I need to find the length of the crease L when a rectangular piece of paper, 6 inches wide, is folded so that one corner touches the opposite side. The angle θ is given, and I need to express L in terms of θ.

First, I visualize the setup. There's a rectangle that's 6 inches wide, and I assume it's long enough since we're folding it. One corner is folded over to touch the opposite side. The crease is the line along which the paper is folded, and it's straight because it's a fold.

When we fold it, the crease will be the perpendicular bisector of the line segment joining a point to its image after folding. That's a key property of folds.

Let me denote the rectangle. Let’s say it's ABCD, with AB = CD = width = 6 inches. Assume AB is the bottom side, CD the top, AD and BC the sides. Let’s say we fold corner A to some point on the opposite side. Opposite side to A would be CD, I think.

The problem says "one corner touches the opposite side." Opposite side probably means the side not adjacent. For corner A, adjacent sides are AB and AD, so opposite side is CD.

So, we fold corner A to a point, say P, on side CD. Then, the crease is the perpendicular bisector of the segment AP.

Since it's a fold, every point on the crease is fixed, and the fold maps points to their symmetric counterparts with respect to the crease line.

Specifically, for point A being mapped to P, the crease is the set of points equidistant to A and P, and it's a straight line perpendicular to AP at its midpoint.

But in this case, since the paper is being folded, the crease must lie within the rectangle, and it's the line such that the distance to A and to P is equal for points on it.

Now, in this fold, when we fold A to P on CD, the crease will intersect the sides of the rectangle.

The problem gives an angle θ, which is probably the angle that the crease makes with some side.

Looking at the diagram, it shows a rectangle with a crease, and an angle θ marked. From the diagram, it seems that θ is the angle between the crease and the side of the rectangle.

In the image, there's a rectangle, width 6 inches, and a crease L, and an angle θ at the top side. It seems that θ is the angle between the crease and the top side or something.

Let me sketch this.

Assume the rectangle has width w = 6 inches. Let’s say it's oriented with width horizontal, so let’s say bottom and top sides are horizontal.

We fold one corner, say bottom-left corner, to a point on the top side.

So, let’s say corner A (bottom-left) is folded to point P on top side CD.

Then, the crease is a line segment within the rectangle, and it will intersect the left side and the bottom side, or perhaps other sides.

Typically, for such a fold, the crease will intersect the two adjacent sides.

Let me recall a standard result or derive it.

I could use coordinate geometry to model this.

Place the rectangle on the coordinate plane.

Set point A at (0,0), B at (6,0), C at (6,h), D at (0,h), where h is the height, which is not given, and probably it will cancel out or we don't need it.

The height might be relevant, but since we're folding to the opposite side, and the crease depends on the height, but in the end, we need L in terms of θ, and w is given, so h might be involved, but perhaps it will cancel.

The angle θ is given, so probably h is related or will be eliminated.

Set A(0,0), B(6,0), C(6,h), D(0,h).

We fold A(0,0) to some point P(x, h) on CD. Since CD is from (0,h) to (6,h), so P has coordinates (p, h) for some p between 0 and 6.

Now, the crease is the perpendicular bisector of the segment AP.

A is (0,0), P is (p,h), so midpoint M of AP is ((0+p)/2, (0+h)/2) = (p/2, h/2).

The slope of AP is (h-0)/(p-0) = h/p.

So, the slope of the perpendicular is -p/h.

Therefore, the crease is the line passing through M(p/2, h/2) with slope -p/h.

So, its equation is:

y - h/2 = (-p/h) (x - p/2)

This crease must lie within the rectangle, and it will intersect the boundaries.

Typically, for such a fold, the crease will intersect the left side and the bottom side, or left and right, but in this case, since we're folding A to P on top, and A is bottom-left, the crease should intersect the bottom side and the left side, or perhaps the top side, but let's see.

The problem is that the crease is a line segment inside the rectangle, and it stops where it meets the boundaries, but in this case, since it's a straight line, and it passes through (p/2,h/2), and has slope -p/h, which is negative, so it slopes downwards to the right if p>0.

For example, if p=6, P is (6,h), then M is (3,h/2), slope of AP is h/6, so perpendicular slope is -6/h.

Equation: y - h/2 = (-6/h)(x - 3)

This line will intersect the bottom side (y=0) and the left side (x=0).

When x=0, y - h/2 = (-6/h)(0 - 3) = (-6/h)(-3) = 18/h

So y = h/2 + 18/h

Similarly, when y=0, 0 - h/2 = (-6/h)(x - 3) => -h/2 = (-6/h)(x - 3)

Multiply both sides by h: - (h^2)/2 = -6 (x - 3)

So (h^2)/2 = 6(x - 3)

x - 3 = h^2 / 12

x = 3 + h^2 / 12

Since the rectangle is only up to x=6, if h is large, this might be greater than 6, but in that case, the crease might intersect the right side instead. But for the fold to be possible with the crease inside, we need to ensure that the crease is within the rectangle.

In the problem, it's given that it's folded and there's a crease, so we assume that for the given θ, it's valid.

But in this case, for general p and h, the crease might not be entirely within the rectangle, but in standard problems, we assume that the crease intersects the two sides adjacent to A, i.e., left and bottom.

In the diagram, it shows the crease going from near the bottom-left to the top, but at an angle.

The angle θ is given, and in the diagram, it seems that θ is the angle between the crease and the horizontal or vertical.

Looking back at the image, it has an angle θ marked near the top, so probably θ is the angle that the crease makes with the side.

In the description, it says "the length in inches of the crease L in terms of angle θ", and width is 6, so L is a function of θ only, but that seems odd because for different heights, the crease length might be different, but the angle θ might depend on height as well.

Perhaps for a given fold, θ is determined, but here θ is given as a parameter, so probably we can express L in terms of θ without h, meaning that θ determines the configuration.

I need to relate p and h.

The angle θ is probably defined in the diagram. Since I don't have the diagram, I need to infer what θ is.

From the image, it shows a rectangle, width 6, and a crease, and at one end, there's an angle θ marked, and it looks like θ is the angle between the crease and the vertical side or horizontal.

In many such problems, θ is the angle that the crease makes with the horizontal or with the side.

Let me assume that θ is the angle that the crease makes with the horizontal. Or with the vertical.

In the diagram, it might be specified. Since the crease is slanted, and θ is marked at the top, but in the fold, the crease may not reach the top.

Perhaps θ is the angle at which the crease is inclined.

Another thought: in such folds, the crease is perpendicular to the line joining A to P, but we can use the fact that the distance from any point on the crease to A and to P is equal.

But for the crease to be within the rectangle, it must be that when we fold, the crease is where the paper is folded, and it intersects the boundaries.

I recall that for a rectangle of width w, folding a corner to the opposite side, the crease length can be found using similar triangles or properties.

Let me denote the crease as line segment EF, where E is on one side, F on another.

In this case, for folding A to P on CD, the crease will intersect AB (bottom) and AD (left side), say.

Let me assume that the crease intersects AB at E and AD at F.

Since it's the perpendicular bisector, and A is at (0,0), and the crease is perpendicular to AP.

But E and F are on the crease, so for any point on EF, distance to A and to P is equal.

In particular, for E on AB, say E(e,0), and F on AD, say F(0,f).

Then, since E is on the crease, distance from E to A is equal to distance from E to P.

Similarly for F.

So, for E(e,0): distance to A(0,0) is |e - 0| = e (since e>0).

Distance to P(p,h) is sqrt( (e - p)^2 + (0 - h)^2 )

So, e = sqrt( (e - p)^2 + h^2 )

Similarly, for F(0,f): distance to A(0,0) is f.

Distance to P(p,h) is sqrt( (0 - p)^2 + (f - h)^2 ) = sqrt(p^2 + (f - h)^2)

So f = sqrt(p^2 + (f - h)^2)

Also, since the crease is a straight line, and E and F are on it, and the line is the set of points equidistant to A and P, so the line EF should be the perpendicular bisector, which we already have.

But from the equations, we can solve for e and f in terms of p and h.

From E: e = sqrt( (e - p)^2 + h^2 )

Square both sides: e^2 = (e - p)^2 + h^2

e^2 = e^2 - 2 e p + p^2 + h^2

0 = -2 e p + p^2 + h^2

2 e p = p^2 + h^2

e = (p^2 + h^2)/(2p)

Similarly, for F: f = sqrt(p^2 + (f - h)^2)

f^2 = p^2 + (f - h)^2

f^2 = p^2 + f^2 - 2 f h + h^2

0 = p^2 - 2 f h + h^2

2 f h = p^2 + h^2

f = (p^2 + h^2)/(2h)

Notice that e and f are both expressed in terms of p and h.

The crease is the line segment from F(0,f) to E(e,0).

The length of the crease L is the distance between E and F.

