Craziest IIT JEE Advanced Geometry problem

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[@jeesimplified-subject]

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[@aayushchhajed2630]

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

[@xyz2915]

​@@aayushchhajed2630Woww 😮 Maine coordinate se karne ka toh socha hi nahi... Kitna easy tha! Bhai aap kamaal ho 🔥

[@adityajha2889]

Bhaiya Coordinate op Boht easily hogya usse

[@shreyashsingh3520]

​@@aayushchhajed2630hm same approach se mainne v Kiya.......

[@Its_me_1729]

​@@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?

[@ishaanroy2436]

People in comment section 🗿 People in exam 🤡

[@drdsouza5285]

Fr

[@reddropgamingyt4965]

Fr

[@wdivyankop]

fr

[@mightycannon1512]

It's because time pressure in jee adv and multiple topics

[@shalini.9872]

Time pressure

[@rishabhjain728]

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

[@ayushmauryars]

I but he is an IIT Professor inside

[@aryajur]

AC=2 sqrt(10) would assume that AB is perpendicular to BC

[@Who_vibesTALKS]

0:45 i am feeling bad for this circle

[@Mr-.neutro9]

I solve it very easily

[@uranium879]

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) !!!!

[@youcuber3237]

OP brooo🔥

[@unnati_hulke]

Wait, How did you get the value of OF and OG?

[@shresthsuraiya3469]

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

[@mohitgoel4894]

Jao rp sir ki class dekho

[@JaadoowithOo]

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

[@RohanDutta-v9l]

Calculation tough nahi tha bas construction predict nahi ho rha tha🫤

[@s.koteshwar6984]

I just subtracted BC value from the radius of the circle which is √50-2=5.071... And the answer was √26=5.099.....

[@harshit3130]

😂

[@shashankshekharsingh2912]

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.

[@Niche_internet_micro_celebrity]

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

[@AyushGautam-lb2uk]

prepare for ioqm

[@Niche_internet_micro_celebrity]

@@AyushGautam-lb2uk maths is interesting... But not that ,also olympiads are for early starters mainly

[@AyushGautam-lb2uk]

@@Niche_internet_micro_celebrity then teachers wont prepare you for top ranks they just prepare you to get selected in iits brother

[@Niche_internet_micro_celebrity]

@@AyushGautam-lb2uk hahahahha!

[@rudrathakur6253]

Koi na ye sab bakch*di advanced me nahi aati

[@arunredddy]

Wrong answer and wrong solution…. Angle OMB CANNOT BE 90 degrees

[@anujgupta7880]

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

[@anujgupta7880]

Hmm i missed that.Thanks for pointing that out

[@sky47136]

This is question oF AIME

[@joohiyadav2847]

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

[@susantparida8369]

Pro!

[@hirenkavad-xs9zs]

same but i took center as origin (it become more hard by it than your method )

[@DineshSahu-dz9dr]

​@@hirenkavad-xs9zsSame bruh i also took centre as orgin then i came to this man's approach

[@randomreality9925]

Damnnn bro !

[@RachitAryanAsthana]

Same Bro, Same method, Isi method se kia. Upar se quadratic bhi bahut easy wala aayega, nice roots.

[@TCshivamarmy]

Actually OB^2 is not 26 it's 26.8629...

[@rishicricstar]

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😊

[@truptilodh6895]

Oh nice method..tq for sharing

[@rishicricstar]

@@truptilodh6895your welcome😊

[@Sah-tc5pr]

NICE !!

[@rishicricstar]

@@Sah-tc5pr thank you..☺️

[@BullsEye2.0]

I cleared JEE in 2014 I have forgotten all the formulae, so I simply looked at those formulae in google then solved it quickly.

[@rv_enemy4367]

My shortcut I took geometry from my box drawn it and OB approx 5.1

[@wbdhdbnfjdj]

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.

[@Gourav.Nishad]

By sandeep : aasan hai, ha aasan hai😂

[@less5715]

Commerce student here solved this without even a pen and paper

[@musicandpoetry8131]

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.

[@jeesimplified-subject]

indeed

[@Mr-.neutro9]

Mera halka sa fig alag bana tha

[@Anime_ki_duniya950]

These questions are basic question in triangles

[@badetisitarambabu8527]

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

[@phymo4135]

I did exactly this, coordinate geometry makes the job really easy

[@hanshalghag2394]

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)

[@L_Ratio_01]

Literally solved it with one hand while eating maggi Edit: I am in post nut clarity and cringing on my comment rn

[@Yash-537]

Bro 🫡🫡

[@drsantoshsingh9891]

Your profile says it all

[@TechnobladeNeverDies-ok]

True sigma

[@SiddhantVerma-n4k]

Bhai sb ek hi hand se solve krte hai 😂

[@vinayaksharmaclass7thcroll832]

​@@SiddhantVerma-n4k😂😂😂

[@FardeenTheDeveloper]

✅I solved it without using cosine rule 🤟🏻. Just two right angled triangles🔺️ and pythagoras theorem.

