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)

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## Comments

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TopNewest4th one is 32C3 - 16.26 - 32.30

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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.

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You are talking about which question ?

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About 6th one

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How many marks would I get in the 3rd question if I forgot to write 2 out of 8 solutions

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How are the marks divided in every question ?

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I think on steps and approach

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How to do the first one

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

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

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but that would only happen when B' coincides with B and C' coincides with C.

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

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What is the answer to 4th and 6th....

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

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I was getting 3 more answers

3+(1.5)^0.5, 3+(2)^0.5, 3+(2.5)^0.5, 3+(3)^0.5 and 3+(3.5)^0.5

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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...

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