Since the previous board was overcrowded with comments , I decided to create a new note so as to keep this step lively. If this also gets crowded , eventually we will again have new RMO board.What you have to do:

1) Propose a problem in the comments. Then I will add that problem here in this note content by giving the respective credit. And after I add the problem , the comment of proposing the problem must be deleted by the problem poster.

2) The problem poster must not post solution to his/her own problem unless someone from our community posts or no one gets it right even after considerable period of time.

3) Inappropriate / trivial comments are not allowed and must be deleted if posted by chance.An enthusiastic discussion is expected.

4) Please reshare this note so that we can reach most of the Brilliantians.

5)The problems to which solutions are posted will be accompanied by a checkmark \(\large(\checkmark)\) at the end.

**Problems:**

**Q1)** Find all prime numbers \(p\) for which there are integers \(x,y\) satisfying \(p+1=2x^2\) and \(p^2+1=2y^2\).\(\large\checkmark\)

**Q2)** 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 \(b\).(Shared by Mycobacterium Tuberculae) \(\large\checkmark\)

**Q3)** In an equilateral \(\Delta ABC\) , \(P\) is a point inside the triangle such that \(PA^2=PB^2+PC^2\).Prove that \(m\angle BPC = 150^\circ\).(Shared by Raven Herd) \(\large\checkmark\)

**Q4)** Evaluate \(\large\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \binom{n}{9} + ... \binom{n}{n}\) with respect to \(n\). (Shared by Svatejas) \(\large\checkmark\)

**Q5)** Consider a cyclic quadrilateral \(ABCD\) such that \(AC\) is the diameter of its circumcircle. Construct points \(A'\) and \(C'\) on \(BD\) such that \(AA' \perp BD\) and \(CC' \perp BD\) . Show that \(DA'=C'B\).(Shared by Karthik V)

**Q6)** Prove that polynomial \(az^n+z+1\) has at least one root in \(|z|\leq 2\). (Shared by Shivamani P)

**Q7)** If \(a,b,c\) are positive reals , determine the minimum value of the expression \(ab(a-c)+bc(b-a)+ca(c-b)\) with proof. (Posed by Nihar M) \(\large\checkmark\)

**Q8)a)** How many ways are there to represent a natural number \(n\) as a sum of \(k\) natural numbers?

**Q8)b)** How many ways are there to represent a natural number \(n\) as a sum of \(k\) non-negative numbers?

**Q9)** Find all ordered pairs of integers \((m,n)\) satisfying: \(3\times 2^m + 1= n^2\).

**Q10)a)** Prove that if \(x^2+px-q\) and \(x^2-px+q\) both factorize into linear factors with integral coefficients, then the positive integers and are respectively the hypotenuse and area of a right angled triangle sides of integer lengths.

**Q10)b)** Show further that if \(x^2+px-q = (x-\alpha)(x-\beta) \ and \ x^2-px+q = (x-\gamma)(x-\delta)\) where \(p,q,\alpha,\beta,\gamma,\delta\) are integers, \(\alpha,\beta,\gamma,\delta\) then are numerically the radii of the incircle and the three excircles of the triangle. (Shared by Svatejas S)

## Comments

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TopNewestQ4) Let \(f(x) = (1+x)^{n} ={n \choose 0}+ {n \choose 1} x + \ldots + {n \choose n} x^{n} \)

Now, \(f(1) + f(\omega) +f(\omega ^2) = {n \choose 0} (1+ \omega + \omega ^2) + {n \choose 1} (1+ \omega ^{2} + \omega ^{4}) + \ldots + {n \choose n} (1+ \omega ^n + \omega ^2n)\) \(= 3 \left( {n \choose 0} + {n \choose 3}+ {n \choose 6} + \ldots \right)\)

So, \({n \choose 0} + {n \choose 3}+ {n \choose 6} + \ldots = \dfrac{f(1) + f(\omega) +f(\omega ^2)}{3} = \dfrac{2^n + (- \omega ^2)^n + ( - \omega)^n}{3}\).

So, we can manipulate the value of this expression by knowing the value of \(n\). – Surya Prakash · 1 year, 3 months ago

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@Surya Prakash Great job!! Can you simply this further to eliminate \(\omega\)? – Svatejas Shivakumar · 1 year, 3 months ago

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For the \(2nd\) question,

Let the roots of first equation be \(m-d,m,m+d\)

Therefore we get that, \(m-d+m+m+d=3a=3m\). Hence \(m=a\)

Then by applying Vieta's formula for other two, we get that, \(3{m}^{2}-{d}^{2}=b\) and \( {m}^{3}-m{d}^{2}=-18c \)

Let the roots for the second equation be \(n/r,n,nr \)

Therefore by applying Vieta's fomula,we get that, \(n=c\) and \(n(1/r+1+r)=-b\) and \( {n}^{2}(1/r+1+r)=1\)

From here we get that, \(n=-1/b=c\)

Substituting these values in the previously acquired equations we get that, \({m}^{3}-m(3{m}^{2}-b)=18/b\)

Simplifying we get that, \(a{b}^{2}-2{a}^{3}b-18=0\)

Applying formula for quadratic roots, \(b=\frac{{a}^{3}\pm \sqrt{{a}^{6}+18a}}{a} \)

For \(b\) to be an integer, \({a}^{6}+18a={k}^{2} \)

We see that, \( {a}^{3}<k<{a}^{3}+3 \)

From here we see that, \(a=2\) only satisfies. Therefore the required solutions are \(a=2\) and \(b=9\). – Saarthak Marathe · 1 year, 3 months ago

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Q1) It is clear that \(p(p-1) = 2(y+x)(y-x)\). If \(p|y-x\) it implies that \(p \leq y-x\), then it should imply that \(p-1 \geq 2(y+x)\). But it is not possible. So, \(p\) does not divide \(y-x\). So, \(p|y+x\) it implies that \(p \leq y+x\), but this implies that \(p-1 \geq 2(y-x)\). So, we get \(p +1 \leq 4x\). Which reduces to \(2x^2 \leq 4x\), \(x \leq 2\). If \(x=1\), then \(p=1\), a contradiction. And \(x=2\) implies that \(p=7\). Therefore, \(p=\boxed{7}\) is the only solution. – Surya Prakash · 1 year, 3 months ago

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My solution for the \(3rd\) one is quite big. I will just give some hints to solve it.

