NMTC 2015 Ramanujan Contest Final Round

1. ABC is a triangle in which $R$ represents the circumradius and $r$ represents the inradius and $d$represents the distance between circumcentre and incentre . Then find $d$ in terms of $R,r$

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

3. If $a,b,c$are real numbers such that $a^{2}=bc , a+b+c=abc$, then prove that $a^{2}\geq3$.

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

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 $1,2,3....,121$ in an $11 \times 11$ 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. $ω=\sum_{n=1}^{2014}\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}$. Find $ω$.

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 $\color{#D61F06}{''novel..and...elegant...solutions.. will ..get ..extra ...credits.''}$.

4 years, 4 months ago

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

4) If $x$ odd, $y$ even, then $y=2\implies x(x+2)=120+z$$\implies x^2-2x-120=z$$\implies (x-12)(x+10)=z$$\implies x=13, z=23$

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

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

6) No. Just write the squares $1^2, 2^2, \ldots , 11^2$ in a column and start filling in squares from $1$. You will find that all moves are forced, and will not create a $11\times 11$ shape.

- 4 years, 4 months ago

- 4 years, 4 months ago

- 4 years, 4 months ago

Here's an even simpler solution.

WLOG let the numbers $1^2, \ldots 11^2$ be going from top to bottom. Consider the stretch of numbers from $k^2$ to $(k+1)^2=k^2+2k+1$. There are $2k$ numbers from these two, which gives $k$ numbers away and $k$ numbers back as the farthest distance this sequence of numbers can go. However, that means that no matter what $k$ 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.

- 4 years, 4 months ago

I don't think the "WLOG" part is legal, it could be possible for $1^2$ to start in the middle and swiggle its way through. I posted a solution for it.

- 4 years, 4 months ago

It's given that $1^2, 2^2, \ldots , 11^2$ all lie on a column. I'm just saying to order them from top to bottom without loss of generality.

- 4 years, 4 months ago

Q 4)

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

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

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

So, letting $k=2l$.

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

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

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

- 4 years, 4 months ago

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 $2(1+3+...+9)=2(5^2)=50$, which cannot happen since it must be divisible by $11$. Therefore the answer is no.

- 4 years, 4 months ago

Can you solve the second question?

- 4 years, 4 months ago

what is r3 in the second question

- 3 years, 5 months ago

r3=r1

- 3 years, 4 months ago

@Calvin Lin
@niharmahajan

- 4 years, 4 months ago

- 4 years, 4 months ago

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

- 4 years, 4 months ago

You need to prove it.

- 4 years, 4 months ago

I have proved it.

- 4 years, 4 months ago

Where?

- 4 years, 4 months ago

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

- 4 years, 4 months ago

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

- 4 years, 4 months ago

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

- 4 years, 4 months ago

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

- 4 years, 4 months ago

@Shivam Jadhav here is the proof for Q1

- 4 years, 4 months ago

- 4 years, 4 months ago

One question. Is Ramanujan the senior round?

- 3 years, 5 months ago