L = sqrt( (e - 0)^2 + (0 - f)^2 ) = sqrt( e^2 + f^2 )

Now, e = (p^2 + h^2)/(2p), f = (p^2 + h^2)/(2h)

So e f = [ (p^2 + h^2)^2 ] / (4 p h)

But L = sqrt(e^2 + f^2) = sqrt{ [ (p^2 + h^2)^2 / (4p^2) ] + [ (p^2 + h^2)^2 / (4h^2) ] } = (p^2 + h^2) / (2 \sqrt{p h} ) * sqrt{ 1/p^2 + 1/h^2 } better to factor.

L = sqrt( e^2 + f^2 ) = sqrt{ \left( \frac{p^2 + h^2}{2p} \right)^2 + \left( \frac{p^2 + h^2}{2h} \right)^2 } = \frac{p^2 + h^2}{2} \sqrt{ \frac{1}{p^2} + \frac{1}{h^2} } = \frac{p^2 + h^2}{2} \sqrt{ \frac{h^2 + p^2}{p^2 h^2} } = \frac{p^2 + h^2}{2} \frac{\sqrt{p^2 + h^2}}{p h}

Simplify:

L = \frac{p^2 + h^2}{2} \cdot \frac{\sqrt{p^2 + h^2}}{p h} = \frac{ (p^2 + h^2)^{3/2} }{2 p h }

Now, this is in terms of p and h, but we have the width w=6, and the angle θ.

The angle θ is probably the angle that the crease makes with the horizontal or vertical.

The crease is from F(0,f) to E(e,0), so the vector from F to E is (e, -f), so the slope is -f/e.

The angle with the horizontal: the slope is rise over run, from F to E, delta x = e, delta y = -f, so slope is -f/e.

So the angle φ with the horizontal is arctan(slope) = arctan(-f/e), but usually we take acute angle.

In the diagram, θ might be the angle with the vertical or horizontal.

Notice that in the expression, we have p and h, but we need to relate to θ.

The angle θ might be given as part of the configuration.

When we fold, the line AP makes an angle with the horizontal.

The line AP from A(0,0) to P(p,h), so its slope is h/p, so angle α with horizontal is arctan(h/p).

But θ is probably the angle of the crease.

The crease has slope -p/h, as we had earlier.

So the angle β that the crease makes with the horizontal is arctan(slope) = arctan(-p/h), so the magnitude is arctan(p/h).

Since it's downward, but the angle is probably taken as acute.

So let’s define θ as the acute angle that the crease makes with the horizontal. In the diagram, it might be specified, but likely it's with the horizontal.

Assume that θ is the angle between the crease and the horizontal direction.

Since the crease has slope m = -p/h, so |m| = p/h, so tanθ = |slope| = p/h.

Is that correct? Slope is rise over run, so for a line with slope m, the angle θ with horizontal is such that tanθ = |m|.

In this case, m = -p/h, so tanθ = p/h.

Yes.

So, tanθ = p / h.

Therefore, p = h tanθ.

Now, we can express L in terms of h and θ, but we have the width w=6.

In this setup, the point P is on CD, which has length w=6, so p must be between 0 and 6.

But also, for the crease to be within the rectangle, we need that E and F are on the sides, which they are, but also that the crease doesn't go outside, but since it's a straight line from (0,f) to (e,0), and both are on the boundaries, it should be fine as long as e <= w and f <= h, but f is on AD, which is height h, so f must be between 0 and h.

Similarly, e between 0 and w.

Now, f = (p^2 + h^2)/(2h)

But p = h tanθ, so f = ( (h tanθ)^2 + h^2 ) / (2h) = (h^2 tan^2 θ + h^2)/(2h) = h (tan^2 θ + 1)/(2) = h sec^2 θ / 2

Since 1 + tan^2 θ = sec^2 θ.

Similarly, e = (p^2 + h^2)/(2p) = (h^2 tan^2 θ + h^2)/(2 h tan θ) = h (tan^2 θ + 1)/(2 tan θ) = h sec^2 θ / (2 tan θ)

Now, sec^2 θ / tan θ = (1 / cos^2 θ) / (sin θ / cos θ) = 1 / (cos θ sin θ) = 2 / sin(2θ)

Since sin(2θ) = 2 sinθ cosθ, so 1/(sinθ cosθ) = 2 / sin(2θ)

So e = h * (2 / sin(2θ)) / 2? Let's see:

e = h * sec^2 θ / (2 tan θ) = h * (1 / (cos^2 θ)) / (2 sinθ / cosθ) = h * (1 / (cos^2 θ)) * (cosθ / (2 sinθ)) = h / (2 cosθ sinθ) = h / (sin(2θ)) * (1/2)? No

1 / (cosθ sinθ) = 1 / ( (1/2) sin(2θ) ) = 2 / sin(2θ)

So e = h * [2 / sin(2θ)] / 2? From above:

e = h * sec^2 θ / (2 tan θ) = h * [1 / (cos^2 θ)] / [2 (sinθ / cosθ)] = h * [1 / (cos^2 θ)] * [cosθ / (2 sinθ)] = h * [1 / (cosθ sinθ)] / 2? No:

[1 / (cos^2 θ)] * [cosθ / (2 sinθ)] = [1 / (cosθ)] * [1 / (2 sinθ)] = 1 / (2 sinθ cosθ) = 1 / (sin(2θ))

I think I messed up.

From:

e = h * \frac{ \sec^2 \theta }{ 2 \tan \theta } = h \cdot \frac{ \frac{1}{\cos^2 \theta} }{ 2 \cdot \frac{\sin \theta}{\cos \theta} } = h \cdot \frac{1}{\cos^2 \theta} \cdot \frac{\cos \theta}{2 \sin \theta} = h \cdot \frac{1}{\cos \theta \cdot 2 \sin \theta} = \frac{h}{2 \sin \theta \cos \theta} = \frac{h}{\sin 2\theta}

Since sin(2θ) = 2 sinθ cosθ, so 1/(sinθ cosθ) = 2 / sin(2θ), so:

e = h / (sin 2θ) ? No:

\frac{h}{2 \sin \theta \cos \theta} = h \cdot \frac{1}{2} \cdot \frac{1}{\sin \theta \cos \theta} = \frac{h}{2} \cdot \frac{2}{\sin 2\theta} = \frac{h}{\sin 2\theta} ? Let's compute:

1 / (sinθ cosθ) = 2 / (2 sinθ cosθ) = 2 / sin(2θ)

So \frac{h}{2} \cdot \frac{1}{\sin \theta \cos \theta} = \frac{h}{2} \cdot \frac{2}{\sin 2\theta} = h / \sin 2\theta

But earlier I have e = \frac{h}{2 \sin \theta \cos \theta} = \frac{h}{2} \cdot \frac{1}{\sin \theta \cos \theta} = \frac{h}{2} \cdot \frac{2}{\sin 2\theta} = \frac{h}{\sin 2\theta}

Yes, e = \frac{h}{\sin 2\theta}

But that can't be right because sin(2θ) ≤ 1, so e ≥ h, but e is on the bottom, which is width, and h is height, they can be different, but in this case, for example if θ small, e large, which might be possible.

Similarly, f = \frac{h}{2} \sec^2 \theta = \frac{h}{2} (1 + \tan^2 \theta) = \frac{h}{2} (1 + \frac{p^2}{h^2}) but earlier we had f = h \frac{\sec^2 \theta}{2} = \frac{h}{2} \sec^2 \theta

Now, the width of the rectangle is w = 6, and e must be less than or equal to w, but in this expression, e = h / sin(2θ), which depends on h, but h is not fixed.

The problem is that for a given θ, and width w, the height h is not specified, but in the fold, when we fold, for a fixed rectangle, h is fixed, but here θ is given, so probably for a given θ, the fold determines p and thus h, but h is part of the rectangle.

I think I missed something. The rectangle has fixed width w=6, but height h is not given, and it's not specified, but in the fold, when we fold, for a given rectangle, we choose P, but here θ is given, so for each θ, there might be a corresponding fold, but for a fixed rectangle, h is fixed, so θ would be determined, but the problem asks for L in terms of θ, implying that θ is the independent variable, so probably the rectangle height is not fixed, or perhaps it is, but we need to express in terms of θ only, which suggests that h is determined by θ and w.

In the configuration, when we fold, the point P is on the opposite side, but for the crease to be defined, and θ given, we need to see what is fixed.

Perhaps for the fold to be possible with a given θ, the height h must be such that the crease is valid, but in this case, since the rectangle is given with width 6, but height is not specified, so probably the height is part of it, but the problem doesn't give height, so likely for the fold to be made, with a given θ, it imposes a relation between h and p, but we have w=6, which is the length of CD.

In our setup, P is at (p,h), and CD is from x=0 to x=6, so p must be between 0 and 6.

