[Guest]

A) you have a regular n-gon of side-length 's'. find the area of the region that contains all the points inside the n-gon that are closer to the center of the n-gon than they are to any of the sides. (You may find it easier to start with a square.)

note:depending on how you do it, you may wind up with an integral that cannot be evaluated. in that case, just leave it as it is.

B) you are given a horizontal line 'a' units above the x axis. Find the area of the region that contains all the points above the x axis that are closer to the origin than they are to any of the horizontal line.

note: this one is actually easier than A, so you may want to do it first.

C)you have a line given in the form of y=mx+b. Find the area of the region that contains all the points between the given line and the x axis that are closer to the origin than they are to any of the horizontal line.

[k-man]

The area of interest will by a regular 2n-sided polygon that has a circumradius of s/4. Using the formula for the area we get A= ns²/16 * sin(pi/n)

[k-man]

Ignore the previous post. I messed up.

The area of interest will by a regular 2n-sided polygon that has a circumradius of a/2, where a is the apothem of the given n-sided polygon.

a = s/(2*tan(pi/n))

Using the formula for the area we get A = ns² * sin(pi/n) / (4 * tan²(pi/n))

[Guest]

Ignore the previous post. I messed up.

The area of interest will by a regular 2n-sided polygon that has a circumradius of a/2, where a is the apothem of the given n-sided polygon.

a = s/(2*tan(pi/n))

Using the formula for the area we get A = ns² * sin(pi/n) / (4 * tan²(pi/n))

Sorry, but that's not it. in fact, the "area of interest" is not a polygon.

[Guest]

I'll attempt to integrate the curve that represents points equidistant to the origin and the line y=a. All points above the x-axis are a maximum of a units from the line y=a so the bounds of integration will be -a to a. The distance of any arbitrary point (x,y) to the line a is |a-y|. For (x,y) below the line a, this is simply a-y. The distance between that point and the origin is (x² + y²)^1/2. Equating these two distance equations yields a curve that encloses points that are closer to the origin than to the line a. so algebra is done:

a-y = (x² + y²)^1/2

a² - 2ay + y²= x² + y²

y = a/2 - x²/(2a²)

integrating y from -a to a yields a²

[Guest]

I'll attempt to integrate the curve that represents points equidistant to the origin and the line y=a. All points above the x-axis are a maximum of a units from the line y=a so the bounds of integration will be -a to a. The distance of any arbitrary point (x,y) to the line a is |a-y|. For (x,y) below the line a, this is simply a-y. The distance between that point and the origin is (x² + y²)^1/2. Equating these two distance equations yields a curve that encloses points that are closer to the origin than to the line a. so algebra is done:

a-y = (x² + y²)^1/2

a² - 2ay + y²= x² + y²

y = a/2 - x²/(2a²)

integrating y from -a to a yields a²

very close and you definitely have the right idea, but you should recheck your work...particularly towards the end.

[Guest]

very close and you definitely have the right idea, but you should recheck your work...particularly towards the end.

whoops

I meant a²-a/3

[unreality]

A) you have a regular n-gon of side-length 's'. find the area of the region that contains all the points inside the n-gon that are closer to the center of the n-gon than they are to any of the sides. (You may find it easier to start with a square.)

note:depending on how you do it, you may wind up with an integral that cannot be evaluated. in that case, just leave it as it is.

B) you are given a horizontal line 'a' units above the x axis. Find the area of the region that contains all the points above the x axis that are closer to the origin than they are to any of the horizontal line.

note: this one is actually easier than A, so you may want to do it first.

C)you have a line given in the form of y=mx+b. Find the area of the region that contains all the points between the given line and the x axis that are closer to the origin than they are to any of the horizontal line.

If you take the polygon and divide it into isoceles triangles (also equilateral in the case of the hexagon) with the narrow point at the center and the wide face (length s) being one of the faces of the polygon. The two equal lines going from the center to this edge (the isoceles legs) are of length R.

