1) Let ABC be a triangle . Let P and Q denote respectively the reflection of B and C in the internal angle bisector of angle A.Show that the triangles ABC and APQ have the same incentre .

2) Let P(x) = x^2 + a * x + b be a quadratic polynomial with real coefficients. Suppose there are real numbers s not equal to t such that P(s) = t and P(t) = s .Prove that( b-st )is the root of the equation x^2 +a * x + b - st = 0.

3) Find all integers a,b,c such that a^2 = bc +1 and b^2= ac+1

4)Suppose 32 objects are placed along a circle at equal distances.In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite .?

5)Two circles X and Y in the plane intersect at two distinct points A and B, and the centre of Y lies on X .Let points C and D be on X and Y ,respectively ,such that C,B and D are collinear .Let point E on Y be such that DE is parallel to AC. Show that AE=AB.

6)Find all real numbers a such that 4<a<5 and a(a -3{a}) is an integer .(Here {a} denotes the fractional part of a.For example {1.5} = 0.5 ; {-3.4} = 0.6)

## Comments

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TopNewest4th one is 32C3 - 16.26 - 32.30 – Easha Manideep D · 9 months, 3 weeks ago

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I got very complicated answer. That is a = 4 + 1/n where n is the +ve root of mn^2 -4n-2 = 0. Ya actually it works. – Easha Manideep D · 9 months, 3 weeks ago

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– Raven Herd · 9 months, 3 weeks ago

You are talking about which question ?Log in to reply

– Easha Manideep D · 9 months, 3 weeks ago

About 6th oneLog in to reply

How many marks would I get in the 3rd question if I forgot to write 2 out of 8 solutions – Samarth Agarwal · 9 months, 3 weeks ago

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– Raven Herd · 9 months, 3 weeks ago

How are the marks divided in every question ?Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

I think on steps and approachLog in to reply

– Raven Herd · 9 months, 3 weeks ago

As for the third one I wrote the method but didn't get the time to write the final solution .What was your approach? I formed the equation a^2+b^2 = 1-ab and used the concept of modulus 4 .Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

After working for long I got : a+b+c=0 a^2+b^2+c^2=2 ab+bc+ca=-1.... but there is also a condition that c=0 and a=b=1,-1.Log in to reply

– Raven Herd · 9 months, 3 weeks ago

Are you sure about this one if yes then post the complete solution.Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

I am sure that I did it wrong.....All my friends got 8 solutions an dI got only 6 :(Log in to reply

– Raven Herd · 9 months, 3 weeks ago

It does have 8 solutions my friends also said the same.Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

What you think they would penalize on 6 solutions?Log in to reply

– Raven Herd · 9 months, 3 weeks ago

I don't think so and I did not say that infact it is better than mine . I am in no position to demoralize you.Log in to reply

How to do the first one – Samarth Agarwal · 9 months, 3 weeks ago

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– Raven Herd · 9 months, 3 weeks ago

I wasted nearly an hour thinking on that one it looked easy but wasn't .Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

I wrote a foolish solution which said that triangles ABC and AB'C' are congurent so incentre will coincide.... :PLog in to reply

– Raven Herd · 9 months, 3 weeks ago

I thought that I will prove that angle B'IC'= 90+ B'AC' because angle BIC = 90 + BAC but solution never came.Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

I was terrified on seeing the figure :PLog in to reply

– Raven Herd · 9 months, 3 weeks ago

Did you do the fifth one?Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

yesLog in to reply

– Raven Herd · 9 months, 3 weeks ago

easiest !Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

Yeah ... it consumed only ten minutesLog in to reply

– Raven Herd · 9 months, 3 weeks ago

I took 3Log in to reply

– Raven Herd · 9 months, 3 weeks ago

but that would only happen when B' coincides with B and C' coincides with C.Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

I know that I am wrong but only hope to get some step marks....Log in to reply

– Raven Herd · 9 months, 3 weeks ago

How many questions did you do in total ? (I mean attempt )Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

I attemted all out of which I am sure of 3.Log in to reply

– Raven Herd · 9 months, 3 weeks ago

I attempted 5 out of which I am hoping for 3.Log in to reply

What is the answer to 4th and 6th.... – Samarth Agarwal · 9 months, 3 weeks ago

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– Raven Herd · 9 months, 3 weeks ago

I think that the answer to the last one is 3+3^1/2 and 3+2^1/2 .How did the paper go ?Log in to reply

3+(1.5)^0.5, 3+(2)^0.5, 3+(2.5)^0.5, 3+(3)^0.5 and 3+(3.5)^0.5 – Samarth Agarwal · 9 months, 3 weeks ago

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– Shubhendra Singh · 9 months, 3 weeks ago

I too got 5 answersLog in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

I have checked question 6 using wolfram alpha there are 5 solutions only :)Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

Someone was saying 6 answers ....:(Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

he said that 6-sqrt(3) is also a solutionLog in to reply

– Raven Herd · 9 months, 3 weeks ago

Then yours maybe correct .Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

Dont know ...thats why waiting for answer keysLog in to reply

– Raven Herd · 9 months, 3 weeks ago

When will it come and how did you approach the problem ?Log in to reply

My approach was that I formed a quadratic in fractional part and found its value by Shir Dharacharya and subjected it to condition 0<f<1 and the result was that there are 5 such integers possible for each I found the value of f and got these as answer... – Samarth Agarwal · 9 months, 3 weeks ago

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– Raven Herd · 9 months, 3 weeks ago

Post the complete solution .Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

Please let me confirm the answer..then I will post a complete solutionLog in to reply

– Raven Herd · 9 months, 3 weeks ago

What about the third one?Log in to reply

– Samarth Agarwal · 9 months, 3 weeks ago

errr....I made a silly mistake in that....I got six solutions but there are 8 solutionsLog in to reply