The crease length L must be in terms of θ only, so probably for a given θ, the height h is determined by the width or something.

From the angle, we have tanθ = p/h, but p can vary, and h can vary, but for the crease to be within the rectangle, we need that the crease intersects the sides as we assumed, which requires that f ≤ h and e ≤ w.

f = (p^2 + h^2)/(2h) = h/2 + p^2/(2h)

Since p = h tanθ, f = h/2 + (h^2 tan^2 θ)/(2h) = h/2 + (h tan^2 θ)/2 = (h/2)(1 + tan^2 θ) = (h/2) sec^2 θ

f must be less than or equal to h, since it's on the left side AD.

f ≤ h ⇒ (h/2) sec^2 θ ≤ h

Assuming h > 0, divide both sides: (1/2) sec^2 θ ≤ 1 ⇒ sec^2 θ ≤ 2 ⇒ |sec θ| ≤ √2 ⇒ |cos θ| ≥ 1/√2 ⇒ |θ| ≤ 45 degrees, since θ is acute probably.

Similarly, e = (p^2 + h^2)/(2p) = (h^2 tan^2 θ + h^2)/(2 h tan θ) = h (tan^2 θ + 1)/(2 tan θ) = h sec^2 θ / (2 tan θ)

e must be less than or equal to w = 6.

e ≤ 6 ⇒ h sec^2 θ / (2 tan θ) ≤ 6

But sec^2 θ / tan θ = 1/(sinθ cosθ) = 2/sin(2θ), as before.

So e = h * 2 / sin(2θ) / 2? Earlier we had e = h / sin(2θ) ? No.

From earlier calculation: e = h * sec^2 θ / (2 tan θ) = h * (1/(cos^2 θ)) / (2 sinθ / cosθ) = h * 1/(cos^2 θ) * cosθ/(2 sinθ) = h / (2 sinθ cosθ) = h / sin(2θ)

Yes, e = \frac{h}{\sin 2\theta}

Similarly, f = \frac{h}{2} \sec^2 \theta

Now, f ≤ h, which as above, requires |θ| ≤ 45°.

Also, e ≤ w = 6.

But e = h / sin(2θ), so h / sin(2θ) ≤ 6.

Moreover, since P is on CD, 0 ≤ p ≤ 6, and p = h tanθ, so 0 ≤ h tanθ ≤ 6.

For the crease to be inside, we also need that the crease doesn't go beyond, but in this case, since it's a straight line from (0,f) to (e,0), and the rectangle is from x=0 to x=6, y=0 to y=h, and since e ≤ 6 and f ≤ h, and the line is in the first quadrant, it should be within the rectangle as long as e ≥ 0, f ≥ 0, which is true for θ acute.

The key point is that for a given θ, the height h can vary, but it must satisfy the constraints from e ≤ 6 and p ≤ 6.

Specifically, p = h tanθ ≤ 6, and e = h / sin(2θ) ≤ 6.

Also, since f ≤ h is already satisfied for |θ| ≤ 45°, and for the crease to be defined, we need f > 0, e > 0, which is true.

But for different h, L is different, but the problem asks for L in terms of θ only, so probably for a given θ, there is a specific h that makes the fold valid, or perhaps the rectangle is fixed, but height is not given, so likely, in the context, the height is determined by the fold or something.

Perhaps I misinterpreted the angle θ.

Another thought: in the diagram, the angle θ might be the angle that the line AP makes, not the crease.

Let me check the diagram description. The user said "the diagram", but I don't have it, but in the text, it says "the length in inches of the crease L in terms of angle θ", and from the context, θ is given in the diagram.

Perhaps θ is the angle at P or something.

Another idea: when they say "the opposite side", and the crease, and angle θ, perhaps θ is the angle between the crease and the side it meets.

In many such problems, for a rectangle of width w, when you fold a corner to the opposite side, the crease length can be related to the angle.

I recall that the length of the crease is w / sinθ or something, but let's think.

Perhaps for the crease to be fully within, and for the fold to be possible, the height must be such that the crease starts and ends on the sides, but in this case, for a given w and θ, we can choose h, but L depends on h.

Unless the height is determined by the condition that the crease is tangent or something, but in this case, it's not.

Let's look back at the expression for L.

We have L = \frac{ (p^2 + h^2)^{3/2} }{2 p h }

And p = h tanθ, so p^2 + h^2 = h^2 tan^2 θ + h^2 = h^2 (tan^2 θ + 1) = h^2 sec^2 θ

So L = \frac{ [ h^2 sec^2 θ ]^{3/2} }{2 (h tanθ) h } = \frac{ (h^3 sec^3 θ) }{2 h^2 tanθ } = \frac{ h sec^3 θ }{2 tanθ }

Now sec^3 θ / tanθ = (1/cos^3 θ) / (sinθ / cosθ) = 1/(cos^2 θ sinθ) = sec^2 θ / sinθ

So L = \frac{h}{2} \frac{ \sec^2 \theta }{ \sin \theta } = \frac{h}{2} \frac{1}{\cos^2 \theta \sin \theta}
From earlier, we have e = h / sin(2θ) = h / (2 sinθ cosθ)

And e must be less than or equal to w=6, but for the crease to be within the rectangle, and for the fold to be possible, typically in such problems, the crease will intersect the adjacent sides, but for that, we need e ≤ w, but w is fixed, so for a given θ, h is chosen such that e = w or something? But that doesn't make sense.

Perhaps for the fold to be made, and the crease to be the one shown, it must be that the crease also touches the top or something, but in the diagram, it might be that the crease starts from the bottom and goes to the top side.

In the user's text, it says "folded as in the diagram", and "the crease L", and angle θ is shown, so probably in the diagram, the crease is not from bottom to left, but perhaps from bottom to right or something.

Let's think differently. When we fold A to P on CD, the crease is the perpendicular bisector, which is a line, and it may intersect the top and bottom or left and right.

In the diagram, from the description, it seems that the crease is slanting across, and θ is marked, likely at the top or side.

Perhaps for the crease to be completely within and to have a defined angle, but in our case, it is within.

Another constraint: when we fold, the paper is folded, and the crease is where it is folded, and for the fold to be flat, the crease must be such that the folded part doesn't overlap or something, but usually it's fine.

But the problem is that L depends on h, but h is not given, so for L to be in terms of θ only, h must be determined by w and θ.

From the width, and the fold, we have p between 0 and 6, and p = h tanθ, and also from e ≤ 6, h / sin2θ ≤ 6.

But also, since P is on CD, p = h tanθ ≤ 6.

Moreover, for the crease to be valid, f must be between 0 and h, which is ok for θ ≤ 45°, as we have.

But for a given θ, there is a range of h that satisfy the inequalities, but L is proportional to h, so it can be different for different h, but the problem asks for a specific L in terms of θ, so probably there is a specific configuration intended.

Perhaps in the diagram, the crease intersects the top side or the right side, not the left side.

For example, in some folds, when you fold A to P on CD, the crease may intersect AB and BC, or AD and CD, etc.

In this case, since A is bottom-left, and P on top, if P is close to A, crease may intersect left and top, but if P is close to C, it may intersect bottom and top or bottom and right.

In the diagram, it likely intersects the bottom and the top side, or bottom and right side.

Let's assume that the crease intersects the bottom side and the top side.

So, let's say the crease intersects AB (bottom) at E(e,0) and CD (top) at G(g,h).

Then, since both E and G are on the crease, distance to A and to P is equal.

For E(e,0): dist to A(0,0) is e, dist to P(p,h) is sqrt((e-p)^2 + (0-h)^2)

So e = sqrt( (e-p)^2 + h^2 ) , same as before.

For G(g,h): dist to A(0,0) is sqrt(g^2 + h^2), dist to P(p,h) is |g - p|, since same y-coordinate.

So |g - p| = sqrt(g^2 + h^2)

Since g and p are both between 0 and 6, and likely g ≠ p, so g - p = - sqrt(g^2 + h^2) or g - p = sqrt(g2 + h2), but since sqrt(g2 + h2) > |g| > |g - p| if p is not g, it must be that g - p = - sqrt(g2 + h2) , because if g - p = + sqrt(g2 + h2), then |g - p| = sqrt(g2 + h2), but for g>0, sqrt(g2 + h2) > g, so if g > p, g-p >0, but g-p < g < sqrt(g2 + h2), contradiction, similarly if g < p, g-p <0, |g-p| = p-g, and p-g >0, and sqrt(g2 + h2) > g, but p-g could be larger or smaller.

Set |g - p| = sqrt(g^2 + h^2)

Square both sides: (g - p)^2 = g^2 + h^2

g^2 - 2g p + p^2 = g^2 + h^2

-2g p + p^2 = h^2

2g p = p^2 - h^2

g = (p^2 - h^2)/(2p)

This must be between 0 and 6, and real, so |p| > |h| or something, but p and h positive, so p^2 - h^2 > 0 for g>0, so p > h.

