Solution:
(Sorry i have zero idea how to make drawings)
Draw a good diagram!
Let $AB \cap DC$, $CD \cap FE$, and $BA \cap EF$ be P, Q, and R, respectively. Let $QR=200, RP=300, PQ=240$. Notice that all smaller triangles formed are all similar to the larger $(200,240,300)$ triangle. Let the side length of the hexagon be x. Triangle $\triangle BCP \sim \triangle RQP$, so $\frac{BC}{BP} =\frac{x}{BP} =\frac{200}{300} \implies BP=\frac{3x}{2}$. Triangle $\triangle AFR \sim \triangle PQR$, so $\frac{AF}{AR}=\frac{x}{AR} = \frac{240}{300} \implies AR=\frac{5x}{4}$. We know $RA+AB+BP=300$, so $\frac{5}{4}x + x + \frac{3}{2}x = 300$. Solving, we get $x=\boxed{080}$. -westwoodmonster
Draw an accurate diagram using the allowed compass and ruler: Draw a to-scale diagram of the $(200,240,300)$ triangle (e.g. 10cm-12cm-15cm). Because of the nature of these lengths and the integer answer needed, it can be assumed that the side length of the hexagon will be divisible by 10. Therefore, a trial-and-error method can be set up, wherein line segments of length $n\cdot 10$, scaled to the same scale as the triangle, can be drawn to represent the sides of the hexagon. For instance, side $BC$ would be drawn parallel to the side of the triangle of length 300, and would have endpoints on the other sides of the triangle. Using this method, it would be evident that line segments of length 80 units, scaled proportionally (4cm using the scale above), would create a perfect equilateral hexagon when drawn parallel to the sides of the triangle. $x=\boxed{080}$. - lackolith

Answer: 080

<｜begin▁of▁sentence｜><｜User｜>Let ABCDEF be a convex equilateral hexagon in which all pairs of opposite sides are parallel. The triangle whose sides are extensions of segments AB, CD, and EF has side lengths 200, 240, and 300. Find the side length of the hexagon.<｜Assistant｜><think>

Okay, so I have this problem about a convex equilateral hexagon where all pairs of opposite sides are parallel. The triangle formed by extending the sides AB, CD, and EF has side lengths 200, 240, and 300. I need to find the side length of the hexagon. Hmm, let me try to visualize this first.

First, an equilateral hexagon means all its sides are equal in length, right? But it's not necessarily regular because the angles might not be equal. Also, the fact that all opposite sides are parallel suggests some symmetry. Maybe it's similar to a regular hexagon but scaled differently in some directions?

The problem mentions a triangle formed by extending AB, CD, and EF. Since opposite sides are parallel, extending these sides should meet at some points to form a triangle. The sides of this triangle are given as 200, 240, and 300. I need to relate these lengths to the side length of the hexagon.

Let me recall that in such a hexagon with opposite sides parallel, the distances between the opposite sides might be related to the triangle's sides. Maybe the triangle is formed by the intersections of these extended sides, and the sides of the triangle correspond to the sums or differences of the hexagon's sides and some other distances?

Wait, perhaps I can model this using vectors or coordinate geometry. Let me try setting up coordinates. Let me assume the hexagon is labeled in order: A, B, C, D, E, F, with each pair of opposite sides parallel. Since it's convex, all the internal angles are less than 180 degrees.

Let me denote the side length of the hexagon as 's'. Since all sides are equal and opposite sides are parallel, the hexagon can be thought of as a "stretched" regular hexagon. In a regular hexagon, opposite sides are parallel and the distance between them is 2 times the short apothem. But here, since it's equilateral but not necessarily regular, the distances between opposite sides might vary.

Alternatively, maybe I can use the concept of similar triangles here. The triangle formed by extending AB, CD, and EF – let me denote the intersection points of these extensions as forming triangle PQR, where PQ = 200, QR = 240, and RP = 300. Wait, but I need to figure out how the sides of the triangle relate to the hexagon's sides.

Alternatively, perhaps the triangle's sides are proportional to the sides of the hexagon. Since all sides of the hexagon are equal, maybe the triangle's sides are multiples of 's'? But the triangle sides are 200, 240, 300, which are all different, so that can't be directly proportional. Hmm, so maybe there's a different relationship.

Alternatively, maybe the triangle is formed such that each side of the triangle is the sum of two sides of the hexagon? But since the hexagon is equilateral, that would mean each side of the triangle is 2s, but again, the triangle sides are different. So that's not it.

Wait, perhaps the triangle is formed by the extensions of AB, CD, and EF. Let me think: if I extend AB, CD, and EF until they meet, the triangle formed by these three lines will have sides of lengths 200, 240, 300. The key here is that the sides of the triangle are the lengths between the intersection points of these extended sides.

Let me try to draw this mentally. Let's say we have the hexagon ABCDEF, with AB, CD, EF extended until they meet. Since opposite sides are parallel, AB is parallel to DE, BC parallel to EF, and CD parallel to FA. Wait, actually, in a hexagon with opposite sides parallel, each pair of opposite sides are parallel. So AB || DE, BC || EF, CD || FA.

Therefore, if I extend AB and DE, they are already parallel, so they won't meet. Wait, but the problem says the triangle is formed by extensions of AB, CD, and EF. So maybe extending AB, CD, and EF until they meet each other?

Let me think step by step. Let me denote the three lines: the extension of AB, the extension of CD, and the extension of EF. Each pair of these lines should intersect at a vertex of the triangle. Since the hexagon is convex, these extensions should form a triangle outside the hexagon.

Let me try to assign variables. Let me suppose that the side length of the hexagon is 's'. Since all sides are equal and opposite sides are parallel, the hexagon can be considered as a centrally symmetric hexagon? Wait, not necessarily, but with opposite sides parallel. Maybe it's affine-regular?

Alternatively, perhaps using vectors. Let me place the hexagon in the coordinate system such that point A is at the origin, and side AB is along the x-axis. Let me denote vectors for each side.

Let me denote the sides as vectors. Since it's a closed polygon, the sum of the vectors is zero. Let me denote each side as vectors **AB**, **BC**, **CD**, **DE**, **EF**, **FA**. Since it's equilateral, each of these vectors has the same magnitude 's', but different directions.

Also, opposite sides are parallel, so **AB** is parallel to **DE**, **BC** parallel to **EF**, and **CD** parallel to **FA**. Therefore, the vectors can be written as:

**AB** = **v₁**

**BC** = **v₂**

**CD** = **v₃**

**DE** = **v₁** (since parallel and same direction? Wait, but direction might be opposite. Hmm, need to be careful here.)

Wait, actually, if AB is parallel to DE, but since the hexagon is convex, the direction of DE would be opposite to AB? Wait, not necessarily. Let me think.

Suppose the hexagon is labeled in order, so moving from A to B to C to D to E to F and back to A. If AB is parallel to DE, then the direction of DE should be the same as AB if the hexagon is regular, but in a general convex hexagon with opposite sides parallel, they might be in the same or opposite directions? Hmm, actually, in a regular hexagon, opposite sides are parallel and in the same direction. Wait, no, in a regular hexagon, opposite sides are parallel but pointing in opposite directions. Wait, actually, in a regular hexagon, each pair of opposite sides are parallel and separated by 180 degrees, so their direction vectors are negatives of each other.

Wait, perhaps in this problem, the opposite sides are parallel but not necessarily in the same direction. So, for example, vector DE would be equal to vector AB but possibly in the opposite direction? Hmm, maybe.

Alternatively, perhaps all opposite sides are both parallel and equal in direction? The problem just says "all pairs of opposite sides are parallel," so direction could be same or opposite. Hmm, this is a bit ambiguous. Maybe I need to assume they are parallel but could be in any direction as long as they are parallel.

Alternatively, perhaps the hexagon can be thought of as a translation of opposite sides. Hmm, maybe coordinate system approach is better.

Let me try to set coordinates. Let me place point A at (0,0). Let me let AB be along the x-axis, so point B is at (s, 0), since the side length is s.

Now, since opposite sides are parallel, DE is parallel to AB. Since AB is along the x-axis, DE must also be horizontal. Let me denote point D as some point, but since the hexagon is convex, the order is important.

Wait, perhaps I need to define all the points step by step. Let me try:

Let me denote the six vertices as A, B, C, D, E, F in order. Since it's a convex hexagon, all the turns are to the same side.

Starting at A(0,0). Moving to B(s,0). Now, from B, we move to C. The direction from B to C is some vector. Since BC is opposite to EF, so BC is parallel to EF. Similarly, CD is opposite to FA, so CD is parallel to FA. And DE is opposite to AB, so DE is parallel to AB, which is along the x-axis.

Therefore, DE must be horizontal. Since DE is opposite to AB, which is along the x-axis from A(0,0) to B(s,0), then DE should also be a horizontal vector. Let me suppose that DE is from point D to E, so the vector DE should be parallel to AB, so it's also along the x-axis. Since the hexagon is convex, the direction of DE should be the same as AB? Or opposite?

Wait, in a regular hexagon, opposite sides are parallel but in opposite directions. For example, if AB is to the right, then DE would be to the left. But since the hexagon is convex, maybe DE is to the right as well? Hmm, not sure. Let me think.

Alternatively, perhaps the vectors for opposite sides are equal? Wait, but in a regular hexagon, opposite sides are in opposite directions. Hmm, maybe in this problem, since it's convex, the opposite sides are parallel but could be in the same direction? Hmm, perhaps.

Alternatively, maybe I can assign coordinates step by step with variables.

Let me try:

Let me set point A at (0,0). Let me let AB be along the x-axis, so point B is (s, 0).

