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https://i.imgur.com/FZ3XYyI.png
I believe this is the correct answer. Basically:
* The ellipse is equal to A[sub]1[/sub]+A[sub]2[/sub]+A[sub]3[/sub]+A[sub]4[/sub]+2X, where 2X is the little eye-shaped thing in the center of the ellipse.
* A[sub]1[/sub]+A[sub]2[/sub]+A[sub]3[/sub]+A[sub]4[/sub] is just the top half of the big ellipse, so it's equal to πr[sup]2[/sup]/2.
* The line that divides X appears to just be the side of a square you'd inscribe in the smaller circle. So you can get the area of x by taking the smaller circle fragment and subtracting the inscribed square fragment.
So assuming my math is right, this is the correct answer.
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Xhin
I believe this is the correct answer. Basically:

The ellipse is equal to A1+A2+A3+A4+2X, where 2X is the little eye-shaped thing in the center of the ellipse.

A1+A2+A3+A4 is just the top half of the big ellipse, so it's equal to πr2/2.

The line that divides X appears to just be the side of a square you'd inscribe in the smaller circle. So you can get the area of x by taking the smaller circle fragment and subtracting the inscribed square fragment.

So assuming my math is right, this is the correct answer.

Posted August 17th
by Xhin

Please bear with me since I don't know the math syntax here.
To start with, the area of an ellipse is defined as:
[y](pi)*a*b[/y]
where [y]a[/y] and [y]b[/y] are the two axes of the ellipse. We will define [y]a[/y] as the longer axis and [y]b[/y] as the shorter axis for this explanation.
To find [y]a[/y], we realize that it spans two of the radius of the smaller circle, or [y]2*.5r = r[/y]; to get the rest of [y]a[/y], we need to find the hypotenuse of the triangle formed if you draw a straight line from corner to corner of one of those sections. The triangle formed will be a 45-45-90 triangle, which will yield a hypotenuse of [y]sqrt(2) * leg length[/y], or in this case [y]sqrt(2)*.5r = r(sqrt(2))/2[/y]. This yields [y]a = r((1+sqrt(2))/2.
Axis [y]b[/y] is somewhat more complicated: we know again that we start with [y]r[/y] given that the inner circle is half the radius of the outer circle, but the middle bit is a bit tricky. To find it, we need to focus on one of the smaller sections (e.g. Section 3), and the aforementioned triangle. We know that the hypotenuse is [y]r*sqrt(2).2[/y]; to get the remaining bit, the distance between the hypotenuse and the edge of the circle, we subtract that from .5r:
[y](r/2 - (r*sqrt(2)/4) = (2r-(r*sqrt(2)))/4 = r(2-sqrt(2))/4[/y]
Now, back to our original formula:
[y]Area(ellipse) = (pi)*a*b = (pi) * r((1+sqrt(2))/2 * r(2-sqrt(2))/4 = (pi) * (r^2)*sqrt(2)/8[/y]
I could be wrong (and please for the love of God someone help me with the LaTEX, I think that's what it's called), but my answer is [y]A=pi*(r^2)*sqrt(2)/8[/y]

Black Yoshi
Please bear with me since I don't know the math syntax here.

To start with, the area of an ellipse is defined as:

(pi)*a*b
where

a and

b are the two axes of the ellipse. We will define

a as the longer axis and

b as the shorter axis for this explanation.

To find

a, we realize that it spans two of the radius of the smaller circle, or

2*.5r = r; to get the rest of

a, we need to find the hypotenuse of the triangle formed if you draw a straight line from corner to corner of one of those sections. The triangle formed will be a 45-45-90 triangle, which will yield a hypotenuse of

sqrt(2) * leg length, or in this case

sqrt(2)*.5r = r(sqrt(2))/2. This yields

a = r((1+sqrt(2))/2.
Axis

b is somewhat more complicated: we know again that we start with

r given that the inner circle is half the radius of the outer circle, but the middle bit is a bit tricky. To find it, we need to focus on one of the smaller sections (e.g. Section 3), and the aforementioned triangle. We know that the hypotenuse is

r*sqrt(2).2; to get the remaining bit, the distance between the hypotenuse and the edge of the circle, we subtract that from .5r:

(r/2 - (r*sqrt(2)/4) = (2r-(r*sqrt(2)))/4 = r(2-sqrt(2))/4
Now, back to our original formula:

Area(ellipse) = (pi)*a*b = (pi) * r((1+sqrt(2))/2 * r(2-sqrt(2))/4 = (pi) * (r^2)*sqrt(2)/8
I could be wrong (and please for the love of God someone help me with the LaTEX, I think that's what it's called), but my answer is

A=pi*(r^2)*sqrt(2)/8
Posted August 17th
by Black Yoshi

I got the same answer that Xhin did. When I have more time, I'll look more carefully at your answer Black Yoshi to see what happened.

EN
I got the same answer that Xhin did. When I have more time, I'll look more carefully at your answer Black Yoshi to see what happened.