Then g = (p^2 - h2)/(2p)

From E, we have e = (p^2 + h^2)/(2p), as before.

The crease is from E(e,0) to G(g,h), so length L = sqrt( (g - e)^2 + (h - 0)^2 )

g - e = [ (p^2 - h^2)/(2p) ] - [ (p^2 + h^2)/(2p) ] = [ (p^2 - h^2) - (p^2 + h^2) ] / (2p) = (-2h^2)/(2p) = -h^2 / p

So |g - e| = h^2 / p

Then L = sqrt( (g - e)^2 + h^2 ) = sqrt( (h^4 / p^2) + h^2 ) = h sqrt( h^2 / p^2 + 1 ) = h sqrt( (h^2 + p^2)/p^2 ) = (h / p) sqrt(p^2 + h^2)

Now, the angle θ. In this case, the crease has slope (h - 0)/(g - e) = h / (g - e) = h / (- h^2 / p) = - p / h

So slope is -p/h, same as before.

So tanθ = |slope| = p/h, same as before.

Then L = (h / p) sqrt(p^2 + h2) = (1 / (p/h)) sqrt(p2 + h2) = (1 / tanθ) sqrt(p2 + h2)

But p2 + h2 = h2 (tan2θ + 1) = h2 sec2θ

So L = (1 / tanθ) * h secθ = (cosθ / sinθ) * h / cosθ = h / sinθ

So L = h / sinθ

Now, this is for the case where the crease intersects bottom and top.

But we have h unknown, and p = h tanθ, and p must be between 0 and 6, and also g = (p2 - h2)/(2p) must be between 0 and 6.

From g = (p2 - h2)/(2p) >0, so p > h.

Also g ≤ 6.

But L = h / sinθ

Now, to have L in terms of θ only, we need another equation.

The width is 6, but how is it used? P is on CD, so p ≤ 6, but also, the crease must be within, but we have g and e.

e = (p2 + h2)/(2p)

And e must be between 0 and 6, since on bottom.

Also, the point G is on top, g between 0 and 6.

For the fold to be possible, and crease to be in the rectangle, we need 0 ≤ e ≤ 6, 0 ≤ g ≤ 6, and 0 ≤ p ≤ 6.

From g = (p2 - h2)/(2p) ≥0, so p ≥ h.

From g ≤ 6, (p2 - h2)/(2p) ≤ 6

From e = (p2 + h2)/(2p) ≤ 6

Also p ≤ 6.

Now, with p = h tanθ, and L = h / sinθ

From p = h tanθ ≤ 6

From g = (p2 - h2)/(2p) = (h2 tan2θ - h2)/(2 h tanθ) = h (tan2θ - 1)/(2 tanθ)

g ≥0, so tan2θ - 1 >0 if h>0, so tan2θ >1, so 2θ > 45°, θ > 22.5° approximately.

Also g ≤ 6: h (tan2θ - 1)/(2 tanθ) ≤ 6

e = (p2 + h2)/(2p) = h (tan2θ + 1)/(2 tanθ) ≤ 6

But L = h / sinθ, so h = L sinθ

Then from p = h tanθ = L sinθ tanθ = L sinθ (sinθ / cosθ) = L sin2θ / (2 cosθ) ? sinθ * sinθ / cosθ = L sin^2 θ / cosθ

But p ≤ 6, so L sin^2 θ / cosθ ≤ 6

Also from g = h (tan2θ - 1)/(2 tanθ) = (L sinθ) ( \frac{ sin2θ / cos2θ - 1 }{2 sinθ / cosθ} ) tan2θ = sin2θ / cos2θ, so tan2θ - 1 = (sin2θ - cos2θ)/cos2θ

2 tanθ = 2 sinθ / cosθ

So g = L sinθ * [ (sin2θ - cos2θ)/cos2θ ] / [2 sinθ / cosθ] = L sinθ * (sin2θ - cos2θ)/cos2θ * cosθ / (2 sinθ) = L (sin2θ - cos2θ) cosθ / (2 cos2θ)

Simplify: sin2θ - cos2θ = 2 sinθ cosθ - (2 cos^2 θ - 1) = 2 sinθ cosθ - 2 cos^2 θ + 1? Better: sin2θ = 2 sinθ cosθ, cos2θ = cos^2 θ - sin^2 θ = 2 cos^2 θ - 1 = 1 - 2 sin^2 θ, but perhaps keep as is.

sin2θ - cos2θ = √2 sin(2θ - 45°) or something, but let's leave.

cos2θ = 2 cos^2 θ - 1 = 1 - 2 sin^2 θ, but perhaps:

g = L * [ (sin2θ - cos2θ) cosθ ] / (2 cos2θ)

And this must be ≤ 6.

Similarly for e.

But this is messy, and for the crease to be fully within, we need that the line from E to G is within the rectangle, which it is if 0≤e≤6, 0≤g≤6, and since it's a straight line, and the rectangle is convex, it should be fine.

But still, for a given θ, L can vary as long as the inequalities are satisfied.

For example, if θ is fixed, h can be chosen, as long as p = h tanθ ≤ 6, and g = h (tan2θ - 1)/(2 tanθ) ≥0, which is true for θ > 22.5°, and g ≤ 6, e ≤ 6.

Then L = h / sinθ, so it depends on h.

But the problem likely assumes that the crease is such that it is the only crease or something, but in this case, for each h, there is a L.

Perhaps in the diagram, the crease is not intersecting the sides at arbitrary points, but is tangent or something, but that doesn't make sense.

Another possibility: perhaps "the opposite side" means the side opposite in the sense of not adjacent, but when we fold, the point P is chosen such that the crease is as shown, and perhaps for the crease to be of extremal length or something, but the problem doesn't say that.

Perhaps for the fold to be made, and the crease to be within the rectangle, the height must be such that the crease is exactly on the boundary or something.

Let's look back at the first approach.

In the first approach, with crease from left to bottom, we had L = \frac{ (p^2 + h^2)^{3/2} }{2 p h } = \frac{ (h^2 sec^2 θ)^{3/2} }{2 (h tanθ) h } = \frac{ h^3 sec^3 θ }{2 h^2 tanθ } = \frac{h sec^3 θ}{2 tanθ}

And with e = h / sin2θ ≤ 6, and p = h tanθ ≤ 6, and for θ ≤ 45°, f = (h/2) sec^2 θ ≤ h, which is true.

But still L depends on h.

Perhaps in that configuration, the crease must also not extend beyond, but it is within.

Unless for the fold to be proper, the folded part doesn't go outside, but usually it's ok.

Perhaps the angle θ is not the angle of the crease, but the angle at A or something.

Let's think of the folded position.

When we fold A to P on CD, the crease is the perpendicular bisector, and the line from A to P is perpendicular to the crease at its midpoint.

The angle between AP and the crease is 90 degrees, but that's not helpful.

Perhaps θ is the angle between AP and the side.

In the diagram, it might be labeled at P or at A.

Another idea: in some problems, θ is the angle that the crease makes with the horizontal, and for a given rectangle, it is fixed, but here it is given as a parameter, so probably the rectangle is not fixed, but the width is fixed, and for each θ, there is a corresponding height that makes the crease have that angle, but then L would depend on h, but h is not given.

Unless the height is determined by the condition that the crease starts at the corner or something, but in this case, it does.

Let's calculate the length when the crease is as short as possible or something, but the problem doesn't say that.

Perhaps for the fold to be made, and the crease to be the one shown, it must be that the crease is parallel to the diagonal or something, but not necessarily.

Let's try to assume that in the diagram, the crease intersects the left side and the bottom side, as in my first approach, and perhaps for the fold, the height is such that the crease is exactly at 45 degrees or something, but not specified.

Perhaps the width is 6, and when we fold, the point P is chosen, but for the angle θ to be defined, and for the crease to be unique for that θ, but it's not.

I recall that in some sources, for a rectangle of width w, folding a corner to the opposite side, the length of the crease is w / (2 sin^2 θ) or something, but let's search my memory.

Perhaps θ is the angle at which the crease is inclined, and for the standard configuration, the height is related.

Another thought: in the diagram, after folding, the corner touches the opposite side, and the crease is shown, and θ is the angle between the crease and the side, and perhaps for the rectangle, the height is not specified, but in the fold, when you fold, the distance from A to the opposite side is the height, but when you fold to a point P, the distance along the side is p, but the vertical distance is h, but h is the height.

Perhaps the height is not given, but in the expression, it cancels when we have the constraint.

Let's go back to the first approach, and use the width constraint.

