Not enough geometry here. Can you find the radius of circle gamma ?

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>>8400263

>>8400263
Yes, it's trivial. Pick up a textbook and learn basic geometry.

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>>8400269
>>8400268

warmed up? good.

what can you tell about a chord of a circle?
prove what you tell.

>>8400275
The angle between the radii which touch the edges of a chord is x such that

x-sin(x) = 2pi*F

where F is the fraction of the area the chord cuts off the circle.

But what if you want to know how two such chords can intersect to cut off a certain fraction of the circle? Then first we calculate the above angle for the two chords:

y-sin(y) = 2pi*F1
z-sin(z) = 2pi*F2

Then the intersection of these two chords can be found by calculating the angle between the radii which touch the endpoints of the chords (specifically, the ones which form the intersection in the middle of the 3 sections which do not contain the center of the circle). Let us call this angle x such that

cos(z/2) cos(x-y/2) csc(x-y/2-z/2) (1-cos(y/2) sec(x-y/2))^2-sin(x) cos(y/2) sec(x-y/2)+x = 2pi F

Where F is the fraction of the area of this intersection.

The proof is left as an exercise for the reader.

>>8400275
Defining a diameter of a circle as any chord which is incident with the center of the circle then the perpendicular bisector of any chord must be a diameter.
Proof:
The perpendicular bisector of a line segment is the set of points that are equidistant from the two points defining that segment, in this case that would be the intersection between the chord and the circle. the center of the circle is equidistant from these two points therefore it is incident with the perpendicular bisector and the bisector must be a diameter.

>>8400306
>x-sin(x) = 2pi*F
>x
>F
>sin
>sec
>pi
with such a simple question you use such impressively useless calculation tools

>>8400322
now that's what I call a statement and its demonstration. thank you anon.

what can we say then about the length of a chord?

>>8400275

If two chords AB and CD cut each other at X, then ||AX||*||XB|| = ||CX||*||XD||.

Let M be the midpoint of AB and N be the midpoint of CD. Let O be the centre nof the circle.

By Euclid III.3, MO is perpendicular to AB.

By Euclid II.5, ||AM||^2 = ||MX||^2 + ||AX||*||XB||

By Euclid I.47, ||MX||^2 + ||OM||^2 = ||OX||^2 and
||MA||^2 + ||OM||^2 = ||AO||^2

Therefore ||AX||*||XB|| = ||AM||^2 - ||MX||^2 = ||AM||^2 + ||OM||^2 - ||OX||^2 = ||AO||^2 - ||OX||^2

Since ||AO|| is simply the radius of the circle, ||AX||*||XB|| only depends on X, thus the statement holds.

>>8400351
Tell me then at what height water one third the volume of a tubular tank will lie if the tank is placed on its side so that it can be rolled?

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>>8400263
What can you tell about a chordal n-dimensional hyperplane through an (n+1)-dimensional sphere?

Prove what you tell.

>>8400351
>>8400374
This sure is taking a while anon... Surely you don't need those "useless calculation tools" for such a simple question?

>>8400409
This is a (an sounds wrong) euclidean geometry thread anon.

>>8400425
Everything I described is Euclidean. Do you know what the terms you're using mean?

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>>8400434
>Transcendental functions
>Euclidean

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>>8400440
>He doesn't know what Euclidean means but still posts as if he does

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Here's another (easy) euclidean problem: find the shortest way from A to B with a mandatory stop by a point of (d)

>>8400381
First, define what n is.
Then define the space you're talking about.
Then an hyperplane of this space don't need to be mentioned its dimension, because of its mere definition.
Then define a (whateverdimension) sphere is.

Then, maybe, we can answer your unclear question that looks more like a brainlet fan of complicated statements from which they don't understand the core.

>>8400440
Euclidean geometry is a geometry in which the parallel postulate holds. It has nothing to do with whether a function is transcendental or not.

