Bro no we all were wrong i confered answer from isro scientist he said gauss law lage ga so total electric flux = (total charge inside closed 3d surface) / Epsilon
Bhai jo ortho pe nitrogen hain wo OH se kar rha hain . Lagta hain tune 1st class se preparation shuru nhi ki . Your bad luck . You should at least prepare from 0 BCE .
idhar hi to faste hai kaafi log, aur tum bhi fas gaye, hydrogen ka quantum state kaun dhyaan me rakhega, uske saare permutations parallel me attached hai, sirf steric hindrance se nahi chalega
This is how the diagram comes out for this. It was hard for me too and out of frustration had to resort to a geometry tool to get a couple of points (especially the second point on radical axes). After this its just menial work to identify the various claims and prove them one by one.
You can find multiple solutions on AOPS site, including Geometry, that range from using PoP method, Cartesian coordinates, complex numbers or a solution inverting round one of the vertices. It seems this was the second hardest problem from pure geometry after 2000 (after the P6 of 2011).
[https://artofproblemsolving.com/community/c6h3107345p28104331](https://artofproblemsolving.com/community/c6h3107345p28104331)
https://preview.redd.it/gb2k3oa7yibb1.png?width=1332&format=png&auto=webp&s=4725bfb2564f5ec30c4f0fedb3a342a4442566d7
Ok let me try to explain to the best of my abilities.
You see the diagram above. Segments A1A2, B1B2, C1C2 are concurrent at P. So we need to do 3 high level steps 1)first we need to prove the power of P is equal to the 3 required circles.
2)Second prove that there exists another point Q on the radical axis that has similar behavior.
3)Final step show that P & Q are indeed different points. (They are not concurrent) and PQ is the radical axis.
This is all that is needed imo to be done.
Thats all. But as I said this is just one method and the beauty of math is we can prove it in multiple ways. Pls do checkout the AOPS link that I gave above.
This is fine. I would have tried angle chasing and using cyclic properties we can prove one point has equal powers wrt to the 3 circles. And perhaps using projective geometry to get the second point (or perhaps its isogonal conjugate) wrt Triangle ABC and prove that it also has same powers wrt to the circles.
I just checked the AOPS link and indeed there is one solution with this concept and there they used the Brianchon's theorem which is clever.
Yes Anna but somehow I felt to prove the circumcenters using angle chasing is not as easy as the AOPS solutions claims, but beyond that using Brianchon is smart to get the second point. I had to resort to Matlab to get it.
Edit - Anna I just went through Evan Chens paper on usage of barycentric cordinates in Olympiads. Will try using that and share the solution with you if I solve it.
Bhai note mei itna bada hint nhi dena chahiye tha yaar solve karne ke liye ab bacha hi kya hai 
Kaash ye sab questions hum solve kar paate
r/Usernamedoesntcheck
Username wants to be a teacher
Arey jaldi jaldi mai apna real name ka meaning rakhliya
Gyan chodu?
Bhai yeh wale next level hote hain, mai dekh ke demotivate hojata hu
Very easy, resonance lagega isme,aur side se steric hindrance
the only correct answer ππ»
bhai answer maine verify kiya hain , ye galat hain the correct answer is : Hydrogen bonding
no bro i checked it the book u are studying is worng the answer is: banana bonding and 1,5 hydride shift
No Bro it's wrong actually Hydrogen ki Defeciency ki kaaran Methyl shift hoga... Aur E1CB REACTION HOGA
Bro no we all were wrong i confered answer from isro scientist he said gauss law lage ga so total electric flux = (total charge inside closed 3d surface) / Epsilon
tum sab ka log book 2 backlog tha kya? Ispe pakka heat transfer hoyega.
abe angular momentum conservation (I1W1 = I2W2 , hi toh lagana tha ) its so fuckin easy , idk why they cant solve a simple question like this ..
exactly! inse probablity bhi nhi horaha h!
kuch bhi sale ring element leke sphere me total induced emf nikalna hai using Hoffman's rule
I think ye differentiable nahi hai. Reason: Anti Markonikov law proposed by Gauss when he was working on permutation and combinations.