In the first approach, with crease from F(0,f) to E(e,0), and e = h / sin2θ, f = (h/2) sec^2 θ

Now, the point P is at (p,h) = (h tanθ, h)

P is on CD, so 0 ≤ p ≤ 6, so 0 ≤ h tanθ ≤ 6

Also, the crease is within, so 0 ≤ e ≤ 6, 0 ≤ f ≤ h.

f ≤ h is satisfied for θ ≤ 45°, as sec^2 θ ≤ 2 for θ ≤ 45°.

e = h / sin2θ ≤ 6

Also, since the crease is a line from (0,f) to (e,0), and the rectangle is from x=0 to x=6, y=0 to y=h, we need to ensure that for x between 0 and e, y is between 0 and h, which it is since the line is in the first quadrant with negative slope.

But additionally, the folded part must be within the rectangle or something, but usually it's not a problem.

For the crease to be the only one, but for each θ, there is a range of h.

However, in the problem, likely it is assumed that the crease is such that it is the shortest or the longest, but not specified.

Perhaps for the fold to be made, and the crease to be straight, and for the angle to be θ, but in this case, for a given rectangle, θ is determined by the choice of P, but here θ is given, so for a fixed rectangle, we can choose P, so θ can be chosen, and then L is determined.

For a fixed rectangle, with width 6 and height h, we can choose P on CD, so choose p between 0 and 6, then the crease is determined, and θ = arctan(p/h), and L = \frac{ (p^2 + h^2)^{3/2} }{2 p h } from first approach.

But L is a function of p and h, but for fixed h, we can express L in terms of θ, since p = h tanθ, so L = \frac{ h sec^3 θ}{2 tanθ} as before.

But this depends on h, and h is not given, so for the problem, perhaps the height is not specified, but in the context, it is part of the rectangle, but not given, so probably we need to assume that the rectangle is square or something, but it's not said.

Perhaps " the opposite side" means the side parallel and opposite, but when we fold, the distance is the height, but in the crease length, it should be independent of height if we choose P appropriately, but not.

Let's calculate for a specific example.

Suppose the rectangle is square, say w = h = 6.

Then fold A(0,0) to P(p,6) on CD.

Then e = (p^2 + 36)/(2p)

f = (p^2 + 36)/(2*6) = (p^2 + 36)/12

Then L = sqrt(e^2 + f^2) = sqrt[ ((p^2+36)/(2p)) ^2 + ((p2+36)/12)^2 ] = (p2+36) sqrt( 1/(4p2) + 1/144 ) = (p2+36) sqrt( (36 + p2)/(144 p2) ) = (p2+36) /(12 p) sqrt(36 + p2) = (p2+36)^{3/2} / (12 p)

Then tanθ = p/h = p/6, so p = 6 tanθ

Then L = [ (6 tanθ)^2 + 36 ]^{3/2} / (12 * 6 tanθ) = [ 36 tan^2 θ + 36 ]^{3/2} / (72 tanθ) = [36 (tan^2 θ + 1)]^{3/2} / (72 tanθ) = (36^{3/2} sec^3 θ) / (72 tanθ) = (6^3 sec^3 θ) / (72 tanθ) = (216 sec^3 θ) / (72 tanθ) = 3 sec^3 θ / tanθ = 3 sec^2 θ / sinθ

= 3 / (cos^2 θ sinθ)
But this depends on θ, but for a square, it's not in terms of θ only.

But the problem has width 6, but height not specified, so probably not square.

Perhaps in the diagram, the height is such that when folded, the crease is parallel to the diagonal or something.

Another idea: perhaps " the opposite side" means that we fold so that the corner touches the opposite side at a specific point, but not specified.

Perhaps for the crease to be of length L, and angle θ, and the width is 6, but the height is determined by the fold.

Let's think about the geometry.

Let me denote the crease as a line with angle θ to the horizontal.

So the crease has slope m = - tanθ, if θ is with horizontal.

Then the crease is a line with slope - tanθ.

Now, when we fold A to P on CD, the crease is the perpendicular bisector of AP.

So the line AP must be perpendicular to the crease.

The crease has slope - tanθ, so its normal has slope 1/(tanθ) = cotθ.

So the line AP has slope cotθ.

Since A is at (0,0), so AP has equation y = (cotθ) x.

P is on CD, which is y = h, so at P, y = h, so h = cotθ * p, so p = h tanθ.

Same as before.

P is on CD, so x=p, y=h, and h = cotθ * p, so p = h tanθ, same as before.

So we are back to the same thing.

Now, the crease passes through the midpoint of A and P.

A(0,0), P(p,h) = (h tanθ, h), so midpoint M( (h tanθ)/2, h/2 )

The crease has slope - tanθ, so its equation is y - h/2 = - tanθ (x - (h tanθ)/2 )

This line must be within the rectangle, and it will intersect the boundaries.

The length of the crease within the rectangle is L, but it is the distance between the intersection points.

To find where it intersects the boundaries.

The line: y - h/2 = - tanθ (x - h tanθ /2)

Let me write as y = - tanθ x + (h tanθ) (tanθ /2) + h/2 = - tanθ x + (h tan^2 θ)/2 + h/2 = - tanθ x + (h/2) (tan^2 θ + 1) = - tanθ x + (h/2) sec^2 θ

So y = - (tanθ) x + (h/2) sec^2 θ

Now, this line has y-intercept when x=0, y = (h/2) sec^2 θ

X-intercept when y=0, 0 = - (tanθ) x + (h/2) sec^2 θ, so (tanθ) x = (h/2) sec^2 θ, so x = (h/2) sec^2 θ / tanθ = (h/2) sec^2 θ * cotθ = (h/2) secθ / sinθ = (h/2) / (sinθ cosθ) = h / (2 sinθ cosθ) = h / sin2θ

Same as e and f earlier.

So the intercepts are at x = e = h / sin2θ, y = f = (h/2) sec^2 θ

For the crease to be within the rectangle, we need 0 ≤ x ≤ w, 0 ≤ y ≤ h when on the boundaries, but since it's a line segment between the intercepts, and for it to be inside, we need that the x-intercept e ≤ w, and y-intercept f ≤ h, and also that the line is in the first quadrant, which it is for θ acute.

So e ≤ w, f ≤ h.

f ≤ h is (h/2) sec^2 θ ≤ h, so sec^2 θ ≤ 2, so |θ| ≤ 45°.

e = h / sin2θ ≤ w = 6.

Also, since P is on CD, and CD has length w=6, and p = h tanθ ≤ 6.

Now, for a given θ, and w=6, to have the crease within the rectangle, we need h / sin2θ ≤ 6, and h tanθ ≤ 6, and h >0, θ acute.

For the fold to be possible, these must be satisfied, but for a given θ, there is a range of h that satisfy.

However, in the problem, likely it is assumed that the crease is exactly filling the width or something, but not specified.

Perhaps for the fold to be made, and the crease to be the longest possible or shortest, but not said.

Another possibility: in the diagram, the crease is not between the intercepts on the axes, but may intersect other sides, but in this case, for the line to be within, if e > w or f > h, it may intersect the top or right side, but in that case, the crease length would be different.

For example, if f > h, then the crease may intersect the top side before the y-axis, but in this case, for θ < 45°, f < h, so it intersects the left side first.

If θ > 45°, then f > h, so the line may intersect the top side first.

For example, if θ > 45°, then f = (h/2) sec^2 θ > h, since secθ > √2 for θ > 45°.

Then the y-intercept is above the rectangle, so the crease will intersect the top side y=h and the bottom side or left side.

So let's consider that case.

Suppose θ > 45°, so f > h.

Then the line y = - (tanθ) x + (h/2) sec^2 θ

When y = h, then h = - (tanθ) x + (h/2) sec^2 θ

so (tanθ) x = (h/2) sec^2 θ - h = h ( sec^2 θ /2 - 1) = h ( (1/cos^2 θ)/2 - 1) = h (1/(2 cos^2 θ) - 1) = h (1 - 2 cos^2 θ)/(2 cos^2 θ) = h (- cos 2θ)/(2 cos^2 θ) since 1 - 2 cos^2 θ = - cos 2θ

So x = [ h (- cos 2θ) / (2 cos^2 θ) ] / tanθ = h (- cos 2θ) / (2 cos^2 θ) * cotθ = h (- cos 2θ) / (2 cos^2 θ) * cosθ / sinθ = h (- cos 2θ) / (2 cosθ sinθ) = h (- cos 2θ) / ( sin 2θ) = - h cot 2θ

Since sin2θ = 2 sinθ cosθ, so 2 cosθ sinθ = sin2θ, so 2 in denominator is for sin2θ? From above:

2 cosθ sinθ = sin2θ, so 2 cosθ sinθ in denominator, but we have 2 cosθ sinθ in the denominator? From:

x = h (- cos 2θ) / (2 cosθ sinθ) = h (- cos 2θ) / ( sin 2θ) because 2 sinθ cosθ = sin2θ, so 2 cosθ sinθ = sin2θ.