[@mr._base_]

If my brain explodes due to over info it's his mistake I'll file for a brain insurance

[@apoorvgautxm]

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

[@AshrafulIslam-gp4rm]

Did it after giving the hint of perpendicular thing

[@HarshRaj-yj5gb]

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.

[@ranjithapremanand8922]

Exactly! This is what I did and it was much easier

[@reddropgamingyt4965]

Ngl I solved it in one Fermi second 🤡

[@prinshutripathi8044]

This is an AIME problem

[@John293._-]

This is a simple question no where jee advanced level

[@pritpalsingh1312]

Solved by applying Pythagoras theorem multiple times

[@Your_Study_Buddy_SD]

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.

[@TonyStark-30001]

😢

[@NirbhayJEE2025]

Ditto same method I used

[@Crazy-Boy0999]

🥲itna padhai kei se kar lete ho yaar 😵‍💫

[@aaravarora2009]

Same approach halwa sawal

[@kashishbindra]

bhai pq 3x kaise likha ?

[@Sharpshootertanishk.]

Bhai mera 11 min :13 sec me ho gaya , bhot hi badhiya question tha agr ban jata na to maje hi aajate 🎉

[@eyw9528]

Answer aa gaya, without hint, but alag bohot ghuma ke aaya.

[@nikulchavda4674]

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😭

[@BadReputation-do8ob]

OB = √(54-8√(10))

[@kinogamerz3380]

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 🎉😂

[@itsinfinity163]

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

[@Dubey161]

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.

[@BhagwanLal-pl4kh]

Aree mori Maiya 😵‍💫

[@TonyStark-30001]

😂😂

[@coolsam2005]

malkhan from bhabhiji ghar par hai 🤣🤣

[@Huzaifa-zd8wi]

Me to NEET wala hu Pr mza aa rha h😅.

[@bhaveshsinghchaudhary2674]

I am in class 10 not able to solve full but still Acquired that figure and able to find the area of oab

[@shauryagupta3644]

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

[@bigbrainsingh9410]

Bhai literally Maine is question ko sirf similarity se solve kar Daya

[@AnkitKunal-fn4yk]

∆OAC=∆OAM HOW it's possible. Please reply

[@Praxprix69]

My brother who is in 9th tried solving it using circles property, he left it midway itself 😂

[@ChannelTerminatedbyYouTube]

PUTNAM MATH COMPETITION 🗿 HARVARD MIT MATH TOURNAMENT 🗿 UCBerkley Math Tournament 🗿🗿 JEE ADVANCED MATHS🐒🙉🙈🐵🦍

[@neelamyadav3936]

I did got the answer right but I had to use calculator for simplifying(my method was different)

[@aditya-1734]

Cosine rule se banya OB=√26, 3min

[@manavbakshi5669]

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.

[@soumitsenapati5612]

Thanks man!

[@ABDxLM]

Ya did same but failed in calculation 😢 Btw using coordinate here is actually easier

[@Chunnumanchu]

Can you please tell how did you solve after getting relation of triangle oma

[@manavbakshi5669]

@@Chunnumanchu Two equations, two variables, just substitute y in terms of x or x in terms of y in the last second degree equation.

[@Chunnumanchu]

@@manavbakshi5669 I got it thanks 😊

[@rednatureyt0786]

literally,I DONT LIE ,I SOLVE THIS QUESTION IN 1 MINUTE AND MY ANSWER IS 2.6 SMALL MISTAKE

[@Mayank-gr9oy]

Bhaiya simple geometry se go raha H done in 7 mins

[@AyushVerma-ui7re]

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.

[@minvsmax9244]

Hard problem ❌ 3rd class problem ☑️

[@anshlohani]

bhai AB perpendicular to BC ko pehle btana chahiye kyuki mai general lekr solve krne baith gya tha

[@weird8599]

bhai pehli baar is channel pe koi ques hua hai

[@niteshanandd]

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

[@niteshanandd]

I've taken the circle as x² + y² = 50 and line AB parallel to y axis

[@Ayush-mg6xw]

Yes same coardinate lagao jab geometry na aye ✌

[@preetbansal1749]

​@@niteshanandd same x=-7 reject karna tha bas

[@lexus_bkl]

Yep, solved using this method in just 5 mins lol. Coordinate Geometry is goated fr

[@Ayush-mg6xw]

@@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

[@CuRiOuS--MEHRAN]

Me applying P.G.T and making a 8th standard question 🗿

[@ViploveTyagi]

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).

[@mathematicalphysicist7576]

It is from AIME not JEE Advance.

[@nothing8497]

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).

[@jeesimplified-subject]

🥲 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

[@ashokkawdikar1976]

I required 40 min without cosine rule

[@jaybawaskar124]

Solved it very easy

[@jaybawaskar124]

I solved it with other method by taking cases which redirect you to only one case where ob comes √26

[@rishicricstar]

​@@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😅

[@rishicricstar]

​@@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😊

[@vahaha1175]

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

[@nathanosmisan2179]

The teacher needs to understand there are far better people than him and needs to get out of his superiority complex.

