NMTC 2015 Ramanujan Contest Final Round

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 $R,r$
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 $\frac{1}{r_{3}}+\frac{1}{r_{2}}-\frac{r}{r_{2}r_{3}}=\frac{2}{h_{a}}$
If $a,b,c$ are real numbers such that $a^{2}=bc , a+b+c=abc$ , then prove that $a^{2}\geq3$ .
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.

Prove that : If three lines from vertices of a triangle are concurrent then their isogonals are also concurrent
$ω=\sum_{n=1}^{2014}\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}$ . 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 $\color{#D61F06}{''novel..and...elegant...solutions.. will ..get ..extra ...credits.''}$ .

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Comments

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.

Daniel Liu - 5 years, 7 months ago

Please help in solving problem number 2.

Shivam Jadhav - 5 years, 7 months ago

Daniel liu can you explain your answer to problem 6 ?

Surekha Jadhav - 5 years, 7 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.

@Daniel Liu – 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.

Xuming Liang - 5 years, 6 months ago

@Xuming Liang – 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.

Daniel Liu - 5 years, 6 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 $(2,59,2)$ .

Surya Prakash - 5 years, 7 months ago

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.

Can you solve the second question?

Shivam Jadhav - 5 years, 6 months ago

what is r3 in the second question

A Former Brilliant Member - 4 years, 7 months ago

@A Former Brilliant Member – r3=r1

Shivam Jadhav - 4 years, 7 months ago

@Calvin Lin
@niharmahajan

@Nihar Mahajan @Surya Prakash

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.

Nihar Mahajan - 5 years, 7 months ago

You need to prove it.

I have proved it.

@Nihar Mahajan – Where?

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

Kushagra Sahni - 5 years, 7 months ago

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

Aakash Khandelwal - 5 years, 7 months ago

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

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

@Shivam Jadhav here is the proof for Q1

Sualeh Asif - 5 years, 7 months ago

Please try second question

One question. Is Ramanujan the senior round?

Swapnil Das - 4 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$