Recent content by Akash47

I've got the problem well. I want to close the thread. How can I do that?

But how is it applied? It would be applicable if ##2n+1## numbers were to be distributed over ##n## buckets. But that would just give that 'at least one bucket contains more than 2 numbers'.But what about the sum of them? Pigeonhole principle deals with the numbers not with the sum of them. Then...

I've got the problem well. The sum of the given numbers is 2n(2n+1)/2 = n(2n+1). And there are n buckets, so the average sum of numbers of each bucket is 2n+1.I've come to this so far. But I'm not sure whether the numbers are to be distributed over all 'n' buckets or some of the buckets can...

Thanks

I know that both 3^x and e^x can't be 0 for real x. Then x^2-4=0 is the only choice and we get x=2,-2. Am I right? Or should I add something?

Your example is not reasonable.I have just guessed that 'AD ' is radius.Then clearly ∠ AUV=∠AVU=45°.But how can you say that ∠DAC=45°?This is totally unreasonable. Have I ever said that ∠ BCA=45°?Please have a good look on the problem.

No,it's not the statement. Please now look in the problem. I've said that I've just guessed that The radius should be AD,but I'm not sure.

I have tried a lot by angle chasing e.g. let ∠ABC=x° then ∠ACB=90°-x°. As AU=AV=radius of circle so ∠AUV=∠AVU=45°. I've connected U,D and V,D. Then ∠UDV=135° etc. But I haven't found any way to get near of proving AE=DE. I have also tried to prove 'the area of triangle AEU= area of triangle...

ehild as well as LCKurtz have helped me much:bow:.Thanks to all of you that you have tried to help me a lot 🥰.

Thanks to all.I have solved the problem.

No,I haven't abondoned the thread. BvU has given me some clues like these: '1+2+4 = ... 2x1 + 3 = 4x1 + 2x2 + 2x4 + 3 = ... manipulate these and see if you can end up with 3 by itself ... ' And: 'If you now can show that, 'area of '4' = area of '2' you're almost done !' But still I can't solve...

I know that f(1) < f(2) < f(3) , then f(f(1)) <f(f(2)) <f(f(3)) and that implies, 3<6<11! Who doesn't know this?I had tried this method and that didn't help me that's why I have posted the problem? Can you please help me a little bit more?

Through symmetry of parallelogram,I have come to this: Here 1,2,3,4 denotes the area of the particular regions.Then I am stuck.Please help what to do next or whether there is any other way.

It is obvious that the function f is not injective. From the given equation, we get f(f(2))=6.And since,there is an inequality given in the problem, I think we can use that to find f(6).But I have got stuck here and can't move.Do I have to find what is f(x) first?Then how?

But isn't my opinion right what I've said in my first post?