Now, from point B, we go to point C. Let me denote the vector BC as (a, b). Since the side length is s, the length of BC must be s, so sqrt(a² + b²) = s.

Then, from point C, we go to point D. The vector CD must be parallel to FA. Since FA is the vector from F to A, which is (-x, -y) if F is at (x,y). Hmm, this might get complicated. Maybe another approach.

Alternatively, since opposite sides are parallel, the vectors AB, CD, EF are all translates of each other? Wait, no, they are opposite sides. Wait, AB is opposite to DE, BC opposite to EF, CD opposite to FA.

Therefore, vector DE is parallel to AB. Since AB is along the x-axis, DE must also be horizontal. Let me suppose DE is in the same direction as AB (to the right). Then, since DE is a side of length s, the coordinates of D and E can be related accordingly.

Wait, perhaps it's better to use vectors for each side.

Let me denote the sides as vectors:

Let me call the sides as vectors:

**AB** = **v₁** = (s, 0) since it's along the x-axis.

**BC** = **v₂** = (a, b), with |**v₂**| = s.

**CD** = **v₃**, which must be parallel to **FA**. Since **FA** is the vector from F to A, which is -(**AB** + **BC** + **CD** + **DE** + **EF**). Hmm, this might get too convoluted.

Alternatively, since opposite sides are parallel, **DE** is parallel to **AB**, so **DE** = k₁**v₁**, where k₁ is a scalar. But since DE has length s, then |k₁**v₁**| = s, so |k₁|*s = s ⇒ |k₁|=1. Therefore, **DE** is either **v₁** or -**v₁**. Since the hexagon is convex, the direction of DE must be such that the polygon doesn't intersect itself. Since AB is from A(0,0) to B(s,0), then DE should be in the opposite direction? Wait, maybe.

Wait, if **DE** is in the same direction as **AB**, then DE would be to the right, but since the hexagon is convex, the points D and E must be placed such that the overall shape is convex. Maybe it's better to suppose that **DE** = -**AB**, so pointing to the left. Let me try that.

So **DE** = -**v₁** = (-s, 0). Then, since DE is from point D to E, so E = D + **DE** = D + (-s, 0). So if I can find coordinates for D, then E is determined.

Continuing, **EF** is opposite to **BC**, so **EF** is parallel to **BC**. Let me denote **EF** = k₂**v₂**, with |**EF**| = s. Since |**v₂**| = s, then |k₂| must be 1. So **EF** is either **v₂** or -**v₂**. Again, convexity might determine the direction.

Similarly, **FA** is opposite to **CD**, so **FA** is parallel to **CD**. Let me denote **FA** = k₃**v₃**, with |**FA**| = s, so again |k₃|=1.

This is getting a bit abstract. Maybe I can write equations based on the polygon closure.

The sum of all vectors must be zero:

**AB** + **BC** + **CD** + **DE** + **EF** + **FA** = 0

Substituting the expressions in terms of **v₁**, **v₂**, **v₃**, etc.:

**v₁** + **v₂** + **v₃** + (-**v₁**) + (k₂**v₂**) + (k₃**v₃**) = 0

Simplify:

(**v₁** - **v₁**) + (**v₂** + k₂**v₂**) + (**v₃** + k₃**v₃**) = 0

So:

(1 + k₂)**v₂** + (1 + k₃)**v₃** = 0

Hmm, so this equation must hold. Since **v₂** and **v₃** are vectors in the plane, unless they are colinear, the coefficients must be zero. But since the hexagon is convex, **v₂** and **v₃** are not colinear. Therefore, the coefficients must be zero:

1 + k₂ = 0 ⇒ k₂ = -1

1 + k₃ = 0 ⇒ k₃ = -1

Therefore, **EF** = -**v₂**, and **FA** = -**v₃**

Therefore, the vectors are:

**DE** = -**v₁**

**EF** = -**v₂**

**FA** = -**v₃**

So the sides are paired as **AB** and **DE** = -**AB**, **BC** and **EF** = -**BC**, **CD** and **FA** = -**CD**

Therefore, the hexagon is centrally symmetric? Wait, because each pair of opposite sides are not only parallel but also equal in magnitude and opposite in direction. That would make the hexagon centrally symmetric, right? Because if you take the midpoint between A and D, B and E, C and F, they should all coincide. So the center of symmetry is the midpoint of AD, BE, CF.

Therefore, this hexagon is a centrally symmetric equilateral hexagon with opposite sides parallel. That simplifies things.

Therefore, in such a hexagon, the opposite sides are equal and opposite vectors, so the hexagon can be thought of as a parallelogram "stretched" into a hexagon. Wait, but how exactly?

Alternatively, since it's centrally symmetric, the coordinates can be defined such that the center is at the origin, but maybe it's easier to place point A at (0,0). Wait, but with central symmetry, if the center is point O, then for any point P, there is a point P' such that O is the midpoint. So if A is at some point, D is the opposite vertex, so the center is the midpoint of AD, similarly for the others.

Alternatively, perhaps I can model the hexagon as follows:

Let me denote the center of symmetry as point O. Then, vectors from O to each vertex are such that opposite vertices are negatives. For example, if vector OA = **a**, then vector OD = -**a**, similarly for the others.

But since the hexagon is equilateral, all sides have the same length. The sides are between consecutive vertices, so the vectors between vertices must all have the same length.

Alternatively, maybe using complex numbers could help here. Let me think of the vertices as complex numbers. Let me denote the center as the origin for simplicity. Then, the vertices can be represented as complex numbers A, B, C, D, E, F such that D = -A, E = -B, F = -C. Because of central symmetry.

Therefore, the hexagon is A, B, C, -A, -B, -C, back to A. Then, the sides are:

AB: from A to B

BC: from B to C

CD: from C to -A

DE: from -A to -B

EF: from -B to -C

FA: from -C to A

Wait, but then CD is from C to -A, so the vector CD is (-A - C). Similarly, FA is from -C to A, which is (A + C). Hmm, but in this case, the sides must all have the same length. Let me check the length of AB: |B - A|, BC: |C - B|, CD: |-A - C| = | - (A + C)| = |A + C|, DE: |-B + A| = |A - B|, EF: |-C + B| = |B - C|, FA: |A + C|.

Wait, so for all sides to be equal:

|B - A| = |C - B| = |A + C| = |A - B| = |B - C| = |A + C|

Hmm, this seems restrictive. Let me see:

First, |B - A| = |A - B|, which is the same.

Similarly, |C - B| = |B - C|.

So the conditions are:

|B - A| = |C - B| = |A + C|

So let me set variables in complex plane. Let me let A and B be complex numbers. Let me suppose A is a real number for simplicity, since we can rotate the coordinate system. Let me set A = a (real), and B = b + ci, where b, c are real numbers. Then, C is another complex number, say C = d + ei.

But this might get too involved. Alternatively, maybe choosing coordinates such that the center is at the origin, and A is at (p, q), then D is at (-p, -q), etc. Hmm, perhaps better to use vectors.

Alternatively, since the problem mentions a triangle formed by extending AB, CD, and EF. Let me think about how those lines form a triangle.

Given the central symmetry, perhaps the triangle is also related symmetrically. Let me try to find equations for the lines AB, CD, EF and find their intersections.

Let me first define coordinates with the center at the origin. Let me denote the center as O(0,0). Then, the vertices are A, B, C, -A, -B, -C in order.

So the hexagon is A, B, C, -A, -B, -C, back to A.

Therefore, the sides are:

AB: from A to B

BC: from B to C

CD: from C to -A

DE: from -A to -B

EF: from -B to -C

FA: from -C to A

Now, let's find the equations of the lines AB, CD, and EF.

First, line AB goes from A to B. Let me denote coordinates:

Let me assign coordinates to points A, B, C. Let me suppose point A is (x_A, y_A), B is (x_B, y_B), and C is (x_C, y_C). Then, the opposite points are -A = (-x_A, -y_A), etc.

The line AB can be parametrized as A + t(B - A), where t is a real number.

Similarly, line CD goes from C to -A, so its direction vector is (-A - C). The parametric equation is C + s(-A - C).

Line EF goes from -B to -C, so direction vector is (-C + B). Wait, from -B to -C is (-C - (-B)) = ( -C + B). So parametric equation is -B + r(-C + B).

We need to find the intersection points of these three lines: AB, CD, EF.

Let me denote the three lines:

Line AB: P(t) = A + t(B - A)

Line CD: Q(s) = C + s(-A - C)

Line EF: R(r) = -B + r(B - C)

We need to find the intersection points of AB and CD, CD and EF, EF and AB, which will form the triangle.

Wait, actually, the triangle is formed by the three lines AB, CD, EF. So each pair of lines intersect at a vertex of the triangle.

So first, find intersection of AB and CD: let's set P(t) = Q(s):

A + t(B - A) = C + s(-A - C)

Similarly, intersection of CD and EF: set Q(s) = R(r):

C + s(-A - C) = -B + r(B - C)

Intersection of EF and AB: set R(r) = P(t):

-B + r(B - C) = A + t(B - A)

These three equations will give the three vertices of the triangle. The sides of the triangle are the distances between these intersection points, which are given as 200, 240, 300. The goal is to find the side length s of the hexagon, which is |B - A| = |C - B| = | -A - C| etc.

This seems quite involved. Maybe there's a better approach using properties of centrally symmetric hexagons.

Alternatively, since the hexagon is centrally symmetric and equilateral, perhaps the triangle formed is similar to the triangle formed by the vectors of the sides?

Alternatively, maybe using the concept of similar triangles or homothety.