>>8400463
Are you trolling me right now? They sure as shit didn't allow trig functions to be used in proofs.

>>8400463
well
you're not being totally honest
we all know what a "beautiful" proof is in euclidean (or in any) geometry. it is a proof that use the less technical tools.
have you read the elements of euclide?
is there any trace of any sin function, of indeterminate, or anything?

now, I think we all agree on what euclidean geometry is.
it's all a question of elegance and style then.

>>8400449
>draw line ab
>construct perpendicular bisector
>extend bisector to line d
>draw a point c where bisector and line d intersect
>a to c to b

>>8400471
Euclidean geometry is an object, not a toolbox. You are confusing how Euclid studied this object with the object itself. Euclid was unaware of many useful theorems that are necessary for a full understanding of Euclidean geometry.

>>8400475
That's fine if you can actually prove what's being asked without trigonometry. But you can't calculate what I asked without it. It's very beautiful to whittle down a proof into the simplest form possible. It's not beautiful to not be able to answer a question.

>>8400471
Euclidean =/= constructible

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>>8400482
counterexample
when A is close to (d)
shortest way is clearly not passing through intersection of perpendicular bisector and (d)

nice try though
keep thinking about perpendicular bisectors

>>8400486
>It's not beautiful to not be able to answer a question
I don't agree.
Thinking about the very minimal conditions under which a question can be addressed is one of the essence of mathematical reasoning.

All minds doing more and more just by technical enjoyment are not mathematicians. They are at most clever users of mathematical techniques.

A is the center of the circle.
Since the triangles ABC and ABD are isosceles we have
1.ABC=ACB
2.ABD=ADB
We can see (I know its not rigorous to say that) that
3.CAD+BAC+BAD=360 therefore
4.CAD=360-BAC-BAD
We know that the sum of the angles in a triangle is 180 so
5.BAC+ABC+ACB=180 therefore
6.BAC=180-ABC-ACB
7.BAD+ABD+ADB=180 therefore
8.BAD=180-ABD-ADB
substituting eq. 6 and 8 into 4 we get
CAD=360-180+ABC+ACB-180+ABD+ADB
which simplifies into
CAD=ABC+ACB+ABD+ADB
using eq. 1 and 2 we get
CAD=2ABC+2ABD
Using the fact that CBD=ABC+ABD we arrive at
CAD=2CBD
Since the measure of an arc is equal to the angle subtending it
We have mArc(CD)=2CBD

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>>8400530
fug forgot pic

>>8400530
legit proof of the inscribed angle

>>8400551
is >>8400358 not legit? It seems pretty neat.

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A solid is constructed by taking a cube of side lengths S and "cutting out" a sphere of radius R < S, as pictured.

The distance between the center of the sphere and the nearest corner of the cube is L. The angle between the distance between the center of the cube and the center of the sphere and the distance between the center of the sphere and the nearest corner is [math]\phi[/math]. The center of the sphere always intersects with a side of the cube.

What is the volume of the solid?

>>8400551
disregard >>8400562 I confused myself.

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>>8400530
anyway I proved that so I could prove this

BFE=DFC
since DCB and DEB subtend the same arc they have the same angle measure (I feel like there was an easier way to go about this)
since two angles of the triangles DCF and BFE are equal the third one must be as well
since all three angles are equal these triangles are similar
therefore there exists some constant k, such that
k*DC=BE, k*DF=BF, and k*CF=EF
DF*k*CF=CF*k*DF is trivially true and by making the substitutions above we get
DF*EF=CF*BF

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>>8400514
We're trying to minimize AC+CB with C as a free variable constrained to the line d. Reflect B over line d to make B'
The length of the path from A to B' is AC+CB'
Its easily proven that CB=CB'
so we have AC+CB'=AC+CB which is the length of the path we're concerned with
so the length of our path is minimized when the path of A to B' is minimized
By definition (or I think as a consequence of the triangle inequality)
The shortest path between two points is a line segment with endpoints at those two points. so the length of AC+CB' is minimized when C is fixed as the intersection of line d and AB'. Since AC+CB' is minimized so is AC+CB.
Point F is our minimizing point.