[@VRZ1105]

I took 35 min to solve this problem after hint

[@pankajchaniara9723]

Bhai yeh to Maine 10th mai IOQM ki taiyari ke liye solve kiya tha

[@utsavdudhatra7674]

"Jitna dimaag marna hai maro"___😂😂

[@ayubshaik2907]

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😅

[@omkardalvi575]

I messed up in the first step for taking AB hypotenuse

[@AaravKadam-k2x]

real bro, us...

[@Yuvi1313]

can be directly solved by intersection chord theorem..

[@solunke.pranav]

Trying before solution: got √26 ,method plotting a rough sketch and brut forcing,circle of radius √50 with centre at origin, trying different chords by taking lines parallel to y axis x=1,2,3,4,5 and calculated distance of P(point of intersection that chord and x axis )and point Q (point of intersection of circle and x axis) which nearly slightly greater than 2 ,then tried to find B on line x=5,by taking various lines y=-1..,got Point B in first try ,so now it was easy peasy just applied Pythagoras,ob²=op²+pb² ,so got 25²+1²=26 hence √26

[@SudhirSingh-ez1wf]

Mene question ka answer sectors, segment ki help se nikala

[@SSN4781]

nice i was able to solve this without any help. I love math

[@dgaming5439]

I saw this question in my ioqm book

[@ishmamrakinkhan6541]

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

[@paramawesome2043]

Literally solved in 2 mins

[@sainiksarkar6722]

It's really interesting but easy than the double disk problem of Mechanics on JEE Advanced 2016 paper.

[@Genus-Homo_Species-Sapiens]

Maths ka comparison Physics se Kya g@dha h

[@ClarkKent-bz9tf]

​@@Genus-Homo_Species-Sapiensbro solved 1 mechanics ques on the hardest year and thinks that every other ques is easy LMAO

[@ardwaj9902]

Real​@@ClarkKent-bz9tf

[@purplehawkyt]

​@@ClarkKent-bz9tfFun Fact: That question was not even that tough as people made it I solved that in my first try 💀

[@ClarkKent-bz9tf]

@@purplehawkyt fr bro it was just a random ass comparison of a math ques with physics lol, things people do to prove how smart they are XD

[@aashishgupta2207]

Mine was an easy approach with only cosine formula Join O to C Let angle(OBA) =x Compute it's cosine in triangle(OAB) Then angle(OBC) = 90+x Compute it's cosine (=-sinx) in triangle OBC Use sin²x + cos²x =1 Solve it you will get OB as √26......

[@reenagupta1636]

Nice and easy approach

[@anikasfunwithart2196]

👍👍👍👍👍

[@anikagupta8235]

Nice👍👍👍👍

[@aashish-anika6556]

Amazing😊😊😊

[@aashishgupta2207]

Thank you

[@Messh-s2y]

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!!

[@Shivam-sx4bx]

Same way bro😂

[@reekhilchawla3197]

bro this is basic ioqm question in aakash reference book given to us in class 9

[@xmartyprince]

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 ❤

[@mainakmanna7431]

Bhai coordinate geometry lagake aramse ho gaya

[@SiddharthSingh-pq1ry]

nahi bana honestly...lekin phir cosine rule lagana tha yeh dekh kar ban gaya....

[@mokshjain7403]

It was easy no kidding. and for OAC I instead used sine rule rest my process was the same as yours

[@sayanghosh9321]

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

[@hailniggas]

ha bhai adv ke result ke din dikh jana gayab mat hona

[@zak-zv8pf]

😂😂 mai tenth me hi hu firvi smjh nahi aya mereko

[@hailniggas]

hes balveer@@zak-zv8pf

[@sayanghosh9321]

@@zak-zv8pf don't know brother Mere boards me ye tha naam bhul Gaya is dono theorem ka Abi to jee mains hai samne

[@sayanghosh9321]

@@hailniggas not a problem bro

[@Bewussteinslage-nc9ik]

easy af question

[@editprov869]

I use only normal geometry theory

[@Gagan1237]

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)

[@ClarkKent-bz9tf]

same bro having solved inmo and usamo geo this was like nothing

[@uhm_yeah398]

Bro really thought i would sleep Lekin me ye soke hi krunga

[@abhinavbhutada2991]

*karke hi souga

[@uhm_yeah398]

​@@abhinavbhutada2991 nhi bhai, soke hi krunga

[@truth1549]

​@@abhinavbhutada2991 sapne m krega bol rh

[@kaustavbhowal2109]

3.5 mins me solved (asan h bro)

[@prathamjain5989]

5 min me hogiya tan(A+B) ke formula se 😂

[@bdash16]

Me after using cosine rule two times:

[@SURAJ_ULTRA_213]

I am a neet student and attempting thi question