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# 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!!!!!!!!

Note by Pranshu Gaba
2 years, 9 months ago

## Comments

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Did anyone get the proofs for question5 &6 · 2 years, 9 months ago

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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) · 2 years, 9 months ago

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Positive integer solution of 1/a+1/b+1/c+1/d =1 · 1 year, 11 months ago

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6 is easy · 1 month, 3 weeks ago

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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.

· 7 months, 1 week ago

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Nicely done :) · 7 months, 1 week ago

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What is the meaning of a non constant arithmetic and geometric progression · 11 months ago

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Pranshu, can you tell me which latex you used to separate the questions by a horizontal line? · 1 year, 9 months ago

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 1 --- 

Priyanshu, this is not latex; it is markdown. Enter three hyphens (shown above) in a new line to get a horizontal line. · 1 year, 9 months ago

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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 · 2 years, 7 months ago

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i am selected for inmo · 2 years, 9 months ago

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Comment deleted Mar 15, 2015

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i didn't get selected. did you? · 2 years, 9 months ago

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in q3 triangle abc is isoceles can be proved.so the perpendicular bisector of side ac is a cevian so BO perpendicular to AC · 2 years, 9 months ago

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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? · 2 years, 9 months ago

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The cutoff must be somewhere around 40. · 2 years, 9 months ago

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Did you get the 3rd one? I tried but I could not get anywhere in that problem. · 2 years, 9 months ago

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No, I tried but I didn't get it. · 2 years, 9 months ago

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5th is so easy. Basic CS/AM-HM. · 2 years, 9 months ago

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whats the answer for 4th question · 2 years, 9 months ago

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I don't have the official answers right now, but when I solved it I got 462 distinct paths. (I may be wrong) · 2 years, 9 months ago

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How ??? · 2 years, 9 months ago

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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? · 2 years, 9 months ago

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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 · 2 years, 9 months ago

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why 11C5 · 2 years, 9 months ago

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or you can do this. if you select 5 diagonals out of the possible diagonal paths, there exists a unique path. · 2 years, 9 months ago

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Haven't thought about it. Will think about such a solution. · 2 years, 9 months ago

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i guess the person could step on the red arrows 5 times · 2 years, 9 months ago

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How ? The question states that the person moves right exactly three times, so three red arrows. · 2 years, 9 months ago

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my bad. i thought the red arrows were the ones when he goes diagonally. he can go diagonally five times. sorry. · 2 years, 9 months ago

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Positive integer solution of 1/a+1/b+1/c+1/d =1 · 1 year, 11 months ago

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what is the ans to question 2 i m getting a =2 b=9 there could be more values · 2 years, 9 months ago

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$$a=2$$ and $$b=9$$ are the only positive integer solutions. · 2 years, 9 months ago

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ya even i got that. and for question 1 i got that a,b and c can never form the sides of a triangle. · 2 years, 9 months ago

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a, b, c are in ratio 4:2:1, hence they cannot be the sides of a triangle. · 2 years, 9 months ago

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yes. i got the same. how did you do it? i factorized the equation. · 2 years, 9 months ago

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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. · 2 years, 9 months ago

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i attended PACE in 9th an 10th and even attended the fast-track batch lectures. · 2 years, 9 months ago

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I made perfect squares. The equation becomes $$(a-2b)^2 + (b-2c)^2 = 0$$. · 2 years, 9 months ago

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how many did you get? · 2 years, 9 months ago

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1, 2 and 4. Which ones did you get? · 2 years, 9 months ago

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i too got 1,2 and 4 but i am not sure about my answer to the fourth question. · 2 years, 9 months ago

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