God I hat geometry

>>8400650
yo thats pretty tricky

>>8400668
Really? The idea behind it is pretty intuitive I think. What do you find tricky about it?

>>8400525
>Thinking about the very minimal conditions under which a question can be addressed is one of the essence of mathematical reasoning.
That doesn't respond to my point though. The question I asked cannot be answered without transcendental functions.

Also, it is not true that the set of beautiful questions is limited to the set of questions with beautiful answers. The prime example of this is Fermat's Last Theorem.

>All minds doing more and more just by technical enjoyment are not mathematicians. They are at most clever users of mathematical techniques.
It's ridiculous to argue that the use of trigonometry makes you a non-mathematician.

>>8400676
no its very intuitive. just never would have thought of it cause dumb

>>8400727
I didn't think of it first time I saw this problem either. Good thing is once you've seen it you can recognize other situations where this kind of trick can be applied.

>>8400680
But what is the point? Yes the solution is possible but getting it is just trivial algebra and then you have to use a calculator to solve it. There's no satisfaction.

>>8400771
The point is that it answers the question. If all you care about are simple answers then you are not a mathematician.

>>8400775
The answer isn't complex. Its boring. There's hardly any thought put into getting it.

God damn I fucking hate geometry

>>8400844
give it a chance

Since no one's posting anymore I'm just gonna be posting basic theorems until I think of something fun.

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We are given a right triangle with sides a,b known and c to be solved for
Create a square with side length of a+b
Partition the square as shown in the picture
since each triangle has 2 congruent sides and a congruent angle between the two sides they are all congruent to each other and to our given triangle. So they each have a hypotenuse equal to c
since the sum of the angles in a triangle is 180 degrees and a right angle is 90 degrees, a right triangle has acute angles that sum to 90 degrees
the sum of angles at each point of the figure inside of our square must be 180 degrees
two of those angles are the acute angles of our right triangle so the remaining angle is a right angle
So the figure inside of our constructed square is itself a square with side length c and area c^2
each triangle has area a*b/2, in total the triangles has area 2ab
the area of the larger square is (a+b)^2=a^2+2ab+b^2
the larger square is decomposable to the 4 triangles and the smaller square so we have that the area of the larger square is equal to sum of the triangles and the smaller square areas
(a+b)^2=2ab+c^2
a^2+2ab+b^2=2ab+c^2
a^2+b^2=c^2

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DCA, ACB, BCE form a straight angle.
DCA+ACB+BCE=180
Since CD is parallel to AB the alternate interior angles created by AC are equal
DCA=BAC
likewise the alternate interior angles created by CB are equal
BCE=ABC
substituting the above two equations into the first we get
ABC+ACB+BAC=180

>>8400564
/sci/ brainlets can't even solve this. Sad.

>>8402732
3DPD 2D only

x^n + y^n != z^n, if n > 2

Prove it. It should be simple to do; 4chan text boxes are larger than any book margin.

>>8402850
>not restricting x, y, z to integers.

(3^(2/3))^3 + (4^(2/3))^3 = (5^(2/3))^3

>>8402885
Fermat btfo

>>8400475
> Elements of Euclid
> relevant

Are there still non-plebs who read Elements? I was going to get a copy but a lot of people say that it's mostly for high school math teachers.

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Tfw hop on /sci/ after a full 2 days of suffering through babby's first Calc 2 optimization problems and see this thread.