Alternatively, since the triangle sides are 200, 240, 300, which is a scaled 20-24-30 triangle (divided by 10), which is a multiple of a 10-12-15 triangle, but actually 200-240-300 is a multiple of 20-24-30, which is a right triangle? Wait, 200² + 240² = 40000 + 57600 = 97600, while 300² is 90000. Not a right triangle. Wait, 200:240:300 simplifies by dividing by 20: 10:12:15. Let me check 10² +12² = 100 + 144 = 244, which is not 225. So not a right triangle. Hmm.

Alternatively, perhaps the triangle's sides are proportional to the distances between the opposite sides of the hexagon?

Wait, in a centrally symmetric hexagon, the distance between two opposite sides can be related to the side length and the angles between the sides.

Alternatively, maybe the triangle formed by the extensions is the outer triangle that circumscribes the hexagon, and the sides of the triangle are each equal to twice the distance between two opposite sides of the hexagon? Not sure.

Alternatively, perhaps the side length of the hexagon can be found using the formula related to the triangle sides. Since the triangle sides are 200,240,300, maybe the hexagon's side is related through some ratio.

Alternatively, maybe the triangle is the outer Napoleon triangle or something similar, but I'm not sure.

Alternatively, let me think of the hexagon as a combination of triangles. Since it's centrally symmetric, maybe splitting it into triangles from the center?

Alternatively, perhaps using the area. The area of the triangle is related to the area of the hexagon. But without more info, maybe not.

Alternatively, let me consider that in such a centrally symmetric hexagon, the triangle formed by those three extended sides would have sides proportional to the sums of the vectors.

Wait, let me think in terms of vectors again. Let me suppose the center is at the origin, so the vertices are A, B, C, -A, -B, -C.

The lines AB, CD, EF can be expressed parametrically as before.

Let me try to compute the intersection points.

First, let me denote vectors:

Let me let vector A = **a**, B = **b**, C = **c**. All vectors are in the plane.

The line AB can be parametrized as **a** + t(**b** - **a**).

The line CD goes from C to -A, so direction vector is -**a** - **c**, so parametrized as **c** + s(-**a** - **c**).

The line EF goes from -B to -C, direction vector is -**c** + **b**, so parametrized as -**b** + r(**b** - **c**).

Now, to find the intersection of AB and CD:

Set **a** + t(**b** - **a**) = **c** + s(-**a** - **c**).

Rearranged:

**a** - t**a** + t**b** = **c** - s**c** - s**a**

Grouping terms:

(1 - t)**a** + t**b** = (1 - s)**c** - s**a**

Bring all terms to left:

(1 - t + s)**a** + t**b** - (1 - s)**c** = 0

This is a vector equation. Since **a**, **b**, **c** are vectors in the plane, unless they are colinear, the coefficients must be zero. But since the hexagon is convex and equilateral, they are not colinear. Therefore:

Coefficient of **a**: (1 - t + s) = 0 ⇒ 1 - t + s = 0

Coefficient of **b**: t = 0

Coefficient of **c**: -(1 - s) = 0 ⇒ 1 - s = 0 ⇒ s = 1

Wait, but if t = 0 from the coefficient of **b**, then from the first equation:

1 - 0 + s = 0 ⇒ s = -1

But from the third equation, s must be 1. Contradiction. Hmm, that suggests that my assumption is wrong, or perhaps the vectors are linearly dependent?

Hmm, maybe I made a mistake in setting up the equations. Let me check again.

Wait, the equation is:

(1 - t)**a** + t**b** = (1 - s)**c** - s**a**

So moving all terms to the left:

(1 - t + s)**a** + t**b** - (1 - s)**c** = 0

Yes, that's correct. So for this to hold, since **a**, **b**, **c** are linearly independent (assuming the hexagon is non-degenerate), their coefficients must be zero. Therefore:

1 - t + s = 0,

t = 0,

-(1 - s) = 0 ⇒ s = 1.

But substituting t = 0 gives s = -1 from first equation, conflicting with s =1. Therefore, this implies that the vectors are linearly dependent, so the equation can be satisfied with some relation between **a**, **b**, **c**.

Hmm, this suggests that my approach might not be the best. Maybe I need to consider specific coordinates.

Let me try to assign coordinates with some symmetry.

Let me suppose that the center of the hexagon is at the origin. Let me set point A at (a, 0). Due to central symmetry, point D is (-a, 0). Since the hexagon is equilateral, the length AB = BC = CD = DE = EF = FA = s.

Starting from point A(a, 0), moving to B. Let me denote point B as (a + s*cosθ, s*sinθ), where θ is the angle between AB and the x-axis. Since AB is a side of length s.

Then, point C can be found by moving from B in some direction, but since BC must also have length s and the hexagon is convex. However, due to central symmetry, point -B must be the point opposite to B, which is point E. Wait, actually, in central symmetry, the opposite of B is -B, so point E is -B, and opposite of C is -C, which is point F.

Wait, let me clarify the order of the hexagon. The vertices are A, B, C, D, E, F. Since it's centrally symmetric, D is opposite A, E opposite B, F opposite C. So:

D = -A,

E = -B,

F = -C.

Therefore, the coordinates are:

A(a, 0),

B(b_x, b_y),

C(c_x, c_y),

D(-a, 0),

E(-b_x, -b_y),

F(-c_x, -c_y).

Now, the sides:

AB: from A(a,0) to B(b_x, b_y), length s.

BC: from B to C, length s.

CD: from C to D(-a,0), length s.

DE: from D to E(-b_x, -b_y), length s.

EF: from E to F(-c_x, -c_y), length s.

FA: from F to A, length s.

Therefore, all these vectors must have length s.

Let me write equations for the lengths:

1. |B - A|² = (b_x - a)^2 + (b_y)^2 = s²,

2. |C - B|² = (c_x - b_x)^2 + (c_y - b_y)^2 = s²,

3. |D - C|² = (-a - c_x)^2 + (0 - c_y)^2 = s²,

4. |E - D|² = (-b_x + a)^2 + (-b_y - 0)^2 = s²,

5. |F - E|² = (-c_x + b_x)^2 + (-c_y + b_y)^2 = s²,

6. |A - F|² = (a + c_x)^2 + (0 + c_y)^2 = s².

Looking at equations 1 and 4:

Equation 1: (b_x - a)^2 + b_y² = s²,

Equation 4: (-b_x + a)^2 + (-b_y)^2 = (a - b_x)^2 + b_y² = same as equation 1. So they are the same equation.

Similarly, equations 2 and 5:

Equation 2: (c_x - b_x)^2 + (c_y - b_y)^2 = s²,

Equation 5: (-c_x + b_x)^2 + (-c_y + b_y)^2 = (b_x - c_x)^2 + (b_y - c_y)^2 = same as equation 2.

Equations 3 and 6:

Equation 3: (-a - c_x)^2 + (-c_y)^2 = ( - (a + c_x))² + c_y² = (a + c_x)^2 + c_y²,

Equation 6: (a + c_x)^2 + (0 + c_y)^2 = same as equation 3. So equations 3 and 6 are the same.

Therefore, we have three unique equations:

1. (b_x - a)^2 + b_y² = s²,

2. (c_x - b_x)^2 + (c_y - b_y)^2 = s²,

3. (a + c_x)^2 + c_y² = s².

So three equations with variables a, b_x, b_y, c_x, c_y, and s. But we need more relations.

Additionally, the triangle formed by lines AB, CD, EF has sides 200,240,300. Let me find the equations of these lines and compute their intersections.

First, let's find the equations of lines AB, CD, EF.

Line AB: passes through A(a,0) and B(b_x, b_y). The parametric equations can be written as:

x = a + t(b_x - a),

y = 0 + t(b_y - 0) = t b_y,

for t ∈ ℝ.

Line CD: passes through C(c_x, c_y) and D(-a, 0). The direction vector is D - C = (-a - c_x, -c_y). Parametric equations:

x = c_x + s(-a - c_x - c_x)? Wait, direction vector is (-a - c_x, -c_y). So:

x = c_x + s*(-a - c_x),

y = c_y + s*(-c_y),

where s ∈ ℝ.

Line EF: passes through E(-b_x, -b_y) and F(-c_x, -c_y). The direction vector is F - E = (-c_x + b_x, -c_y + b_y). Parametric equations:

x = -b_x + r*(-c_x + b_x),

y = -b_y + r*(-c_y + b_y),

where r ∈ ℝ.

Now, we need to find the intersection points of these three lines.

First, find intersection of AB and CD.

Set the parametric equations equal:

For AB: (x, y) = (a + t(b_x - a), t b_y),

For CD: (x, y) = (c_x - s(a + c_x), c_y(1 - s)).

Set equal:

a + t(b_x - a) = c_x - s(a + c_x),

t b_y = c_y(1 - s).

We have two equations with variables t and s.

Similarly, we can solve for t and s.

From the second equation: s = 1 - (t b_y)/c_y, assuming c_y ≠ 0.

Substitute into first equation:

a + t(b_x - a) = c_x - [1 - (t b_y)/c_y]*(a + c_x)

Expand the right side:

c_x - (a + c_x) + (t b_y / c_y)(a + c_x)

= c_x - a - c_x + (t b_y / c_y)(a + c_x)

= -a + (t b_y (a + c_x))/c_y

Therefore, left side: a + t(b_x - a) = right side: -a + (t b_y (a + c_x))/c_y

Bring all terms to left:

a + t(b_x - a) + a - (t b_y (a + c_x))/c_y = 0

2a + t [ (b_x - a) - (b_y (a + c_x))/c_y ] = 0

Solving for t:

t [ (b_x - a) - (b_y (a + c_x))/c_y ] = -2a

This is getting very complicated. Maybe I need to make some assumptions to simplify.

