Craziest IIT JEE Advanced Geometry problem
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- Опубліковано 22 бер 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 🤡
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It's because time pressure in jee adv and multiple topics
Time pressure
I solve it very easily
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)*
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Can be solved only using Pythagoras theorem. Assuming AB is perpendicular to BC which was not given in the question but i assumed it and solved it like that. Later I see all solutions are assuming it.
Put a perpendicular on AB at X. You get 3 equations:
AX^2+OX^2=50
AX+BX=6
for hypoteneuse OC we have:
BX^2+(OX+2)^2=50
very easy to solve for AX=5, BX=1,OX=5. And now we have:
OX^2+BX^2=OB^2
Here the simplest solution
AC is radius ,, so AC -BC= root50-2=5.07
let the cordinates of C be (x, y). Then cordinates of A will be (x-2, y+6). imagine the circle to be centered at origin solve both equations and u will get x, y. its straightforward but messy from there
The real correct answer will be {86-24(50^.5 - 1)^0.5}^0.5 that is approxily root 26
0:45 i am feeling bad for this circle
Very easy question
I am in class 9
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
My shortcut
I took geometry from my box drawn it and OB approx 5.1
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😭
Commerce student here solved this without even a pen and paper
I saw this question in my ioqm book
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
Done without hint
By sandeep : aasan hai, ha aasan hai😂
Me to NEET wala hu
Pr mza aa rha h😅.
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 🎉😂
Ngl I solved it in one Fermi second 🤡
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
Wrong answer and wrong solution…. Angle OMB CANNOT BE 90 degrees
I cleared JEE in 2014
I have forgotten all the formulae, so I simply looked at those formulae in google then solved it quickly.
This is a simple question no where jee advanced level
I am in class 10 not able to solve full but still
Acquired that figure and able to find the area of oab
Isko solve karne me 5 min bhi nahi lage mujhe😂😂😂
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😅
Did it after giving the hint of perpendicular thing
It can be done only using Pythagoras theoram
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
Bhai mera 11 min :13 sec me ho gaya , bhot hi badhiya question tha agr ban jata na to maje hi aajate 🎉
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
Thanks
Answer aa gaya, without hint, but alag bohot ghuma ke aaya.
Thanks bhaiya for questions
Mera halka sa fig alag bana tha
Solved in first attempt 🗿,but time laga 😅
I did got the answer right but I had to use calculator for simplifying(my method was different)
OB = √(54-8√(10))
Solved by applying Pythagoras theorem multiple times
Actually OB^2 is not 26 it's 26.8629...
These questions are basic question in triangles
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.
My approach was almost same just didnt think abt cosine rule........it really is a nice video and such videos never fail to maintain my courage towards jee advance.......thnks to u i feel adv. Is still doable
Hard problem ❌ 3rd class problem ☑️
Calculation tough nahi tha bas construction predict nahi ho rha tha🫤
Bhaiya simple geometry se go raha H done in 7 mins
✅I solved it without using cosine rule 🤟🏻.
Just two right angled triangles🔺️ and pythagoras theorem.
This is an AIME problem
3.5 mins me solved (asan h bro)
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?
My brother who is in 9th tried solving it using circles property, he left it midway itself 😂
Mene question ka answer sectors, segment ki help se nikala
Me applying P.G.T and making a 8th standard question 🗿
This is question oF AIME
bhai AB perpendicular to BC ko pehle btana chahiye kyuki mai general lekr solve krne baith gya tha
5 min me hogiya tan(A+B) ke formula se 😂
Finally you uploaded 😂🎉
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
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.
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
Bahit accha question tha aisa hee aur laiye please
It was easy no kidding. and for OAC I instead used sine rule rest my process was the same as yours
Bhai literally Maine is question ko sirf similarity se solve kar Daya
Cosine rule se banya OB=√26, 3min
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 😊
literally,I DONT LIE ,I SOLVE THIS QUESTION IN 1 MINUTE AND MY ANSWER IS 2.6 SMALL MISTAKE
I required 40 min without cosine rule
I use only normal geometry theory
Bhai yeh to Maine 10th mai IOQM ki taiyari ke liye solve kiya tha
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)
Literally solved in 2 mins
"Jitna dimaag marna hai maro"___😂😂
easy af question
I took 35 min to solve this problem after hint
bro this is basic ioqm question in aakash reference book given to us in class 9
Anyone prepare putnam competition 🎉🎉🎉
Two words
Just two words
"Pythagoras Theorem"
I was so close but i solved it on my mind my answer was √25 (class 9 wbbse)
Me after using cosine rule two times:
It is from AIME not JEE Advance.
10:41 😂😂epic
Got the qs after hint... creativity at its peak
If my brain explodes due to over info it's his mistake I'll file for a brain insurance
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
bhai pehli baar is channel pe koi ques hua hai
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😂😂😂
Hacker hai bhai hacker
I am a neet student and attempting thi question
I am able to solve this problem by my self
Bhai coordinate geometry lagake aramse ho gaya
Literally solved it with one hand while talking on my phone with my girlfriend.... 😂
nahi bana honestly...lekin phir cosine rule lagana tha yeh dekh kar ban gaya....
Advanced se bhi advanced 😮
It can be easily solved using basics of circle like parametric points on circle as a function of any angle
This question is simple just use intersection chord theorm by extending AB AND CB to meet ag circle
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
Mene to Pythagoras se 2 min me solve krdiya