>>8400263
I solved this by assuming that the top corner is the origin point and the 6unit lines are the x and y coordinate axes

The larger fourth of a circle then is represented by the function y = -sqrt(36-x^2) for 0<x<6

The tangent point of the circle Gamma and the larger circle is the intersection point between the larger circle and the line of -45degrees extending down from the origin. That line is represented by the function y=-x for x>0

set equal and solve:

-x = -sqrt(36-x^2)
x^2 = 36-x^2
x^2-18 = 0
=> x = +-3sqrt2 but x>0 by our choice of x so x=3sqrt2

Since y = -x, the tangent point is (3sqrt2, -3sqrt2)

I then realized the inscribed circle could only be tangent to each coordinate axes if it was tangent where x=r and y=-r where r is the radius of Gamma. Since the center of circle Gamma is not the origin, we can model its function with a simple transform:

(x-r)^2+(y+r)^2=r^2

=> x^2-2xr+r^2+y^2+2yr+r^2=r^2
= x^2+y^2-2xr+2yr+r^2=0

We can solve for r because we know that circle Gamma contains (3sqrt2, -3sqrt2). So since in the case that x=3sqrt2 and y=-3sqrt2 then y=-x which implies y^2=x^2, and we can simplify further:

2x^2-4xr+r^2=0

So now note that r is the variable we're solving for and using a specific value for x so use the quadratic formula

r = [4x+-sqrt(16x^2-4*2x^2)]/2
= r = [4x+-sqrt(8x^2)]/2
= r = [4x+-2xsqrt2]/2
= r = 2x+-xsqrt2 = x(2+-sqrt2)

plug in x = 3sqrt2

=> r = 3sqrt2(2+-sqrt2)
Note that the way the circle is inscribed is such that diameter<6 => r<3
so r=3sqrt2(2-sqrt2) =~ 2.4853

Suck a dick OP

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Allow the vertices of an 'inscribed' triangle to fall on the extended sides of the outer triangle, so in the picture shown the blue triangle is 'inscribed' in the red triangle.

Then, given any two triangles such that the first is 'inscribed' in the second, show that there exists a third triangle which completes a 'loop' of triangles, each one 'inscribed' in the next. How many triangles can complete the 'loop'?

>>8403412
I put in all this work before I opened the thread and then realized that this guy >>8400268
is much smarter than me

>>8403336
The struggle never ends fellow brainlet

>>8403323
>Euclide's Elements one of the most published books ever
>made a whole school of maths until 18th century
Are you taking for granted statements from friends of friends of faggots? Get out and go buy a brain. Or kill yourself (better option).

>think of Euler time travelling in 2016
who he even understand what we do with integration or derivation ?

Of course, not. So your statement about comparing objective "levels" of maths is totally irrelevant.

So please fuck off.

Now.

Right now.

>>8403447
Haven't quite understood your question.
The proof of the alignment of those three intersections (see figure) is a classical of the beginning of idea of projective geometry though.

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>>8403447

>>8400263
OP masterfully gets /sci/ to do his grade 7 homework

>>8404175
You have obscenely high standards for 7th graders.

>>8404175
OP here

So you're feeling you're too smart for problems? You probably are no mathematician. We like to play about with any kind of problem. Easy or not.

Even if you claim your mathematical skills to be way ahead of the skills of a 7th grader, you wouldn't have liked me to ask you a proof of the reciprocal Pythagorea theorem.

It is an interesting proof. And I challenge you to give it here without looking at any written resources.

Go on, faggot. Even if I doubt you are even able to state what this theorem and its reciprocal theorem is. Go on.

Peace and love.

>>8404175

>can't solve any of the problems
>calls it 7th grade because it's geometry

nice try brainlet

>>8400650
Holy shit why didn't I think of that? That's pretty much how virtual images in plane mirrors operate.

I was thinking it would just be the perpendicular bisector because I knew that the angles ACD and BCE had to be equal (which in your answer they are due to vertical angle congruence, but in retrospect wouldn't be true in my case unless line AB was parallel to d), but I never thought of just reflecting point B along line d

>>8404059

Point I doesn't seem to be a special point at all.

>>8405693
what do you call 'special' ?

>>8405693
I is the intersection of GH and AC, but it isn't really anything else, and that intersection is almost always going to exist anyways.