Let me assume that the hexagon is such that point A is along the x-axis, and the center is at the origin. Let me also suppose that the vectors are arranged such that the triangle formed is the outer triangle with sides 200,240,300. Maybe the triangle is the outer triangle circumscribing the hexagon, and the sides of the triangle are each twice the distance from the center to the sides of the hexagon? Not sure.

Alternatively, perhaps the side length s of the hexagon is related to the sides of the triangle through the formula s = (200 + 240 + 300)/something? But that might not be precise.

Alternatively, maybe using the concept of similar triangles. If the triangle sides are 200,240,300, perhaps the hexagon's side is the harmonic mean or something? Not sure.

Alternatively, perhaps the triangle is formed such that each side is the sum of two sides of the hexagon projected in some direction. Since the hexagon is equilateral, maybe each side of the triangle is 2s times some trigonometric function. Hmm.

Alternatively, let me think of the problem in terms of the outer triangle. The triangle is formed by extending AB, CD, EF. Since the hexagon is convex and equilateral with opposite sides parallel, the triangle is larger than the hexagon, and the sides of the triangle are each composed of two sides of the hexagon? Wait, but the triangle sides are 200,240,300, which are all different, so maybe each side corresponds to a different pair of sides of the hexagon?

Alternatively, perhaps the triangle's sides are each equal to 2 times the side length of the hexagon multiplied by some factor based on angles. Since in a regular hexagon, extending the sides would form a larger regular hexagon, but here it's a triangle.

Alternatively, maybe the triangle is the result of the three extended sides forming a triangle where each side is the sum of the lengths from the hexagon's sides. But since all sides of the hexagon are equal, maybe each side of the triangle is 2s, but scaled by some factor. However, the triangle sides are different, so that can't be.

Alternatively, perhaps the triangle is formed such that each side is the distance between two non-adjacent vertices of the hexagon? Not sure.

Alternatively, maybe using homothety. If the hexagon is inscribed in the triangle, then the triangle is a homothetic image of the hexagon. The homothety ratio could relate the sides.

Alternatively, let me think of the problem in terms of the Routh's theorem. Routh's theorem gives the ratio of areas for a triangle formed by lines extended from the sides of another triangle. But this might not directly apply here.

Alternatively, since the hexagon is centrally symmetric, the triangle formed by AB, CD, EF might also be related through central symmetry. Wait, but a triangle can't be centrally symmetric unless it's degenerate. Hmm.

Alternatively, perhaps the triangle's sides are each twice the distance between opposite sides of the hexagon. The distance between two parallel sides in a centrally symmetric hexagon would be twice the distance from the center to a side. If the triangle's sides are proportional to these distances, then maybe the side length s can be found via some relation.

Let me denote the distances from the center to each pair of opposite sides as d1, d2, d3. Then the distance between the opposite sides would be 2d1, 2d2, 2d3. Perhaps the sides of the triangle are proportional to these distances?

Alternatively, in a centrally symmetric hexagon, the triangle formed by those three lines might have sides proportional to 2d1, 2d2, 2d3. But how exactly?

Alternatively, the sides of the triangle could be equal to 2d1 + 2d2 + 2d3? Not sure.

Alternatively, perhaps the triangle's sides are each equal to the sum of two of the distances between opposite sides. For example, if the triangle is formed by extending AB, CD, EF, then each side of the triangle is the sum of the distances between the opposite sides?

Alternatively, maybe the sides of the triangle are each equal to twice the side length of the hexagon divided by the sine of some angle. Hmm.

Alternatively, let me think of the problem in terms of the affine regular hexagon. An affine regular hexagon is a hexagon that can be obtained by applying an affine transformation to a regular hexagon. Since affine transformations preserve ratios and parallelism, maybe this problem's hexagon is an affine image of a regular hexagon, and the triangle is the affine image of some triangle related to the regular hexagon.

In a regular hexagon, extending every other side forms a larger regular hexagon, but here it's a triangle. Hmm, perhaps in the affine transformation, the regular hexagon is scaled differently in different directions, turning the extended sides into a triangle.

Alternatively, in a regular hexagon, the triangle formed by extending three alternate sides (like AB, CD, EF) would form an equilateral triangle, but in this problem it's a 200-240-300 triangle, which is scalene. So the affine transformation must have scaled the regular hexagon into a non-regular one, turning the equilateral triangle into a scalene one.

The key idea is that affine transformations preserve ratios along lines and parallelism, so the side length of the hexagon can be found by scaling the regular hexagon's side length according to the triangle's sides.

In a regular hexagon with side length s, the triangle formed by extending AB, CD, EF would be an equilateral triangle with side length 3s. Wait, let me check.

Wait, in a regular hexagon, each side is length s. If you extend the sides AB, CD, EF, which are every other side, then the distance between the extended sides would form a larger equilateral triangle. The side length of that triangle would be 3s. Because each side of the triangle is formed by three sides of the hexagon. Wait, actually, in a regular hexagon, extending those sides would form a triangle where each side is twice the side length of the hexagon. Wait, maybe not.

Alternatively, let me consider coordinates for a regular hexagon centered at the origin with a vertex at (s, 0). The vertices are at angles 0°, 60°, 120°, 180°, 240°, 300°. Let me compute the equations of lines AB, CD, EF.

Let me take a regular hexagon with side length s. Let me label the vertices as A, B, C, D, E, F in order.

Coordinates:

A: (s, 0)

B: (s/2, (s√3)/2)

C: (-s/2, (s√3)/2)

D: (-s, 0)

E: (-s/2, -(s√3)/2)

F: (s/2, -(s√3)/2)

Now, lines AB, CD, EF.

Line AB: from A(s,0) to B(s/2, (s√3)/2). The slope is [( (s√3)/2 - 0 ) / (s/2 - s)] = [ (s√3)/2 ] / (-s/2 ) = -√3.

Equation: y - 0 = -√3(x - s)

Similarly, line CD: from C(-s/2, (s√3)/2) to D(-s, 0). The slope is [0 - (s√3)/2] / [ -s + s/2 ] = [ - (s√3)/2 ] / ( -s/2 ) = √3.

Equation: y - (s√3)/2 = √3(x + s/2 )

Line EF: from E(-s/2, - (s√3)/2) to F(s/2, - (s√3)/2). This is a horizontal line at y = - (s√3)/2.

Wait, so line EF is horizontal. Let me find where these lines intersect.

First, intersection of AB and CD:

AB: y = -√3 x + √3 s

CD: y = √3 x + √3*( -s/2 + s/2 )? Wait, let me re-derive CD's equation properly.

Wait, point C is (-s/2, (s√3)/2), and point D is (-s, 0). The slope is √3 as above.

Equation using point C:

y - (s√3)/2 = √3(x + s/2 )

So y = √3 x + (√3)(s/2) + (s√3)/2 = √3 x + s√3

Wait, but when x = -s, y should be 0:

Plug x = -s: y = √3*(-s) + s√3 = 0. Correct.

So CD's equation is y = √3 x + s√3.

AB's equation is y = -√3 x + √3 s.

Set equal:

-√3 x + √3 s = √3 x + s√3

Bring terms together:

-√3 x - √3 x = s√3 - √3 s ⇒ -2√3 x = 0 ⇒ x = 0

Then y = -√3*0 + √3 s = √3 s.

So intersection point is (0, √3 s).

Intersection of CD and EF:

Line CD: y = √3 x + s√3

Line EF is horizontal at y = - (s√3)/2.

Set equal:

√3 x + s√3 = - (s√3)/2 ⇒ √3 x = - (s√3)/2 - s√3 = - (3s√3)/2 ⇒ x = - (3s/2 )

Thus, intersection point is (-3s/2, - (s√3)/2 )

Intersection of EF and AB:

Line AB: y = -√3 x + √3 s

Line EF: y = - (s√3)/2

Set equal:

-√3 x + √3 s = - (s√3)/2 ⇒ -√3 x = - (s√3)/2 - √3 s ⇒ -√3 x = - (3s√3)/2 ⇒ x = (3s)/2

Thus, intersection point is ( 3s/2, - (s√3)/2 )

Therefore, the three vertices of the triangle are:

1. (0, √3 s )

2. (-3s/2, - (s√3)/2 )

3. ( 3s/2, - (s√3)/2 )

Now, let's compute the lengths of the sides of this triangle.

First, between (0, √3 s ) and (-3s/2, - (s√3)/2 ):

Distance squared:

( (-3s/2 - 0 )^2 + ( - (s√3)/2 - √3 s )^2 )

= ( (9s²/4) + ( - (s√3)/2 - 2s√3/2 )^2 )

= (9s²/4 + ( - (3s√3)/2 )^2 )

= 9s²/4 + (27s²/4 ) = 36s²/4 = 9s² ⇒ distance is 3s.

Second, between (-3s/2, - (s√3)/2 ) and (3s/2, - (s√3)/2 ):

Distance squared:

( (3s/2 + 3s/2 )^2 + ( - (s√3)/2 + (s√3)/2 )^2 )

= ( (3s)^2 + 0 ) = 9s² ⇒ distance 3s.

Third, between (3s/2, - (s√3)/2 ) and (0, √3 s ):

Same as the first side, distance 3s.

So in a regular hexagon, the triangle formed is an equilateral triangle with side length 3s. Therefore, if in this problem the triangle has sides 200,240,300, which is a scalene triangle, the hexagon is an affine transformation of the regular hexagon.

An affine transformation can scale differently along different axes, turning the equilateral triangle into a scalene one. The key is that affine transformations preserve ratios along lines and parallelism, so the side length of the hexagon can be found by scaling appropriately.

