Craziest IIT JEE Advanced Geometry problem
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- Опубликовано: 5 сен 2024
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Sab isko itna complicate kyu kar rahe hai
X²+Y²=50
(X+2)² + (Y-6)² = 50
Seedha x and y ki value aa jati hai
A(5,5) C(7,-1) AND B ka distance pata hai to B(5,-1) ab bas OB nikalna hai √(25+1)=√26
@@aayushchhajed2630Woww 😮
Maine coordinate se karne ka toh socha hi nahi... Kitna easy tha!
Bhai aap kamaal ho 🔥
Bhaiya
Coordinate op
Boht easily hogya usse
@@aayushchhajed2630hm same approach se mainne v Kiya.......
@@aayushchhajed2630 BHAI YEH SAB GALAT HAI
AAPKO ITNI INFORMATION HAI HI NAHI KI AAP -2,6 PE CENTRE LE KAR EK CIRCLE BANA DO ROOT50 RADIUS KA
AAPNE -2,6 PE CIRCLE KYON BANAYA?
People in comment section 🗿
People in exam 🤡
Fr
Fr
fr
It's because time pressure in jee adv and multiple topics
Time pressure
I solve it very easily
0:45 i am feeling bad for this circle
My shortcut
I took geometry from my box drawn it and OB approx 5.1
Simply solved using Pythagoras theorem only! Took two variables x and y and applied pythagoras twice to get two equations of circles and one of the intersection point gave positive x and y and OB² = x²+y², this was how I assumed x and y, this was not even close to advanced it was jee mains level stuff, just complicated it by using cosine rule 😂
(x,y) came out to be (5,1) hence √26 took 2 mins
I think there is no need to make it complex
Method 1:
By extending AB let it meets the circle at point E and by extending CB it meets the circle at point F
Join point A and C
Draw OD perpendicular bisector of the chord AC , meet O with point A
AO=R, AC =2√10 and AD=√10
Let angle AOD =x ,
tanx=1/2 (in triangle AOD)
Draw CE then same chord AC will construct angle x (1/2 of the angle made at the centre) at point E
Now in triangle BEC
Angle BEC=x
And tanx=1/2=BC/BE=2/BE
BE=4
AB×BE=BC×BF
BF=6×4/2=12
Draw OP perpendicular to AE and OQ perpendicular to CF
We get OP=6-5=1 and OQ=12-14/2=5
OB²=OP²+OQ²=1²+5²=26
Method 2:
Let the coordinates of point B : (0,0)
and coordinates of the centre of the circle be: (-g, -f)
Where g is +ve and f is -ve
OB²=g²+f²
the equation of the family of circles passing through the points (2,0) and (0,6) would be
x(x-2)+y(y-6)+k(3x+y-6)=0
g=(3k-2)/2 , f=(k-6)/2 , c=-6k
R=√(g²+f²-c)
by putting value and then solve we get k=4 ✓
g=5 , f=-1
OB=√26
I but he is an IIT Professor inside
AC=2 sqrt(10) would assume that AB is perpendicular to BC
sir i have a very simple solution,, extend AB to meet the circle say at D let BD=x similarly extend CB to meet the circle at E
Using POWER OF POINT AB.BD=CB.BE we get BE=3x
Drop perpendicular from center to BE and AB at F and G
we get OF=3-x/2 and OG=3x/2-1
Using pythagoras we get x=4
THEREFORE OB^2=OF^2+OG^2
OB=sqrt(26) !!!!
OP brooo🔥
Wait, How did you get the value of OF and OG?
I am in class 10 not able to solve full but still
Acquired that figure and able to find the area of oab
Mene question ka answer sectors, segment ki help se nikala
I'm glad that I'm through with all this geometry calculus physics shit, and chemistry too. I'm happy with where I am and thinking I've did this back then feels soo amazing about me. Abhi toh 34+45 ke liye bhi calci lagrahi😅
I just subtracted BC value from the radius of the circle which is √50-2=5.071...
