Problem 21
In the figure, two unit(radius 1) circle are externally tangent. A square is touching both the circle and their common tangent. If the square side length is \frac{a}{b} then what is a+b ?
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Bdmo প্রস্তুতি
আমি noob তাই আমি পারি না
Someone, please post a new problem…
\triangle ABC is a right triangle with \angle ABC=90°. P is a point inside the triangle such that PA=10, PB=6 and \angle APB = \angle BPC = \angle CPA. Find the measure of PC.
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∠APB=∠BPC=∠CPA = 120
\triangle APB এ কোসাইন সূত্র অনুযায়ী
AB^2 = AP^2 + BP^2 - 2BP \times PA cos(APB) = 10^2 + 6^2 - 2.10.6.cos(120)= 196
AB^2 = 196....(i)
আবার, \triangle BPC এ
BC^2 = BP^2 + PC^2 - 2.BP.PC.cos(BPC)= PC^2+6^2-2.6.PC.cos(120)
BC^2=PC^2 +6PC + 36 .....(ii)
আবার \triangle APC এ
CA^2 = AP^2 + PC^2-2.PA.PC.cos(APC) = 10^2 + PC^2 - 2.10.PC.cos(120)
CA^2 = PC^2 + 10PC+100.........(iii)
আবার , \triangle ABC এ
CA^2 = AB^2 + BC^2
PC^2 + 10PC+100=196 +PC^2 +6PC + 36
4PC = 132
PC = 33
(no problem to submit)
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Next problem, plz! 
(Problem 23)
কোনো ত্রিভুজের অন্তঃকেন্দ্র ও পরিকেন্দ্র যথাক্রমে I, O. এর অন্তঃবৃত্তের ব্যাসার্ধ ও পরিবৃত্তের ব্যাসার্ধ যথাক্রমে r, R. প্রমান করো যে,
OI^2 = R^2-2rR
I think we are dead 
The calibration round #15 is tomorrow. It’s gonna be alive again no worries.
Let O be the circumcenter of \triangle ABC and I be the incenter of it. Ray AI intersects the circumcircle at L. Ray LO intersects the circumcircle at M. ID is perpendicular to AB. ID=r, r be the radius of incircle. IO=d, d be the distance between O and I. OQ = OL = OM = OP = OQ = R, R be the radius of circumcircle.
Let ∠A=α,∠B=β.
In \triangle ADI and \triangle MBL, ∠ADI=∠MBL=90∘ and ∠DAI=∠BML (they lie on same cord). Hence both the triangles are similar.
So, \frac{ID}{BL} = \frac{AI}{ML} \Rightarrow ID×ML=AI×BL.
Therefore, AI×BL=2Rr ……………….1
Because, ID= r, and ML= diameter of circumcircle =2R.
From the figure, Again, ∠BIL=\frac{∠A}{2}+\frac{∠B}{2}=\fracα2+\fracβ2. And ∠IBL=\frac{∠A}{2}+\frac{∠B}{2}=\fracα2+\fracβ2. Hence, ∠IBL=∠BIL. So, IL=BL
From equation 1 we have, AI×BL=2Rr
As OI intersects circumcircle on both the sides at P and Q then,
PI×QI=AI×BL=2Rr ………….2
The distance PI=R+d, The distance QI=R-d
Putting this in equation 2 we get, (R+d) (R-d)= 2Rr.
Solving the above equation we get,
R^2−d^2=2Rr. Putting radius terms in one side, d^2=R^2−2Rr hence the prove is complete.
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Problem: 24
In isosceles trapezoid ABCD , parallel bases \overline{AB} and \overline{CD} have lengths 500 and 650 respectively, and AD=BC=333. The angle bisectors of \angle A and \angle D meet at P, and the angle bisectors of \angle{B} and \angle{C} meet at Q. Find PQ.
ABCD একটি সমদ্বিবাহু ট্রাপিজিয়াম যেখানে AB=500 এবং CD=650 দুটি সমান্তরাল বাহু, এবং AD=BC=333 । \angle A এবং \angle D এর সমদ্বিখণ্ডকদ্বয় পরস্পর P এবং \angle B এবং \angle C এর সমদ্বিখণ্ডকদ্বয় পরস্পর Q বিন্দুতে ছেদ করে। PQ এর মান বের কর।
Source
AIME 2022
Here EF and AB parallel for what we can say that \angle BAP = \angle APE. On the other hand, \angle EAP = \angle APE. So AE=PE. In the same way, DE=PE. So AE=DE=PE=\frac{333}{2}. In the same way, FQ=\frac{333}{2}.
Now, it’s a isosceles trapezoid and AE=DE=BF=CF. So EF=\frac{500+650}{2}=575.
So PQ=575-(\frac{333}{2}+\frac{333}{2})=242.
Problem 25
O is the center of AB semicircle. Let O′ be the circle’s center that is tangent to AB at D and the semicircle at C. Find \angle ACD.
Problem 25
\angle{ACD}={23⁰}
I think it is the ans
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Problem 26
In triangle ABC the medians \overline{AD} and \overline{CE} have lengths 18 and 27, respectively, and AB=24. Extend \overline{CE} to intersect the circumcircle of ABC at F. The area of triangle AFB is m\sqrt{n}, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n
Source
AIME 2002
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Nope, you are wrong.
Try again…
ok I will again try
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