Let me denote the affine transformation matrix as:

[ a  b ]

[ c  d ]

Applied to the regular hexagon, transforming the equilateral triangle with sides 3s into the given triangle with sides 200,240,300.

However, affine transformations can complicate the relationship. Alternatively, since the problem's hexagon is equilateral and the triangle sides are given, perhaps the side length s of the hexagon is related by the formula 3s = (200 + 240 + 300)/3? No, that would be 280, but that's probably not correct.

Alternatively, since in the regular case, the triangle sides are 3s, so s = (triangle side)/3. But here, the triangle is not regular, so maybe the hexagon's side is related to the triangle's sides through some harmonic mean or other relation.

Alternatively, since affine transformations preserve the ratio of areas and lengths along directions, perhaps the side length s can be found by considering the triangle sides as scaled versions of the regular triangle's sides.

Suppose that the affine transformation scales the x-axis by a factor of k_x and y-axis by k_y. Then, the equilateral triangle with sides 3s would become a triangle with sides scaled by sqrt( (k_x)^2 + (k_y)^2 ) in each direction, but since the triangle sides are along different directions, the scaling factors would vary per side.

Alternatively, perhaps the triangle sides correspond to the three different axes of the affine transformation. Let me think of the regular hexagon's triangle sides of 3s along three different directions (each 120 degrees apart). After affine transformation, these directions are scaled differently, leading to the triangle sides 200,240,300.

Let me denote the scaling factors along three directions separated by 120 degrees as k1, k2, k3. Then, the sides of the transformed triangle would be 3s * k1, 3s * k2, 3s * k3. But the problem states the sides are 200,240,300. However, without knowing the angles between the sides, it's hard to relate k1, k2, k3.

Alternatively, perhaps the triangle's sides correspond to the three different axes of the affine transformation, so that each side is scaled by a different factor. If the original triangle was equilateral with sides 3s, then the affine transformation scales each side to 200,240,300. Therefore, the scaling factors are 200/(3s), 240/(3s), 300/(3s). But affine transformations have consistent scaling factors across the plane, so this might not hold.

Alternatively, perhaps the area of the triangle is scaled by the determinant of the affine transformation. But without more information, this might not help.

Alternatively, since the problem's hexagon is equilateral, the affine transformation must preserve the equality of the hexagon's side lengths. In an affine transformation, lengths are not preserved unless it's a similarity transformation. To keep all sides of the hexagon equal, the affine transformation must be a similarity transformation (i.e., scaling, rotation, translation). But a similarity transformation would preserve the shape, so the triangle would remain equilateral, which contradicts the given triangle sides. Therefore, this approach might be invalid.

Hmm, maybe I need to think differently. Let me recall that in the regular hexagon case, the triangle side is 3s. Here, the triangle sides are 200,240,300. If the hexagon is a scaled version in different directions, perhaps the side length s is such that 3s is the harmonic mean or something of the triangle sides?

Alternatively, maybe the side length s is given by 200 + 240 + 300 divided by some factor. Wait, but that seems arbitrary.

Alternatively, perhaps using the formula for the side length of a hexagon inscribed in a triangle. Wait, but I don't recall such a formula.

Alternatively, let me consider that in the regular case, the triangle's area is (sqrt(3)/4)*(3s)^2. The given triangle has sides 200,240,300. Let me compute its area using Heron's formula and set up a ratio.

Compute the area of the given triangle:

sides a=200, b=240, c=300.

s = (200+240+300)/2 = 740/2 = 370.

Area = sqrt(370*(370-200)*(370-240)*(370-300)) = sqrt(370*170*130*70).

Compute this:

First, compute the product inside the sqrt:

370*170 = 62,900

130*70 = 9,100

Then 62,900 * 9,100 = ?

Wait, actually, 370*170*130*70 = (370*70)*(170*130) = (25,900)*(22,100) = ?

Hmm, this is getting too big. Alternatively, factor terms:

370 = 10*37

170 = 10*17

130 = 10*13

70 = 10*7

Thus, product is 10^4 * 37*17*13*7

Thus, Area = sqrt(10^4 * 37*17*13*7) = 10^2 * sqrt(37*17*13*7)

Compute 37*17 = 629, 13*7=91, then 629*91.

Compute 629*91:

629*90 = 56,610

629*1 = 629 → total 57,239

Thus, Area = 100 * sqrt(57,239)

Hmm, sqrt(57,239) ≈ 239.247, so area ≈ 100 * 239.247 ≈ 23,924.7.

In the regular hexagon case, the area of the triangle would be (sqrt(3)/4)*(3s)^2 = (sqrt(3)/4)*9s² = (9sqrt(3)/4)s².

The area of the hexagon itself is (3sqrt(3)/2)s².

The ratio of the triangle area to the hexagon area is (9sqrt(3)/4 s²) / (3sqrt(3)/2 s²) ) = (9/4)/(3/2) = (9/4)*(2/3) = 3/2. So the triangle area is 1.5 times the hexagon area in the regular case.

If in the given problem, the triangle area is approximately 23,924.7, then the hexagon area would be (2/3)*23,924.7 ≈ 15,949.8. But how does this help find s?

Alternatively, perhaps the ratio of the triangle's side to the hexagon's side is preserved in some way. In the regular case, triangle side is 3s. Here, the triangle has sides 200,240,300. Maybe the hexagon's side is the harmonic mean of these divided by 3? Not sure.

Alternatively, since the triangle is formed by extending the sides, perhaps each side of the triangle is equal to the sum of two sides of the hexagon in some direction. Since all sides of the hexagon are length s, each side of the triangle would be 2s times some factor based on angles between sides.

Alternatively, let me think of the triangle's sides as the result of the three extended sides of the hexagon. Each side of the triangle is formed by two sides of the hexagon. For example, the side of length 200 might be formed by extending AB and CD until they meet, and the distance between those extensions is 200. Since AB and CD are opposite sides, separated by some distance, the length of the triangle's side would relate to that distance and the angle between the sides.

Wait, in the regular hexagon, the distance between opposite sides is 2*(apothem). The apothem is (s*sqrt(3))/2, so the distance between opposite sides is s*sqrt(3). The triangle's side is 3s, so the ratio is 3s / (s√3) ) = sqrt(3). So the triangle's side is sqrt(3) times the distance between opposite sides.

In the given problem, if the triangle's sides are 200,240,300, then the distances between the opposite sides of the hexagon would be 200/sqrt(3), 240/sqrt(3), 300/sqrt(3). But since the hexagon is equilateral, maybe these distances are related to the side length s through some formula.

Alternatively, in a centrally symmetric hexagon, the distance between two opposite sides can be expressed in terms of the side length and the angles between the sides. If the hexagon is equilateral, all sides are length s, but the angles can vary. The distances between opposite sides would then depend on those angles.

Let me denote the three distances between opposite sides as d1, d2, d3. Then, the sides of the triangle would be proportional to these distances. In the regular case, all d_i are equal to s*sqrt(3), and the triangle sides are 3s, so the proportionality is 3s = k * d_i, where k = sqrt(3). Wait, actually in regular case, triangle side is 3s and d_i = s√3, so 3s = sqrt(3) * d_i ⇒ d_i = 3s / sqrt(3) = s√3, which matches.

Therefore, in general, the sides of the triangle would be equal to (d_i) * sqrt(3). Wait, no, in regular case, triangle side is 3s = sqrt(3) * (distance between sides). So triangle side = sqrt(3) * distance between opposite sides.

Therefore, in the given problem, the distances between the opposite sides of the hexagon would be:

d1 = 200 / sqrt(3),

d2 = 240 / sqrt(3),

d3 = 300 / sqrt(3).

But in a centrally symmetric equilateral hexagon, the distances between opposite sides can be related to the side length s and the angles between the sides.

Let me consider the hexagon as composed of three pairs of opposite sides, each pair separated by a distance d1, d2, d3. The hexagon can be thought of as a parallelogram stretched in three different directions, but since it's a hexagon, maybe it's formed by three rhombuses?

Alternatively, in a centrally symmetric hexagon with opposite sides parallel and equal, the distance between each pair of opposite sides can be expressed as 2 times the height of the hexagon in that direction. Since it's equilateral, the height in each direction would relate to the side length and the angle between adjacent sides.

Let me consider one pair of opposite sides, say AB and DE. The distance between them is d1. Since AB is of length s, and the hexagon is equilateral, the distance d1 can be expressed as s * sin(theta1), where theta1 is the angle between AB and the direction perpendicular to AB. Wait, maybe more precisely, if the angle between two adjacent sides is theta, then the distance between the opposite sides can be calculated based on theta.

Alternatively, in a centrally symmetric hexagon, the distance between two opposite sides is twice the distance from the center to the side. Let me denote the distance from the center to side AB as h1, then the distance between AB and DE is 2h1. Similarly for the others.

In that case, the triangle sides would be related to 2h1, 2h2, 2h3. From the regular case, in a regular hexagon, the distance from center to a side is (s√3)/2, so 2h1 = s√3, and the triangle side is 3s, so 3s = sqrt(3) * (2h1) ⇒ 3s = sqrt(3)*(s√3) ⇒ 3s = 3s, which holds.

Therefore, in general, the sides of the triangle are equal to sqrt(3) times the distances between the opposite sides. Wait, no, in the regular case, triangle side = sqrt(3) * (distance between sides). Wait, triangle side = 3s, distance between sides = s√3, so 3s = (sqrt(3))*(s√3) = 3s. So yes, triangle side = sqrt(3) * distance between sides.

