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stumped tbh ngl
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Based Post
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I think i may have a solution soon
jk im still stumped
jk im still stumped
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Tbh we need more high IQ threads like this.
Yes, that’s correct.
Could you give me a brief outline of the method you used to arrive at that conclusion? I’d love to know, since I only have two different solutions I plan on presenting tomorrow
This is an interesting one.
I think it can be solved by using the angle MLK as an unknown parameter x. Then you solve for the distance MJ to be 5 cos(x).The distance MJ is four times the sine of the angle MLJ, which is x plus the angle KLJ, which is known. PS. No it's not, my bad.
I think it can be solved by using the angle MLK as an unknown parameter x. Then you solve for the distance MJ to be 5 cos(x).
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Are the triangles KML and JKL congruent? I assumed it, but I'm not sure.
Will do tommorow
Nah don't pay attention to what I wrote, I was trying to be quick but I ended up writing something wrong.
Hint for problem #0004
Try looking at isosceles JML
Try looking at isosceles JML
I used Mathematica, ngl.
Apparently MK is half a radius. If one can prove that then it's easy.
PS. Found the most elegant way. See below in the thread.
Apparently MK is half a radius. If one can prove that then it's easy.
PS. Found the most elegant way. See below in the thread.
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Is the solution supposed to be that complex? Wow.
Probably not. This is kind of the brute force approach, I'm sure there is a shortcut somehow.
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I am brain dead at math
math is for faggots
Oh I think I see the smarter way to do it.
Guys, next time you propose a geometry problem, I suggest you name your points with letters A, E, R, J, F, L, N, G, H, T, B and make it so the demonstration involves segments ER, AF, or triangles TBH, JFL and NGL 
We can consider the whole circle and the point L' with whom L forms a diameter. Then everything is much simpler, especially since L', K and J are aligned (not obvious, but probably not difficult to prove either)
Then the squared diameter is (5+3)² + 4² = 64 + 16 = 80. We can divide by 4 to get the squared radius, which gives 20, and by four again and multiply by pi to get the quarter of the disk area. This gives 5pi.
PS. To prove that L', K and J are aligned, we just need to see that the triangle L'JL is right-angled in J because this angle intercepts a diameter. Thus the line L'J is orthogonal to JL, and since KJ is orthogonal to JL too, these lines are the same because by one point only one line can intersect an other line with a right angle (Euclide's axiom).
Then the squared diameter is (5+3)² + 4² = 64 + 16 = 80. We can divide by 4 to get the squared radius, which gives 20, and by four again and multiply by pi to get the quarter of the disk area. This gives 5pi.
PS. To prove that L', K and J are aligned, we just need to see that the triangle L'JL is right-angled in J because this angle intercepts a diameter. Thus the line L'J is orthogonal to JL, and since KJ is orthogonal to JL too, these lines are the same because by one point only one line can intersect an other line with a right angle (Euclide's axiom).
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How do you figure this?
I have considered this but still I cannot find a way to determine the length of the radius.
L'K = KL, and KL = 5 by hypothesis, so L'K = 5.
L'J = L'K +KJ = 5+3 = 8
KJ = 3 and JL = 4 by hypothesis
Beautiful solution, well done! I did not solve the problem this way, but this is brilliant. Kudos to you, man.Oh I think I see the smarter way to do it.
Guys, next time you propose a geometry problem, I suggest you name your points with letters A, E, R, J, F, L, N, G, H, T, B and make it so the demonstration involves segments ER, AF, or triangles TBH, JFL and NGLWe can consider the whole circle and the point L' with whom L forms a diameter. Then everything is much simpler, especially since L', K and J are aligned (not obvious, but probably not difficult to prove either)
Then the squared diameter is (5+3)² + 4² = 64 + 16 = 80. We can divide by 4 to get the squared radius, which gives 20, and by four again and multiply by pi to get the quarter of the disk area. This gives 5pi.
PS. To prove that L', K and J are aligned, we just need to see that the triangle L'JL is right-angled in J because this angle intercepts a diameter. Thus the line L'J is orthogonal to JL, and since KJ is orthogonal to JL too, these lines are the same because by one point only one line can intersect an other line with a right angle (Euclide's axiom).
Also, I will totally use segments ER, AF and triangles TBH, JFL. Even maybe quadrilateral GOER next tme.
One of my solutions involves the cosine addition identity. Namely cos(a+b)=cos(a)cos(b)-sin(a)sin(b).