Yes, x = - h cot 2θ

Since θ > 45°, 2θ > 90°, so cot2θ < 0, so x > 0.

So at y=h, x = - h cot 2θ >0.

Similarly, when x=0, y = (h/2) sec^2 θ > h, so it intersects the top side at say G(x_g, h) with x_g = - h cot 2θ

Then it will intersect the bottom side or the right side.

When y=0, 0 = - (tanθ) x + (h/2) sec^2 θ, so x = [(h/2) sec^2 θ] / tanθ = (h/2) sec^2 θ cotθ = (h/2) secθ / sinθ = h / (2 sinθ cosθ) = h / sin2θ, same as e.

But if e > w, it may intersect the right side.

For the crease to be within, and for it to be the line segment from the intersection with one side to another.

In the diagram, likely for the given angle, it is configured so that the crease intersects the bottom and top sides.

So let's assume that, and for that, we need that it intersects bottom and top, which requires that when x=0, y > h, which is f > h, so θ > 45°, and when y=0, x >0, which is true, and also that the x-interect e = h / sin2θ ≤ w, and the x-coordinate at y=h is x_g = - h cot 2θ, and it must be between 0 and w.

From earlier, x_g = - h cot 2θ = h / ( - tan 2θ) since cot2θ = 1/tan2θ, and for θ > 45°, 2θ > 90°, tan2θ <0, so - tan2θ >0, so x_g = h / ( - tan 2θ) >0.

And x_g ≤ w.

Also, since it intersects bottom at E(e,0) with e = h / sin2θ, and top at G(x_g, h) with x_g = - h cot 2θ.

For the line segment to be within the rectangle, we need 0 ≤ e ≤ w, 0 ≤ x_g ≤ w, and since y from 0 to h, it should be fine.

Then the length of the crease is the distance between E(e,0) and G(x_g,h).

So L = sqrt( (x_g - e)^2 + (h - 0)^2 )

From earlier, in the second approach, when we assumed it intersects bottom and top, we had g and e, and L = (h / p) sqrt(p^2 + h^2) = h / (h tanθ) * h secθ = (1/ tanθ) * h / cosθ = h / ( sinθ cosθ) = 2h / sin2θ

From earlier calculation, L = (h / p) sqrt(p^2 + h2) and p = h tanθ, so L = (h / (h tanθ)) * sqrt( (h tanθ)^2 + h2) = (1/ tanθ) * h sqrt( tan^2 θ + 1) = (cosθ / sinθ) * h secθ = (cosθ / sinθ) * h / cosθ = h / sinθ

Yes, L = h / sinθ

Now, with the constraints.

P is at (p,h) = (h tanθ, h), and 0 ≤ p ≤ w = 6, so 0 ≤ h tanθ ≤ 6

Also, e = h / sin2θ = h / (2 sinθ cosθ) ≤ 6

x_g = - h cot 2θ = h / ( - tan 2θ) = h * ( - 1 / tan 2θ) since tan2θ <0 for θ>45°, so -1/tan2θ >0.

And 0 ≤ x_g ≤ 6

Also, since the crease is from (x_g,h) to (e,0), and for it to be in the rectangle, and for the fold to be proper, but mainly the range.

But still, for given θ, h can vary.

However, in this case, for the crease to be fully within, and for the fold to be made, but there is no additional constraint.

Perhaps in the problem, it is assumed that the crease is symmetric or something, but not specified.

Another idea: perhaps " the opposite side" means that we fold to a point on the opposite side, but in the diagram, it might be that the fold is such that the crease is parallel to the width or something, but not.

Let's look at the answer or think differently.

Perhaps the angle θ is not the angle of the crease, but the angle at which the folded part is, but that might be complicated.

Perhaps when they say "the crease L" and "angle θ", and from the diagram, θ is the angle between the crease and the side, but in the folded position.

Let's try to consider the unfolded position.

I recall that in such folds, the length of the crease can be given by L = w / (2 sin^2 (θ/2)) or something, but let's calculate for a specific value.

Suppose we take a rectangle with width w=6, and height h, and fold A to P on CD.

Suppose we choose P such that the crease is at 45 degrees.

So slope of crease is -1, so tanθ = 1, so θ = 45°.

Then from first approach, p = h tanθ = h *1 = h

e = h / sin(90°) = h / 1 = h

f = (h/2) sec^2 45° = (h/2) (√2)^2 = (h/2)*2 = h

Then L = sqrt(e^2 + f^2) = sqrt(h^2 + h^2) = h√2

Now, p = h ≤ 6, e = h ≤ 6, so h ≤ 6, so L = h√2 ≤ 6√2

But for different h, L is different.

If we choose h=6, then L = 6√2 ≈8.485

If we choose h=3, then p=3, e=3, f=3, L=3√2≈4.24

So different L for same θ.

But the problem asks for L in terms of θ, so for a given θ, L is not unique, unless there is a specific choice.

Perhaps for the fold to be made, and the crease to be the one that is shortest or longest, but not specified.

Perhaps in the diagram, the point P is chosen such that the crease is as long as possible or as short as possible.

For example, for fixed h, what P maximizes or minimizes L.

From L = \frac{ (p^2 + h^2)^{3/2} }{2 p h }

Let me write as L = \frac{ (p^2 + h^2)^{3/2} }{2 p h }

For fixed h, this is a function of p.

Let me set u = p, then L(u) = k (u^2 + c)^{3/2} / u, where c=h^2, k=1/(2h)

Then dL/du = k [ (3/2) (u^2 + c)^{1/2} * 2u * u - (u^2 + c)^{3/2} * 1 ] / u^2 = k (u^2 + c)^{1/2} [ 3u^2 - (u^2 + c) ] / u^2 = k (u^2 + c)^{1/2} (2u^2 - c) / u^2

Set dL/du = 0, so 2u^2 - c = 0, u^2 = c/2, u = h / √2

Then p = h / √2

Then tanθ = p/h = 1/√2, θ = 35.264° approximately.

Then L = \frac{ ((h^2/2) + h^2)^{3/2} }{2 (h/√2) h } = \frac{ (3h^2/2)^{3/2} }{2 h^2 / √2 } = \frac{ (3/2)^{3/2} h^3 }{2 h^2 / √2 } = (3/2)^{3/2} h / 2 * √2 = (3/2)^{3/2} h √2 / 2

(3/2)^{3/2} = (3/2) sqrt(3/2) = (3/2) sqrt(3)/sqrt(2) = (3 sqrt(6)) / 4? Let's calculate.

(3/2)^{3/2} = ((3/2)^3)^{1/2} = sqrt(27/8) = (3sqrt(3))/(2sqrt(2)) = (3 sqrt(6)) / 4 after rationalizing.

Then L = [ (3 sqrt(6)) / 4 ] * h * √2 / 2 = (3 sqrt(6) sqrt(2) h) / (42) = (3 sqrt(12) h) / 8 = (32 sqrt(3) h) / 8 = (6 sqrt(3) h) / 8 = (3 sqrt(3) h) / 4

Then with w=6, and p = h / √2 ≤ 6, so h ≤ 6√2, but L = (3 sqrt(3) / 4) h, which still depends on h.

So not good.

Perhaps for the maximum crease length.

For fixed h, L(u) = k (u^2 + c)^{3/2} / u

As u->0+, L-> ∞, as u->∞, L~ k u^{3/2} / u = k u^{1/2} -> ∞, and at u= h/√2, minimum.

So minimum at p = h/√2, L = (3 sqrt(3) / 4) h as above.

But still depends on h.

Perhaps in the problem, the rectangle is such that the height is equal to the width, but not specified.

Another idea: perhaps " the opposite side" means that we fold to a point on the opposite side, but the distance is fixed, but not.

Let's read the problem again: "one corner touches the opposite side" and "the perimeter of the small rectangle is 60 cm" but that was for a different problem. In this problem, it's "a rectangular piece of paper 6 inches wide" and "folded so that one corner touches the opposite side", and " the length of the crease L in terms of angle θ".

But in the earlier part, there was a diagram with 126 sticks, but that was for a different purpose.

For this problem, only the last part is relevant.

Perhaps for this fold, with width w=6, and angle θ, the height is not specified, but in the fold, when we fold, the vertical distance from A to CD is h, but when we fold to P, the straight-line distance is sqrt(p^2 + h^2), but not helpful.

Perhaps the angle θ is the angle that the line AP makes with the horizontal.

In many problems, θ is the angle that the line from the corner to the point on the opposite side makes with the horizontal.

Let me try that.

So assume that θ is the angle that AP makes with the horizontal.

So A(0,0), P on CD, so P(p,h), and angle that AP makes with horizontal is θ, so tanθ = h/p, so h = p tanθ.