And the answer was √26=5.099.....
😂
Here's a solution using "Power of a Point".
Extend line AB to meet the circle again at D. By Pythagoras on ABC we have AC=sqrt40. Let angle ADC = x. Then, by sine rule, we have sin(x) = AC/2r = 1/sqrt5, which means tanx = 1/2 = BC/BD = 2/BD which gives BD = 4. Hence, |Power of B w.r.t. circle| = |BA*BD| = |r^2-OB^2| = 6*4 = 24 = |50-OB^2| and hence OB^2 = 26.
In general, the answer is r^2 - BC^2 *sqrt(4r^2-AC^2)/AC
Jao rp sir ki class dekho
Solved it very easy
I solved it with other method by taking cases which redirect you to only one case where ob comes √26
@@jaybawaskar124maine center ko origin manke sare points ke coordinate likhe kiya to uss easily ho gya, maine uska explanation bhi comment me likh diya h,ruko m copy paste krke likh hi deta hu yaha bhi😅
@@jaybawaskar124Mera method chhota aur easy h,
Maine O ko origin let kr liya aur A points ke coordinate likh diya theta variable ke term me √50cos theta,√50sin theta. Phir A coordinate, AB ki length and BC ki help se C point ke coordinate likh diya √50cos theta + 2,√50sin theta-6, ab C point to circle pe lie kr rha h, to C point ke coordinate ko circle ki equation (x²+y²=50) me satisfy krake theta ki 2 values aayi sin theta = 1/√2 or 1/√50. Ab B point ke coordinate usi tarah se likh diya √50cos theta,√50sin theta-6. Ab origin se iski distance hi question me pucha h, wo nikal liya theta ki dono value dalke, ussd OB √26 aur √74 aaya. Ab √74 ho nhi skta kyuki isse B point ki distance center se radius (√50) se bhi jyada ho rhi thi, so the ans is √26.
Wese ye likhne me bada h kyuki Maine pura explain kiya h, lekin krne pe bahut Chhota sa h😊
These questions are basic question in triangles
Did it after giving the hint of perpendicular thing
bhai AB perpendicular to BC ko pehle btana chahiye kyuki mai general lekr solve krne baith gya tha
Actually OB^2 is not 26 it's 26.8629...
The approach that you took is a complex and tedious approach. This problem can be solved using elementry geometry(Pythagoras theorem) in 2 steps. This is a moderate level PRMO question which we teach to 9th Grade students.
Note:- PRMO is the gateway for RMO and then INMO which 9th, 10th and 11th grade students take.
indeed
Wrong answer and wrong solution…. Angle OMB CANNOT BE 90 degrees
✅I solved it without using cosine rule 🤟🏻.
Just two right angled triangles🔺️ and pythagoras theorem.
I cleared JEE in 2014
I have forgotten all the formulae, so I simply looked at those formulae in google then solved it quickly.
Time taken: 8 minutes (Genuinely, see my approach below, jhoot bolne se mera koi fayda nahi hota :D)
Method: Pythagoras theorem and properties of circles (Class 9 level math)
Not a difficult question, just a bit imaginative.. Bhai aapki videos mujhe bohot acchi lagti hai, but ye starting ka statement galat laga.. 'hona toh nahi hai' bolne se aapne waise hi aadhe logo ka confidence gira diya, and unko genuine try dene se rok diya.. please apni phrasing soch kar kiya karo, many of us look up to you 🙏
My approach:
1) Mark center O, and draw the diameter parallel to the line AB. Let this diameter be PQ, where P is near A and Q is near B.
2) Extend line AB to touch the circle at D.
3) Extend line CB to meet diameter PQ at R.
4) Drop perpendicular from center O to AB at E.
5) Let BD = 2x, BR = y. Thus, OE is also y.
