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.

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 pp for which there are integers x,yx,y satisfying p+1=2x2p+1=2x^2 and p2+1=2y2p^2+1=2y^2.\large\checkmark

Q2) The roots of the equation x33ax2+bx+18c=0 x^3-3ax^2+bx+18c=0 form a non-constant arithmetic progression and the roots of the equation x3+bx2+xc3=0 x^3+bx^2+x-c^3=0 form a non-constant geometric progression. Given that a,b,ca,b,c are real numbers, find all positive integral values aa and bb.(Shared by Mycobacterium Tuberculae) \large\checkmark

Q3) In an equilateral ΔABC\Delta ABC , PP is a point inside the triangle such that PA2=PB2+PC2PA^2=PB^2+PC^2.Prove that mBPC=150m\angle BPC = 150^\circ.(Shared by Raven Herd) \large\checkmark

Q4) Evaluate (n0)+(n3)+(n6)+(n9)+...(nn)\large\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \binom{n}{9} + ... \binom{n}{n} with respect to nn. (Shared by Svatejas) \large\checkmark

Q5) Consider a cyclic quadrilateral ABCDABCD such that ACAC is the diameter of its circumcircle. Construct points AA' and CC' on BDBD such that AABDAA' \perp BD and CCBDCC' \perp BD . Show that DA=CBDA'=C'B.(Shared by Karthik V)

Q6) Prove that polynomial azn+z+1az^n+z+1 has at least one root in z2|z|\leq 2. (Shared by Shivamani P)

Q7) If a,b,ca,b,c are positive reals , determine the minimum value of the expression ab(ac)+bc(ba)+ca(cb)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 nn as a sum of kk natural numbers?

Q8)b) How many ways are there to represent a natural number nn as a sum of kk non-negative numbers?

Q9) Find all ordered pairs of integers (m,n)(m,n) satisfying: 3×2m+1=n23\times 2^m + 1= n^2.

Q10)a) Prove that if x2+pxqx^2+px-q and x2px+qx^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 x2+pxq=(xα)(xβ) and x2px+q=(xγ)(xδ)x^2+px-q = (x-\alpha)(x-\beta) \ and \ x^2-px+q = (x-\gamma)(x-\delta) where p,q,α,β,γ,δp,q,\alpha,\beta,\gamma,\delta are integers, α,β,γ,δ\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
4 years ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Q4) Let f(x)=(1+x)n=(n0)+(n1)x++(nn)xnf(x) = (1+x)^{n} ={n \choose 0}+ {n \choose 1} x + \ldots + {n \choose n} x^{n}

Now, f(1)+f(ω)+f(ω2)=(n0)(1+ω+ω2)+(n1)(1+ω2+ω4)++(nn)(1+ωn+ω2n)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((n0)+(n3)+(n6)+)= 3 \left( {n \choose 0} + {n \choose 3}+ {n \choose 6} + \ldots \right)

So, (n0)+(n3)+(n6)+=f(1)+f(ω)+f(ω2)3=2n+(ω2)n+(ω)n3{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 nn.

Surya Prakash - 4 years ago

Log in to reply

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

Log in to reply

Q1) It is clear that p(p1)=2(y+x)(yx)p(p-1) = 2(y+x)(y-x). If pyxp|y-x it implies that pyxp \leq y-x, then it should imply that p12(y+x)p-1 \geq 2(y+x). But it is not possible. So, pp does not divide yxy-x. So, py+xp|y+x it implies that py+xp \leq y+x, but this implies that p12(yx)p-1 \geq 2(y-x). So, we get p+14xp +1 \leq 4x. Which reduces to 2x24x2x^2 \leq 4x, x2x \leq 2. If x=1x=1, then p=1p=1, a contradiction. And x=2x=2 implies that p=7p=7. Therefore, p=7p=\boxed{7} is the only solution.