Wait, actually, triangle side = sqrt(3) multiplied by the distance between the opposite sides. Therefore, in the given problem, the distances between the opposite sides are:

d1 = 200 / sqrt(3),

d2 = 240 / sqrt(3),

d3 = 300 / sqrt(3).

But in a centrally symmetric equilateral hexagon, the three distances d1, d2, d3 must satisfy some relationship due to the hexagon being closed.

Let me think of the hexagon as being formed by three pairs of vectors. Since it's centrally symmetric, the vectors can be considered as **a**, **b**, **c**, -**a**, -**b**, -**c**. The distances between the opposite sides can be related to the lengths of these vectors and the angles between them.

Alternatively, the distance between two opposite sides (e.g., AB and DE) can be calculated as the length of the projection of the vector from AB to DE onto the direction perpendicular to AB.

Since the hexagon is centrally symmetric, the vector from AB to DE is twice the vector from the center to the midpoint of AB. Wait, perhaps this is getting too abstract.

Alternatively, in a centrally symmetric polygon, the distance between a pair of opposite sides is twice the distance from the center to the side.

Let me denote the distance from the center to side AB as h1, so the distance between AB and DE is 2h1. Similarly for the others.

The area of the hexagon can be calculated as the sum of the areas of the six triangles formed with the center. Each pair of opposite sides contributes two triangles with area (base * height)/2. So total area is 6*( (s * h)/2 ) for each pair? Wait, no, actually for each pair of opposite sides, the area contributed would be (perimeter/2 * distance between sides)/2? Wait, maybe better to think:

The area of a centrally symmetric polygon can be calculated as the sum over each pair of opposite sides: (distance between the sides) * (length of the sides) / 2 * 2 (since two sides). Wait, actually, for a pair of opposite sides each of length s, separated by distance d, the area contributed by them is d*s. Since there are three pairs, total area is d1*s + d2*s + d3*s = s*(d1 + d2 + d3).

Alternatively, in a parallelogram (which is a 4-sided centrally symmetric polygon), the area is base * height, which is similar to this idea.

Therefore, for the hexagon, area = s*(d1 + d2 + d3).

In the regular hexagon case, each d_i = s√3, so area would be s*(3s√3) = 3√3 s², which matches the known formula ( (3√3)/2 s² )? Wait, no, wait, regular hexagon area is (3√3 / 2 ) s². Hmm, discrepancy here.

Wait, perhaps my assumption is wrong. Let me recalculate.

In a regular hexagon with side length s, the distance between opposite sides is 2*(apothem). The apothem is (s√3)/2, so the distance between opposite sides is 2*(s√3)/2 = s√3. There are three pairs of opposite sides, so according to the formula above, area would be s*(d1 + d2 + d3) = s*(3*s√3) = 3√3 s². But the actual area is (3√3)/2 s². So my formula is off by a factor of 2. Maybe the correct formula is (d1 + d2 + d3)*s / 2 ?

Let me check: (3*s√3)*s /2 = (3√3/2)s², which matches. So the correct formula is Area = (d1 + d2 + d3)*s / 2.

Therefore, in the regular case, that works.

Now, in the given problem, the distances between opposite sides are d1 = 200/√3, d2 = 240/√3, d3 = 300/√3. Therefore, the area would be ( (200/√3 + 240/√3 + 300/√3 ) * s ) / 2 = ( (740/√3 ) * s ) / 2 = (740 s ) / (2√3 ).

But we also need another expression for the area of the hexagon to relate it to s. However, without more information, this might not help.

Alternatively, perhaps the triangle's area can also be expressed in terms of the hexagon's area. In the regular case, the triangle's area is (sqrt(3)/4)*(3s)^2 = (9√3/4)s², and the hexagon's area is (3√3/2)s². The ratio is (9√3/4)/(3√3/2) ) = (9/4)/(3/2) = 3/2. So the triangle's area is 1.5 times the hexagon's area.

In the given problem, the triangle's area is sqrt(370*170*130*70) as computed earlier, which is sqrt(370*170*130*70). Let me compute that more accurately.

Wait, let me recompute the area using Heron's formula properly.

Given sides a=200, b=240, c=300,

s = (200 + 240 + 300)/2 = 370,

Area = sqrt(370*(370-200)*(370-240)*(370-300)) = sqrt(370*170*130*70).

Let me factor each term:

370 = 2 * 5 * 37,

170 = 2 * 5 * 17,

130 = 2 * 5 * 13,

70 = 2 * 5 * 7.

Thus, multiplying together:

2^4 * 5^4 * 37 * 17 * 13 * 7.

Therefore, sqrt(2^4 *5^4 *37*17*13*7) = 2^2 *5^2 * sqrt(37*17*13*7) = 4*25*sqrt(37*17*13*7) = 100*sqrt(37*17*13*7).

Compute 37*17 = 629,

13*7 = 91,

then 629*91:

Compute 629*90 = 56,610,

plus 629*1 = 629,

Total = 56,610 + 629 = 57,239.

Thus, Area = 100*sqrt(57,239).

Compute sqrt(57,239). Let me see:

239^2 = 57,121,

240^2 = 57,600,

So sqrt(57,239) is between 239 and 240. Let me compute 239.2^2:

239 + 0.2:

(239 + 0.2)^2 = 239² + 2*239*0.2 + 0.2² = 57,121 + 95.6 + 0.04 = 57,216.64

Still less than 57,239.

239.3^2 = 239.2^2 + 2*239.2*0.1 + 0.1^2 ≈ 57,216.64 + 47.84 + 0.01 ≈ 57,264.49, which is over.

Thus sqrt(57,239) ≈ 239.24.

Thus, Area ≈ 100 * 239.24 ≈ 23,924.

So the triangle's area is approximately 23,924.

If the ratio of triangle area to hexagon area is 3/2 as in the regular case, then the hexagon's area would be (2/3)*23,924 ≈ 15,949.

But using the formula Area = (d1 + d2 + d3)*s / 2,

where d1 = 200/√3 ≈ 115.47,

d2 = 240/√3 ≈ 138.56,

d3 = 300/√3 ≈ 173.20,

so sum ≈ 115.47 + 138.56 + 173.20 ≈ 427.23,

Thus, Area ≈ (427.23 * s)/2 ≈ 213.615 * s.

Setting this equal to 15,949,

s ≈ 15,949 / 213.615 ≈ 74.66.

But this is an approximate value, and the problem likely requires an exact answer. So this approach might not be the right one.

Alternatively, perhaps the key is that in the regular case, the triangle sides are 3s, so s = triangle_side / 3. Here, since the triangle is not regular, but the hexagon is equilateral, maybe the side length s is the harmonic mean of the triangle sides divided by 3?

Harmonic mean of 200,240,300 is 3/(1/200 + 1/240 +1/300) = 3/( (6/1200 + 5/1200 +4/1200 )) = 3/(15/1200) = 3*(1200/15)= 3*80=240. Then s=240/3=80. Hmm, that gives s=80. Maybe that's the answer?

Alternatively, the side length is 80. Let me see if that makes sense.

Wait, if s=80, then in the regular case the triangle sides would be 240, which is one of the given sides. But the given triangle has sides 200,240,300. So maybe 80 is the answer? Alternatively, maybe the side length is 40? Let me think.

Alternatively, perhaps the triangle sides are proportional to the distances between the opposite sides, which are related to the hexagon's side through some trigonometric relation.

Let me consider that in the regular hexagon, each distance between opposite sides is s√3, and the triangle side is 3s. So the ratio of triangle side to distance is 3s / (s√3) ) = √3. Therefore, in the given problem, each distance between opposite sides is triangle_side / √3. Thus, the three distances are 200/√3, 240/√3, 300/√3.

In a centrally symmetric equilateral hexagon, the three distances must satisfy some relationship. Since the hexagon is closed, the sum of vectors must be zero. Let me consider the vectors again.

Let me denote the three distances as d1, d2, d3. The hexagon can be thought of as having three pairs of opposite sides, each separated by d1, d2, d3. The side length s is related to these distances and the angles between the sides.

Alternatively, in such a hexagon, the side length s can be found using the formula derived from the three distances:

1/(d1) + 1/(d2) + 1/(d3) = 1/s

Wait, not sure. Alternatively, perhaps using the formula for the area in two different ways.

Earlier, Area = (d1 + d2 + d3)*s / 2,

Also, the triangle's area is related to the hexagon's area by a factor. In the regular case, the factor was 3/2 (triangle area is 1.5 times hexagon area). If that factor holds here, then:

Triangle Area = (3/2) * Hexagon Area,

Thus,

Hexagon Area = (2/3)*Triangle Area,

Therefore,

(d1 + d2 + d3)*s / 2 = (2/3)*Triangle Area,

Plugging in the values,

( (200 + 240 + 300)/√3 ) * s / 2 = (2/3)*23924,

Wait, but 200/√3 + 240/√3 + 300/√3 = (740)/√3,

Thus,

(740/√3)*s / 2 = (2/3)*23924,

Solving for s,

s = ( (2/3)*23924 * 2 ) / (740/√3 )

= ( (4/3)*23924 ) / (740/√3 )

= ( (4*23924)/(3*740) ) * √3

Calculate numerator: 4*23924 ≈ 95,696,

Denominator: 3*740 = 2,220,

Thus,

≈ (95,696 / 2,220) * 1.732 ≈ (43.11) * 1.732 ≈ 74.6,

Which matches the approximate value earlier. But since the problem likely has an exact answer, perhaps the exact value is 80.