6) Since perpendicular from center bisects the chord, AE = DE = AD/2 = 3+x [This is why I took BD as 2x, not as x, since it would be halved]
7) Also, since ED = 3+x, BD = 2x, thus BE = 3-x, and OR = 3-x
8) In triangle AOE, by pythagoras theorem: OE² + AE² = OA², so, y² + (3+x)² = r² ...........(1)
9) In triangle OBR, by pythagoras theorem: OR² + BR² = OB², so, y² + (3-x)² = OB² ...........(2)
10) In triangle ORC, by pythagoras theorem: OR² + RC² = OC², so, (3-x)² + (y+2)² = r² ...........(3)
11) Now, we have 3 equations, and 3 variables (since r is given), rest is simple algebraic manipulations
By sandeep : aasan hai, ha aasan hai😂
From pure geometry,
Extend AB to meet the circle at X and BC to meet the circle at Y. BY/BX = AB/BC = 3 from similarity
Let BX = x and BY = 3x
Now draw perpendicular from O to BY, name M and O to AX name N.
MY = (BY + BC)/2 = (3x + 2)/2
ON = MB = BY - MY = (3x - 2)/2
AN = (AB + BX) / 2 = (6 + x)/2
ON² + AN² = OA² = r²
This will give the value of x as 4.
Then,
NB = AB - AN = 6 - (6 + 4)/2 = 1
ON = (3 × 4 - 2)/2 = 5
OB² = ON² + NB²
OB = √26
The solution seems long but it takes less time to process each step.
Bhai mera 11 min :13 sec me ho gaya , bhot hi badhiya question tha agr ban jata na to maje hi aajate 🎉
Solved by applying Pythagoras theorem multiple times
I did got the answer right but I had to use calculator for simplifying(my method was different)
It becomes simple when you take obas origin A(√50 cosx,√50sinx)
Thus C(√50cosx + 2,√50sinx -6) are also on the circle, get the ratios and answer what ever you want
Meanwhile I am confused that value of radius is greater than value of "AB"
Since, ✓50 > ✓36
than how the diagram shows
AB greater than radius of circle.
By this conclusion, all the scholars who has taken origin as B(0,0).
And considered cente "O" to lie in second quadrant will be false. As then O will lie in third quadrant.
Also those who has taken O as origin and considered B to lie in Fourth quadrant is false.
As in this it will lie in first quadrant.
Hope so, you can understand that diagram is wrong.
So prepare diagram yourself and then solve.
It was easy no kidding. and for OAC I instead used sine rule rest my process was the same as yours
The issue with many teachers is that before starting cordinate geometry they don't teach euclid's / that basic geometry that's why question like these seems impossible
prepare for ioqm
@@AyushGautam-lb2uk maths is interesting... But not that ,also olympiads are for early starters mainly
@@Niche_internet_micro_celebrity then teachers wont prepare you for top ranks they just prepare you to get selected in iits brother
@@AyushGautam-lb2uk hahahahha!
Koi na ye sab bakch*di advanced me nahi aati
Answer aa gaya, without hint, but alag bohot ghuma ke aaya.
It is totally up to you, how much you can manipulate PEOPLE. That's what this guy did. This problem could easily be done but he chosen the difficult way. Bro you are a JEE Aspirant, I mean you have done coordinate geometry in your syllabus💀 and a circle problem is given to you which even doesn't have any constraints like center coordinates, then why not just let it be on graph with center at origin? (because it could be easily done that way).
🥲 you are right, That solution didn’t click when I made it first. But indeed, considering that point as origin would make it a mains type question
Mjaa aagya
My method was to take b as origin and find center of circle knowing that the circle passes through (0,6) and (2,0)
We get a quadratic but we can neglect one value knowing that the center of circle lies in 2nd quadrant
Using equations for geometry makes the job a hell lot easier
And the ob is basically dist of center from origin
Pro!
same but i took center as origin (it become more hard by it than your method )
@@hirenkavad-xs9zsSame bruh i also took centre as orgin then i came to this man's approach
Damnnn bro !