Then the crease is the perpendicular bisector of AP.

Midpoint M(p/2, h/2) = (p/2, (p tanθ)/2)

Slope of AP is tanθ, so slope of perpendicular is -1/(tanθ) = - cotθ.

So crease: y - h/2 = - cotθ (x - p/2)

Now, it will intersect the boundaries.

Assume it intersects bottom and left sides.

So at F(0,f), E(e,0).

Then for F(0,f): dist to A and P equal.

Dist to A: f

Dist to P: sqrt((0-p)^2 + (f-h)^2) = sqrt(p^2 + (f-h)^2)

So f = sqrt(p^2 + (f-h)^2)

As before, f = (p^2 + h^2)/(2h)

Similarly for E(e,0): e = sqrt((e-p)^2 + (0-h)^2) , so e = (p^2 + h^2)/(2p)

Then L = sqrt(e^2 + f^2) = sqrt[ ((p^2 + h^2)/(2p)) ^2 + ((p^2 + h2)/(2h)) ^2 ] = (p^2 + h^2) / 2 * sqrt(1/p^2 + 1/h^2) = (p^2 + h2) / 2 * sqrt((h^2 + p^2)/(p^2 h^2)) = (p^2 + h2)^{3/2} / (2 p h)

Now with h = p tanθ, so p^2 + h2 = p^2 + (p^2 tan^2 θ) = p^2 (1 + tan^2 θ) = p^2 sec^2 θ

So L = [ (p^2 sec^2 θ)^{3/2} ] / (2 p * p tanθ) = [ p^3 sec^3 θ ] / (2 p^2 tanθ) = (p sec^3 θ) / (2 tanθ) = p sec^2 θ / (2 sinθ) since 1/tanθ = cotθ = cosθ/sinθ, so / (2 sinθ) * sec^2 θ = p / (2 sinθ cos^2 θ)
sec^3 θ / tanθ = sec^2 θ / (sinθ cosθ) * secθ / secθ = sec^2 θ / (sinθ cosθ) = 1/( sinθ cos^2 θ) = 4 / (2 sinθ cos^2 θ) = 4 / ( sin 2θ cosθ)
But still has p.

Now, p is the x-coordinate of P, and it must be between 0 and 6, but also the crease must be within, so e = (p^2 + h2)/(2p) = (p^2 sec^2 θ)/(2p) = (p sec^2 θ)/2 ≤ 6

f = (p2 + h2)/(2h) = (p2 sec2 θ)/(2 p tanθ) = (p sec2 θ)/(2 tanθ) = p / (2 sinθ) since sec2θ / tanθ = 1/(sinθ cosθ) = 2/ sin2θ, so f = p * 2 / sin2θ / 2 = p / sin2θ? Let's see:

f = (p2 + h2)/(2h) = (p2 sec2 θ)/(2 p tanθ) = (p sec2 θ)/(2 tanθ) = p / (2 sinθ) because sec2θ / tanθ = 1/(sinθ cosθ) = 2/ sin2θ, so f = p * 2 / (2 sin2θ) = p / sin2θ? No:

(p sec2 θ) / (2 tanθ) = p * (1/cos2 θ) / (2 sinθ / cosθ) = p * 1/(cos2 θ) * cosθ/(2 sinθ) = p / (2 cosθ sinθ) = p / sin2θ

Yes, f = p / sin2θ

Then f ≤ h = p tanθ, so p / sin2θ ≤ p tanθ

Assume p>0, so 1 / sin2θ ≤ tanθ

sin2θ = 2 sinθ cosθ, so 1/(2 sinθ cosθ) ≤ sinθ / cosθ

Multiply both sides by sinθ cosθ (assuming θ acute, so >0): 1/2 ≤ sin^2 θ

so sin^2 θ ≥ 1/2, |sinθ| ≥ 1/√2, so θ ≥ 45°.

Then e = (p sec2 θ)/2 ≤ 6

But p can be from 0 to 6, and for each p, L is different.

To have L in terms of θ only, we need to eliminate p.

From e = (p sec2 θ)/2 ≤ 6, but this is an inequality, not equality.

Perhaps for the crease to be exactly at the edge, but not specified.

Perhaps in the diagram, the crease passes through the corner or something, but unlikely.

Another thought: perhaps " the opposite side" means that we fold so that the corner touches the opposite side at the corner, but then it's not a crease, or if to the opposite corner, then it's the diagonal, but then the crease would be the perpendicular bisector of the diagonal, which is a line, but in a rectangle, it may not be within.

For example, if we fold A to C, then AP is the diagonal, midpoint is (3, h/2) if w=6, then perpendicular bisector, etc.

But then the angle is fixed, not given as θ.

So not.

Perhaps for this fold, with width 6, and angle θ, the height is 6, i.e., it's a square, but not specified.

Let's assume it is a square, w=h=6.

Then from earlier, with θ being the angle of the crease, L = 3 / (cos^2 θ sinθ) for the first approach, but that was for the square.

From earlier, for square, L = 3 sec^2 θ / sinθ = 3 / (cos^2 θ sinθ)

But this is for the square, and it depends on θ, but for a given fold, θ is determined, but the problem has θ as given, so for each θ, there is a L, but in a square, for each P, there is a θ, and L, so it is a function, but the width is 6, but for a square, it is defined.

But the problem doesn't say it's a square, so probably not.

Perhaps the height is such that the crease is horizontal or something, but not.

Let's look at the answer or think of the diagram.

In the diagram, there is an angle θ marked, and from the drawing, it seems that θ is at the top, between the crease and the top side.

So perhaps at the top side, the crease makes an angle θ with the top side.

So with the horizontal.

So same as before.

Perhaps with the vertical.

Let's assume that at the point where it intersects the top side, the angle with the top side is θ.

So if it intersects CD at G(g,h), then the slope of the crease is tan of the angle with horizontal.

If it makes angle θ with the top side, which is horizontal, so same as with horizontal.

So slope m, and tan of the angle with horizontal is |m|.

So same as before.

Perhaps it makes angle θ with the vertical, so with horizontal it is 90-θ.

But same thing.

I think I need to assume that for the crease to be valid, it must be that the height h is such that the crease is tangent to the top or something, but that doesn't make sense.

Perhaps in the fold, when we fold, the crease is such that it is the line that is fixed, and for the corner to touch the opposite side, it must be that the distance is correct, but in this case, it is.

Another idea: perhaps " the opposite side" means the side parallel and opposite, but when we fold, the point P is not on the side, but the corner touches the side at a point, but the side is not the top, but the side wall.

For example, fold A to a point on BC, the right side.

Then "opposite side" might mean the side not adjacent, which for A is BC or CD, but CD is opposite, BC is adjacent in a way.

In a rectangle, for corner A, adjacent sides are AB and AD, opposite side is CD.

But if we fold to a point on BC, then it's not on the opposite side.

The problem says "the opposite side", so likely CD.

Perhaps in some interpretations, but probably not.

Let's try to search for a standard formula.

Upon thinking, I recall that for a rectangle of width w, when you fold a corner to a point on the opposite side, the length of the crease is given by L = w / (2 sin^2 (α)) or something, but let's see.

Perhaps the angle θ is related to the fold.

Let's consider the small rectangle in the earlier problem, but that was different.

Perhaps for this problem, the small rectangle mentioned in the diagram is not relevant, but in this case, it's separate.

Another thought: in the diagram for this problem, there is a small rectangle or something, but from the image, it's just the folded paper, and a crease, and angle θ.

Perhaps the "small rectangle" is the crease or something, but not.

Let's read the user's message: "Bridget folds a square piece of paper 6 inches wide so that one corner touches the opposite side. The length in inches of the crease L in terms of angle θ is"

And "see image 1", but no image, but from the text, it's a square paper, 6 inches wide, so it is a square! I missed that.

In the problem, it is a square paper! " a square piece of paper 6 inches wide" so since it's square, width = height = 6 inches.

I missed that. So it is a square.

So w = h = 6 inches.

Then we can proceed.

So let to the square 6x6.

A(0,0), B(6,0), C(6,6), D(0,6).

Fold A(0,0) to P(p,6) on CD, so 0 ≤ p ≤ 6.

Then crease is perpendicular bisector of AP.

Midpoint M(p/2, 3)

Slope of AP is (6-0)/(p-0) = 6/p

So slope of perpendicular is -p/6

Equation: y - 3 = (-p/6) (x - p/2)

Now, it will intersect the boundaries.

Assume it intersects bottom AB and left side AD.