Look at isosceles JML and see if you can find a sneaky way to use this
.If we call P the midpoint of JL, then MP is also orthogonal to JL. Isn't cos(x)=adjacent/hypotenuse?
I'm looking forward to itAlso, I will totally use segments ER, AF and triangles TBH, JFL. Even maybe quadrilateral GOER next tme.
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Why this IQ mog thread was pinned?
@Divergent_Integral @WorldYamataizer, why aren't you studymaxxxing so that you can betamaxxx later?
You have hopes, unlike braindead cels like me.
@Divergent_Integral @WorldYamataizer, why aren't you studymaxxxing so that you can betamaxxx later?
You have hopes, unlike braindead cels like me.
First, this is not an IQ mog thread. This is simply a thread where incels who enjoy racking their brain on a challenging math problem can come together and do so.
I assure you none of us here were gifted with an “extraordinary talent in mathematics” as I was unable to solve problem #0003, and I stumped a few users with #0004. We just like a challenging puzzle!
Maybe mods pinned it because this thread will be thriving for a while? Not sure about that. I am studymaxxing btw. I’m going into my first year of Uni this fall for math. Im not planning on betabux. I just have to get my degree so my rice parents can have “peace of mind” then I can move on with my life.
Ewwww sounds bittersweet, but at least you are being educated and doing something with your lifetime unlike rotting neets like me who can't study in uni.
Still mogs me
Misery loves company! Dw brocel, everyone here on .co got ur back. I hope you find some really sweet, zen copes.Ewwww sounds bittersweet, but at least you are being educated and doing something with your lifetime unlike rotting neets like me who can't study in uni.
Still mogs me
MATHmxxing rather METHmaxx
Despite two Degrees i still feel like a cavemen everytime i'm looking at Math problems.
I was recently at an international airport, trying to get from one end of a very long terminal to another. It inspired in me the following simple maths puzzle, which I thought I would share here:
Suppose you are trying to get from one end A of a terminal to the other end B. (For simplicity, assume the terminal is a one-dimensional line segment.) Some portions of the terminal have moving walkways (in both directions); other portions do not. Your walking speed is a constant
, but while on a walkway, it is boosted by the speed
of the walkway for a net speed of
. (Obviously, given a choice, one would only take those walkways that are going in the direction one wishes to travel in.) Your objective is to get from A to B in the shortest time possible.
Suppose you are trying to get from one end A of a terminal to the other end B. (For simplicity, assume the terminal is a one-dimensional line segment.) Some portions of the terminal have moving walkways (in both directions); other portions do not. Your walking speed is a constant
- Suppose you need to pause for some period of time, say to tie your shoe. Is it more efficient to do so while on a walkway, or off the walkway? Assume the period of time required is the same in both cases.
- Suppose you have a limited amount of energy available to run and increase your speed to a higher quantity (or, if you are on a walkway). Is it more efficient to run while on a walkway, or off the walkway? Assume that the energy expenditure is the same in both cases.
Ngl this problems gave me a hard time. I guess I'm too out of touch with normal eulerian geometry.
Maby after the exams I might give a look into it, but too much mathceling might actually kill me so Idk
Maby after the exams I might give a look into it, but too much mathceling might actually kill me so Idk
Nice solutions, @WorldYamataizer. Very ingenious, fellow mathcel.
Here's
Problem #0005 (a bit easier than the previous ones)
At which times during the day do the hour and minute hands of an ordinary analog clock make exactly equally sized angles on both sides of the vertical line joining the 12h and 6h marks? Give your answer to the nearest second. You may assume that both hands move at a perfectly continuous rate, in such a way that every time the minute hand hits the 12h mark, the hour hand exactly hits an hour mark.
Bonus problem: See post #88 by @nihility
@Diocel @grondilu @mental_out @SelfCrucified @your personality @Irredeemable @Caesercel @mNFwTJ3wz9 @WorldYamataizer @nihility
Here's
Problem #0005 (a bit easier than the previous ones)
At which times during the day do the hour and minute hands of an ordinary analog clock make exactly equally sized angles on both sides of the vertical line joining the 12h and 6h marks? Give your answer to the nearest second. You may assume that both hands move at a perfectly continuous rate, in such a way that every time the minute hand hits the 12h mark, the hour hand exactly hits an hour mark.
