Geometric challenges or puzzles (can include math)

[Anonymous No.5879148]

Any geometry (or any other) challenges or puzzles that you want to share?

[Anonymous No.5879158]

>>5879148
alright
theres 5 circles in a square, the circles touch edges perfectly, with no gaps between them.
all circles are the same size, with radius of 6.5cm.

find the dimensions of the square that they are in

[Anonymous No.5879163]

>>5879158
me again.
puzzle 2: using the digit '8' exactly 8 times, along with any mathematical symbols you like, make it total 1000.

go

[Anonymous No.5879177]

>>5879148
any1 else doin this thread or wat?

someone else post some

[Anonymous No.5879179]

>>5879158
the dimension can be several things, but I guess they are divided by 45 degrees out of the center?

[Anonymous No.5879184]

Move 3 coins (and don't touch coins with other coins etc), and make the triange point upwads

[Anonymous No.5879190]

>>5879179
>dimension
Size you fucking moron.

[Anonymous No.5879193]

>>5879179
well obviously, seeing as it's a fucking square!
also, solve my second puzzle first, seeing as it's easier

i got more ;)

also, the first puzzle only has one answer. there might be multiple methods on the second puzzle

[Anonymous No.5879197]

>>5879184
easy as fuck, hun. im pretty sure i have the answer to this saved in gif form somewhere, but i cant be assed to fucking find it, so it's just easier to do this
<<<

the hexagon piece inside the line stays static. move the other 3 coins around in a mirror image.

[Anonymous No.5879204]

>>5879197
you cant solve it without the answer... lol even knowing the middle stays the same

[Anonymous No.5879217]

>>5879163
unsolvable

[Anonymous No.5879215]

>>5879204
i CAN fucking solve it pindick!
theres only 3 other coins to fucking move! and any of them can go anywhere, it still fucking works.

i knew this years ago, i practically fucking invented that puzzle

[Anonymous No.5879223]

>>5879217
no it fucking isnt you short-sighted little spastic!

if you were more cheatingly inclined and less mind-numbingly fucktarded then you could have fucking googled that shit to find out how wrong you are.

but you're lazy and stupid, huh?

good to know

[Anonymous No.5879226]

>>5879215
>>5879197
No need to be a cunt dude.

[Anonymous No.5879227]

>>5879226
no need to be a dude, cunt

[Anonymous No.5879230]

>>5879226
no need to be a dude, cunt

[Anonymous No.5879237]

>>5879158
Diagonal is 6 times the radius = 39, divided by sqrt(2) gives side a=39sqrt(2)/2=25,58 and area of 760,5.

[Anonymous No.5879240]

>>5879163
>>5879223
888+88+8+8+8 =1000

[Anonymous No.5879241]

>>5879230
no need to be a picard, dicktard

[Anonymous No.5879242]

>>5879230
no cunt to be a dude, need.
>>>/b/

[Anonymous No.5879243]

Guys I need your help and it is thread related. My mom wants to make a pie but the recipe she has is for a plate with a diameter of 23 cm. Her plate is 28cm. I thought I had to multiply the ingredients by 28/23, but apparently the ingredients come short. By what do I need to multiply the ingredients?

[Anonymous No.5879245]

>>5879237
No, the diagonal is 70,3848...

[Anonymous No.5879246]

>>5879243
Salt and pepper.

[Anonymous No.5879252]

>>5879237
first of all, dont use fucking commas as decimal points, cunt.

secondly, you cant fucking read because i said SIX POINT FUCKING FIVE!

[Anonymous No.5879248]

>>5879245
No idea how are you getting that, but seems wrong.

>>5879243
Look at it as area of circle. Calculate the ratio of areas instead of diameters.

[Anonymous No.5879259]

>>5879242
no penis to penetrate my mingus, have

>>>/lgbt/

[Anonymous No.5879261]

>>5879240
yeh, that's right, but i bet you fucking googled it

[Anonymous No.5879262]

>>5879237
When you do that, you don't get the distance from the end of the circle and in to the corners.

[Anonymous No.5879265]

>>5879248
I don't understand. What should the calculation be? For the pie btw

[Anonymous No.5879273]

>>5879262
Oh sorry, I just realized that I'm fucking retarded, I'll have a less quick look at the problem now.