Same Bro, Same method, Isi method se kia. Upar se quadratic bhi bahut easy wala aayega, nice roots.
Hello everyone i think i have an easier way to do this question.
We need to find the length OB
So now join OC
Then we have a right angle triangle i.e OBC
Now use the Pythagoras theorem.
OB sq + BCsq = OC SQ
OB sq +4 = 50
OB = root 46
Correct me if i am wrong
Hmm i missed that.Thanks for pointing that out
me(sees question): arre baap re ye kya hai
also me(looks at quesition for 30 seconds): haha ye to laddu sawaal hai
answer dekhne se pehle ye meri approach thi:
1. AB ke perpendicular ek radius drop kiya. let point of intersection be M and point on ciircumference be P.
2. OM=OP-BC => OM=√50 -2
3. find AM using pythagoras' theorem.
4. MB= 6-AM
5. find OB using pythagoras' theorem.
i solved in just 2 min . i draw a circle of radius 7 unit and all this and measure a line by scale answer is 5 unit 🎉😂
OB = √(54-8√(10))
Cosine rule se banya OB=√26, 3min
So What I did was Assuming the B as Origin (0,0) C(2,0) And A(0,6) . Now Taking the Center O as (X,Y).
Now Using the Distance Formulae I Got.
50= X² + (Y-2)²
Now We Know The Y Coordinate Must Should Between 0 and 6 .
By Hit and Trial you Got Y can be 1,2,5. And X be 5,6,1.
But You Can't take (X,Y) as (6,2).
So Either Taking 1 , 5 or 5 , 1. You Get √26.
Tadaa 🎉 You Got it By Simple Distance Formula😭
bro this is basic ioqm question in aakash reference book given to us in class 9
Mera method chhota aur easy h,
Maine O ko origin let kr liya aur A points ke coordinate likh diya theta variable ke term me √50cos theta,√50sin theta. Phir A coordinate, AB ki length and BC ki help se C point ke coordinate likh diya √50cos theta + 2,√50sin theta-6, ab C point to circle pe lie kr rha h, to C point ke coordinate ko circle ki equation (x²+y²=50) me satisfy krake theta ki 2 values aayi sin theta = 1/√2 or 1/√50. Ab B point ke coordinate usi tarah se likh diya √50cos theta,√50sin theta-6. Ab origin se iski distance hi question me pucha h, wo nikal liya theta ki dono value dalke, ussd OB √26 aur √74 aaya. Ab √74 ho nhi skta kyuki isse B point ki distance center se radius (√50) se bhi jyada ho rhi thi, so the ans is √26.
Wese ye likhne me bada h kyuki Maine pura explain kiya h, lekin krne pe bahut Chhota sa h😊
Oh nice method..tq for sharing
@@truptilodh6895your welcome😊
NICE !!
@@Sah-tc5pr thank you..☺️
This type of problem is actually good for brain excercise but instead of solving this problem i can go with CAD software were i can solve this problem in few seconds
If you can't apply basic geometry and construction in some questions it doesn't mean the question is hard compared to JEE. Please look, think and then speak. It took me 1 min to solve the question after looking at it for 1 min.
It is from AIME not JEE Advance.
This is a simple question no where jee advanced level
∆OAC=∆OAM HOW it's possible. Please reply
Commerce student here solved this without even a pen and paper
I saw this question in my ioqm book
Just join o to c
oa=oc
Calculate ocb angle using basic trigonometry in triangle oac and triangle abc. Obviously you get obc in some form of cos or sine
Now since oc, bc and angle ocb known calculate ob using triangle obc.