@EN: @Xhin: @Black Yoshi:
The radius [i]r[/i] of a circle with the same area as that of an ellipse with semimajor axis [i]a[/i] and semiminor axis [i]b[/i], is the geometric mean of that semimajor axis and that semiminor axis.
[i]r[/i] = squrt([i]a*b[/i]) .
You can easily get that from pi*[i]r*r[/i] = pi*[i]a*b[/i].
———
Working out a compass-and-straightedge construction of such a circle is a different matter; maybe a more difficult and/or more complicated one, certainly a less obvious one.
————
If you use every part of the cut-up circle to fill the constructed figure, leaving no gaps and not covering anything twice, then they have to have the same area exactly.
That’s clearly not what you did.
Black Yoshi is on the right track.
And so is Xhin.
I didn’t check the details of either, though.

chiarizio
@EN: @Xhin: @Black Yoshi:

The radius

*r* of a circle with the same area as that of an ellipse with semimajor axis

*a* and semiminor axis

*b*, is the geometric mean of that semimajor axis and that semiminor axis.

*r* = squrt(

*a*b*) .

You can easily get that from pi*

*r*r* = pi*

*a*b*.

———

Working out a compass-and-straightedge construction of such a circle is a different matter; maybe a more difficult and/or more complicated one, certainly a less obvious one.

————

If you use every part of the cut-up circle to fill the constructed figure, leaving no gaps and not covering anything twice, then they have to have the same area exactly.

That’s clearly not what you did.

Black Yoshi is on the right track.

And so is Xhin.

I didn’t check the details of either, though.

Edited August 17th
by chiarizio

Constructing this ellipse with a compass and straight should be relatively straightforward -- all you have to be able to do is construct both of these circles:
https://i.imgur.com/7xCjwha.png
Then the other two ends of your ellipse are just circles with the centerpoint of each corner of the "eye" that trace one radius out.
The centerpoints of the small circles appear to be exactly one radius horizontally and one radius vertically away from each other, so that's pretty convenient.
It would probably take a lot of trace lines, but it would definitely work.

Xhin
Constructing this ellipse with a compass and straight should be relatively straightforward -- all you have to be able to do is construct both of these circles:

Then the other two ends of your ellipse are just circles with the centerpoint of each corner of the "eye" that trace one radius out.

The centerpoints of the small circles appear to be exactly one radius horizontally and one radius vertically away from each other, so that's pretty convenient.

It would probably take a lot of trace lines, but it would definitely work.

Posted August 17th
by Xhin

@Xhin:
No, that’s an ovalish shape that’s not an ellipse.
No part of an ellipse is any part of any circle, unless the ellipse [b]is[/b] that circle.
OTOH the figure EN asked about isn’t an ellipse. It’s a figure made up out of four pieces of a circle.
It’s not a smooth curve.
OTOH we’ve all been talking about how to calculate its [i]area.[/i]
That’s not what EN asked for; they asked for its [i]volume.[/i]
Its volume is zero.

chiarizio
@Xhin:

No, that’s an ovalish shape that’s not an ellipse.

No part of an ellipse is any part of any circle, unless the ellipse

**is** that circle.

OTOH the figure EN asked about isn’t an ellipse. It’s a figure made up out of four pieces of a circle.

It’s not a smooth curve.

OTOH we’ve all been talking about how to calculate its

*area.*
That’s not what EN asked for; they asked for its

*volume.*
Its volume is zero.

Posted August 17th
by chiarizio

Yeah, it's an oval not a true ellipse. I meant area in the picture.

EN
Yeah, it's an oval not a true ellipse. I meant area in the picture.

Whatever the case, you should be able to construct it.
I recently lucked into a really cheap compass set, so I might do just that.

Xhin
Whatever the case, you should be able to construct it.

I recently lucked into a really cheap compass set, so I might do just that.

Posted August 17th
by Xhin

> Yeah, it's an oval not a true ellipse. I meant area in the picture.
I believe the pieces have the same tangent at the joins, but the curvature of the ends is double that of the sides.
So it’s a smooth join to the first degree, but not the second.
Unless I’m wrong again!

chiarizio
Yeah, it's an oval not a true ellipse. I meant area in the picture.

I believe the pieces have the same tangent at the joins, but the curvature of the ends is double that of the sides.

So it’s a smooth join to the first degree, but not the second.

Unless I’m wrong again!

Posted August 18th
by chiarizio

I spent the better part of the other night going back over my calculations...and only now do I realize that I was assuming the shape was a true ellipse, when in fact it's not. I ran the numbers by Xhin's method and got Xhin's answer, I ran them by my own method and got something that wasn't Xhin's answer and wasn't my original answer either. So Xhin's right in this one, I believe. Sometimes in mathematics you can make assumptions, but this is one of those cases where making assumptions can lead you into trouble.

