# RMO Board Part 2

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.

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)

Note by Nihar Mahajan
2 years, 10 months ago

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Q4) 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 $$n$$.

- 2 years, 10 months ago

@Surya Prakash Great job!! Can you simply this further to eliminate $$\omega$$?

- 2 years, 10 months ago

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 $$b=9$$.

- 2 years, 10 months ago

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.

- 2 years, 10 months ago

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

1. Try it by co-ordinate geometry.

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

3. 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 $${x}^{2}+{y}^{2}+\sqrt{3}y-1/4=0$$

4. 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. $$P=x+iy$$

5. 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 )

- 2 years, 10 months ago

$$Nice one:$$ But Here is a still simpler one :)

Solution to $$3rd$$ stuff

$$Construction$$: Draw $$PX$$ = $$PC$$ such that $$\angle CPX = 60^{\circ}$$. Finally join $$BX$$ and $$CX$$.

Clearly $$\Delta CPX$$ is equilateral. => $$CX$$ = $$PC$$ ------ $$1$$ and Also, $$\angle PCX = \angle ACB$$ = $$60^{\circ}$$ => $$\angle PCX - \angle PCB = \angle ACB - \angle PCB$$ => $$\angle BCX = \angle ACP$$ ------- $$2$$

From the results in $$eq^{n}$$ $$1$$, $$2$$ and that $$BC$$ = $$AC$$, we conclude that $$\Delta BCX \cong \Delta ACP$$. => $$BX$$ = $$PA$$

But $$Given$$ $$PB^{2} + PC^{2}$$ = $$PA^{2}$$

=> $$PB^{2} + PX^{2}$$ = $$BX^{2}$$ ............... [$$PC$$ = $$PX$$ and $$BX = PA$$]

Therefore by the converse of $$Pythagoras... theorem$$, it is obvious that $$\angle BPX$$ = $$90^{\circ}$$

Thus the required angle = $$\angle BPC$$ = $$\angle BPX + \angle CPX$$ = $$90^{\circ}$$ + $$60^{\circ}$$ = $$150^{\circ}$$.

- 1 year, 4 months ago

$$Question 5:$$

$$Const:$$ The constructions are clearly shown in the above $$diagram.$$ Join $$BM$$, $$DM$$, $$C'M$$ ,finally join $$A'M$$ and produce it to meet $$CC'$$ at $$B'$$.

$$AB$$ is the diameter of the $$\odot ABCD.$$

=> $$\angle ABC$$ = $$\angle ADC$$ = $$90^{\circ}$$ => $$BM$$ = $$DM$$ [= $$1/2AC$$ ] => $$\angle MBC'$$ = $$\angle MDC'$$ --- $$1$$

Clearly, $$\Delta AA'M \cong \Delta CB'M$$ => $$A'M$$ = $$B'M$$ => $$M$$ is the midpoint of $$A'B'$$.

But $$\Delta A'B'C'$$ is right angled at $$C'$$ and also $$M$$ is the midpoint of $$A'B'.$$ => $$C'M$$ = $$A'M$$ ---- $$2$$

=> $$\angle MC'B$$ = $$\angle MDA'$$ ----- $$3$$

From the $$eq^{n}$$ $$1$$, $$2$$ and $$3$$: $$\Delta MC'B \cong \Delta MDA'$$ => $$BC'$$ = $$A'D.$$

$$K.I.P.K.I.G.$$

- 1 year, 4 months ago

Hey Nihar plead post solution for question(9). Thanks!

- 2 years, 8 months ago

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

- 2 years, 10 months ago

@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 $$3$$.

- 2 years, 10 months ago

Its correct.

- 2 years, 10 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.

- 2 years, 10 months ago

It satisfies.

- 2 years, 10 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.

- 2 years, 10 months ago

In that way!!!!! Then it is understood that it should be $$3.[\frac{n}{3}]$$

- 2 years, 10 months ago

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

- 2 years, 10 months ago

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 $$a=b=c=0$$.

- 2 years, 10 months ago

1) p = 7

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

comparing of y + x and p gives p = 7

- 2 years, 10 months ago

@Dev Sharma Kindly always post a full solution.

- 2 years, 10 months ago

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

- 2 years, 10 months ago

let me check

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

- 2 years, 10 months ago

- 2 years, 10 months ago

If $$x^{5}-x^{3}+x=\eta$$, the minimum value of $$x^{6}$$ in terms of $$\eta$$.

- 2 years, 10 months ago