[Anonymous No.5879285]

>>5879158
S = side length
D = diagonal length

D=4*6.5+2sqrt(6.52+6.52*tan(pi/4)2)
=26+2sqrt(2*6.52)

D2=2S2
=> S2=D2/2 (approx 985 cm2)

[Anonymous No.5879290]

>>5879262
So yeah, 10 seconds later, you need to add 1 (2 halves on each end) diagonal of a square with a=2r with 4r_circle to make up for the empty space.
D=4*6,5+13sqrt(2)=26+13sqrt(2)=44,38
a_big_square=31,38

Can somebody check up on my math, I'm fucking up a lot tonight.

[Anonymous No.5879292]

>>5879285
..........
D=4*6.5+2sqrt(6.5^2+6.5^2*tan(pi/4)^2)
=26+2sqrt(2*6.5^2)

D^2=2S^2
=> S^2=D^2/2 (approx 985 cm^2)

[Anonymous No.5879294]

Area of square is about 815,47

[Anonymous No.5879307]

>>5879193
>i got more ;)

Give more then!

[Anonymous No.5879310]

>>5879148
find angle X in this isosceles triangle

[Anonymous No.5879314]

>>5879294
>>5879292

Who is right?

[Anonymous No.5879317]

>>5879158
>>5879158
35.1923 x 35.01923

[Anonymous No.5879322]

>>5879310
30 degrees

[Anonymous No.5879331]

>>5879314
Me.
Area=(26+2sqrt(2*(6.5)^2))^2/2

[Anonymous No.5879349]

>>5879310
>>5879322
20 degrees.

[Anonymous No.5879354]

>>5879349
>implying DEC is a right angle

[Anonymous No.5879361]

>>5879331
>>5879314
You are so right.
I made a mistake.. I did it this way:
((6r+2*6,5*(sqrt(2)-1)) / sqrt(2)) ^2

[Anonymous No.5879375]

>>5879354
>DEC is a right angle.

No.
Top is 20 deg. and it is an isoceles => B and C is 80 deg.
This means the small angle above B is 20 deg. and the one above C is 30 deg.
And you can then easily know that the "?" is 20 deg.

[Anonymous No.5879384]

>>5879375
>pathetic trolling attempt

[Anonymous No.5879394]

>>5879384
Prove it wrong

[Anonymous No.5879420]

>>5879417
whose frequency approaches infinity

[Anonymous No.5879417]

>>5879148

make a filled cardioid out of a sine function.

[Anonymous No.5879469]

>>5879375
20 + 80 + 80 = 200

B and C must be 70, faggot.

>1/10 i replied

[Anonymous No.5879491]

>>5879469
80+80=160 + 20 = 180. wow.

[Anonymous No.5879516]

Please don't respond if you know the answer, or google it.

1
11
21
1211
111221
312211
13112221
1113213211

[Anonymous No.5879549]

>>5879516
>asks a question.
>"don't respond if you know the answer"

oh ok, so you only want wrong answers then?
then why did you fucking post it!??

fucking eejit

[Anonymous No.5879619]

>>5879285
omg, wtf are you doin??
no! you're wrong

>>5879317
no, and its a fucking square!
so here's a clue:
both dimensions of a square are exactly the same!

[Anonymous No.5879632]

>>5879375
>And you can then easily know that the "?" is 20 deg.
must have missed a few steps there partner, since you're wrong.

[Anonymous No.5879633]

>>5879619
>>5879619
Tell the answer then. was 815 right?

[Anonymous No.5879679]

>>5879310
50 degrees

[Anonymous No.5879687]

>>5879633
[sarcasm] yes, it was 815. i asked for teh dimensions of the square, and its 815 centimeters by 815 centimeters, even though the radius of one of teh circles is only 6.5 centimeters, and the square cant contain 3 circles in a line, but yeh, somehow it works out as exactly 815 centimeters for the sidelength of the square. congratulations, i am so jelly of your IQ [/sarcasm]

[Anonymous No.5879740]

In easy, simplified picture form (with red being a circle's centre).

[Anonymous No.5879746]

>>5879679
sorry nigga, my bad, it's actually 30 degrees. I've been awake for 26 hours and I'm slightly drunk so I fucked up the math first time.

[Anonymous No.5879747]

>>5879740
yes! finally, a non-retard!
yeh. that's correct.
and somebody already solved the 8's to 1000 one, so i guess you fags are due another puzzle
gimme a minute

[Anonymous No.5879763]

<<<
you are a chess-knight
start anywhere you like, but you must hit each square once and only once.

post solution in a numerical-coded format, like 3-6-11-17-etc so i can check correct answers.