Wait, let me see:

If s = 80,

Then the distances would be d1 = 200/√3, d2=240/√3, d3=300/√3,

Then the area of the hexagon would be ( (200 + 240 + 300)/√3 * 80 ) / 2 = (740/√3 *80)/2 = (740*40)/√3 ≈ (29,600)/1.732 ≈ 17,099,

But the triangle area is ~23,924,

The ratio is 23,924 /17,099 ≈ 1.398, which is close to 1.5 but not exact. Hmm.

Alternatively, maybe the ratio is different. Perhaps in non-regular hexagons, the ratio isn't constant.

Alternatively, perhaps there's a better approach using the concept that in such a hexagon, the side length s is given by the formula s = (a + b + c)/3 where a,b,c are the sides of the triangle. But (200+240+300)/3 = 740/3 ≈ 246.666, which doesn't seem right.

Alternatively, since in the regular case, the triangle sides are three times the hexagon side, maybe here the hexagon side is the geometric mean of the triangle sides divided by 3?

Geometric mean of 200,240,300 is (200*240*300)^(1/3) = (14,400,000)^(1/3) ≈ 243. So 243/3 ≈ 81. Which is close to the approximate 74.6 but not exact.

Alternatively, perhaps the side length is 40. Let me see:

If s=40,

Then distances would be d1=200/√3 ≈115.47,

d2=240/√3≈138.56,

d3=300/√3≈173.20,

Sum is ≈427.23,

Area would be (427.23 *40)/2≈8544.6,

Triangle area is ~23,924,

Ratio is ~2.8, which is not 1.5.

Hmm.

Alternatively, maybe the side length is 60:

Then area would be (427.23 *60)/2≈12816.9,

Ratio ~23924/12816.9≈1.866, still not 1.5.

Hmm.

Alternatively, perhaps my assumption about the relationship between the triangle area and hexagon area is incorrect in non-regular hexagons.

Let me think differently. Let me consider the problem in terms of the affine transformation.

Suppose that the regular hexagon is transformed by an affine transformation that scales the x-axis by a factor of k_x and y-axis by k_y, resulting in the triangle sides becoming 200,240,300.

In the regular hexagon, the triangle sides are all 3s. After affine transformation, the sides along the x-axis would be scaled by k_x, those along y-axis by k_y, and those along the 60-degree directions by sqrt(k_x² + k_y² + 2k_xk_y cos(theta)), depending on the angle.

Alternatively, since the triangle sides are along three different directions separated by 120 degrees, the scaling factors for each direction would be different.

Let me denote the three triangle sides as a=200, b=240, c=300.

In the regular case, all sides are 3s. After affine transformation, each side is scaled by a factor depending on its direction.

Let me suppose that the affine transformation scales the regular hexagon such that the three triangle sides become a, b, c. Since affine transformations can be represented by scaling along principal axes, perhaps the side length s of the hexagon can be found by:

a = 3s * k1,

b = 3s * k2,

c = 3s * k3,

where k1, k2, k3 are scaling factors along three different directions.

However, the hexagon remains equilateral, so the affine transformation must preserve the length of the hexagon's sides. In an affine transformation, this requires that the transformation is a similarity transformation (i.e., scaling, rotation, translation), but then the triangle would remain equilateral, which it's not. Therefore, this is a contradiction, so affine transformation approach might not work.

Alternatively, perhaps the hexagon is not an affine regular hexagon but has a different structure.

Let me try to think of the problem in terms of the system of equations from the coordinates approach earlier.

Recall that in the coordinate system with center at origin, the three equations were:

1. (b_x - a)^2 + b_y² = s²,

2. (c_x - b_x)^2 + (c_y - b_y)^2 = s²,

3. (a + c_x)^2 + c_y² = s².

Additionally, the triangle formed by lines AB, CD, EF has sides 200,240,300. The coordinates of the triangle's vertices were found in terms of a, b_x, b_y, c_x, c_y.

Earlier, in the regular case, the triangle vertices were (0, √3 s), (3s/2, - (s√3)/2 ), (-3s/2, - (s√3)/2 ). The distances between these points gave the triangle sides as 3s each.

In the general case, the triangle vertices would be:

Intersection of AB and CD: let's call this point P,

Intersection of CD and EF: point Q,

Intersection of EF and AB: point R.

The distances PQ, QR, RP are 200,240,300 in some order.

This gives three equations for the distances, which can be written in terms of a, b_x, b_y, c_x, c_y, s.

This system might be solvable, but it's quite complex. Let me see if I can make some assumptions to simplify.

Assume that the hexagon is such that point A is along the x-axis at (a,0), and the hexagon is symmetric with respect to the x-axis. This would mean that point C is the mirror image of point B across the x-axis.

Let me suppose that:

Let me set b_y = -c_y,

and c_x is related to b_x,

and so on.

Wait, if the hexagon is symmetric about the x-axis, then:

Point B has coordinates (b_x, b_y),

then point F, which is opposite to C, should be the mirror of point C.

Wait, maybe this symmetry would impose that the triangle is also symmetric about the x-axis, so two of its sides are equal, but the given triangle has all sides different. Therefore, this symmetry is not possible.

Alternatively, perhaps the hexagon has rotational symmetry of 180 degrees, which it does as it's centrally symmetric, but not reflection symmetry.

Alternatively, let me suppose that the triangle's sides correspond to the three different distances between opposite sides scaled by sqrt(3), as in the regular case.

Thus, if the distances between the opposite sides are d1=200/√3, d2=240/√3, d3=300/√3,

then the side length s of the hexagon can be found using the relation from the area.

The area of the hexagon is (d1 + d2 + d3)*s / 2,

and the area of the triangle is (sqrt(3)/4)*(a^2 + b^2 + c^2), but no, the triangle is scalene.

Alternatively, using the formula for the area of a triangle given three sides (Heron's formula), which we did earlier.

But without another equation relating s and the distances, it's difficult.

Wait, perhaps in a centrally symmetric equilateral hexagon, the following relation holds:

1/d1 + 1/d2 + 1/d3 = 1/s,

where d1, d2, d3 are the distances between opposite sides.

If that's the case, then plugging in d1=200/√3, d2=240/√3, d3=300/√3,

we get:

√3/200 + √3/240 + √3/300 = 1/s,

Factor out √3:

√3 (1/200 + 1/240 + 1/300) = 1/s,

Compute the sum inside:

Find a common denominator for 200,240,300. Let's see:

Factor each:

200 = 2^3 *5^2,

240 = 2^4 *3*5,

300 = 2^2 *3*5^2,

LCM is 2^4 *3*5^2 = 16*3*25 = 1200.

Thus,

1/200 = 6/1200,

1/240 = 5/1200,

1/300 = 4/1200,

Total: 6+5+4 =15/1200 = 1/80.

Thus,

√3*(1/80) = 1/s ⇒ s = 80/√3 ≈46.188,

But this is not a nice number, and the problem likely expects an integer.

Alternatively, maybe the formula is different. Perhaps:

d1*d2*d3 = s^3,

then s = cube root(d1*d2*d3).

Plugging in,

d1=200/√3, d2=240/√3, d3=300/√3,

product = (200*240*300)/( (√3)^3 ) = (14,400,000)/(3√3) = 4,800,000/√3,

s = cube root(4,800,000/√3). This is messy.

Alternatively, perhaps the formula is d1 + d2 + d3 = 3s,

then s = (200 +240 +300)/(3√3) = 740/(3*1.732) ≈ 740/5.196 ≈142.8. Not likely.

Alternatively, maybe the side length s is the radius of the circumscribed circle of the triangle.

The circumradius R of the triangle is given by R = (a*b*c)/(4*A),

where a=200, b=240, c=300, A≈23924,

Thus R = (200*240*300)/(4*23924) ≈ (14,400,000)/(95,696) ≈ 150.5.

If s is related to R, maybe s = R / something. Not sure.

Alternatively, perhaps the side length of the hexagon is the same as the radius of the inscribed circle of the triangle.

The inradius r = A/s, where s is the semi-perimeter,

r = 23924 / 370 ≈64.66.

Hmm, 64.66 is close to 60*sqrt(3)≈103.9, no.

Alternatively, perhaps the side length s is 80, which is a divisor of 200,240,300.

Let me check if s=80 works.

Assume s=80.

Then, in the regular case, the triangle sides would be 240, which is one of the given sides. The other sides 200 and 300 would imply that the hexagon is "compressed" in one direction and "stretched" in another.

Perhaps the hexagon is such that two of the distances between opposite sides are less and more than the regular case.

If s=80,

then the distances would be d1=200/√3 ≈115.47,

d2=240/√3≈138.56,

d3=300/√3≈173.20.

Now, in a centrally symmetric equilateral hexagon, the three distances must satisfy some relationship derived from the geometry.

Let me consider the vectors again. Let me denote vectors **a**, **b**, **c** from the center to vertices A, B, C.

Since the hexagon is centrally symmetric, the vertices are A, B, C, -A, -B, -C.

The vectors must satisfy:

**A** + **B** + **C** = 0,

because the hexagon is closed. Wait, no, the sum of the vectors from the center to the vertices in a centrally symmetric polygon must be zero, since for every vector **v**, there is a -**v**.

Therefore, **A** + **B** + **C** + (-**A**) + (-**B**) + (-**C**) = 0, which is trivial. But the sum of **A** + **B** + **C** must be zero for the hexagon to close properly?

Wait, let me think of the hexagon as a polygon. The sum of the vectors around the polygon must be zero.

The vectors representing the sides are:

**AB** = **B** - **A**,

**BC** = **C** - **B**,

**CD** = -**A** - **C**,

**DE** = -**B** - (-**A**) = **A** - **B**,

**EF** = -**C** - (-**B**) = -**C** + **B**,

**FA** = **A** - (-**C**) = **A** + **C**.

The sum of these vectors must be zero:

(**B** - **A**) + (**C** - **B**) + (-**A** - **C**) + (**A** - **B**) + (-**C** + **B**) + (**A** + **C**) = ?

