Re: Maths- Gym for Brainiacs

by spanky Sun Feb 26, 2012 9:32 pm

Immortal Babun wrote:Too hard or not interested?

by Babun Tue Feb 28, 2012 5:28 pm

spanky489 wrote:
Immortal Babun wrote:Too hard or not interested?

You're lazy. You don't even have to calculate a lot eco smile

by Guest Fri Apr 06, 2012 12:57 am

Cool thread. Bump.

by Babun Fri Apr 06, 2012 10:39 am

Maje10 wrote:Cool thread. Bump.

Solve the last question and I'm in Very Happy

Maths- Gym for Brainiacs - Page 12 400270_282970735084117_100001135798775_740493_1045761260_n

by spanky Fri Apr 06, 2012 7:26 pm

hell, its about time

by Ganso Fri Apr 06, 2012 9:06 pm

LOOOOL i just realized those are the graphs :bow:

i used the X^2,X and |x| to remember it lol

i always have a hard time remebering the graphs for cos and sin

by Guest Sat Apr 07, 2012 4:31 am

Immortal Babun wrote:Next MS Paint question, you guys seem to be interested in geometry

topic: geometry
difficulty: hard
requirements: 2nd year of high school

Spanky invented a brand new triangular merry-go-round for a more bumpy feeling in the corners . It's shape is square with the side length of 10m and the moving part itself is an equilateral triangle ( all sides with the same length) with length of 5m
Spanky wants to demonstrate his baby in action therefore he invites Kiranr and Halamadrid2 for a round. Those two take the first two sits ( black dots) and Spanky the last one ( red dot). Each dot is 2/3 height away from the opposite side. The round ends when Spanky ( red dot) returns into his initial postion.
Babun watched those three in their natural environment with a smirk and thought, how much of a way did Spanky travel?

This is going to sound silly, but whatever...

Using the Pythagorean Theorem, the height of the triangle is 4.330127019? Is this relevant?

by Ganso Sat Apr 07, 2012 12:58 pm

you cannot use the Pythagorean theory because that is not a right triangle,its equilateral

by Ganso Sat Apr 07, 2012 1:51 pm

btw my answer is 40M or around 40M

by spanky Sat Apr 07, 2012 2:29 pm

Ganso wrote:you cannot use the Pythagorean theory because that is not a right triangle,its equilateral

you can, u split the triangle in half and get 2 right angle triangles.

by Babun Sat Apr 07, 2012 2:39 pm

Ganso wrote:btw my answer is 40M or around 40M

The answer isn't 42 Maths- Gym for Brainiacs - Page 12 Trollface-2

The triangle makes 2 rounds until Spanky is in the left corner again...

The answer lies somewhere between 80 < x < 110.

spanky489 wrote:
Ganso wrote:you cannot use the Pythagorean theory because that is not a right triangle,its equilateral

you can, u split the triangle in half and get 2 right angle triangles.

This! If something doesn't fit, make it fitting ( this is also how you bust a cherry) eco smile

Maje10 wrote:

This is going to sound silly, but whatever...

Using the Pythagorean Theorem, the height of the triangle is 4.330127019? Is this relevant?

You need Spanky's position. The height might help you but he sits lower than the total height.

Spanky and co., no new question until you solve this one :lol!:

by kiranr Sat Apr 07, 2012 3:10 pm

In the first move of the triangle, did the red dot move 6.6 units?

by spanky Sat Apr 07, 2012 3:33 pm

ok babun ill try solving this since im bored and its the weekend Very Happy

by spanky Sat Apr 07, 2012 4:12 pm

ok so basically due to the dot being equally far apart from the other vertices means that we have 2 different radiuses of rotation, one is the big radius which is the distance from either of the furthest vertices to the point and the small radius from the point and to the nearest vertex. the small radius is equal to 1/3 of the triangles height. the other radius we can calculate using the euclidean distance of points equation, to calculate the height and distance along the x axis of the point we use trigonometric functions.

height of triangle: sqrt(25-2.5*2.5)=4.33m
so small radius: 1/3 * 4.33 = 1.44m

lets call the point where im sitting point A with the coordinates (1.44*cos30,1.44*sin30)=(1.25,0.72)
big radius: sqrt(0.72*0.72 + 5 - 1.25 * 5 - 1.25)=sqrt(0.52+14.06)= 3.82m

now i drew out the problem on a piece of paper because even though my intuition told me that this point would plot a few circles till the end i could not perceive the amount of degrees it actually rotated. after drawing it on paper i cmae to the conclusion that the point rotated 600 degrees with the small radius and 1200 degrees with the big radius.

now we just use the equation for perimeter for both radiuses

BIG RADIUS LENGTH: 2*pi*3.82 * 1200/360 = 79.98m
SMALL RADIUS LENGTH: 2*pi*1.44*600/360=15.08m
total path length: 79.98m+15.08m= 95.06m

there babun i solved it. next question Very Happy

by Babun Sat Apr 07, 2012 4:21 pm

@spanky, you have to get back into your initial position ( more than 1 round) Very Happy

by spanky Sat Apr 07, 2012 4:25 pm

the triangle goes around 3 times

updated my answer its 93.2 not fifty something.

the thing i did wrong was before i thought the triangle rotated by 60 degrees once per side but it actually did 120 degrees and then the additional 30 degrees to line up with the next side.

by kiranr Sat Apr 07, 2012 4:39 pm

I got 96.5!

by Babun Sat Apr 07, 2012 4:45 pm

Interesting, I lost the solution so I calculated again. My answer is 95.10m :vagi:

My small radius is :

r1= 4.33/3 = 1.444 (m )

Big radius:

r2= r1²+ 5²- 2*r1*5*cos( 30°) = 3.8186 (m)

by spanky Sat Apr 07, 2012 4:50 pm

well it would help if u guys posted how u both got those answers.

by Babun Sat Apr 07, 2012 4:53 pm

spanky489 wrote:ok so basically due to the dot being equally far apart from the other vertices means that we have 2 different radiuses of rotation, one is the big radius which is the distance from either of the furthest vertices to the point and the small radius from the point and to the nearest vertex. the small radius is equal to 1/3 of the triangles height. the other radius we can calculate using the euclidean distance of points equation, to calculate the height and distance along the x axis of the point we use trigonometric functions.

height of triangle: sqrt(25+2.5*2.5)=5.59m
so small radius: 1/3 * 5.59 = 1.86m

lets call the point where im sitting point A with the coordinates (1.86*cos30,1.86*sin30)=(1.61,0.93)
big radius: sqrt(0.93*0.93 + 5 - 1.61 * 5 - 1.61)=sqrt(0.87+11.49)= 3.52m

now i drew out the problem on a piece of paper because even though my intuition told me that this point would plot a few circles till the end i could not perceive the amount of degrees it actually rotated. after drawing it on paper i cmae to the conclusion that the point rotated 600 degrees with the small radius and 1200 degrees with the big radius.

now we just use the equation for perimeter for both radiuses

BIG RADIUS LENGTH: 2*pi*3.52 * 2(720 degrees) = 73.72m
SMALL RADIUS LENGTH: 2*pi*1.86=19.48m
total path length: 44.18m+11.69m= 93.2m

there babun i solved it. next question

It should be sqrt(25-2.5*2.5)= 4.33
There is your fault. You used Pythagoras in a wrong way Very Happy

Otherwise your solution is right.

by Ganso Sat Apr 07, 2012 4:54 pm

spanky489 wrote:
Ganso wrote:you cannot use the Pythagorean theory because that is not a right triangle,its equilateral

you can, u split the triangle in half and get 2 right angle triangles.

damn,didnt think about that hmm

by spanky Sat Apr 07, 2012 4:54 pm

Immortal Babun wrote:
spanky489 wrote:ok so basically due to the dot being equally far apart from the other vertices means that we have 2 different radiuses of rotation, one is the big radius which is the distance from either of the furthest vertices to the point and the small radius from the point and to the nearest vertex. the small radius is equal to 1/3 of the triangles height. the other radius we can calculate using the euclidean distance of points equation, to calculate the height and distance along the x axis of the point we use trigonometric functions.

height of triangle: sqrt(25+2.5*2.5)=5.59m
so small radius: 1/3 * 5.59 = 1.86m

lets call the point where im sitting point A with the coordinates (1.86*cos30,1.86*sin30)=(1.61,0.93)
big radius: sqrt(0.93*0.93 + 5 - 1.61 * 5 - 1.61)=sqrt(0.87+11.49)= 3.52m

now i drew out the problem on a piece of paper because even though my intuition told me that this point would plot a few circles till the end i could not perceive the amount of degrees it actually rotated. after drawing it on paper i cmae to the conclusion that the point rotated 600 degrees with the small radius and 1200 degrees with the big radius.

now we just use the equation for perimeter for both radiuses

BIG RADIUS LENGTH: 2*pi*3.52 * 2(720 degrees) = 73.72m
SMALL RADIUS LENGTH: 2*pi*1.86=19.48m
total path length: 44.18m+11.69m= 93.2m

there babun i solved it. next question

It should be sqrt(25-2.5*2.5)=5.59m

yea i just saw this as well.

by Babun Sat Apr 07, 2012 4:56 pm

You both solved the question, kiranr and spanky Very Happy

Next question

Disappearing Squares!

topic: goemetry
difficulty: medium
requirements: last year of high school

Maths- Gym for Brainiacs - Page 12 2q247ja

Maths- Gym for Brainiacs - Page 12 2q247ja

I'm lazy to explain, see the picture above. A square area is divided into 9 smaller square units. All of the small squares possess identical area. Now, I remove the one in the middle( see the white part in the pic above). Afterwards, I divide the small squares into 9 smaller squares again and remove the middle part again. Repeat...( see the pic ).

Questions:

1. How much area of the initial square will remain ( black area) if we repeat the process 10 times?
2. How much area of the initial square will remain ( black area) if we repeat the process infinite numbers of times?
3. How long are the boders of the removed areas for each step ( white area)? Calculate the version for infinite numbers of removals as well.

This whole question is an appetizer to the real one which I'll post after eco smile

by spanky Sat Apr 07, 2012 5:01 pm

ok i updated my original post so everything is correct there now so the other guys will know how we solved it.

by kiranr Sat Apr 07, 2012 5:04 pm

Well, i did not do it using a circle. I just got two formations of moves that keep repeating when the triangle is moving along the side of the square and three formations that keep repeating when the triangle is moving from one side to the adjacent line. I got the numbers of each moves. Along the side, the two sets of moves are 6.6 and 2.5 and from side to side, the three sets of moves are 2.33, 2 and 0.76. Then i just added it up for all the three rounds.

In the first round it moves 29.3 +7.09. In the third round, 29.3 + 5.52 and the second round 18.2 + 7.42. The number before the plus for the movement along the sides and the number after the plus for movement from side to side.

The total comes out to 96.5.

by Ganso Sat Apr 07, 2012 5:05 pm

what are we supposed to do on that question?

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