Surya Prakash - 4 years ago

Log in to reply

For the 2nd2nd question,

Let the roots of first equation be md,m,m+dm-d,m,m+d

Therefore we get that, md+m+m+d=3a=3mm-d+m+m+d=3a=3m. Hence m=am=a

Then by applying Vieta's formula for other two, we get that, 3m2d2=b3{m}^{2}-{d}^{2}=b and m3md2=18c {m}^{3}-m{d}^{2}=-18c

Let the roots for the second equation be n/r,n,nrn/r,n,nr

Therefore by applying Vieta's fomula,we get that, n=cn=c and n(1/r+1+r)=bn(1/r+1+r)=-b and n2(1/r+1+r)=1 {n}^{2}(1/r+1+r)=1

From here we get that, n=1/b=cn=-1/b=c

Substituting these values in the previously acquired equations we get that, m3m(3m2b)=18/b{m}^{3}-m(3{m}^{2}-b)=18/b

Simplifying we get that, ab22a3b18=0a{b}^{2}-2{a}^{3}b-18=0

Applying formula for quadratic roots, b=a3±a6+18aab=\frac{{a}^{3}\pm \sqrt{{a}^{6}+18a}}{a}

For bb to be an integer, a6+18a=k2{a}^{6}+18a={k}^{2}

We see that, a3<k<a3+3 {a}^{3}<k<{a}^{3}+3

From here we see that, a=2a=2 only satisfies. Therefore the required solutions are a=2a=2 and b=9b=9.

Saarthak Marathe - 4 years ago

Log in to reply

My solution for the 3rd3rd one is quite big. I will just give some hints to solve it.

  1. Try it by co-ordinate geometry.

  2. Take BB as (1/2,0)(-1/2,0) and CC as (1/2,0)(1/2,0).Therefore,we get AA as (3/2,0)(\sqrt{3}/2,0).

  3. Then take PP as(x,y)(x,y) use the provided equation of PA2=PB2+PC2{PA}^{2}={PB}^{2}+{PC}^{2} to get the equation of locus of PP.The equation comes out to be x2+y2+3y1/4=0{x}^{2}+{y}^{2}+\sqrt{3}y-1/4=0

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

  5. By simplifying the equation obtained in the 4th4th step, we get that BPC=150\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 - 4 years ago

Log in to reply

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

Solution to 3rd3rd stuff ConstructionConstruction: Draw PXPX = PCPC such that CPX=60\angle CPX = 60^{\circ}. Finally join BXBX and CXCX.

Clearly ΔCPX\Delta CPX is equilateral. => CXCX = PCPC ------ 11 and Also, PCX=ACB\angle PCX = \angle ACB = 6060^{\circ} => PCXPCB=ACBPCB\angle PCX - \angle PCB = \angle ACB - \angle PCB => BCX=ACP\angle BCX = \angle ACP ------- 22

From the results in eqneq^{n} 11, 22 and that BCBC = ACAC, we conclude that ΔBCXΔACP\Delta BCX \cong \Delta ACP. => BXBX = PAPA

But GivenGiven PB2+PC2PB^{2} + PC^{2} = PA2PA^{2}

=> PB2+PX2PB^{2} + PX^{2} = BX2BX^{2} ............... [PCPC = PXPX and BX=PABX = PA]

Therefore by the converse of Pythagoras...theoremPythagoras... theorem, it is obvious that BPX\angle BPX = 9090^{\circ}

Thus the required angle = BPC\angle BPC = BPX+CPX\angle BPX + \angle CPX = 9090^{\circ} + 6060^{\circ} = 150150^{\circ}.

The almighty knows it all. - 2 years, 6 months ago

Log in to reply

1) p = 7

p(p1)=2(y+x)(yx)p(p - 1) = 2(y + x)(y - x)

comparing of y + x and p gives p = 7

Dev Sharma - 4 years ago

Log in to reply

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

Nihar Mahajan - 4 years ago

Log in to reply

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

Dev Sharma - 4 years ago

Log in to reply

@Dev Sharma Include this in your previous comment and complete your solution. Don't make parts of your solution.