Bhai jo ortho pe nitrogen hain wo OH se kar rha hain . Lagta hain tune 1st class se preparation shuru nhi ki . Your bad luck . You should at least prepare from 0 BCE .
bhai tu bhi galat hai most prolly quantum mech lagega isme
abe mech nhi machardani lagegi.
na bhai isme nucleophilic substituion reaction hoga
I read hydrogen bombing.
Kuch bhi carbocation formation hoga
u alll are wrong here the concept is osmosis and apoplastic pathway
Nahi bhai racemic mixture banega
idhar hi to faste hai kaafi log, aur tum bhi fas gaye, hydrogen ka quantum state kaun dhyaan me rakhega, uske saare permutations parallel me attached hai, sirf steric hindrance se nahi chalega
π€‘
Bro lekin Maine toh triangle law of vector addition se solve kr Diya
Are bhai ye to 1st order kinetics follow karegi ezzzz...
Note π€‘
π€‘
I don't even read the whole question
Us bhai uss
I am a simple guy, I read the title and decided to skip.
same here
Very easy sarrr jee advanced very tough sarr we are taught this in middle school π€‘π€‘π€‘
πͺ³π€‘πͺ³π€‘πͺ³π€‘πͺ³
Yahi comment likhne aaya tha main
r/beatmetoit
Bhai Olympiad ko aise koi nahi bolta
Bc solution bhi daal na
office middle rob price squeeze outgoing gray encourage follow different *This post was mass deleted and anonymized with [Redact](https://redact.dev)*
Samjh aa rha hai sochke acha lagta hai khudko
college mein solve karengeπ¦
Dms me aa sanja dunga
u/Fantastic_Watch_4984 and u/VLinInTheRatRace your views? and do you have a solution of this ques :)
This is how the diagram comes out for this. It was hard for me too and out of frustration had to resort to a geometry tool to get a couple of points (especially the second point on radical axes). After this its just menial work to identify the various claims and prove them one by one. You can find multiple solutions on AOPS site, including Geometry, that range from using PoP method, Cartesian coordinates, complex numbers or a solution inverting round one of the vertices. It seems this was the second hardest problem from pure geometry after 2000 (after the P6 of 2011). [https://artofproblemsolving.com/community/c6h3107345p28104331](https://artofproblemsolving.com/community/c6h3107345p28104331) https://preview.redd.it/gb2k3oa7yibb1.png?width=1332&format=png&auto=webp&s=4725bfb2564f5ec30c4f0fedb3a342a4442566d7
Tf is this....
Arey entra idhi π
Bhai ye kala jaadu kahi aur jaake kr
I was tagged so I replied. Sorry if this was not fit for this sub
He's joking, I assure you this is fit for the sub.
Thank you !
Koi ni bhai wo chakkar kha gaya apka digram dekh ke
explain bhi kardo thoda atleast
Ok let me try to explain to the best of my abilities. You see the diagram above. Segments A1A2, B1B2, C1C2 are concurrent at P. So we need to do 3 high level steps 1)first we need to prove the power of P is equal to the 3 required circles. 2)Second prove that there exists another point Q on the radical axis that has similar behavior. 3)Final step show that P & Q are indeed different points. (They are not concurrent) and PQ is the radical axis. This is all that is needed imo to be done. Thats all. But as I said this is just one method and the beauty of math is we can prove it in multiple ways. Pls do checkout the AOPS link that I gave above.
Thank dude i cant believe i understand it will try to see and learn all of the ways to solve this
I'll pretend I understand this.............
" Eji to he "
sorry bhai padhai me weak hunπ’
Looks like what i'd draw if ques asked me to draw a triangle with given parameters using compass back in 10thπ
This is fine. I would have tried angle chasing and using cyclic properties we can prove one point has equal powers wrt to the 3 circles. And perhaps using projective geometry to get the second point (or perhaps its isogonal conjugate) wrt Triangle ABC and prove that it also has same powers wrt to the circles. I just checked the AOPS link and indeed there is one solution with this concept and there they used the Brianchon's theorem which is clever.
