Only 1% could solve this insanely difficult problem

If this question is easy for you, I see great things for your future. Thanks to Rahul and Battu for the suggestion! This problem appeared on the 2022 JEE Advanced Paper 1 Mathematics section as Question 8. The JEE Advanced is an extremely difficult exam for admission into India’s prestigious IIT schools. I think this is one of the easier questions of the paper. I give credit to the Unacademy Atoms video about the paper which helped me understand how to solve the problem.

2022 JEE Advanced paper 1 answer sheet (see Q.8)

Unacademy Atoms video solutions to paper 1 (Q8 is around 59:00)

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[anil]

According to JEE Advanced 2022 report, only 0.95% of the candidates have answered this question correctly.

[pas]

Damn, who would have guessed 4 years ago that a question from the exam which decided my college would appear in this channel? I scored barely 35% in this Maths paper, and 50% overall, and am now in a good IIT!

[sweepingtime]

Setting up the diagram is the key to getting the solvable equations. It was tedious enough to be difficult, and oh boy this was the easiest question on that paper...

[MarieAnne.]

Using coordinate geometry:
A = (0, 0), B = (1, 0), C = (0, 3), M (midpoint of BC) = (1/2, 3/2), BC = √(AB²+AC²) = √(1²+3²) = √10
Larger circle has center (1/2, 3/2) and radius R = BC/2 = √10/2
Smaller circle has center (r, r) and radius r (since it is tangent to both the x-axis and y-axis).
For circles that are internally tangent, distance between centers = R - r
√((1/2 - r)² + (3/2 - r)²) = √10/2 - r
(1/2 - r)² + (3/2 - r)² = (√10/2 - r)²
1/4 - r + r² + 9/4 - 3r + r² = 10/4 - r√10 + r²
r² = 4r - r√10
Since r ≠ 0, divide both sides by r
r = 4 - √10

The calculations are pretty much the same as in the video, but I find the setup a little simpler.

[bdot]

I can confidently say there was no way I could have solved this.

[itsOnlyPIYUSH]

I also gave this year 2022 Jee Advanced and I am happy to tell you @MindYourDecisions that this questions has taken my 3 mins of Precious time and the result was Amazing . LoL 😁

[mihaiio]

I am impressed that you used the module, |r - 1.5|. Very nice.

[cblpu]

I solved this using coordinate geometry. It is less elegant however very little brainpower is needed which can be helpful in nervous exam situations.

Let vertices of the right angled triangle be (3, 0), (0, 0), (1, 0).
The circumcenter simply by its definition is (1/2, 3/2). The circumcircle also passes through (0, 0). So we get that the equation of the circle is x^2+y^2-x-3y=0
Now let the circle whose radius we want be x^2+y^2+2gx+2fy+c=0

Now it touches the coordinate axes at two points say (x_1, 0), (0, y_1). Putting these points we get two quadratic equations in x_1 and y_1 whose discriminant is 0. Therefore we get that g^2=c=f^2 => g=f
Now the circles touch internally. This means the distance between their centres c_1 and c_2 is equal to the difference of the radius of the larger circle and the smaller circle.i.e, C_1C_2= r_1-r where r_1 is the radius of the circumcircle. [(1/2-g)^2+(3/2-f)^2]^(1/2) = √10/2 - √(-c+g^2+f^2).
But, g^2-c=0 and g^2 = f^2
This allows us to completely simplify the expression into the quadratic equation
f^2+f(√10-4)=0
From here we obtain that f = 4-√10. since the radius of the circle is √(-c+f^2+g^2) and -c+g^2 =0, we have the radius = 4-√10

[ayushrajput]

We can use coordinate geometry in this question two sides of triangle as y and x axis and the center of smaller circle will lie on y=x
Result will be 4- √10

[Aditisoukar]

I gave jee advanced this year and skipped all maths questions except 4 which i attempted just to qualify 😂

[rcht]

Ah yes, me watching JEE videos after JEE

[lapaget]

There is also a formula for this problem. It's r = a + b - c. Applying with a=1, b=3, c=√10, then r= 1+3-√10 = 4-√10

[gen]

I haven't normally encountered such type of qs in pyqs, this does however remind me a lot of ioqm pattern papers. Certainly using those techniques helped in solving the question.

[anuragsingha]

I scored only 15 marks in math.Physics and chemistry saved my rank

[moisesbarrera]

Beautiful solution sir. Thank you. Greetings from México City.

[tmpt]

i remember solving this question in JEE ADVANDED this year
took a little time... but eventually did it

[jasnoorsingh]

Tbh using coordinate geometry is made it pretty easy than thinking about all this stuff

[ThanksThanks-enno]

You could have did it like assuming a as origin and writing coordinates of both centres then c1c2=r1-r3

[piyushkumarjha]

Remeber the stuff, this has a another approach by mixtilinear incircles, a general problem was in rajeev manocha and similar came in AIME

[mr_angry_kiddo]

Hi Sir iam Battu, thank for providing the solution ☺️