Exactly! This is what I did and it was much easier
PUTNAM MATH COMPETITION 🗿
HARVARD MIT MATH TOURNAMENT 🗿
UCBerkley Math Tournament 🗿🗿
JEE ADVANCED MATHS🐒🙉🙈🐵🦍
Bro I solved it using coordinate geometry
Taking O as (0,0) and taking A as (x,y) then B is (x,y-6) and C is (x+2,y-6)
And both and A and C lie on circle so I solved it using general equation
It only took me 3 minutes
Please reply wether I am correct or wrong
Let centre be origin and assume point a be (a,b) and then b will be(a,b-6) and c point will be(a+2, b-6) then the equation of circle will be x^2 + y^2 = 50 and point a and c lies on circle so substitute and solve you will get a value5 and b value 5
I did exactly this, coordinate geometry makes the job really easy
bhai thoda elaborate kar.....
a^2 + b^2 + 4a - 12b + 40 = 50 kiya toh jaake
12b - 4a = 40 or 3b - a = 10 aaya........
lekin isse sol kaise nikla ki (a,b) = (5,5)
Where is it given that b is right angle if it’s given never presume unless it gets proven from somewhere basic rule for advanced jee
Bhai exact 13 minutes mai solve ho gaya tha bas thodi inverse trigonometry lagakar and cosine rule use kar kar
Alternate solution: Extend AB to D on the circle. (better complete the circle for clear visualisation)
Assume DB to be some x.
drop a perpendicular from the center on the chord AD at Y.
then first of all find OY by Pythagorean theorem => (3+x/2)^2 + OY^2 = 50.
from here we will get OY in terms of x.
Also BY=3-x/2. now that we know OY and BY we will get OB=sqrt(50-4x) i.e. in terms of x
Now I used a bit of coord geo. WLOG let the center be 0,0. We can write the coordinates of the point C that will be [ -(OY+2), -(BY) ]. We have already calculated OY and BY above in terms of x
Now using the equation of circle we will get ---> (OY+2)^2 +(BY)^2 = 50 since the point C lies on the circle and from here on solving we will get x=4.
Now OB=sqrt(5-4*6)=sqrt(26)
problem I faced --> solving the (OY+2)^2 + (BY)^2 =50 will be a bit of challenge, not because its something very difficult but calculative. I had to use wolfram alpha to reach to x=4.
I will be happy to receive suggestions to shorten my solution
Thank you!!
Same way bro😂
Bhai yeh to Maine 10th mai IOQM ki taiyari ke liye solve kiya tha
4 MINUTE SOLUTION:
Let the centre of the circle be the origin O(0, 0}, and the coordinates of B be B(x, y).
Then the coordinates of A and C are A(x, y+6) and C(x+2, y). Both these points lie on the circle and therefore the coordinates must satisfy the equation of the circle. We thus get 2 equations in 2 unknowns:
x^2 +(y+6)^2 = 50
(x+2)^2 + y^2 = 50
Subtracting second from the first, we get
x = 3y+8
Putting this back in any of the equations gives a quadratic equation in y:
y^2 + 6y + 5 = 0
so y = -1 or -5
so x = 5 or -7
The first solution seems right for the given picture (the second is when C is on the left of O).
The coordinates of B are B(5, -1)
so OB = sqrt(26).
ABC right angle kese ho gya jb question me nhi diya hai? figure dikhne se kkuch nhi hota
Question direct circles ke property se ban jayega.
Extend AB to P and BC to Q to make chords AP and CQ. Now using the theorem for two perpendicular chords in a circle, AB/BC = PB/BQ.
Let PB= x then PQ = 3x.
Now for perpendicular chords, we have formula 4r² = a²+b²+c²+d², where a,b,c,d are legths of segments of chord. On putting in this formula, we get x=4. So length of AP= 6+4=10.
Now draw chord OM perpendicular to AP. Then AM=5. From Pythagoras OM=5. Now join OB to make right traingle OMB. Here MB = AB-AM = 6-5=1
And now apply pythagoras, OB = sgrt(26).
Moderate question if someone knows properties of circle well.
