2017 IMO Problem #4

[kennethshi]

Took me 2 hours to solve this problem, but it was worth the time. I think they reduced the difficulty of IMO this year. Anyways, I'll share my solution here. It is easy to show by angle chasing that triangle RSK is similar to triangle TAR. and since angle ATS=angle KRT, we conclude that triangle ATS is similar to triangle TRK--->angle RTK=angle STK=angle SAT. Q.E.D.

[vanquyennguyen]

Bài toán tương đương:
Cho tam giác RKS có RK=a;KS=b;SR=c;J là điểm sao cho SJ=d và tứ giác RKSJ nội tiếp;T là điểm sao cho S là trung điểm RT;A là điểm sao cho KT^2=KJ.KA; hãy chứng minh RA là tiếp tuyến của đường tròn (RKSJ)(bài toán ngược với bài toán gốc).
Lời giải:
Biết cosKJS=cosKRS=>JK;Biết RT=2SR, KR và cos KRS=>KT;mà KT^2=JK.KA=>KA
=>RA^2=JA.KA(đpcm).

[solosolo]

great video series about IMO problems, how did you reach that level of math that you can solve all this problems by yourself?

[EssentialsOfMath]

I follow the proof up until the last part. How does KTS = SAT imply tangency?

[meetukumar]

Sir plzz tell if this method is correct
We know AR is tangent to the small circle and join AT then from centre of smaller circle we would construct a line on AT parallel to AR then it would be a rectangle and join T to centre of larger circle and extend it so it would be parallel to KA by proving angle TKA to be 90 and then ad they are parallel KT is tangent

[ShahriarNafiz]

angle PTR=TRA but PTR=TRA=SKR how?? someone please explain!!

[FUkOffMAFFiA]

You don't have to prove that RK||AT. Instead, draw SP such that AS=SP and AP is diagonal then you get that the diagonals bisects each other at S ==> ARPT is a paralellogram.

[akashsudhanshu]

Please send me the theorem according to which you said angle sat and angle STK make it a tangent