NMTC 2015 Ramanujan Contest Final Round

  1. ABC is a triangle in which RR represents the circumradius and rr represents the inradius and ddrepresents the distance between circumcentre and incentre . Then find dd in terms of R,rR,r

  2. ABC is a triangle . Point P is on side BC . r1,r2,rr_{1},r_{2},r are inradii of Triangle ABP,ACP,ABC respectively. hah_{a} is the altitude of ABC from A . Prove that 1r3+1r2rr2r3=2ha\frac{1}{r_{3}}+\frac{1}{r_{2}}-\frac{r}{r_{2}r_{3}}=\frac{2}{h_{a}}

  3. If a,b,ca,b,care real numbers such that a2=bc,a+b+c=abca^{2}=bc , a+b+c=abc, then prove that a23a^{2}\geq3.

  4. If (x,y,z)(x,y,z) are prime numbers then find all solutions of x(x+y)=120+zx(x+y)=120+z.

5.Let ABC be an acute angled triangle with BC > AC. Let O be the circumcentre , H be the orthocentre of the triangle ABC . CF is the altitude . The perpendicular to OF at E meets the side CA at P . Prove that angle FHP = angle BAE.

6.Is it possible to write the numbers 1,2,3....,1211,2,3....,121 in an 11×1111 \times 11 table so that any two consecutive numbers be written in cells with a common side and all perfect squares lie in a straight column?

7.The positive integers are seprated into two subsets with no common elements. Show that one of these subsets must contain a three term A.P.

  1. Prove that : If three lines from vertices of a triangle are concurrent then their isogonals are also concurrent

  2. ω=n=120141+1n2+1(n+1)2ω=\sum_{n=1}^{2014}\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}. Find ωω.

Please reshare this note so that more people get to know about it. This note contain all the questions asked in the exam . Enjoy solving these questions. Remember the guidelines novel..and...elegant...solutions..will..get..extra...credits.\color{#D61F06}{''novel..and...elegant...solutions.. will ..get ..extra ...credits.''}.

Note by Shivam Jadhav
3 years, 11 months ago

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3) a3=abc=a+b+ca+2bc=3a    a23a^3=abc=a+b+c\ge a+2\sqrt{bc}=3a\implies a^2\ge 3

4) If xx odd, yy even, then y=2    x(x+2)=120+zy=2\implies x(x+2)=120+z    x22x120=z\implies x^2-2x-120=z    (x12)(x+10)=z\implies (x-12)(x+10)=z    x=13,z=23\implies x=13, z=23

Else x(x+y)x(x+y) is even so z=2z=2 and x(x+y)=122=2×61    x=2,y=59x(x+y)=122=2\times 61\implies x=2, y=59

Only solutions are (x,y,z)=(13,2,23),(2,59,2)(x,y,z)=(13, 2, 23), (2, 59, 2)

6) No. Just write the squares 12,22,,1121^2, 2^2, \ldots , 11^2 in a column and start filling in squares from 11. You will find that all moves are forced, and will not create a 11×1111\times 11 shape.

Daniel Liu - 3 years, 11 months ago

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Please help in solving problem number 2.

Shivam Jadhav - 3 years, 11 months ago

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Daniel liu can you explain your answer to problem 6 ?

Surekha Jadhav - 3 years, 11 months ago

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Here's an even simpler solution.

WLOG let the numbers 12,1121^2, \ldots 11^2 be going from top to bottom. Consider the stretch of numbers from k2k^2 to (k+1)2=k2+2k+1(k+1)^2=k^2+2k+1. There are 2k2k numbers from these two, which gives kk numbers away and kk numbers back as the farthest distance this sequence of numbers can go. However, that means that no matter what kk we choose, we cannot reach either the top left or top right square, depending on where the column is. Thus, that square cannot be labelled by a square, and we are done.