😢
Ditto same method I used
🥲itna padhai kei se kar lete ho yaar 😵💫
Same approach halwa sawal
bhai pq 3x kaise likha ?
Bhai literally Maine is question ko sirf similarity se solve kar Daya
How can you find cos(angleOAB) when ∆OAB is not a right angled triangle? Isn't it that trig ratios like cos and sin are only for right angled traingles? (I am a 10th grade student so i don't have much knowledge abt it 🤧)
You can split it into 2 triangles by drawing a perpendicular and then you can verify the formula
Ngl I solved it in one Fermi second 🤡
I am a class 12 jee aspirant. Bhaiya aapne jo solution bataya maine nhi dekha kyuki mujhe pata tha mujhe samajh nhi aayega par bhaiya ye question Sirf trigo aur cosine rule ka use karke kiya maine aur ans match hone ki khushi to kya hi batau. Gave 1 hr to the question
Bhai mae 10th ka new session mae hu litrally mere dost ne ye question bheja tha mane algebra lagake solve kardiya bhai im litrally supprised
Literally solved it with one hand while eating maggi
Edit: I am in post nut clarity and cringing on my comment rn
Bro 🫡🫡
Your profile says it all
True sigma
Bhai sb ek hi hand se solve krte hai 😂
@@user-xu6en5ed4l😂😂😂
Well 😢 I thaught about this idea to solve but 😂 u know this was first
I didn't knew chord rule , didn't knew cosine rule 😊
😂 So I stopped and watch video in 4x speed 😂
Now I am happy 😊
Thanks brother ❤
Bro what are you saying I have practically just finished 10th and some how I know cosine rule and cos(A±B) expansion it's the only thing required to solve this question once you told angle ABC WAS 90 Degrees
I was so close but i solved it on my mind my answer was √25 (class 9 wbbse)
Me to NEET wala hu
Pr mza aa rha h😅.
√31.25
My approach. Let the coordinate of B be (a,b) so y coordinate a would be (-√(50-a²)) and x coordinate of C would be √(50-b²). So b + √50-a² = 6 and √(50-b²) - a = 2. On solving we get a = 5 and b = 1. So distance = √26
I've taken the circle as x² + y² = 50 and line AB parallel to y axis
Yes same coardinate lagao jab geometry na aye ✌
@@niteshanandd same x=-7 reject karna tha bas
Yep, solved using this method in just 5 mins lol. Coordinate Geometry is goated fr
@@lexus_bkl nah bro real champs(olympiad kids) do it by geometry its actually a pity for we don't know much geometry it reflects how we didn't study in our junior years but none the less solving is more important iit>>being a star kid
We can also use perpendicular chord theorem here. Extend CB and AB to meet the circle at D and E respectively. Now drop perpendiculars OM and ON to chords AE and CD and let the lengths of the perpendiculars be x and y. AM=AB-y=6-y and NC=NB+BC=x+BC=x+2. A perpendicular from the centre to any chord bisects it, so BE=ME-BM=AM-BM=6-2y and BD=NB+ND=NB+NC=2x+2. Using perpendicular chord theorem we get that BD•BC=BA•BE. So 2(2x+2)=6(6-2y). We can now use the radius to get another relation in x and y. In triangle OMA, x^2 + (6-y)^2 = 50. After solving we get x=5 and y=1. OB^2 = x^2 + y^2 = 26.
Edit: This is intersecting chord theorem and works for all intersecting chords, not only perpendicular chords.
Thanks man!
Ya did same but failed in calculation 😢
Btw using coordinate here is actually easier
Can you please tell how did you solve after getting relation of triangle oma
@@Chunnumanchu Two equations, two variables, just substitute y in terms of x or x in terms of y in the last second degree equation.
@@manavbakshi5669 I got it thanks 😊
You can solve purely geometrically.