[Anonymous No.5879773]

>>5879516
31131211131221

fuck the man

[Anonymous No.5879779]

>>5879763
be more specific nigga.
I guess I can go 1-4-5 but not 1-12-13.

[Anonymous No.5879778]

>>5879158
answer is the height and width =
here is my working for the question.
r = 6.5
6.5*2 = 13
13*3 = 39
a squared+ b squared = c squared
c squared = 39*39 = 1521
a squared = b squared
1521/2 = 760.5 = a or b squared
square root of 760.5 = 27.577
height and width = 27.557
point me out if I am wrong

[Anonymous No.5879784]

>>5879773
>please dont respond if you know the answer
and you obviously do know the answer, because they didnt even actually ask a fucking question!
for all we know, it could be some random pyramid of numbers
(obviously it's not, and i know the trick to it as well, but he never actually said "what is the next line?" or something)

[Anonymous No.5879788]

>>5879784
it's so heavily implied that he didn't have to

[Anonymous No.5879789]

>>5879778
>point me out if I am wrong
you are wrong.
same mistake as that other spaz who DOESNT FUCKING REALISE THAT 3 DIAMETERS DONT REACH THE EXACT CORNERS OF THE SQUARE! SO YOU LOSE DISTANCE AND GET THE WRONG FUCKING ANSWER!

and correct answer has already been posted anyway, so GTFO

[Anonymous No.5879799]

>>5879779
>be more specific
to someone who isnt actually retarded, it's pretty fucking clear
idiotic nonces that are ignorant as to how knight move in chess need not apply

[Anonymous No.5879815]

>>5879789
I forgot about the corners thanks for the correction

[Anonymous No.5879832]

>>5879815
...srisly /sci/, do the knight chess one
it aint even that hard, wtf.

[Anonymous No.5879859]

>>5879310
40 degrees.

[Anonymous No.5879862]

>>5879859
stop fucking guessing and POST A METHOD!

[Teacup !obkpuSWh5M No.5879921]

>>5879310
110º

[Anonymous No.5879925]

et X be a projective complex manifold. Show that every Hodge class on X is a linear combination with rational coefficients of the cohomology classes of complex subvarieties of X.

[Anonymous No.5879927]

>>5879763
here's one:
3-10-16-9-1-7-5-12-14-17-11-2-8-15-6-4-13

[Anonymous No.5879931]

>>5879927
yep, winrar

[Anonymous No.5879934]

>>5879925
*Let

[Anonymous No.5881347]

MOAR PUZZELS!

>those aylossi

[Anonymous No.5881392]

>>5879925
>implying there have been more than a handful of people on /sci/ who even know what any of those things are

[Anonymous No.5881393]

Just watched a video, surprisingly a little tough after a while.

0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

You can put as many symbols as you like in between them, but it has to equal six. No writing of numbers, constants and variables. No operations which involves writing symbols with numbers on them. For example, I can write a square root, but not a cube root, as that would require writing a three.

I'll give the first one free:

6 + 6 - 6 = 6

[Anonymous No.5881398]

>>5881393
>in between them
Or before or after, sorry if that was a little confusing.

[Anonymous No.5881404]

>>5881393
2 + 2 + 2 = 6
3 * 3 - 3 = 6
5 / 5 + 5 = 6
7 - 7 / 7 = 6

Give me a second for the others

[Anonymous No.5881414]

>>5881404
(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
sqrt(4) + sqrt(4) + sqrt(4) = 6

Still not sure about 8 and 9

[Anonymous No.5881417]

>>5881414
8-sqrt(sqrt(8+8))=6
sqrt(9)*sqrt(9)-sqrt(9)=6

[Anonymous No.5881425]

>>5881393
4+4-sqrt(4)=6

(9-sqrt(9)) % 9=6
% is modulus operator

[Anonymous No.5881426]

>>5881414
8 - sqrt(sqrt(8 + 8))
sqrt(9) * sqrt(9) - sqrt(9) = 6

[Anonymous No.5881436]

more geometric shapes!

[Anonymous No.5881509]

>>5879158
it has a lenght of 17.5cm

[Anonymous No.5881512]

[Anonymous No.5882298]

>>5881436
nobody has proven a solution for this one yet:
>>5879310
you only need three things: triangle sums to 180, property of isosceles, property of equilateral.

[Anonymous No.5882726]

>>5882298
I fuckin give up man.
Could you at least post a solution to Pastebin or something, so those of us that aren't good enough can get some closure?