Black Yoshi
I spent the better part of the other night going back over my calculations...and only now do I realize that I was assuming the shape was a true ellipse, when in fact it's not. I ran the numbers by Xhin's method and got Xhin's answer, I ran them by my own method and got something that wasn't Xhin's answer and wasn't my original answer either. So Xhin's right in this one, I believe. Sometimes in mathematics you can make assumptions, but this is one of those cases where making assumptions can lead you into trouble.

Posted August 18th
by Black Yoshi

@Xhin:
> Whatever the case, you should be able to construct it. I recently lucked into a really cheap compass set, so I might do just that.
Let us know what progress you make!

chiarizio
@Xhin:

Whatever the case, you should be able to construct it. I recently lucked into a really cheap compass set, so I might do just that.

Let us know what progress you make!

Posted August 19th
by chiarizio

I too got the same answer as Xhin for the total area of the quasi-elliptical region. Black Yoshi is correct that it is it not an actual ellipse by comparing the area found by Xhin with that of an ellipse with semi-minor and semi-major axes a and b (which are easy to find from the diagram). I got a different answer from BY, but I might have not been careful. The area of the circumscribing ellipse is somewhat bigger than the quasi-ellipse considered in the problem - difference being around \(0.777r^2\).
In any case, Xhin has the correct total area. Nice problem, EN!
By the way, the circumference of an ellipse is tricky and not so easy to find as the circle (which is \(2\pi\) times the radius). It's given by a non-elementary integral.

The Fly
I too got the same answer as Xhin for the total area of the quasi-elliptical region. Black Yoshi is correct that it is it not an actual ellipse by comparing the area found by Xhin with that of an ellipse with semi-minor and semi-major axes a and b (which are easy to find from the diagram). I got a different answer from BY, but I might have not been careful. The area of the circumscribing ellipse is somewhat bigger than the quasi-ellipse considered in the problem - difference being around \(0.777r^2\).

In any case, Xhin has the correct total area. Nice problem, EN!

By the way, the circumference of an ellipse is tricky and not so easy to find as the circle (which is \(2\pi\) times the radius). It's given by a non-elementary integral.

Edited August 22nd
by The Fly

I was successful.
https://i.imgur.com/fNSa6dr.png
My tablet camera isn't the best thing in the world. Even with some post-processing it's hard to see the detail here.
My steps were basically:
1. Construct a circle
2. Construct its diameter (there's a bunch of ways to do this, idk what the correct way is -- I just ran a circle from the outer edge to the center, creating a line where that circle crossed the original cirlcle, then bisected the line.
3. Construct its diameter perpendicular to the other diameter (for me, this meant bisecting the first diameter)
4. Create a smaller circle with radius r/2 of the first circle. In this case I just bisected one of the radiuses and then created a circle between the original center and this new point.
This gives you the original figure EN drew.
5. Create a "second center" perpendicular to radius of the smaller circle and one radius away. There's a bunch of ways to do this too -- I opted for creating a circle from one edge and creating another circle from an edge perpendicular to it (via the X I drew in steps 2 and 3), then where they cross is the "second center"
6. Use this second center to construct a second small circle and a second large circle. I cheated a bit here and just used other shapes to find r and r/2 and then built circles from them and the second center.
7. Construct diameter lines which are perpendicular to the old diameter lines. Again, easy to cheat here -- just draw lines between the second center and the heavily-used points of the first small circle's edge.
This creates this:
https://i.imgur.com/7xCjwha.png
From there, you go on either corner of the "eye" and trace out one small circle radius (or r/2) to make those sections of EN's shape. Then thicken the lines accordingly.

Xhin
I was successful.

My tablet camera isn't the best thing in the world. Even with some post-processing it's hard to see the detail here.

My steps were basically:

1. Construct a circle

2. Construct its diameter (there's a bunch of ways to do this, idk what the correct way is -- I just ran a circle from the outer edge to the center, creating a line where that circle crossed the original cirlcle, then bisected the line.

3. Construct its diameter perpendicular to the other diameter (for me, this meant bisecting the first diameter)

4. Create a smaller circle with radius r/2 of the first circle. In this case I just bisected one of the radiuses and then created a circle between the original center and this new point.

This gives you the original figure EN drew.

5. Create a "second center" perpendicular to radius of the smaller circle and one radius away. There's a bunch of ways to do this too -- I opted for creating a circle from one edge and creating another circle from an edge perpendicular to it (via the X I drew in steps 2 and 3), then where they cross is the "second center"

6. Use this second center to construct a second small circle and a second large circle. I cheated a bit here and just used other shapes to find r and r/2 and then built circles from them and the second center.

7. Construct diameter lines which are perpendicular to the old diameter lines. Again, easy to cheat here -- just draw lines between the second center and the heavily-used points of the first small circle's edge.

This creates this:

From there, you go on either corner of the "eye" and trace out one small circle radius (or r/2) to make those sections of EN's shape. Then thicken the lines accordingly.

Posted August 22nd
by Xhin