Choose origin at O = (0, 0)
Assume without loss of generality that AB is vertical (parallel to Y axis), which means that BC is horizontal (parallel to X axis)
A = (a, b)
B = (a, b-6)
C = (a+2, b-6)
Equation of circle is
x^2 + y^2 = 50
A is on the circle, which means that
a^2 + b^2 = 50
C is on the circle, which means that
(a+2)^2 + (b-6)^2 = 50
a^2 + 4a + 4 + b^2 - 12b + 36 = 50
(a^2 + b^2) + 4a - 12b + 40 = 50
(50) + 4a - 12b + 40 = 50
a - 3b + 10 = 0
a = 3b - 10
(3b - 10)^2 + b^2 = 50
9b^2 - 60b +100 + b^2 - 50 = 0
10b^2 - 60b + 50 = 0
b^2 - 6b + 5 = 0
(b-5)(b-1) = 0
b = 5 or b = 1
(a, b) = { (5, 5), (-7, 1) }
OB = sqrt((a-0)^2 + (b-6-0)^2)
OB^2 = 5^2 + (-1)^2
OB^2 = 26
OB = sqrt(26)
*or*
OB = sqrt((a-0)^2 + (b-6-0)^2)
OB^2 = (-7)^2 + (-5)^2
OB^2 = 74
OB = sqrt(74)
OB is required to be less than r = sqrt(50), since B is a point inside the circle. Hence this is an extraneous solution.
Therefore the answer is just *OB = sqrt(26)*
Bhai B right angle hey bola hi nahi apney wahi hogai confusion figure mey to dikhi nahi perpendicular sign or itni choti figure mey perpendicular kese manley 😅
Bhai isko resultant nikal kai bhi kiya ja sakta tha
With due respect I would like to say that you are exaggerating a lot by saying that it's beyond IIT-JEE level or whatsoever... Yaa it's a good one but please don't disrespect IIT-JEE by saying that the given question is it's limit... Honestly speaking the problem is quite simple for any IIT JEE aspirants who have studied Solution of Triangles (SOT) chapter properly. Just apply cosine rule in Traingle OBC and in Traingle OAB and solve the set of equations u will get a Biquadratic equation in terms of OB from where u get two solutions of it one is √74 and other is √26 and from common sense u can say that it can't be larger than radius itself that is √50, Hence the answer is √26...
Ye aapne jo bhi lambi si geometry lagayi hai pta nhi kya kya naye naye extensions uski koi need nahi sir...
Was just scrolling through the RUclips and ye video dikha jisme kuch bhi fizool sa likha hua thaa ki IITJEE aspirants se nhi hoga wagare wagera... No offence but Please don't give any caption to ur videos to get Views(Harder than Hardest of IIT JEE advance... Huhh kuch bhi feek dete hain log Aaj kal ... )
yup what u said is right, question is a lot easier using other solutions as well.
Maybe at first glance i felt like its harder than it is, My bad 😅
@@jeesimplified-subject it's ok Sir.
literally,I DONT LIE ,I SOLVE THIS QUESTION IN 1 MINUTE AND MY ANSWER IS 2.6 SMALL MISTAKE
Maine hint pane ke baad ek alag method se kiya .. lekin mera answer √50 - 2 aa raha hai .. jiska value 5.079 hai or aapke answer ka numerical value 5.099 , almost same !!
Lekin mera 3-4 line mai ho gaya 😂😂
Me applying P.G.T and making a 8th standard question 🗿
Hard problem ❌ 3rd class problem ☑️
nahi bana honestly...lekin phir cosine rule lagana tha yeh dekh kar ban gaya....