It's possible to figure out R if you know n (the degree of the polygon) because the angle in this triangle is 2pi/n, and the sine of half of that (pi/n) equals the "opposite" (s/2) over the "hypotenuse" ®. so SIN(pi/n) = s/(2R)... consequently R = s/(2sin(pi/n))

Anyway, my point with dividing this up into triangles is that for any point in the triangle, the side of the polygon that it's closest to is the one formed by the base of the triangle (the leg with length s). This makes a kind of rotational symmetry so that whatever shape gives it best for this triangle is repeated rotationally to form the overall region of points. In other words I can find the area of the points closer to the center than the edge, then multiply it by n for the total area.

So if you think about an arbitrary point in the triangle, it forms a line between the "center" (which is actually the apex of the triangle) and the edge. Say that this line is of length L, and the point is at position x. If x > L/2, the point is outside of the region. If x < L/2, the point is within the region.

If the line happens to be along the edge of the triangle, L=R. Beyond that, it changes as the angle changes, from 0 radians away from the L=R line, up until halfway across, so pi/n radians. After that L increases again back to R.

So in fact, dividing the triangle in half is even better, then multiply that area by 2, then by n for the whole polygon.

The formula for polar area (easily derivable from the formula of a circular sector then applied to this current situation) will be the integral from theta=alpha to theta=beta, of (L(theta)/2)^2 / 2 (dtheta)

Multiplying all of that by 2n yields:

n/4 * integral (from 0 to pi/n) of L(theta)^2 dtheta

Now to find L(theta).

We know L(0) = R = s/(2sin(pi/n))

As theta increases, we have a right triangle, with hypotenuse of length R, angle of theta, and we want the "adjacent" leg. cos(theta) = L / R

Duh. so L simply equals R*cos(theta). That's why, when theta=0, L is just R.

In other words, L(theta) = s/(2sin(pi/n))*cos(theta)

and L(theta)^2 = (s^2)/(4sin^2(pi/n)) * cos^2(theta)

Pulling out the R^2 constant in front, we get:

Area of Region = n/4 * (s^2)/(4sin^2(pi/n)) * integral (from 0 to pi/n) of cos^2(theta) dtheta

That's solvable with the half angle formula cos(x)^2 = 1/2 + cos(2x)/2

Doing the math leads to the indefinite integral theta/2 + sin(2*theta)/4. Then evaluate it from our bounds, theta=0 to theta=pi/n and get:

pi/(2n) + sin(2pi/n)/4 - 0 - 0

multiply that by the (n/4)R^2 constant pulled out and get:

(n*s*s)/(16sin(pi/n)^2) * (pi/(2n) + (1/4)sin(2pi/n))

[Guest]

My final answer was the integral from t= pi/n to t= (1-1/n)pi of [na/(1+ sin(t))]dt

To arrive at this conclusion I took a similar path as I used for part B. The difference is that I started by breaking the n-gon into n congruent triangles. The reason for this is that if you treat the n-gon like a curve, then finding the distance from the n-gon to any point is the same as finding the distance between the point and the part of the curve closest to the point. The part of the curve closest to the point has a gradient pointing directly at the point. In the case of a polygon this just means that the shortest line that connects the point and the polygon will be perpendicular to the point on the curve where it touches. I hope I’ve made it obvious enough at this point why the shortest line will stay within one triangle.

At this point I used the fact that the triangles are congruent to simplify the matter to just finding the desired area of one triangle and multiplying by the number of triangles. This is where it becomes similar to my answer for part B.

I decided to use polar coordinates this time since the bounds are not end points of a line, but edges of a triangle. The distance from the origin to point (r,t) is simply r. To find the distance from that point to the n-gon, that is, the distance from that point to the side of the triangle opposite the origin, I rotated the triangle to be parallel to the x-axis. Then the equation was simply

d = a-y = a- rsin(t)

equating the two to get the boundary of the desired area I got

r = d = a- rsin(t)

r= a/(1+ sin(t)

so now that I knew what I was integrating the only tricky part left was determining the bounds. I would be integrating from some angle t₀ to the angle t₀ + t₁. Due to the orientation of the triangle on the axis, 2t0 + t1= pi in radians. So t0 = (pi-t1)/2 and t0 + t1 = (pi+t1)/2. These are my bounds of integration but they aren’t in terms of n. t₁ is simply t_n/n where t_n is the sum of the interior angles of an n-gon. tn = (n-2)pi so upon substitution of (1-2/n)pi for t₁, my bounds of integration became pi/n and pi-pi/n. I then moved the n inside the integral.

