Messages posted by hsbhatt

This is the locus of a circle when k is not 1. Specifically such circles are known as Appollonius circles.

You can read more about them at http://jwilson.coe.uga.edu/emt725/Apollonius/Cir.html

i.e.

Now

It immediately follows that

You have missed the roots where is a cube root of unity. So, the quadratic also has this property

all the given numbers are of the form 4k+3 and no square is of this form

It looks good. Yet I would suggest that

(1) you enable latex which makes math and physics posts very readable. You can include a link to a latex tutor and/or provide a keypad with most commonly used commands like \frac \implies etc.

(2) Have teachers who post regularly. Encourage them to post good concept based questions themselves. The presence of some good teachers will give confidence to your students

(3) Have a moderator look around for answers provided by a person that has not been acknowledged. One of the reasons for my not coming to the site often is that I post and then I dont know if any one let alone the OP has seen it. its like talking to an audience of zero - not tempting at all.

Rewrite the problem by substituting 1 by (a+b+c) as having to prove that

Setting and dividing by each factor on both sides by 2, we can further rewrite as

given x+y+z=1, to prove that which is well known and easy as

Multiplying the above three inequalities concludes the solution

quite obviously (0,1).

If b and c are equal, you can make a as small as you please (as we only need a>b-c)

If a is very large then . But again from triangle inequality a<b+c. Hence the range is (0,1)

Congratulations Vichar!

Well as you have very little time left, I would suggest visiting mathlinks.ro

This is an international forum is most used by students preparing for math olympiads across the world

There is a large bank of questions from contests across the world over several years which you will find very useful

@rishabh.

The angles A,B,C of the triangle are

Hence

So the questions wants us to prove that

But this is true as

Let the number of terms be n. Then the sum of terms of the AP is

Suppose the common ratio of the terms is r, then

Now the sum of terms of the GP can be written as

The typical bracketed term is

We will now prove that this is less than or equal to a+b =

WLOG we may assume that r>1 (otherwise we read the sequence backwards)

So we are to prove that

Hence, each bracketed term is less than or equal to a+b.

Hence the entire sum is less than or equal to thus proving the statement

Since when did we abandon the convention that the remainder is to be less than the divisor?

The correct answer is 9

Hence the given expression equals . Hence the remainder is 1

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