Let me compute term by term:

1. (B - A)

2. + (C - B) → total so far: C - A

3. + (-A - C) → total: C - A - A - C = -2A

4. + (A - B) → total: -2A + A - B = -A - B

5. + (-C + B) → total: -A - B - C + B = -A - C

6. + (A + C) → total: -A - C + A + C = 0.

So it checks out. Therefore, the condition is automatically satisfied.

However, the distances between the opposite sides can be related to the vectors.

The distance between sides AB and DE is twice the distance from the center to side AB.

The distance from the center (origin) to the line AB can be calculated using the formula for distance from a point to a line.

The line AB is defined by points A and B. The distance from origin to this line is |(**A** × **B**)| / |**B** - **A**|,

where × denotes the cross product (in 2D, the magnitude is |A_x B_y - A_y B_x|).

Similarly for the other sides.

Let me denote:

Let me let **A** = (a, 0), since we can place A along the x-axis.

Let **B** = (b_x, b_y),

**C** = (c_x, c_y).

Since the hexagon is centrally symmetric, the sum **A** + **B** + **C** must equal zero? Wait, no, that's not necessarily true. Wait, the center is the origin, so the midpoint of A and D is the origin, so D = -A, similarly for others. The vectors **A**, **B**, **C** can be any vectors as long as the opposite vertices are negatives.

However, the condition for the hexagon to close is already satisfied as shown earlier.

The distance from the origin to side AB is |**A** × **B**| / |**B** - **A**|.

Similarly, the distance between AB and DE (which is opposite to AB) is twice this distance.

Therefore, d1 = 2 * |**A** × **B**| / |**B** - **A**|,

Similarly,

d2 = 2 * |**B** × **C**| / |**C** - **B**|,

d3 = 2 * |**C** × **A**| / |**A** - **C**|.

Given that all sides have length s,

|**B** - **A**| = s,

|**C** - **B**| = s,

|**-A** - **C**| = s,

and so on.

This is getting very involved, but perhaps with the assumption that **A** is (a, 0), we can write equations.

Let me set **A** = (a, 0).

Then, **B** = (b_x, b_y),

**C** = (c_x, c_y).

The first side AB has length s:

sqrt( (b_x - a)^2 + (b_y)^2 ) = s. (Equation 1)

Second side BC has length s:

sqrt( (c_x - b_x)^2 + (c_y - b_y)^2 ) = s. (Equation 2)

Third side CD has length s:

sqrt( (-a - c_x)^2 + (0 - c_y)^2 ) = s ⇒ sqrt( (a + c_x)^2 + c_y^2 ) = s. (Equation 3)

Fourth side DE is from -A to -B, which has length same as AB, so it's s.

Similarly for others.

Now, the distances between opposite sides:

Distance between AB and DE is d1 = 200/√3,

Distance between BC and EF is d2 = 240/√3,

Distance between CD and FA is d3 = 300/√3.

Compute d1:

The line AB is through points A(a,0) and B(b_x, b_y).

The distance from the origin to line AB is |(a*b_y - 0*b_x)| / sqrt( (b_x - a)^2 + b_y^2 )

= |a b_y| / s,

since the denominator is the length of AB, which is s.

Therefore, the distance between AB and DE (opposite side) is twice this:

d1 = 2*( |a b_y| / s ) = 200/√3,

Similarly, for d2:

The line BC is through B(b_x, b_y) and C(c_x, c_y).

The distance from origin to BC is | (b_x c_y - b_y c_x) | / sqrt( (c_x - b_x)^2 + (c_y - b_y)^2 )

The denominator is s,

so distance is |b_x c_y - b_y c_x| / s,

thus d2 = 2*( |b_x c_y - b_y c_x| / s ) = 240/√3,

For d3:

The line CD is through C(c_x, c_y) and D(-a,0).

The distance from origin to CD is | c_x*0 - c_y*(-a) + ... | Wait, better to compute using the formula.

The line CD goes from C(c_x, c_y) to D(-a, 0).

The vector along CD is (-a - c_x, -c_y).

The distance from origin to line CD can be calculated as | (c_x * 0 - c_y*(-a) + ... ) | Wait, better to use the formula for distance from origin to line through two points.

The line through C(c_x, c_y) and D(-a,0):

The distance from origin is | (c_x * 0 - c_y*(-a) + (-a)c_y - 0*c_x ) | / sqrt( ( -a - c_x )^2 + ( -c_y )^2 )

Wait, perhaps better to use the formula:

Distance = | (c_x)(0) - c_y*(-a) + (-a - c_x)(0) - ... | Hmm, getting confused.

Alternatively, the distance from origin to line CD is | (c_x * 0 - c_y*(-a) + ... ) | Wait, let me use the standard formula:

The line through points (x1,y1) and (x2,y2) is:

(y2 - y1)x - (x2 - x1)y + x2 y1 - y2 x1 = 0

So for CD: points C(c_x, c_y) and D(-a,0):

The line equation is:

(0 - c_y)x - (-a - c_x)y + (-a)c_y - 0*c_x = 0,

Simplify:

- c_y x + (a + c_x)y - a c_y = 0

The distance from origin (0,0) to this line is | -0 + 0 - a c_y | / sqrt( c_y² + (a + c_x)^2 )

= | -a c_y | / sqrt( (a + c_x)^2 + c_y^2 )

The denominator is the length of CD, which is s,

Thus distance is |a c_y| / s,

Therefore, the distance between CD and FA (opposite side) is twice this:

d3 = 2*( |a c_y| / s ) = 300/√3,

So now we have three equations:

1. 2*( |a b_y| / s ) = 200/√3 ⇒ |a b_y| = (200/√3)*(s/2) = (100 s)/√3,

2. 2*( |b_x c_y - b_y c_x| / s ) = 240/√3 ⇒ |b_x c_y - b_y c_x| = (240/√3)*(s/2) = (120 s)/√3,

3. 2*( |a c_y| / s ) = 300/√3 ⇒ |a c_y| = (300/√3)*(s/2) = (150 s)/√3,

Additionally, the three length equations:

Equation1: (b_x - a)^2 + b_y² = s²,

Equation2: (c_x - b_x)^2 + (c_y - b_y)^2 = s²,

Equation3: (a + c_x)^2 + c_y² = s².

This is a system of six equations with variables a, b_x, b_y, c_x, c_y, s.

Assuming all variables are positive (since the hexagon is convex and we can choose coordinates accordingly), we can drop the absolute value:

1. a b_y = (100 s)/√3,

2. b_x c_y - b_y c_x = (120 s)/√3,

3. a c_y = (150 s)/√3,

Equation1: (b_x - a)^2 + b_y² = s²,

Equation2: (c_x - b_x)^2 + (c_y - b_y)^2 = s²,

Equation3: (a + c_x)^2 + c_y² = s².

Let me see if I can express variables in terms of s.

From equation 3: a c_y = (150 s)/√3 ⇒ c_y = (150 s)/(√3 a).

From equation 1: a b_y = (100 s)/√3 ⇒ b_y = (100 s)/(√3 a).

From equation 2: b_x c_y - b_y c_x = (120 s)/√3.

Let me substitute c_y and b_y from above into equation2:

b_x*(150 s)/(√3 a) - (100 s)/(√3 a)*c_x = (120 s)/√3,

Multiply both sides by √3 a / s:

b_x*150 - 100 c_x = 120 a,

Thus:

150 b_x - 100 c_x = 120 a ⇒ divide by 10:

15 b_x -10 c_x = 12 a ⇒ 3 b_x - 2 c_x = (12/5) a ⇒ Hmm, maybe keep as 15b_x -10c_x =12a.

Now, let's look at equation3: (a + c_x)^2 + c_y² = s²,

We can substitute c_y from above:

c_y = (150 s)/(√3 a),

Thus,

(a + c_x)^2 + [ (150 s)/(√3 a) ]² = s²,

Similarly, equation1: (b_x -a)^2 + b_y² = s²,

Substitute b_y = (100 s)/(√3 a):

(b_x -a)^2 + [ (100 s)/(√3 a) ]² = s²,

Equation2: (c_x - b_x)^2 + (c_y - b_y)^2 = s²,

Substitute c_y and b_y:

(c_x - b_x)^2 + [ (150 s)/(√3 a) - (100 s)/(√3 a) ]² = s²,

Simplify the second term:

[ (50 s)/(√3 a) ]²,

Thus equation2 becomes:

(c_x - b_x)^2 + (50 s / (√3 a))² = s²,

Now, this is a lot of equations, but perhaps we can find ratios.

Let me denote k = s/a ⇒ a = s/k,

Let me see if this substitution helps.

Let me set a = s/k ⇒ k = s/a ⇒ a = s/k.

Then,

From equation1:

b_y = (100 s)/(√3 a) = (100 s)/(√3*(s/k)) ) = (100 k)/√3,

Similarly,

c_y = (150 s)/(√3 a) = (150 s)/(√3*(s/k)) = (150 k)/√3,

Equation3 substitution:

(a + c_x)^2 + c_y² = s²,

Expressed as:

(s/k + c_x)^2 + (150k/√3)^2 = s²,

Equation1 substitution:

(b_x - s/k)^2 + (100k/√3)^2 = s²,

Equation2 substitution:

(c_x - b_x)^2 + (50k/√3)^2 = s²,

Also, from the modified equation2 earlier:

15b_x -10c_x =12a =12*(s/k),

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