I was gonna include images but I'm computerlogically challenged.

[unreality]

B) you are given a horizontal line 'a' units above the x axis. Find the area of the region that contains all the points above the x axis that are closer to the origin than they are to any of the horizontal line.

position of point: (x,y)

distance from origin: D = sqrt(x^2 + y^2)

distance from horizontal line: H = a-y

Additionally we know that y>0 because all points are above the x-axis.

And that y<a because above the line would mean it's definitely closer to the line than to the origin.

To find the region of points closer to the origin than the line, D must be less than H. To find the boundary case, I will set D = H and reduce...

sqrt(x^2 + y^2) = a-y

x^2 + y^2 = a^2 - 2ay + y^2

x^2 = a^2 - 2ay

2ay = a^2 - x^2

y = a/2 - (x^2)/(2a)

Visually, it means y=a/2 when x is 0, and as x goes to the left or right, y drops away. The upper boundary i determined (y<a) is satisfied automatically by this. But the lower boundary (y>0) must be satisfied by finding the points of intersection. If there was no restriction, the area would just be infinity. Here goes:

0 = y = a/2 - (x^2)/(2a)

a^2 - x^2 = 0

a^2 = x^2

x = (+/-) a

So the region starts at (-a,0), curves upward, topping off at (0, a/2) and then hitting the x axis again at (a,0).

Now that I know the lower and upper bounds x=-a and x=a, I can just integrate y to find the area:

integral (from -a to a) of (a/2 - (x^2)/(2a)) dx

Preserving symmetry:

= 2 * integral (from 0 to a) of y(x) dx

pulling out the 1/2:

= 1 * integral (from 0 to a) of (a - x^2/a) dx

= (ax - (x^3)/(3a)) evaluated from 0 to a

= a*a - (a^3)/(3a)

= a^2 - a^2/3

= (2/3)*a^2

[Guest]

If you take the polygon and divide it into isoceles triangles (also equilateral in the case of the hexagon) with the narrow point at the center and the wide face (length s) being one of the faces of the polygon. The two equal lines going from the center to this edge (the isoceles legs) are of length R.

It's possible to figure out R if you know n (the degree of the polygon) because the angle in this triangle is 2pi/n, and the sine of half of that (pi/n) equals the "opposite" (s/2) over the "hypotenuse" ®. so SIN(pi/n) = s/(2R)... consequently R = s/(2sin(pi/n))

Anyway, my point with dividing this up into triangles is that for any point in the triangle, the side of the polygon that it's closest to is the one formed by the base of the triangle (the leg with length s). This makes a kind of rotational symmetry so that whatever shape gives it best for this triangle is repeated rotationally to form the overall region of points. In other words I can find the area of the points closer to the center than the edge, then multiply it by n for the total area.

So if you think about an arbitrary point in the triangle, it forms a line between the "center" (which is actually the apex of the triangle) and the edge. Say that this line is of length L, and the point is at position x. If x > L/2, the point is outside of the region. If x < L/2, the point is within the region.

If the line happens to be along the edge of the triangle, L=R. Beyond that, it changes as the angle changes, from 0 radians away from the L=R line, up until halfway across, so pi/n radians. After that L increases again back to R.

So in fact, dividing the triangle in half is even better, then multiply that area by 2, then by n for the whole polygon.

The formula for polar area (easily derivable from the formula of a circular sector then applied to this current situation) will be the integral from theta=alpha to theta=beta, of (L(theta)/2)^2 / 2 (dtheta)

Multiplying all of that by 2n yields:

n/4 * integral (from 0 to pi/n) of L(theta)^2 dtheta

Now to find L(theta).

