**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!!!!!!!!

## Comments

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TopNewestDid anyone get the proofs for question5 &6 – Nitish Deshpande · 2 years, 1 month ago

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– Mohnish Chakravarti · 2 years, 1 month 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)Log in to reply

Positive integer solution of 1/a+1/b+1/c+1/d =1 – Gebretsadkan Gebereyohannes · 1 year, 2 months ago

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What is the meaning of a non constant arithmetic and geometric progression – Aditya Thomas · 2 months, 2 weeks ago

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Pranshu, can you tell me which latex you used to separate the questions by a horizontal line? – Priyanshu Mishra · 1 year, 1 month 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 – Drishtant Jain · 1 year, 10 months ago

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i am selected for inmo – Nitish Deshpande · 2 years, 1 month ago

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– Sai Prasanth Rao · 2 years, 1 month ago

i didn't get selected. did you?Log in to reply

in q3 triangle abc is isoceles can be proved.so the perpendicular bisector of side ac is a cevian so BO perpendicular to AC – Nitish Deshpande · 2 years, 1 month 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? – Mohnish Chakravarti · 2 years, 1 month ago

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– Pranshu Gaba · 2 years, 1 month ago

The cutoff must be somewhere around 40.Log in to reply

– Mohnish Chakravarti · 2 years, 1 month ago

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

– Pranshu Gaba · 2 years, 1 month ago

No, I tried but I didn't get it.Log in to reply

5th is so easy. Basic CS/AM-HM. – Krishna Ar · 2 years, 1 month ago

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whats the answer for 4th question – Shrihari B · 2 years, 1 month ago

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– Pranshu Gaba · 2 years, 1 month ago

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

– Shivram Badhe · 2 years, 1 month ago

How ???Log in to reply

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 · 2 years, 1 month ago

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– Mohnish Chakravarti · 2 years, 1 month ago

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 11C5Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

why 11C5Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

or you can do this. if you select 5 diagonals out of the possible diagonal paths, there exists a unique path.Log in to reply

– Mohnish Chakravarti · 2 years, 1 month ago

Haven't thought about it. Will think about such a solution.Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

i guess the person could step on the red arrows 5 timesLog in to reply

– Pranshu Gaba · 2 years, 1 month ago

How ? The question states that the person moves right exactly three times, so three red arrows.Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

my bad. i thought the red arrows were the ones when he goes diagonally. he can go diagonally five times. sorry.Log in to reply

Positive integer solution of 1/a+1/b+1/c+1/d =1 – Gebretsadkan Gebereyohannes · 1 year, 2 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 – Nitish Deshpande · 2 years, 1 month ago

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– Pranshu Gaba · 2 years, 1 month ago

\(a=2\) and \(b=9\) are the only positive integer solutions.Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

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

– Pranshu Gaba · 2 years, 1 month ago

a, b, c are in ratio 4:2:1, hence they cannot be the sides of a triangle.Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

yes. i got the same. how did you do it? i factorized the equation.Log in to reply

– Mohnish Chakravarti · 2 years, 1 month ago

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.Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

i attended PACE in 9th an 10th and even attended the fast-track batch lectures.Log in to reply

– Pranshu Gaba · 2 years, 1 month ago

I made perfect squares. The equation becomes \((a-2b)^2 + (b-2c)^2 = 0\).Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

how many did you get?Log in to reply

– Pranshu Gaba · 2 years, 1 month ago

1, 2 and 4. Which ones did you get?Log in to reply

– Sai Prasanth Rao · 2 years, 1 month ago

i too got 1,2 and 4 but i am not sure about my answer to the fourth question.Log in to reply