/sci/ - Science & Math

[Anonymous No.11608812]

http://graphics8.nytimes.com/packages/pdf/education/harvardexam.pdf
Can /sci/ solve this 1899 Harvard entrance exam question?
>prove that circles are to each other as the squares of their radii
No circular reasoning allowed (you can't assume S=pi*r^2 because that's just presupposing the result).

[Anonymous No.11608816]

>you can't assume S=pi*r^2 because that's just presupposing the result
ok then I'm just gonna prove that S = pi*r^2 and then solve it

[Anonymous No.11608822]

>>11608816
Yes, the question exactly asks you to prove that there is some constant c such that S=pi * r^2. (which happens to be pi, but that's beside the point)
Can you prove it?

[Anonymous No.11608823]

>>11608822
yes, it's trivial

[Anonymous No.11608827]

>>11608816
Yes, the question exactly asks you to prove that there is some constant c such that [math]S = c r^2 [/math]. (which happens to be [math]\pi[/math], but that's beside the point)
Can you prove it?

[Anonymous No.11608831]

>>11608823
>yes, it's trivial
Ok then provide a proof.

[Anonymous No.11608833]

>>11608831
sorry kid, there are more important problems for me to solve

[Anonymous No.11608834]

>>11608812
>solve these medium-hard problems in a subject that was taught intensely in schools in 1899 but is barely touched now

[Anonymous No.11608837]

>>11608812
I'm not exactly sure what they want. This seems like a geometry question, do they want you to do the Archimedes thing where you approximate it by the area of the polygon?
I'm almost certain they can't expect you to be doing anything with calculus or limits, because I strongly doubt high schoolers would have known calculus in 1899, but there's no way to get actually get the exact formula with pi without doing some kind of limit somewhere.

[Anonymous No.11608841]

Am I just retarded or isn't it obvious? One can draw a regular n-polygon as a rotation of n-triangles sharing a point at the center. We know the formula for the area of a triangle, so therefore we know the area of an inscribed regular polygon as the multiple by n. We can approximate a circle as an apeirogon, thus by taking the limit we get the area of a circle. The radius squared is because two of the sides of the triangle are the radius, and that quantity persists in the limit.

[Anonymous No.11608844]

>>11608837
The question does not ask you to provide an exact formula for [math]\pi[/math]. It merely asks you to prove that there's a constant [math]c[/math] such that [math]A=cr^2[/math] where A is the area of the circle and r is the radius. Can you do it? No calculus required. You may need to use something similar to epsilon-delta proofs though.
>>11608833
>I can't do it.
Ok mathlet.

[Anonymous No.11608856]

>>11608844
>It merely asks you to prove that there's a constant c such that A=cr2
No it doesn't.
It asks you to show "how the area of a circle may be found". Saying that the area is proportional to some unspecified constant times r^2 does not show you how to find the area.

[Anonymous No.11608860]

>>11608841
You seem to have realized the point of this thread better than other anons and have a right idea in mind (although different from mine) but you still don't have a complete proof.
>thus by taking the limit we get the area of a circle
You have to prove this.
The rest of the post looks fine to me.

[Anonymous No.11608907]

>>11608860
I'm unsure how one can prove an apeirogon has the same area as a circle without going into set theory or calculus proper. The closest I can get is by recalling that the definition of a circle is the set of points equidistant from the origin. Given a Euclidean metric, this produces our familiar roundboi. An apeirogon, by virtue of being regular and inscribed, will therefore share an arbitrarily high number of points with a circle, plus some extra for its sides. But that extra implies this line of thought's very refutation as that extra merely vanishes in the limit.

As for proving that two shapes sharing the same points have the same area, consider that two sets are equal iff they have precisely the same members - no more, no less in either. Which I guess solidly refutes this line of thought. Huh. Surprisingly good question OP.

[Anonymous No.11608916]

>>11608812
solve this problem: have sex
haha you cannot

[Anonymous No.11608917]

>>11608907
You could show that the difference between the are of the inscribed polygon and the circle tends to zero by fitting the circle into another polygon and showing that the difference of the area of the two polygons tends to 0 as the number of sides increases. There are other strategies to prove it though (for example, using squares).

[Anonymous No.11608919]

>>11608916
anime trannies btfo

[Anonymous No.11608932]

>>11608917
While that would work, it leaves open the question why would these two different shapes share the (exact) same area? I guess it depends on construction. If the apeirogon actually has infinite sides (is this even valid? I'm not a mathematician), you can't find a point that is an edge, whereas in the limit it always has such points.