Nihar Mahajan - 4 years ago

Log in to reply

@Dev Sharma Kindly always post a full solution.

Mehul Arora - 4 years ago

Log in to reply

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

Multiplying we get a2b+b2c+c2a3abc=abc(ac+ba+cb3a^2b+b^2c+c^2a-3abc= abc(\frac {a}{c}+ \frac {b}{a} + \frac {c}{b}-3).

By AM-GM inequality, ac+ba+cb3\frac {a}{c}+ \frac {b}{a} + \frac {c}{b} \ge 3.

Therefore, minimum value of the expression ac+ba+cb\frac {a}{c}+ \frac {b}{a} + \frac {c}{b} is 33.

Therefore, minimum value of ab(ac)+bc(ba)+ca(cb)ab(a-c)+bc(b-a)+ca(c-b) is 00 and equality occurs when a=b=c=0a=b=c=0.

Log in to reply

@Surya Prakash @Nihar Mahajan @Svatejas Shivakumar There seems to be a blunder in problem statement of Q4). I guess it meant " Simplify (n0)+(n3)+(n6)+(n9)+...(n3n3) \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 (nn) \binom{n}{n} doesn't make sense at all if n n is not a multiple of 3 3 .

Karthik Venkata - 4 years ago

Log in to reply

Its correct.

Saarthak Marathe - 4 years ago

Log in to reply

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

Karthik Venkata - 4 years ago

Log in to reply

@Karthik Venkata It satisfies.

Saarthak Marathe - 4 years ago

Log in to reply

@Saarthak Marathe Let f(n)=(n0)+(n3)+(n6)+(n9)+...(nn) f(n) = \large\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \binom{n}{9} + ... \binom{n}{n}.Then f(4)=(40)+(43)+(44) f(4) = \binom{4}{0} + \binom{4}{3} + \binom{4}{4} ? I am certainly sure about the last summand being (n3n3) \binom{n}{ 3 \cdot \lfloor \frac{n}{3} \rfloor } , contrary to (nn) \binom{n}{n} mentioned in the question.

Karthik Venkata - 4 years ago

Log in to reply

@Karthik Venkata In that way!!!!! Then it is understood that it should be 3.[n3]3.[\frac{n}{3}]

Saarthak Marathe - 4 years ago

Log in to reply

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

Karthik Venkata - 4 years ago

Log in to reply

1) Solve y3=x3+8x26x+8y^3 = x^3 + 8x^2 - 6x + 8 for positive integer x and y.

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

3) Find all real a for which x42ax2+x+a2ax^4-2ax^2+x+a^2- a has all real root

Dev Sharma - 4 years ago

Log in to reply

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

Harsh Shrivastava - 3 years, 10 months ago

Log in to reply

Question5:Question 5: Const:Const: The constructions are clearly shown in the above diagram.diagram. Join BMBM, DMDM, CMC'M ,finally join AMA'M and produce it to meet CCCC' at BB'.

ABAB is the diameter of the ABCD.\odot ABCD.

=> ABC\angle ABC = ADC\angle ADC = 9090^{\circ} => BMBM = DMDM [= 1/2AC1/2AC ] => MBC\angle MBC' = MDC\angle MDC' --- 11

Clearly, ΔAAMΔCBM\Delta AA'M \cong \Delta CB'M => AMA'M = BMB'M => MM is the midpoint of ABA'B'.

But ΔABC\Delta A'B'C' is right angled at CC' and also MM is the midpoint of AB.A'B'. => CMC'M = AMA'M ---- 22

=> MCB\angle MC'B = MDA\angle MDA' ----- 33

From the eqneq^{n} 11, 22 and 33: ΔMCBΔMDA\Delta MC'B \cong \Delta MDA' => BCBC' = AD.A'D.

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

The almighty knows it all. - 2 years, 6 months ago

Log in to reply

If x5x3+x=ηx^{5}-x^{3}+x=\eta, the minimum value of x6x^{6} in terms of η\eta.

Shivam Jadhav - 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...