Try it by co-ordinate geometry.

Take \(B\) as \((-1/2,0)\) and \(C\) as \((1/2,0)\).Therefore,we get \(A\) as \((\sqrt{3}/2,0)\).

Then take \(P\) as\((x,y)\) use the provided equation of \({PA}^{2}={PB}^{2}+{PC}^{2}\) to get the equation of locus of \(P\).The equation comes out to be \({x}^{2}+{y}^{2}+\sqrt{3}y-1/4=0\)

Then take the \(X-Y\) plane as the Argand plane and apply the Rotation theorem of Complex numbers for \(B,P,C\) in order.Here the complex numbers are, i. \(B=-1/2\), ii. \(C=1/2\) and iii. \(P=x+iy\)

By simplifying the equation obtained in the \(4th\) step, we get that \(\angle{BPC}={150}^{\circ}\)

You may consider this as a solution because I have provided all the steps required.(But this step completely depends on you :P ) – Saarthak Marathe · 1 year, 3 months ago

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Hey Nihar plead post solution for question(9). Thanks! – Harsh Shrivastava · 1 year, 1 month ago

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1) Solve \(y^3 = x^3 + 8x^2 - 6x + 8\) for positive integer x and y.

2) Two positive integer a and b such that \(a^{a}b^{b}\) is divisible by 2000. What is least possible value of ab?

3) Find all real a for which \(x^4-2ax^2+x+a^2- a\) has all real root – Dev Sharma · 1 year, 3 months ago

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@Surya Prakash @Nihar Mahajan @Svatejas Shivakumar There seems to be a blunder in problem statement of

Q4). I guess it meant " Simplify \( \large \binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \binom{n}{9} + ... \binom{n}{ 3 \cdot \lfloor \frac{n}{3} \rfloor } \)." The last summand being \( \binom{n}{n} \) doesn't make sense at all if \( n \) is not a multiple of \( 3 \). – Karthik Venkata · 1 year, 3 months agoLog in to reply

– Saarthak Marathe · 1 year, 3 months ago

Its correct.Log in to reply

– Karthik Venkata · 1 year, 3 months ago

Try substituting \( n = 4 \) in the summation. The link between last term of the summation ( i.e. last summand ) and previous summands is not clear.Log in to reply

– Saarthak Marathe · 1 year, 3 months ago

It satisfies.Log in to reply

– Karthik Venkata · 1 year, 3 months ago

Let \( f(n) = \large\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \binom{n}{9} + ... \binom{n}{n}\).Then \( f(4) = \binom{4}{0} + \binom{4}{3} + \binom{4}{4} \) ? I am certainly sure about the last summand being \( \binom{n}{ 3 \cdot \lfloor \frac{n}{3} \rfloor } \), contrary to \( \binom{n}{n} \) mentioned in the question.Log in to reply

– Saarthak Marathe · 1 year, 3 months ago

In that way!!!!! Then it is understood that it should be \(3.[\frac{n}{3}] \)Log in to reply

– Karthik Venkata · 1 year, 3 months ago

I guess it would have been understood if it was just ..... , but rather it mentioned the last summand wrong 😓.Log in to reply

7) @Nihar Mahajan We have \(ab(a-c)+bc(b-a)+ca(c-b)\).

Multiplying we get \(a^2b+b^2c+c^2a-3abc= abc(\frac {a}{c}+ \frac {b}{a} + \frac {c}{b}-3\)).

By AM-GM inequality, \(\frac {a}{c}+ \frac {b}{a} + \frac {c}{b} \ge 3\).

Therefore, minimum value of the expression \(\frac {a}{c}+ \frac {b}{a} + \frac {c}{b}\) is \(3\).

Therefore, minimum value of \(ab(a-c)+bc(b-a)+ca(c-b)\) is \(0\) and equality occurs when \(a=b=c=0\). – Svatejas Shivakumar · 1 year, 3 months ago

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1) p = 7

\(p(p - 1) = 2(y + x)(y - x)\)

comparing of y + x and p gives p = 7 – Dev Sharma · 1 year, 3 months ago

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@Dev Sharma Kindly always post a full solution. – Mehul Arora · 1 year, 3 months ago

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– Nihar Mahajan · 1 year, 3 months ago

Sorry to say , but you have dropped out some cases. This solution is incomplete.Log in to reply

p can't divide 2 and y - x so p divides only y + x and by some observations, i found x is equal to or smaller than 2, so when x = 1 then p = 1. No. then x = 2 so p = 7 – Dev Sharma · 1 year, 3 months ago

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– Nihar Mahajan · 1 year, 3 months ago

Include this in your previous comment and complete your solution. Don't make parts of your solution.Log in to reply

If \(x^{5}-x^{3}+x=\eta\), the minimum value of \(x^{6}\) in terms of \(\eta\). – Shivam Jadhav · 1 year, 3 months ago

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