Waste less time on Facebook — follow Brilliant.
×

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, 7 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Did anyone get the proofs for question5 &6 Nitish Deshpande · 2 years, 7 months ago

Log in to reply

@Nitish Deshpande 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 · 2 years, 7 months ago

Log in to reply

Positive integer solution of 1/a+1/b+1/c+1/d =1 Gebretsadkan Gebereyohannes · 1 year, 9 months ago

Log in to reply

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.

Rohit Camfar · 5 months ago

Log in to reply

@Rohit Camfar Nicely done :) Pranshu Gaba · 5 months ago

Log in to reply

What is the meaning of a non constant arithmetic and geometric progression Aditya Thomas · 8 months, 3 weeks ago

Log in to reply

Pranshu, can you tell me which latex you used to separate the questions by a horizontal line? Priyanshu Mishra · 1 year, 7 months ago

Log in to reply

@Priyanshu Mishra

1
---
Priyanshu, this is not latex; it is markdown. Enter three hyphens (shown above) in a new line to get a horizontal line. Pranshu Gaba · 1 year, 7 months ago

Log in to reply

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 · 2 years, 4 months ago

Log in to reply

i am selected for inmo Nitish Deshpande · 2 years, 7 months ago

Log in to reply

Comment deleted Mar 15, 2015

Log in to reply

@Mihir Chakravarti i didn't get selected. did you? Sai Prasanth Rao · 2 years, 7 months ago

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, 7 months ago

Log in to reply

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, 7 months ago

Log in to reply

@Mohnish Chakravarti The cutoff must be somewhere around 40. Pranshu Gaba · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba Did you get the 3rd one? I tried but I could not get anywhere in that problem. Mohnish Chakravarti · 2 years, 7 months ago

Log in to reply

@Mohnish Chakravarti No, I tried but I didn't get it. Pranshu Gaba · 2 years, 7 months ago

Log in to reply

5th is so easy. Basic CS/AM-HM. Krishna Ar · 2 years, 7 months ago

Log in to reply

whats the answer for 4th question Shrihari B · 2 years, 7 months ago

Log in to reply

@Shrihari B 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 · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba How ??? Shivram Badhe · 2 years, 7 months ago

Log in to reply

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

image

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, 7 months ago

Log in to reply

@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 · 2 years, 7 months ago

Log in to reply

@Mohnish Chakravarti why 11C5 Sai Prasanth Rao · 2 years, 7 months ago

Log in to reply

@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 · 2 years, 7 months ago

Log in to reply

@Sai Prasanth Rao Haven't thought about it. Will think about such a solution. Mohnish Chakravarti · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba i guess the person could step on the red arrows 5 times Sai Prasanth Rao · 2 years, 7 months ago

Log in to reply

@Sai Prasanth Rao How ? The question states that the person moves right exactly three times, so three red arrows. Pranshu Gaba · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba my bad. i thought the red arrows were the ones when he goes diagonally. he can go diagonally five times. sorry. Sai Prasanth Rao · 2 years, 7 months ago

Log in to reply

Positive integer solution of 1/a+1/b+1/c+1/d =1 Gebretsadkan Gebereyohannes · 1 year, 9 months ago

Log in to reply

what is the ans to question 2 i m getting a =2 b=9 there could be more values Nitish Deshpande · 2 years, 7 months ago

Log in to reply

@Nitish Deshpande \(a=2\) and \(b=9\) are the only positive integer solutions. Pranshu Gaba · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba 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 · 2 years, 7 months ago

Log in to reply

@Sai Prasanth Rao a, b, c are in ratio 4:2:1, hence they cannot be the sides of a triangle. Pranshu Gaba · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba yes. i got the same. how did you do it? i factorized the equation. Sai Prasanth Rao · 2 years, 7 months ago

Log in to reply

@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 · 2 years, 7 months ago

Log in to reply

@Mohnish Chakravarti i attended PACE in 9th an 10th and even attended the fast-track batch lectures. Sai Prasanth Rao · 2 years, 7 months ago

Log in to reply

@Sai Prasanth Rao I made perfect squares. The equation becomes \((a-2b)^2 + (b-2c)^2 = 0\). Pranshu Gaba · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba how many did you get? Sai Prasanth Rao · 2 years, 7 months ago

Log in to reply

@Sai Prasanth Rao 1, 2 and 4. Which ones did you get? Pranshu Gaba · 2 years, 7 months ago

Log in to reply

@Pranshu Gaba i too got 1,2 and 4 but i am not sure about my answer to the fourth question. Sai Prasanth Rao · 2 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...