RMO 2014 Mumbai Region

Regional Mathematical Olympiad 2014 (Mumbai Region)

Instructions

There are six questions in this paper. Answer all questions.
Each question carries 10 points
Use of protractors, calculators, mobile phone is forbidden.
Time alloted: 3 hours

Questions

1 Three positive real numbers $a, b, c$ are such that $a^2 + 5b^2 + 4c^2 - 4ab - 4bc = 0$ . Can $a, b, c$ be the lengths of sides of a triangle? Justify your answer.

2 The roots of the equation

$x^3 -3ax^2 + bx + 18c = 0$

form a non-constant arithmetic progression and the roots of the equation

$x^3 + bx^2 + x - c^3 = 0$

form a non-constant geometric progression. Given that $a, b, c$ are real numbers, find all positive integral values $a$ and $b$ .

3 Let $ABC$ be an acute-angled triangle in which $\angle ABC$ is the largest angle. Let $O$ be its circumcentre. The perpendicular bisectors of $BC$ and $AB$ meet $AC$ at $X$ and $Y$ respectively. The internal bisectors of $\angle AXB$ and $\angle BYC$ meet $AB$ and $BC$ at $D$ and $E$ respectively. Prove that $BO$ is perpendicular to $AC$ if $DE$ is parallel to $AC$ .

4 A person moves in the $x-y$ plane moving along points with integer co-ordinates $x$ and $y$ only. When she is at point $(x,y)$ , she takes a step based on the following rules:

(a) if $x+y$ is even she moves either to $(x+1,y)$ or $(x+1, y+1)$ ;

(b) if $x+y$ is odd she moves either to $(x,y+1)$ or $(x+1, y+1)$ .

How many distinct paths can she take to go from $(0,0)$ to $(8,8)$ given that she took exactly three steps to right $((x,y)$ to $(x+1,y))$ ?

5 Let $a, b, c$ be positive numbers such that

$\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} \leq 1.$

Prove that $(1+a^2 )(1+b^2)(1+c^2) \geq 125$ . When does the equality hold?

6 Let $D, E, F$ be the points of contact of the incircle of an acute-angled triangle $ABC$ with $BC, CA, AB$ respectively. Let $I_1, I_2, I_3$ be the incenters of the triangles $AFE, BDF, CED,$ respectively. Prove that the lines $I_1D, I_2E, I_3F$ are concurrent.

Post your innovative solutions below!! Enjoy!!!!!!!!

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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}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

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}`	$\boxed{123}$

Comments

Did anyone get the proofs for question5 &6

Nitish Deshpande - 6 years, 6 months ago

There are 2 ways to solve this. 1. AM-GM inequality. 2. Without using any theorems, we can prove the square of each individual term to be greater than 4(try figuring out on your own first) and then add 1 to each term. Finally multiply all the terms to get the final result. Equality holds true at a=b=c=2(which is clearly visible)

Mohnish Chakravarti - 6 years, 6 months ago

Positive integer solution of 1/a+1/b+1/c+1/d =1

Gebretsadkan Gebereyohannes - 5 years, 7 months ago

whats the answer for 4th question

Shrihari B - 6 years, 6 months ago

I don't have the official answers right now, but when I solved it I got 462 distinct paths. (I may be wrong)

Pranshu Gaba - 6 years, 6 months ago

How ???

Shivram Badhe - 6 years, 6 months ago

@Shivram Badhe – Here's one way to start. We make a graph of all the possible paths. It looks something like

image

According to the condition in the question, the person must step on exactly 3 of the red arrows. Can you continue now?

@Pranshu Gaba – i guess the person could step on the red arrows 5 times

Sai Prasanth Rao - 6 years, 6 months ago

@Sai Prasanth Rao – How ? The question states that the person moves right exactly three times, so three red arrows.

@Pranshu Gaba – my bad. i thought the red arrows were the ones when he goes diagonally. he can go diagonally five times. sorry.

@Pranshu Gaba – I would say that 462 sounds right.Since the person is allowed to go only 3 steps to the right, she can go only 3 units on the x-axis. And since she needs to reach 8 on the x axis, she must take exactly 5 steps to reach the required point. But now having only options as going upwards and moving diagonally, she needs exactly 5 diagonals as that is the only other step which can take her +1 unit on both axes. But moving right doesn't contibute to moving upwards. And since the 5 diagonal steps contribute only 5 steps upwards, she must take exactly 3 steps upwards. Now fixing the diagonal steps will complementarily fix the steps upwards and towards the right or vice versa. So, the required answer is 462 or 11C5

@Mohnish Chakravarti – or you can do this. if you select 5 diagonals out of the possible diagonal paths, there exists a unique path.

@Sai Prasanth Rao – Haven't thought about it. Will think about such a solution.

@Mohnish Chakravarti – why 11C5

5th is so easy. Basic CS/AM-HM.

Krishna Ar - 6 years, 6 months ago

1 and 5 were so easy. I wonder why they asked them. I was on the right track for the 2nd question however, a shitty mistake while writing the equation led to me getting no solutions :-( I also attempted the 4 and the 6th but I am not sure of the solutions. How much marks do you expect? What is the expected cut-off?

The cutoff must be somewhere around 40.

Did you get the 3rd one? I tried but I could not get anywhere in that problem.

@Mohnish Chakravarti – No, I tried but I didn't get it.

in q3 triangle abc is isoceles can be proved.so the perpendicular bisector of side ac is a cevian so BO perpendicular to AC

i am selected for inmo

I am preparing for RMO and I am in class 10 please someone help me in preparing.and pls help me in the 2nd Question i tried but i dont know where i am wrong

Drishtant Jain - 6 years, 3 months ago

Pranshu, can you tell me which latex you used to separate the questions by a horizontal line?

Priyanshu Mishra - 5 years, 6 months ago

---

Priyanshu, this is not latex; it is markdown. Enter three hyphens (shown above) in a new line to get a horizontal line.

Pranshu Gaba - 5 years, 6 months ago

What is the meaning of a non constant arithmetic and geometric progression

Aditya Thomas - 4 years, 7 months ago

Well the Maharastra is much ahead in the race for IMO in India. I tried the geometry ones(nice). Both Mumbai and Pune are doing well. I couldn't try the non geometric one due to lack of time. Here are the geometric ones.

Vishwash Kumar ΓΞΩ - 4 years, 3 months ago

Nicely done :)

Pranshu Gaba - 4 years, 3 months ago

Ya really a nice solution.Thumbs Up to you.

D K - 2 years, 10 months ago

6 is easy

Shree Ganesh - 3 years, 10 months ago

what is the ans to question 2 i m getting a =2 b=9 there could be more values

$a=2$ and $b=9$ are the only positive integer solutions.

ya even i got that. and for question 1 i got that a,b and c can never form the sides of a triangle.

@Sai Prasanth Rao – a, b, c are in ratio 4:2:1, hence they cannot be the sides of a triangle.

@Pranshu Gaba – yes. i got the same. how did you do it? i factorized the equation.

@Sai Prasanth Rao – I made perfect squares. The equation becomes $(a-2b)^2 + (b-2c)^2 = 0$ .

@Pranshu Gaba – how many did you get?

@Sai Prasanth Rao – 1, 2 and 4. Which ones did you get?

@Pranshu Gaba – i too got 1,2 and 4 but i am not sure about my answer to the fourth question.

@Sai Prasanth Rao – Did you attend Pace in grade 9 and 10? I have a feeling that we have studied together in the "fast-track" batch lectures held in dadar.

@Mohnish Chakravarti – i attended PACE in 9th an 10th and even attended the fast-track batch lectures.

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}`	$\boxed{123}$