Discover the Solution to the World's Hardest Easy Problem on Thinkzone.wlonk.com

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In summary, the conversation discusses a question on a webpage about finding the solution to a diophantine equation involving angles in a triangle. One participant suggests drawing the triangles to scale and using parallel lines as a strategy to solve the problem. Another participant mentions that drawing to scale is not typically taught but might be helpful in this case. The conversation ends with one participant admitting to looking up the solution online and another participant recommending seeking help from teachers and mentors.
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  • #2
I haven't the solution, I get the equation x + y = 130, where x and y are angles. If [itex] x, y \in Z^+[/itex]
This is a diophantine equation. We know: x, y > 0. y > x. And y is greater than 90 degrees.
the most probable angles:
[itex] x=20^o\,\, ,y=110^o[/itex]
or
[itex] x=30^o\,\, ,y=100^o[/itex]
 
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  • #3
I tried a couple of things but didn't solve it yet , but there's probably a way to build your way up to the x angle using the tricks in my sketch (just don't have the patience to construct the triangles properly and doing it now):
 

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  • #4
Simple geometry is all that is required.

Why can't you do this for yourself? Don't give up so easy.
 
  • #5
janhaa said:
diophantine equation
wazz it,
i tried it hard that's why asking askin solution,
well i need an approach
i asked it to one of my faculty at coaching and he straightaway said it's wrong so i needed to check
 
  • #6
janhaa said:
I haven't the solution, I get the equation x + y = 130, where x and y are angles. If [itex] x, y \in Z^+[/itex]
This is a diophantine equation. We know: x, y > 0. y > x. And y is greater than 90 degrees.
the most probable angles:
[itex] x=20^o\,\, ,y=110^o[/itex]
or
[itex] x=30^o\,\, ,y=100^o[/itex]

reenmachine said:
I tried a couple of things but didn't solve it yet , but there's probably a way to build your way up to the x angle using the tricks in my sketch (just don't have the patience to construct the triangles properly and doing it now):
whatss y in equation
x+y= 130

reenmachine i agree myself i even not have patience and doing it (doing angle sum property question) , thanks for tip of parallel lines but what benefit it does
 
  • #7
namanjain said:
reenmachine i agree myself i even not have patience and doing it (doing angle sum property question) , thanks for tip of parallel lines but what benefit it does

I don't know since I didn't solve it.It was just a suggestion of strategy to attack the problem , maybe it's a dead end , but the parallel lines will at least allow you to chase more angles , whether or not they end up being useful to find angle x.

If you don't have the patience to try it , I suggest you do not ask others to do it for you as it's not recommended on the forum to ask for answers without showing that you tried hard before hand.I'm not a moderator , just friendly advice.Also , saying that you tried hard is not showing that you tried hard.
 
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  • #8
till here i ve done (in attchment)
one of image is inverted
so i did the mirror image from webcam toy
 

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  • #9
You should draw the triangles/angles to scale if you want to try that strategy.

It should look like the picture below
 

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  • #10
kingo1979 said:
from the original figure below, y seems equal angle bde

http://thinkzone.wlonk.com/mathfun/triangle.htm
=====
where x + y + 50 = 180
however i got that but i also meant that diophtine (whatever is name) equation

well where is it taught
 
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  • #11
reenmachine said:
you should draw the triangles/angles to scale if you want to try that strategy.

It should look like the picture below

does that help, because doing any of olympiads we were never told to draw to scale and i was always of kind to do geometry of this kind mentally so it's quite difficult for me
though i think i'll print that page
 
  • #12
Uff! After long work with a correct precise figure i bored and saw solution on one of other web pages

is it bad of me not getting it (i'm 15 yrs and preparing rmo and iit)
 
  • #13
namanjain said:
does that help, because doing any of olympiads we were never told to draw to scale and i was always of kind to do geometry of this kind mentally so it's quite difficult for me
though i think i'll print that page

I don't know , you should listen to your teachers before you listen to me that's for sure :tongue2:

You should also listen to mentors here as they are qualified to help you.
 
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  • #14
Might try to prove both of these triangles (or another pair of triangles) are similar to prove x = 20 :
 

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What is the "World's hardest easy problem"?

The "World's hardest easy problem" is a mathematical puzzle created by American mathematician Paul Erdős. It is a simple problem that is notoriously difficult to solve and has puzzled mathematicians for decades.

What is the specific problem involved in the "World's hardest easy problem"?

The specific problem involved in the "World's hardest easy problem" is finding the minimum number of colored regions needed to color a map without any adjacent regions having the same color. This is known as the four color problem.

Why is the "World's hardest easy problem" considered difficult to solve?

The "World's hardest easy problem" is considered difficult to solve because it is easy to understand and state, but finding a proof or solution has proven to be extremely challenging. Many mathematicians have attempted to solve it, but it still remains unsolved.

What is the significance of the "World's hardest easy problem" in mathematics?

The "World's hardest easy problem" is significant in mathematics because it has sparked interest and research in the field of topology and graph theory. It has also led to the development of new techniques and algorithms that have been used in other mathematical problems.

Has anyone been able to solve the "World's hardest easy problem"?

No, the "World's hardest easy problem" has not been officially solved. However, in 1976, two mathematicians, Kenneth Appel and Wolfgang Haken, used a computer to prove that any map can be colored with only four colors without any adjacent regions having the same color. This is considered a solution, but it has not been accepted by all mathematicians due to the use of a computer in the proof.

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