bhai pehli baar is channel pe koi ques hua hai
Co ordinate geometry se A ko 5,5 assume karke easily ho jayega
Aur pure geometry se karne k liye circles k properties pata hone chahiye
Isme do properties use honge AB.BD = BC.BE aur a²+b²+c²+d²=4r² where a,b,c,d are the lengths of line segments
Bohot dino baad ye property use kar Raha hu
Last 9th 10th me use kiya tha
ha bhai adv ke result ke din dikh jana gayab mat hona
😂😂 mai tenth me hi hu firvi smjh nahi aya mereko
hes balveer@@zak-zv8pf
@@zak-zv8pf don't know brother
Mere boards me ye tha naam bhul Gaya is dono theorem ka
Abi to jee mains hai samne
@@hailniggas not a problem bro
Produce AB to meet the circle at X and Produce CB to meet the circle at Y. Let BX = a BY = b. By chord theorem 6a=2b --> b= 3a. We know that 4(radius)^2 = AB^2 + BC^2 + BX^2 + BY^2 --> 4(50) = 6² + 2² + a² + b² --> 200= 36 +4 + a² + 9a² --> 160 = 10a² --> a=4 AX = 10. Construct OM perpendicular AX. Join OX. MX =AX/2 = 10/2 = 5. In triangle OMX by Pythagoras theorem OM = 5. BM = 1. In triangle OMB by Pythagoras theorem OB = sqrt(26)
Did the same thing
What do you mean by we know that 4r²=ab²+a²+b²+ob²
@@lesterfoundcookies😳
bro chill you are 4 years old bruh
hum log ka kaiya hoga
This is old AIME pyq(1983 year) and it's easy for mathematical Olympiad aspirants
Here o is centre,Firstly join b and c and then join a and o and finally join o and b
Now let angle (bac) = alfa
Let angle angle(oab)=t
Now drop perpendicular from O to line AC and let d be the foot of perpendicular then let angle (doa) = x then we get angle (oad)=90-x
Then we have 90-x= t+alfa , t = 90-(x+alfa)
Put cos on bot side to get
Cos(t) = sin(x+ alfa) now observe from the given lengths than sin(x) = 1by√5 and sin(alfa) = 1by√10 then we get cos(t) = 1by√2 hence then in the triangle oab using the cosine rule we get here ob²=26 and ob = √26
( Sorry but this question was not tough)
same bro having solved inmo and usamo geo this was like nothing
That's why I took pencil,eraser,compass,protector and a scale in jee papers
Believe me or not 20+ marks are free by constructing the given questions
The questions like
1:- find the distance of line from a point
2:- find centre of circle
3:- find the angle between two lines
Are easy to solve by constructing and take less than 4 min per question 😅😅😅😂😂
2022 ke jee mains me lekar Gaya tha scale compass no one objected
Par lagta hai NTA ne 2023 me comment padh liya tha sirf pen allow Kiya .
Par NDA me abhi bhi allow hai ye sab assessories
who allowed instruments there
Maine pehlei hi ek glti kii , O se AB kei upper perpendicular Dal diya 🤧🤧🤧🤧
Bhir 2 ghante dimmag daudata raha
This is an AIME problem
5 min me hogiya tan(A+B) ke formula se 😂
Pehle batana chahiye tha na ki angle abc 90 degree hai
My brother who is in 9th tried solving it using circles property, he left it midway itself 😂
As a 10the grade student it sure takes some time for me. I have just found the answer in 2 hours least. Fun fact is tomorrow is my English exam XD
I use only normal geometry theory
Just shift the chord back by 2 units to the right, and assume equation of chord is x = p, try solving for p, question will solve it self 😅
PS: at p = 7, all heavy sq rts become puppies 😂
Two words
Just two words
"Pythagoras Theorem"
If we use some common sense , compass and scale
Than it is not that tough
I have done it in 2 min with answer 5
it's been 3 years since i gave my jee advanced, and let me tell you, this question is still a piece of cake. Easiest method is to calculate the angle OCB. to do so, calculate length AC, then calculate angle ACO, also calculate angle ACB, simple trigonometry, now just subtract ACO from ACB, this gives OCB, now use cosine rule of triangle and get the answer, ~sqrt(26.2). No geometric intuition required.