So on AB (y=0): 0 - 3 = (-p/6) (x - p/2)

-3 = (-p/6) (x - p/2)

Multiply both sides by -6/p: ( -3 * -6/p ) = x - p/2

18/p = x - p/2

x = 18/p + p/2

Similarly, on AD (x=0): y - 3 = (-p/6) (0 - p/2) = (-p/6)(-p/2) = p^2/12

So y = 3 + p^2/12

Now, the crease from F(0, f) with f = 3 + p^2/12

To E(e,0) with e = 18/p + p/2

Then length L = sqrt( (e - 0)^2 + (0 - f)^2 ) = sqrt( e^2 + f^2 )

But f is y-intercept, e is x-intercept, and since the line has negative slope, the distance from (0,f) to (e,0) is sqrt(e^2 + f^2), yes.

So L = sqrt( e^2 + f^2 ) = sqrt{ \left( \frac{18}{p} + \frac{p}{2} \right)^2 + \left( 3 + \frac{p^2}{12} \right)^2 }

Now, the angle θ. In the diagram, likely θ is the angle that the crease makes with the horizontal.

The crease has slope m = -p/6, so tanφ = |m| = p/6, where φ is the angle with horizontal.

So let’s set θ = φ, so tanθ = p/6, so p = 6 tanθ

Then substitute.

e = 18/p + p/2 = 18/(6 tanθ) + (6 tanθ)/2 = 3 / tanθ + 3 tanθ = 3 ( \frac{1}{\tan\theta} + \tan\theta ) = 3 ( \cot\theta + \tan\theta )

f = 3 + p^2/12 = 3 + (36 tan^2 θ)/12 = 3 + 3 tan^2 θ = 3 (1 + tan^2 θ) = 3 sec^2 θ

Then L = sqrt( e^2 + f^2 ) = sqrt{ [3 (\cot\theta + \tan\theta)]^2 + [3 \sec^2 \theta]^2 } = 3 sqrt{ ( \cot\theta + \tan\theta )^2 + \sec^4 \theta }

Now, \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin\theta \cos\theta} = \frac{1}{\sin\theta \cos\theta} = \frac{2}{\sin 2\theta}

And \sec^4 \theta = \frac{1}{\cos^4 \theta}

So inside: \left( \frac{2}{\sin 2\theta} \right)^2 + \frac{1}{\cos^4 \theta} = \frac{4}{\sin^2 2\theta} + \frac{1}{\cos^4 \theta}

Now sin 2θ = 2 sinθ cosθ, so sin^2 2θ = 4 sin^2 θ cos^2 θ

So \frac{4}{4 \sin^2 \theta \cos^2 \theta} + \frac{1}{\cos^4 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} + \frac{1}{\cos^4 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} + \frac{\sin^2 \theta}{\sin^2 \theta \cos^4 \theta} wait no.

Common denominator is sin^2 θ cos^4 θ

First term: \frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta \cos^4 \theta} ? No.

\frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}

Second term: \frac{1}{\cos^4 \theta} = \frac{1}{\cos^4 \theta}

So sum: \frac{1}{\sin^2 \theta \cos^2 \theta} + \frac{1}{\cos^4 \theta} = \frac{1}{\cos^2 \theta} \left( \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} \right) = \frac{1}{\cos^2 \theta} \left( \csc^2 \theta + \sec^2 \theta \right)

\csc^2 θ + sec^2 θ = 1/sin^2 θ + 1/cos^2 θ = (cos^2 θ + sin^2 θ)/(sin^2 θ cos^2 θ) = 1/(sin^2 θ cos^2 θ)

So sum = \frac{1}{\cos^2 \theta} \cdot \frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^4 \theta}

From above:

Sum = \frac{1}{\sin^2 \theta \cos^2 \theta} + \frac{1}{\cos^4 \theta} = \frac{1}{\cos^2 \theta} \left( \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} \right) = \frac{1}{\cos^2 \theta} \cdot \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\cos^2 \theta} \cdot \frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^4 \theta}

Yes.

So inside the sqrt: \frac{1}{\sin^2 \theta \cos^4 \theta}

So L = 3 sqrt{ \frac{1}{\sin^2 \theta \cos^4 \theta} } = 3 / ( |sin θ| cos^2 θ )

Since θ is acute, sinθ >0, so L = 3 / (sinθ cos^2 θ)

We can write as L = 3 / (sinθ (1 - sin^2 θ)) but perhaps leave it.

Note that 1/(sinθ cos^2 θ) = sinθ / (sin^2 θ cos^2 θ) = sinθ / ( (1/4) sin^2 2θ ) = 4 sinθ / sin^2 2θ, but not simpler.

Since the problem likely wants a simplified expression.

But this is in terms of θ.

Now, to confirm, if θ=45°, tanθ=1, p=6*1=6, then e = 18/6 + 6/2 = 3 + 3 =6, f=3 + 36/12=3+3=6, L=sqrt(6^2+6^2)=6√2

From formula, L=3 / (sin45 cos^2 45) = 3 / ( (√2/2) * (1/2) ) since cos45=√2/2, cos^2 45= (2/4)/1? (√2/2)^2 = (2/4)=1/2

sin45=√2/2

So 3 / ( (√2/2) * (1/2) ) = 3 / ( √2 / 4 ) = 3 * 4 / √2 = 12 / √2 = 12 √2 / 2 = 6 √2, correct.

Another check: if p=3, tanθ=p/6=0.5, θ=arctan(0.5)≈26.565°

e=18/3 + 3/2=6+1.5=7.5

f=3 + 9/12=3+0.75=3.75

L=sqrt(7.5^2 + 3.75^2)=sqrt(56.25 + 14.0625)=sqrt(70.3125)=8.385? 7.5^2=56.25, 3.75^2=14.0625, sum 70.3125, sqrt=8.385? 8.385^2=70.308225, close to 70.3125, yes.

From formula, θ=arctan(0.5), tanθ=0.5, sinθ=1/√5? tanθ=0.5, so sinθ= tanθ / sqrt(1+tan^2 θ) = 0.5 / sqrt(1+0.25) = 0.5 / sqrt(1.25) = 0.5 / (√5 / 2) sqrt(1.25)=sqrt(5/4)=√5 / 2? 1.25=5/4, sqrt(5/4)=√5 / 2

So sinθ= (0.5) / (√5 / 2) ? No, sinθ = tanθ / sqrt(1+tan^2 θ) = (1/2) / sqrt(1 + 1/4) = (1/2) / sqrt(5/4) = (1/2) / (√5 / 2) = 1/√5

Similarly, cosθ=1 / sqrt(1+tan^2 θ) = 1 / sqrt(1+0.25) = 1 / (√5 / 2) * 2/2? sqrt(1.25)=√(5/4)=√5 / 2, so cosθ= adjacent/hypotenuse, if opp=1, adj=2, hyp=√5, so cosθ=2/√5

Then L=3 / (sinθ cos^2 θ) = 3 / [ (1/√5) * (4/5) ] = 3 / (4 / (5√5)) = 3 * 5√5 / 4 = 15 √5 / 4

Now √5≈2.236, 15*2.236=33.54, /4=8.385, matches the earlier calculation.

So correct.

Now, we can write it as L = \frac{3}{\sin \theta \cos^2 \theta}

To write it neatly, perhaps as L = 3 \csc \theta \sec^2 \theta

But probably the fractional form is fine.

Since the problem might expect a simplified form.

Note that \frac{1}{\sin \theta \cos^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos^2 \theta} = \frac{\sin \theta}{\cos^2 \theta} + \frac{\cos \theta}{\sin \theta} = \tan \theta \sec \theta + \cot \theta, but not simpler.

So I think \frac{3}{\sin \theta \cos^2 \theta} is acceptable.

But in the expression, it's in terms of θ, and we have it.

Now, the problem is to box the answer.

So for the square paper of side 6, crease length L = \frac{3}{\sin \theta \cos^2 \theta}

We can write as L = 3 \csc \theta \sec^2 \theta, but perhaps the fractional form is better.

To make it a single fraction.

Note that \frac{1}{\sin \theta \cos^2 \theta} = \frac{1}{\sin \theta (1 - \sin^2 \theta)} but not helpful.

So I think it's fine.

So the answer is $\frac{3}{\sin \theta {\cos }^2\theta }$

But let me confirm the angle.

In the diagram, θ is the angle that the crease makes with the horizontal, which is what we assumed, since tanθ = p/6, and p/6 is the slope magnitude.

In the fold, if the crease is sloping down to the right, the angle with horizontal is acute, so θ as defined.

So I think this is correct.

So the length of the crease is $\frac{3}{\sin \theta {\cos }^2\theta }$ inches.

But to box it.

We can write it as $3\csc \theta {\sec }^2\theta$, but perhaps the first form is clearer.

Since the problem gives angle θ, and asks for L in terms of θ, this should be it.

So \boxed{\dfrac{3}{\sin \theta \cos^{2} \theta}}

To match the format.

In the expression, it's positive, and for θ acute, it's fine.

So I think this is the answer.