Daniel Liu - 3 years, 11 months ago

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@Daniel Liu I don't think the "WLOG" part is legal, it could be possible for 121^2 to start in the middle and swiggle its way through. I posted a solution for it.

Xuming Liang - 3 years, 10 months ago

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@Xuming Liang It's given that 12,22,,1121^2, 2^2, \ldots , 11^2 all lie on a column. I'm just saying to order them from top to bottom without loss of generality.

Daniel Liu - 3 years, 10 months ago

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Q 4)

Case 1: If zz is an even prime i.e. z=2z=2. Then x(x+y)=2×61x(x+y) = 2 \times 61. But since x+y>xx+y > x, the only possibility is x=2x=2 and y=59y=59. So, one solution in this case (2,59,2)(2,59,2).

Case 2: If zz is an odd prime. Then x(x+y)x(x+y) is odd, which implies that both xx and x+yx+y giving out that yy is even. But yy is a prime, so y=2y=2. So the equation transforms into x2+2x(120+z)=0x^2 + 2x -(120+z)=0. So, the discriminant should be a perfect square.

    Δ=4z+484=k2;kZ\implies \Delta = 4z + 484 =k^2 \quad ; \quad k \in \mathbb{Z}     2k\implies 2|k

So, letting k=2lk=2l.

    z=(l11)(l+11)\implies z = (l-11)(l+11)

But zz is a prime. So, l11=1l-11=1 and l+11=zl+11=z, which gives us that z=23z=23. Substituting in the above equation gives out that x=11x=11. Therefore (11,2,23)(11,2,23) is only solution in this case.

Therefore, total number of solutions are 22. And they are (11,2,23)(11,2,23) and (2,59,2)(2,59,2).

Surya Prakash - 3 years, 11 months ago

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  1. Suppose the answer is yes, Observe that any path that begins with odd squared number and ends on the next (even) squared number only occurs on one side of the column, vise versa for paths that begin with even squares. We can count the number of squares on the "odd" side to be 2(1+3+...+9)=2(52)=502(1+3+...+9)=2(5^2)=50, which cannot happen since it must be divisible by 1111. Therefore the answer is no.

Xuming Liang - 3 years, 10 months ago

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Can you solve the second question?

Shivam Jadhav - 3 years, 10 months ago

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what is r3 in the second question

Neel Khare - 2 years, 11 months ago

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@Neel Khare r3=r1

Shivam Jadhav - 2 years, 11 months ago

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@Calvin Lin
@niharmahajan

Shivam Jadhav - 3 years, 11 months ago

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@Nihar Mahajan @Surya Prakash

Shivam Jadhav - 3 years, 11 months ago

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I had done Q1 about 1 year ago. The relation is d=R(R2r)d=\sqrt{R(R-2r)}. It is famous and called as Euler Triangle formula. Its proof is based on cyclicity and a bit of trigonometry.

Nihar Mahajan - 3 years, 11 months ago

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You need to prove it.

Shivam Jadhav - 3 years, 11 months ago

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I have proved it.

Nihar Mahajan - 3 years, 11 months ago

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@Nihar Mahajan Where?

Shivam Jadhav - 3 years, 11 months ago

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The answer to Q.4 is (2,59,2) and (11,2,23). So there are two solutions.

Kushagra Sahni - 3 years, 11 months ago

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1)d^2=R^2-2Rr

Aakash Khandelwal - 3 years, 11 months ago

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Q 7) 2) https://brilliant.org/problems/simply-awsome/ Similar problem

Surya Prakash - 3 years, 11 months ago

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@Daniel Liu @Surya Prakash @Calvin Lin @Best of Geometry @Xuming Liang please help me in solving question number 2.

Shivam Jadhav - 3 years, 11 months ago

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@Shivam Jadhav here is the proof for Q1

Sualeh Asif - 3 years, 11 months ago

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Please try second question

Shivam Jadhav - 3 years, 11 months ago

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One question. Is Ramanujan the senior round?

Swapnil Das - 2 years, 11 months ago

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