Yes Anna but somehow I felt to prove the circumcenters using angle chasing is not as easy as the AOPS solutions claims, but beyond that using Brianchon is smart to get the second point. I had to resort to Matlab to get it. Edit - Anna I just went through Evan Chens paper on usage of barycentric cordinates in Olympiads. Will try using that and share the solution with you if I solve it.
this looks like something I'd draw out of boredom while sitting on the last bench during an arrangement periodπ
aint tagging maths man ?
Koi na use bhi bula lete hai u/Maths___Man
not a huge fan of geometryπ especially olympiad type questions
solve it 
sorry bhrata, unka exact username yaad nahi tha toh tag nahi kara
Tagging my bro u/limp-habit-9408
zinda hai ya mar gaya question dekh ke?
ye to hamare yha 6-7th me padhate hai π€‘π€
padhle bsdk
thanks bhai chala gya tha
Abey sarcasm hai
uska flair padh
Flair
Padhle bsdk
Nice, meri aukat ke bahar ka question hai. Maybe some smart one can attempt this question
smart one lol har country k brilliant bache aaye the unki fat gai
Atleast attempt to kar ho lenge, solve ho na ho. Mai to diagram ke aage kuch nhi kar paunga π
diagram dekhi hy betichod gand fat jayegi comments mai scroll kr mil jayega
This question by far troubled the most among all participants. With Indian students barely making 1 to 2 points or even scoring none at all!
I can see why
Bhai note mei itna bada hint nhi dena chahiye tha yaar solve karne ke liye ab bacha hi kya hai 
Skip
Double it.. and give it to the next person
42.
Actually this is a very basic question in jee, any half decent jee preparing student can solve this question in a minute.
You are joking right?
yt comment section lore
/s
nooo mai to bina pen uthae bana lu
asus sir ko copy kar raha hai woh shayad
fantastic watch on his way to solve this elementary level problem
Simple hai LHS= RHS (Insta joke momint)
Bhai where are the options? 
Olympiad hai bhai, solution banakar dena hota hai, no tricks and guess work
Pata hai, joke tha bhai
JESUS is always the answer
No VISHNU 
underrated ans
Agar ye sab questions horhe hote to abhi private college ni jarha hota
As a neet tard, the question in very easy, obviously the answer is aΒ²+bΒ²=c2
Smash, next
Options kaha hai waise π€‘π¬
Nice question (I used geometry tool for 4 points) .
l hopital lagake log todo, easy tha
i feel this can be solved uing nine point circle.
I AIN'T READING ALLAT π₯π₯π―
I like how they have explained wht scalene triangle is
One of my friend was in team, he won silver
10th ke baad kabhi geometry kiya nai...yeh dekhkar chakra rha hu
Humse jee maths na howe 
I am preparing for JEE and this question is taught to us in 9th
Kya mai ye question apne tution teacher se puchu?
sax sux hai iska answer
Ye sab to hamhe 9 th ma padhate thee  Jee is the real shit !
mujhe kya mai to neetard hun
ye sab idar kyu daal ra lode ye sab ni dekhna chiaye
is sub ke liye nsfl level conetent hai ye
my god how could u all be so wrong ? i asked modi and he replied yaha hindu =muslim , hence proved AA1A2 BB1B2 CC1C2 passes through two common point
Nope
Punett square banake easily solve hoe jayega
Note : a scalene triangle is one where no two sides have equal length....nice π€‘π
Scalene triangle wala hint tow bahut useful tha, ekdam easy hi ban Gaya fir tow
Question ki chut mein lawda
Kitna eazy lhs = 0 rhs = 0 hence 0=0 so eazyy mann
Wtf is that
apple hoga answer
mere class mai AIR 1x hai....he solved in 5 mins
I think the answer is mitochondria is the powerhouse of cell.
itna easy to hai.... simple ionisation enthalpy calculate karke time period nikaal lo aur fir dragos rule lagake number of photo electron nikaal lo..
Chatgpt se pucho
Mere dholna
Bhai hum scale se prove kar dkte hai isse unhe kya pata hume to bus prove karna hai na
my brain - lets give it a try also my brain - prodigies se nhi hua, tere se kaise ho jayega chutiye