[Anonymous No.5883358]

>>5882726
Bump for this.

[Anonymous No.5883372]

>>5882726
Let
(1) △ABC be isosceles, with ∠ACB = 20º, AC = BC
(2) D on AC, ∠ABD = 50º
(3) E on BC, ∠BAE = 60º

Find ∠AED.

(4) (1) B. => ∠BAC = ∠ABC = 80º
(5) (2)(4) A. => ∠ADB = 50º
(6) (2)(5) B. => △ADB isosceles, AD = AB

(7) B. Let F on EB, AF = AB, so △ABF isosceles
(8) (4)(7) B. => ∠AFB = 80º, A. => ∠BAF = 20º
(9) (3)(8) A. => ∠EAF = 40º
(10) (3)(4) A. => ∠CAE = 20º
(11) (9)(10) A. => ∠DAF = 60º
(12) (6)(7) B. => AD = AF, so △DAF isosceles
(13) (11)(12) C. => ∠ADF = ∠AFD = 60º, so △DAF equilateral, DF = AF
(14) (3)(4) A. => ∠AEB 40º
(15) (9)(14) B. => △AFE isosceles, EF = AF, A. => ∠AFE = 100º
(16) (13)(15) A. => ∠DFE = 40º
(17) (13)(15) B. => EF = DF, so △DFE isosceles
(18) (16)(17) B. => ∠FDE = ∠FED = 70º
(19) (14)(18) => ∠AED = ∠FED - ∠AEB = 30º

∴ QED
A. three angles of a triangle add to 180º
B. two sides equal <=> two opposite angles equal (= (180º-A)/2) <=> isosceles
C. three sides equal <=> two angles 60º <=> equilateral

[Anonymous No.5883377]

>>5883372
>(2) D on AC, ∠ABD = 50º
>(3) E on BC, ∠BAE = 60º
I can already tell you this is horribly wrong.
Fuck, ∠BAE isn't even an angle.

[Anonymous No.5883408]

>>5883377
oops, this proof was for a diagram with D and E swapped, and C was the 20° angle. How confusing.

Here is a clearer, animated proof:
http://gogeometry.com/LangleyProblem.html

[Anonymous No.5883837]

>>5879158

not too hard

[Anonymous No.5885531]

>>5883837
that's a way to do it too.

[Anonymous No.5885558]

>>5879310
>>5882298
>nobody has proven a solution for this one yet:

i think i've managed to do it. i just filled in every single possible angle. with angles in a straight line = 180, angles around a point are 360, and angles in a triangle are 180, as well as isosceles triangles having 2 identical angles.
<<<

[Anonymous No.5885561]

>>5883837
wow, nice! i actually prefer that method

[Anonymous No.5885566]

>>5885561
Why can't you into middle school geometry, EK?

[Anonymous No.5885585]

>>5885566
firstly, wtf?
oh, checking my images again? w/e
secondly, im the one who actually posted the fucking problem in the first place, i know the answer, and some anon already solved it.
also, this is my answer from a previous thread
<<<

image check that, faggot
you'll see the problem is an old one.

(INb4 EK is a faggot for posting old problems and not being original and making new puzzles)
fuck off, it's old enough that everyone has already forgotten.

[Anonymous No.5885911]

>>5879163
(8888-888)/8=1000

[Anonymous No.5886004]

[Anonymous No.5886303]

>>5886004
Impossible to solve with that amount of data

[Anonymous No.5886335]

>>5886303
It's possible.

[Anonymous No.5886347]

>>5886004

2?

[Anonymous No.5886401]

>>5886347
That's what I'm thinking. I don't know how to go through a rigorous mathematical explanation of it, but logic tells me they're similar triangles, so I guess 2. Can someone explain?

[Anonymous No.5886433]

>>5886004

Don't draw lines as perpendicular when you don't want them to be. That's just being misleading for the sake of it.

Angles ACB = ADB (subtended by common chord AB)
Angles CAD = CBD (subtended by common chord CD)
Angles BXC = AXD (BD and AC are straight lines intersecting at X)
Thus, triangles AXD and BCX are similar. Hence, x/4 = 4/8 and so x = 2.

[Anonymous No.5886490]

>>5886433
angle ACB?

[Anonymous No.5887853]

>>5886490

[Anonymous No.5888763]

>>5886433
How did you find out the angle to start with, or the circumference, or the diameter?

[Anonymous No.5888790]

http://www.wolframalpha.com/input/?i=batman+equation

na na na na na na BATMAN