Bonus problem: See post #88 by @nihility
@Diocel @grondilu @mental_out @SelfCrucified @your personality @Irredeemable @Caesercel @mNFwTJ3wz9 @WorldYamataizer @nihility
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more efficient to tie your shoe off the walkwayI was recently at an international airport, trying to get from one end of a very long terminal to another. It inspired in me the following simple maths puzzle, which I thought I would share here:
Suppose you are trying to get from one end A of a terminal to the other end B. (For simplicity, assume the terminal is a one-dimensional line segment.) Some portions of the terminal have moving walkways (in both directions); other portions do not. Your walking speed is a constant, but while on a walkway, it is boosted by the speedof the walkway for a net speed of. (Obviously, given a choice, one would only take those walkways that are going in the direction one wishes to travel in.) Your objective is to get from A to B in the shortest time possible.
- Suppose you need to pause for some period of time, say to tie your shoe. Is it more efficient to do so while on a walkway, or off the walkway? Assume the period of time required is the same in both cases.
- Suppose you have a limited amount of energy available to run and increase your speed to a higher quantity
(or
, if you are on a walkway). Is it more efficient to run while on a walkway, or off the walkway? Assume that the energy expenditure is the same in both cases.
more efficient to run on the walkway
Efficient to do it on the walkway.
Assume you "race" yourself, and one version of you stops to tie the shoe just before getting on a walkway, while the other version gets on the walkway and starts tying their shoes. The latter has a lead, which cannot be compensated for.
This isn't math tbh.
it is lel
the second one is tbh
not the first.
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first can be answered on a mathematical level and on phically intuitive level
Assuming time of sprint is constant t1, distances for walkway is (u + v') * t1 and distance for non walkway is v' * t1 WLOGSuppose you have a limited amount of energy available to run and increase your speed to a higher quantity(or, if you are on a walkway). Is it more efficient to run while on a walkway, or off the walkway? Assume that the energy expenditure is the same in both cases.
time of sprint in walkway = (v' / v) * t1 + t1
time of sprint in non-walkway = ((u + v')/(u+v)) * t1 + t1
Obviously 2nd one is lower time
Even the second is less but still intuitive tbh
Like if you think of v' is infinity/very high or u is infinity/very high
It would obviously be useful to use the "teleport" to skip the walking (v' is high)
Or it would be obvious to speed up while walking, rather than during the "teleport" (u is high)
Are you asking this ?
if so then timing is h,m
and the equation is h/12 + m/60 = (30-m)/60
h = 0,1,2,3...11
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I said both sides of the line. So like 1:50, but more precisely.
then it will be : h/12 + m/60 = 1 - m/60
h = 0,1,2,3...11
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Basically from midnight, and then every 43200/13 seconds.
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Mathematicians on the forum? Wtf is wrong with society
Our best representative went ER before going ER was even a thing :
I'm having trouble wrapping my head around problem #0005. I'm not sure how to even begin. Gonna go on a walk to ponder it. I'm thinking trig functions.
I'll get back to yall if i get somewhere.
I'll get back to yall if i get somewhere.
Holy fuck that is complex as shit
Also made a retarded error in this lol
then it will be : h/12 + m/(60*12) = 1 - m/60
h = 0,1,2,3...11
Yeah I may have made it look more complex than necessary.
We just have to consider the angles a and b of respectively the minutes hand and the hours hand, as separating them from the midnight position. To satisfy the problem requirement, they must be opposite modulo 2 pi.
a = -b + k 2 pi
By definition we also have :
a = 2 pi t/3600 and b = 2 pi t/(12 * 3600)
so :
2 pi t/3600 = - 2 pi t/ (12 * 3600) + k 2 pi
t /3600 = - t /(12 * 3600) + k
t/3600 (1 + 1/12) = k
t = 43200/13 k
Notice that this cycle lasts precisely twelve thirtheenth of an hour. About 55 minutes.
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I am very pleased with this second solution of yours.
Problem #0006
In the diagram, GOER is a rectangle. Circles of radii 3,4,5 are drawn in as shown. What are the dimensions of the rectangle?
Edit: For clarity, each circle is tangent to anything it touches.
Caution! I think this is a challenging problem. Have fun!
@Divergent_Integral @SelfCrucified @grondilu @mNFwTJ3wz9 @Diocel @mental_out @your personality @Irredeemable @Caesercel @nihility
In the diagram, GOER is a rectangle. Circles of radii 3,4,5 are drawn in as shown. What are the dimensions of the rectangle?
Edit: For clarity, each circle is tangent to anything it touches.
@Divergent_Integral @SelfCrucified @grondilu @mNFwTJ3wz9 @Diocel @mental_out @your personality @Irredeemable @Caesercel @nihility
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