We know L(0) = R = s/(2sin(pi/n))

As theta increases, we have a right triangle, with hypotenuse of length R, angle of theta, and we want the "adjacent" leg. cos(theta) = L / R

Duh. so L simply equals R*cos(theta). That's why, when theta=0, L is just R.

In other words, L(theta) = s/(2sin(pi/n))*cos(theta)

and L(theta)^2 = (s^2)/(4sin^2(pi/n)) * cos^2(theta)

Pulling out the R^2 constant in front, we get:

Area of Region = n/4 * (s^2)/(4sin^2(pi/n)) * integral (from 0 to pi/n) of cos^2(theta) dtheta

That's solvable with the half angle formula cos(x)^2 = 1/2 + cos(2x)/2

Doing the math leads to the indefinite integral theta/2 + sin(2*theta)/4. Then evaluate it from our bounds, theta=0 to theta=pi/n and get:

pi/(2n) + sin(2pi/n)/4 - 0 - 0

multiply that by the (n/4)R^2 constant pulled out and get:

(n*s*s)/(16sin(pi/n)^2) * (pi/(2n) + (1/4)sin(2pi/n))

good thinking, but unfortunately, it's not quite right.

for a square of side-length 2, the area of the region is 4*(4*sqrt(2)-5)/3

[Guest]

If you take the polygon and divide it into isoceles triangles (also equilateral in the case of the hexagon) with the narrow point at the center and the wide face (length s) being one of the faces of the polygon. The two equal lines going from the center to this edge (the isoceles legs) are of length R.

It's possible to figure out R if you know n (the degree of the polygon) because the angle in this triangle is 2pi/n, and the sine of half of that (pi/n) equals the "opposite" (s/2) over the "hypotenuse" ®. so SIN(pi/n) = s/(2R)... consequently R = s/(2sin(pi/n))

Anyway, my point with dividing this up into triangles is that for any point in the triangle, the side of the polygon that it's closest to is the one formed by the base of the triangle (the leg with length s). This makes a kind of rotational symmetry so that whatever shape gives it best for this triangle is repeated rotationally to form the overall region of points. In other words I can find the area of the points closer to the center than the edge, then multiply it by n for the total area.

So if you think about an arbitrary point in the triangle, it forms a line between the "center" (which is actually the apex of the triangle) and the edge. Say that this line is of length L, and the point is at position x. If x > L/2, the point is outside of the region. If x < L/2, the point is within the region.

If the line happens to be along the edge of the triangle, L=R. Beyond that, it changes as the angle changes, from 0 radians away from the L=R line, up until halfway across, so pi/n radians. After that L increases again back to R.

So in fact, dividing the triangle in half is even better, then multiply that area by 2, then by n for the whole polygon.

The formula for polar area (easily derivable from the formula of a circular sector then applied to this current situation) will be the integral from theta=alpha to theta=beta, of (L(theta)/2)^2 / 2 (dtheta)

Multiplying all of that by 2n yields:

n/4 * integral (from 0 to pi/n) of L(theta)^2 dtheta

Now to find L(theta).

We know L(0) = R = s/(2sin(pi/n))

As theta increases, we have a right triangle, with hypotenuse of length R, angle of theta, and we want the "adjacent" leg. cos(theta) = L / R

Duh. so L simply equals R*cos(theta). That's why, when theta=0, L is just R.

In other words, L(theta) = s/(2sin(pi/n))*cos(theta)

and L(theta)^2 = (s^2)/(4sin^2(pi/n)) * cos^2(theta)

Pulling out the R^2 constant in front, we get:

Area of Region = n/4 * (s^2)/(4sin^2(pi/n)) * integral (from 0 to pi/n) of cos^2(theta) dtheta

That's solvable with the half angle formula cos(x)^2 = 1/2 + cos(2x)/2

Doing the math leads to the indefinite integral theta/2 + sin(2*theta)/4. Then evaluate it from our bounds, theta=0 to theta=pi/n and get:

pi/(2n) + sin(2pi/n)/4 - 0 - 0

multiply that by the (n/4)R^2 constant pulled out and get:

(n*s*s)/(16sin(pi/n)^2) * (pi/(2n) + (1/4)sin(2pi/n))

"So if you think about an arbitrary point in the triangle, it forms a line between the "center" (which is actually the apex of the triangle) and the edge. Say that this line is of length L, and the point is at position x. If x > L/2, the point is outside of the region. If x < L/2, the point is within the region."

