance: dx=1.5, dy=√3/2, distance sqrt( (1.5)^2 + (√3/2)^2 ) = sqrt( 2.25 + 0.75) = sqrt(3)

Similarly, D and F: from (-1,0) to (0.5, -√3/2), dx=1.5, dy=√3/2, distance sqrt(2.25 + 0.75) = sqrt(3)

So B, D, F form an equilateral triangle with side length √3.

Now, the center O is inside this triangle.

When we fold the three vertices A, C, E to O, we are essentially folding the hexagon along these three lines to bring the vertices to O.

After folding, the paper might be stacked or the shape is this equilateral triangle.

But the hexagon has area, and after folding, the three corners are folded in, so the visible shape could be the triangle BDF.

With the three vertices at O, which is inside the triangle.

O is the centroid of the triangle BDF.

B (0.5, √3/2), D (-1,0), F (0.5, -√3/2), centroid: x = (0.5 -1 + 0.5)/3 = (0)/3 =0, y=( √3/2 + 0 - √3/2)/3=0, so O (0,0), yes.

So after folding, the figure formed is an equilateral triangle with vertices at B, D, F.

And the three marked vertices are at O.

So the shape is an equilateral triangle.

But is that with one fold or multiple folds?

The problem says "folded", which might imply one action, but in this case, it requires three separate folds.

However, in practice, for paper folding, you can make multiple creases with one folding action, but it's a bit ambiguous.

Perhaps there is a single fold that brings the three vertices to the center by folding along a line that is not a straight line, but a curve, but that seems complicated.

Another possibility: in some interpretations, for a regular hexagon, folding vertices to the center might be done by folding along the lines from the center to the vertices, but that doesn't move the vertices.

I think the multiple fold lines approach is more plausible.

Moreover, in the context, the resulting figure is an equilateral triangle.

Perhaps for the hexagon, with three vertices folded to center, the shape is a smaller hexagon or something.

But in this case, with vertices A, C, E folded to O, the remaining points are B, D, F, which form a triangle.

The original hexagon has six vertices; after folding three to O, we have points B, D, F, and O, but O is not a vertex, it's a point.

The boundary after folding: with the creases, the paper is divided, and when folded, the outer shape is the triangle BDF.

And the folded parts are inside or on top.

But the figure formed is the equilateral triangle.

Perhaps it's a regular triangle.

Now, to confirm, the side length of this triangle is distance between B and F, which is | √3/2 - (-√3/2) | = √3, since same x, dy= √3, so distance √3.

Original side S=1, so side √3 times larger? S=1, distance from center to vertex is 1, so in the triangle, vertices at distance: B to O is 1, but B to D is sqrt( (0.5- (-1))^2 + (√3/2 -0)^2 ) = sqrt( (1.5)^2 + (√3/2)^2 ) = sqrt( 2.25 + 0.75 ) = sqrt(3), as before.

Original hexagon side is 1, so the triangle has side √3.

But in terms of area or something.

I think for the purpose, the shape is an equilateral triangle.

Perhaps it's a triangle with the same orientation.

In the diagram, the three marked vertices might be different.

Suppose the three marked vertices are A, B, C or something.

For example, in the user's message, it says "as shown", and from the text, it might be specific.

But since no diagram, I think A, C, E is a reasonable assumption.

Perhaps it's the vertices at 0°, 60°, 120°, but that might not be symmetric for folding to center.

I think with the multiple perp bisectors, we get a triangle.

Moreover, in some math problems, this is known.

So the figure formed is an equilateral triangle.

Therefore, the shape is a triangle.

But to be precise, it's an equilateral triangle.

Perhaps it's a smaller shape.

Another thought: when you fold the three vertices to the center, the creases form a smaller hexagon or something.

In this case, the creases are the three lines: x=0.5, BD, and FD.

BD is from B to D, which is a line, and x=0.5 from B to F, and FD from F to D.

So they form triangle BDF.

No smaller hexagon.

Perhaps the figure formed is the polygon after folding.

I think equilateral triangle is the answer.

So I'll go with that.