[Anonymous No.11608944]

>>11608932
>is this even valid?
Not really. As far as I can tell, your apeirogon with infinite sides is no different, mathematically, from the circle itself. They're identical.
>it leaves open the question why would these two different shapes share the (exact) same area?
I would not refer to the apeirogon at all. The fact that the difference of the are of the n-sided polygon and the circle tends to zero implies that the are of the n-sided regular polygon tends to the are of the circle as n gets large.
There will be no one polygon inscribed in the circle whose area is equal to the circle but they get closer and closer, and that's exactly how you calculate areas of more complicated shapes: by approximating them with simpler shapes and calculating the areas of the simpler shapes, which you know about.
Though don't get me wrong, you can make the notion of an apeirogon rigorous by, for example, declaring it to be the (closure of) set-theoretic union of all the points in all the different polygons that make it up. But then in your case you would indeed get the exact set-theoretic circle.
The fact that you are able to approximate it by such simple shapes and that scaling the circle just scales the triangles by a factor of r^2, proves that the limits are also scaled by r^2.

[Anonymous No.11608952]

>>11608932
He's not really responding to the idea you're posting, he's just prattling about the standard Wikipedia method for some reason.

Whether an apeirogon is "valid" depends on what context you mean. You can't actually draw a regular apeirogon in the Euclidean plane, unless you're willing to go full retard and accept that an infinitely long straight line is a polygon. You can't draw one "around" a circle.
In other contexts it sometimes makes more sense, but not here.

[Anonymous No.11608973]

>>11608944
>But then in your case you would indeed get the exact set-theoretic circle.
Derp. Good point.

>>11608952
>You can't actually draw a regular apeirogon in the Euclidean plane
Could you elaborate? This thread is obviously a mathlet thread and it's not immediately obvious why this is true. Also, IIRC, hyperbolic space allows this/a similar construction where it has a point at infinity - what changes at the moment the curvature precisely equals zero?

[Anonymous No.11609339]

>>11608812
Easy.
Construct an equilateral triangle of any size. Compute the perpendicular bisectors of the lines, and construct their intersection. Connect the intersection point with the endpoints of the equilateral triangle, and extend inefinitely. Take the radius of the circle you are working with, and construct a circle with that radius at the intersection point. The circle will intersect all three of the indefinitely long lines, forming three isosceles triangles. Because the indefinitely long lines were constructed from an equilateral triangle, the isosceles triangles must all have SAS congruency, hence, you constructed an equilateral triangle inside the circle you want to work with.

Bisecting the edges of this regular, inscribed polygon is simple enough, just connect the edge bisection points to the center point, and extend to the circumference. The new circumference points should be connected in sequence to the previous inscribed polygon's circumference points, constructing an inscribed, regular polygon of 2n sides (n is the number of sides of the previous regular, inscribed polygon).

Now that an arbitrary regular polygon of 3*2^n sides can be inscribed inside of the circle in question, one can simply convert this inscribed polygon to a circumscribed polygon by treating the endpoints of the circumscribed polygon as apothem points of the circumscribed polygon. In other words, you just take those lines emanating from the center, to the 3*2^n circumference points, and construct lines perpendicular to the radii, at the circumference points (this can be trivially done by letting the emanating lines continue for twice the radius of the circle, and then applying line bisection on those line segments). These tangent lines can be extended until the intersect neighboring tangent lines, and those intersections will be the points of the regular, circumscribed polygon of 3*2^n sides.

Computing the areas of the inner and outer polyhedra is trivial.

[Anonymous No.11609392]

>>11609339
The inner, regular polyhedra of 3*2^n sides approach the area of the outer regular polyhedra of 3*2^n sides, and the area of the circle is always between these two values. This means the inner and/or outer polyhedra are a Cauchy sequence of constructible numbers (comeasurable to the radius of the circle), and this Cauchy sequence represents the area of the circle. Because we are using the Real number system, the limit of the Cauchy sequence does exist as a number, so the Cauchy sequence has a limit, and that limit is the area of the circle.

We can use this Cauchy sequence representation of the area of the circle to establish the fact that the limit of the Cauchy sequence must scale with a factor squared, whenever the radius of the original circle is scaled with said factor (because we know the polyhedra in the Cauchy sequence scale with the factor squared).

[Anonymous No.11609613]

>>11608812
That a circle goes to other circles as do squares to other squares can be established geometrically. As one scales up the area of a square, the ratio of the area of square and the circle it circumscribes does not change.

The area of a square is the length of one size multiplied by that of the other side. Thus squares go to each other as the squares of their side length. Thus circles go to each other as the square of the length of the sides of the squares that circumscribe them, which is obviously equal to some constant multiplied by the radius of the circle. Thus circles to other circles as the square of their radii.