Elliptical Room in the White House

[Globex]

I've read somewhere that there is a room in the white house that's built as a perfect ellipse and as a result of the room's shape, two people standing at the foci of the ellipse would be able to hear each others' whispers from across the room due to the fact that all the sound waves emitted from one of the foci will reflect off the walls and converge on the other focus. I forgot the name of this room; I tried to Google it but it came up empty. Does anyone happen to know what this room is called?

 

[Camikaze]

It's actually in the Capitol. It's called National Statuary Hall. It says in the Wikipedia article that John Quincy Adams used the acoustic property you mentioned to eavesdrop.

 

[Perfection]

The Oval Office and Blue Room are both elliptical.

 

[Atticus]

I first thought the Oval Office, and wikied it -I'll have to admit that I didn't know what oval exactly is, and that it might be synonym for ellipse. Wiki said: "the office was inspired by the elliptical Blue Room", so it's the Blue Room.

 

[Perfection]

Oval isn't precisely defined in general parlance. Wikipedia described it as "elliptical", but I now think that's a little suspect in terms of it actually being an ellipse and holding the proper acoustic properties. I'm not really sure what geometric shape the oval office really is.

 

[ParadigmShifter]

Sure it is - it means egg-shaped

Eggs aren't ellipsoids though nor elliptical in cross-section.

 

[peter grimes]

Since we're talking about ellipses, here's something I've always wondered:

Take an ellipse, then add a border to it. What is the name of the shape described by the outer edge of the border? I know it's not an ellipse, but I don't know what it is...

 

[ParadigmShifter]

What sort of border?

A bounding box would be a rectangle

 

[peter grimes]

a border of constant width. For example, a 2 unit wide strip.

 

[ParadigmShifter]

Looks like there is no special name for this for an ellipse.

http://mathworld.wolfram.com/ParallelCurves.html

 

[Padma]

Why would there be a special name? The border around the ellipse merely defines a different, larger ellipse.

 

[Atticus]

Not necessarily: the bigger thing's borders are at constant distance from the smaller ellipse, whereas in the ellipse the sum of distances from focal points should be constant.

For circles it would be bigger circle, but if you consider similar stripe around a square, it wouldn't have sharp corners, but round ones instead.

It might be ellipse if the original figure is ellipse, but I have strong intuition that it is not. (Too lazy to prove at the moment).

 

[ParadigmShifter]

I thought it would be an ellipse at first too (like the circle) but it isn't: consider a really thin ellipse. It's more like a sausage shape than an ellipse in that case.

EDIT: I'm 99.99999% certain it will only be another ellipse if the foci are coincident, i.e it's a circle I can't be bothered with a proof either, maybe someone may want to if you post it in the "let's discuss maths" thread.

 

[Padma]

Yeah, I'm too lazy to get into the math, myself. I'll just take your word for it.

 

[ParadigmShifter]

There is a reason we don't allow ellipse-ellipse collisions in our computer games

As Barbie says: "Math is hard".

EDIT: Basically it involves solving a quartic which although doable isn't really a great thing to spend time doing. We do ellipse-triangle collision only really, if they interact with other ellipses we use a cylinder or a pill (cylinder capped with hemispheres) instead.

EDIT2: I think Unreal Engine has that though, I've certainly seen the solve quartic function in that.

 

[Atticus]

One way to do it would be to choose ellipse whose foci (lets say a and b) are on the x-axis. Now every point on the ellipse has constant sum of distances from points a and b, let's that sum is r.

If you cover this ellipse with a "belt" of width 2, then the point where bigger figure intersects the x-axis has sum of distances r+4. If the bigger figure was ellipse, every other point of it should have the same sum of distances, but this is easily seen to be false: Just take any point that isn't on the x-axis, let's say d. There's one point on the smaller ellipse whose exact distance from d is 2, lets call that point c. Now if you go from point a to c, from there to point d, back to c again and from there to point b, the sum of distances will be r+4. However, there are shorter paths* from both a to d and from b to d, so d's sum distance is less than r+4.

Here's a figure I drawed quickly with paint:

This is the idea, details would require little more effort. Hope I recalled ellipse's definition right...


*This needs argument. It can be doen by noticing that if you draw tangent at point c, the d will be on that tangent's normal.

EDIT: sorry... I tried once to edit images smaller within IMG-tags, but it didn't work...
EDIT2: Nevermind, I wasted tinypic's resources instead.

 

[Perfection]

One way to do it would be to choose ellipse whose foci (lets say a and b) are on the x-axis. Now every point on the ellipse has constant sum of distances from points a and b, let's that sum is r.

If you cover this ellipse with a "belt" of width 2, then the point where bigger figure intersects the x-axis has sum of distances r+4. If the bigger figure was ellipse, every other point of it should have the same sum of distances, but this is easily seen to be false: Just take any point that isn't on the x-axis, let's say d. There's one point on the smaller ellipse whose exact distance from d is 2, lets call that point c. Now if you go from point a to c, from there to point d, back to c again and from there to point b, the sum of distances will be r+4. However, there are shorter paths* from both a to d and from b to d, so d's sum distance is less than r+4.

Here's a figure I drawed quickly with paint:

This is the idea, details would require little more effort. Hope I recalled ellipse's definition right...


*This needs argument. It can be doen by noticing that if you draw tangent at point c, the d will be on that tangent's normal.

EDIT: sorry... I tried once to edit images smaller within IMG-tags, but it didn't work...
EDIT2: Nevermind, I wasted tinypic's resources instead.

I considered that style of proof for awhile. There however is a complicating factor, just because a figure has inconsistent sum of distances between a and b doesn't prove that it isn't an ellipse, only that it isn't an ellipse with foci at a and b.

 

[Atticus]

You're right

 

[ParadigmShifter]

Triangle inequality simplified your argument anyway

EDIT: Nice use of the word foci anyway, well done. Focuses is so vulgar

 

[Atticus]

I thought a while it, foci sounds bit snobish, but focuses little stupid...

Anyway, here's another way how you could do the proof, although not very satisfactory, since I don't want to go into details, and am too lazy to complete it:

Suppose that the foci of original ellipse are (-a,0) and (a,0), and each point on the original ellipse has sum distance r from them.

Lets suppose that the bigger figure is ellipse (and it is the original one with belt of thickness 2 around it). It's fairly obvious that it's foci also lie on the x-axis, and are equidistant from (0,0). So let's call them (-b,0) and (b,0). Now take the point on the left side of the bigger figure where it intersects the x-axis, let's call the point (-c,0). Now calculate the sum distance from the new foci, it should be r+4, because |-b-(-a)| and |b-a| cancel each other out (this is clear if you draw a picture, but pretty hard to explain in words).

So if the bigger figure is ellipse, the sum distance is r+4.

Next calculate where the smaller ellipse intersects the y-axis, it's in point sqrt(r^2/4 -a^2) above the origo (and below too), so the bigger figure intersects it in 2+sqrt(r^2/4-a^2). From this you can calculate the place of new foci, knowing that they have sum distance of r+4 from that point.

Now you should pick some point on the ellipse that doesn't intersect the axis, and show that it has sum distance from the foci different than r+4. I'm too lazy to continue, but I guess there's one point on each quadrant (defined by the axis) that isn't good for this, and all the others are.