TiMartin's Math Corner

It's a little hard.

 

i'm brain dead atm, but i can do that when i'm not tired

 

Do you mean surface area? or total length of the edges

 

surface area i think but idk

 

I shouldn't have stopped doing math D;

 

Perimeter is the sum of all sides. In this case it's P1P2 + S1S2 + P1S2 + P2S1.

 

Answered it. Now to type up the proof

 

Let A(1), A(2), A(3) be the centers of balls K(1), K(2) and K(3) respectively.
A(1), A(2) and A(3) form an equilateral triangle as A(1)A(2) is a straight line length 4, A(2)A(3) is a straight line length 4 and A(1)A(3) is a straight line length 4.
P(1) and P(2) lie on the midpoints of lines A(2)A(3) and A(1)A(3) respectively, as the point of contact between 2 circles which are tangential to each other lies on the line connecting the 2 centers.
angle(P(2)A(3)P(1)) = angle(A(1)A(3)A(2)) (because they are the same angle)
P(2)A(3)/A(1)A(3) = P(1)A(3)/A(2)A(3)
Therefore triangle(A(1)A(2)A(3)) is similar to triangle(P(1)P(2)P(3))
Therefore P(1)P(2) || A(1)A(2)

Let A(4) be the center of K(4)
A(1)A(2)A(3) form a triangle with sides passing through S(1), S(2) and P(3). (Same proof as above)
angle(S(1)A(4)S(2)) = angle(A(1)A(4)A(2)) (because they are the same angle)
S(1)A(4)/A(1)A(4) = S(2)A(4)/A(2)A(4)
Therefore triangle(S(1)A(4)S(2)) is similar to triangle(A(1)A(4)A(2)
Therefore S(1)S(2) || A(1)A(2)
Therefore S(1)S(2) || A(1)A(2) || P(1)P(2)
Therefore S(1)S(2) || P(1)P(2)

P(1)P(2) = 2
S(1)S(2) = 3
P(1)S(2) = P(2)S(1) = sqrt(8) + sqrt (3) (from a triangle using pythagorus)
P(1)P(2) + S(1)S(2) + P(1)S(2) + P(2)S(1) = ~14.12

 

TiMartin1069 what's 9 + 10?

 

-2 i think

 

That has already been asked. I have no idea. Maybe Minus can answer that one.

 

u stupid

 

9=1+1+1+1+1+1+1+1+1
10=1+1+1+1+1+1+1+1+1+1
so 9+10=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
and 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1=19
so 9+10=19

 

Loominarty Questions
You are given a triangle that has sides of 66cm, 73cm, and 94cm. One of the angles is right-angled (meaning that it is possible by trial and error to calculate what each of the angles are). Inside this triangle is a square, so that three corners are in contact with the lines bounding the triangle. One of the sides or the square, which we shall now dub z, is also tangent to a circle, with a radius such that the centre of the circle lies along the side of the triangle with length 73cm. You are also given a regular octagon, which you are told is the same area as the total are of the circle and triangle if they are taken together (i.e. the overlapping area is not counted twice), and one side of this octagon forms another side of equal length belonging to a second square. The area of this square is dubbed x.

1.An isosceles triangle is drawn so that it has the same area as the above square (i.e. x), and with two sides that are equal to the square root of x (henceforth dubbed y). What is the length of the third side?
2.Through cunning use of Pythaogoras' Theorem, prove that aliens do not exist.
3.Give the value, to three significant figures, of x.

 

Very hard. lol no

 

i only read the end so i guess, since it says x.... x files? idk im not good at math

 

I can tell.

 

The proof is good, but the bolded texts are probably mistyped. Underlined are just wrong. You've noticed that triangles (A1A2A4) and (S1S2A4) are similar, use another proportion and you'll have the length of S1S2. For P1S2 and P2S1 you'll need to use the law of cosines in the triangle (A2S2P1). Also notice that A4P1 is the height of an isosceles triangle (A2A3A4).

 

Ahh yes, stupid me i just started assumeing stuff.

 

So, i know that there is infinite numbers of pi but can you tell me all you know?