In this paragraph you seem to assume the distance between the point and the edge is L-x. If you think of a point that's at the very edge of the triangle, with L= 2x then by your assumption this point is on the edge of the desired area. This isn't the case. The distance from that point to the apex of the triangle is indeed x. But if you draw a line from the point to the hexagon such that the line is perpendicular to the edge, this line, the edge, and the side of the triangle form a right triangle. The edge of the triangle is the hypotenuse so it is clearly shorter than the perpendicular line. So the distance from the point to the edge is less than x.

[Guest]

apparently me and magician were on the same wavelength with those last two posts.

[unreality]

"So if you think about an arbitrary point in the triangle, it forms a line between the "center" (which is actually the apex of the triangle) and the edge. Say that this line is of length L, and the point is at position x. If x > L/2, the point is outside of the region. If x < L/2, the point is within the region."

In this paragraph you seem to assume the distance between the point and the edge is L-x. If you think of a point that's at the very edge of the triangle, with L= 2x then by your assumption this point is on the edge of the desired area. This isn't the case. The distance from that point to the apex of the triangle is indeed x. But if you draw a line from the point to the hexagon such that the line is perpendicular to the edge, this line, the edge, and the side of the triangle form a right triangle. The edge of the triangle is the hypotenuse so it is clearly shorter than the perpendicular line. So the distance from the point to the edge is less than x.

Ah there it is, thanks! Back to the drawing board If my answer to B is right though I think I can fairly easily extrapolate it to part C

[Guest]

whoops

I meant a²-a/3

closer this time, but still a little off... If I had to guess where you made your mistake, I would say it was either in your calculus, or you simply had the wrong integrand.

Edited December 23, 2010 by magician

[Guest]

closer this time, but still a little off... If I had to guess where you made your mistake, I would say it was either in your calculus, or you simply had the wrong integrand.

Was my answer for B right? If so then this is probably an algebra mistake.

[Guest]

Was my answer for B right? If so then this is probably an algebra mistake.

no, it was close though.

[Guest]

no, it was close though.

whoops that was a mistype. I didn't mean to ask about my answer to B since that was what you were referencing in the first place. I meant my answer to A.

[Guest]

whoops that was a mistype. I didn't mean to ask about my answer to B since that was what you were referencing in the first place. I meant my answer to A.

It wasn't quite right. Your bounds were off (although I don't know if the integrals are equal) and you forgot that you need to integrate in polar coordinates. when integrating in polar coordinates, you must square the integrand and divide it by 2. Other than that, it seems about right.

wait a sec, the bounds are alright after all . I just used slightly different ones, but they are equal anyway.

[unreality]

what about my answer to B?

[Guest]

what about my answer to B?

I'm sorry, I overlooked that post. Yes, you got it! Nice work!

[Guest]

It wasn't quite right. Your bounds were off (although I don't know if the integrals are equal) and you forgot that you need to integrate in polar coordinates. when integrating in polar coordinates, you must square the integrand and divide it by 2. Other than that, it seems about right.

wait a sec, the bounds are alright after all . I just used slightly different ones, but they are equal anyway.

I felt like I was forgetting a Jacobian multiple but I didn't know for sure what it was so I just ignored it in hopes that it would go away. Thanks! I'll give it another shot.

[Guest]

I felt like I was forgetting a Jacobian multiple but I didn't know for sure what it was so I just ignored it in hopes that it would go away. Thanks! I'll give it another shot.

well, being a high school student, I haven't heard of a Jacobian multiple, but I'll just assume it makes sense in this context...