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 2006-10-29, 12:28 #1 fetofs     Aug 2005 Brazil 2·181 Posts Another easy one Could anyone enlighten me with the solution: Find all integer solutions of the equation $x^3-y^3=3(x^2-y^2)$ and explain why your answer is correct.
2006-10-29, 15:29   #2
victor

Oct 2005
Fribourg, Switzerlan

22×32×7 Posts

Quote:
 Originally Posted by fetofs Could anyone enlighten me with the solution: Find all integer solutions of the equation $x^3-y^3=3(x^2-y^2)$ and explain why your answer is correct.
$\large{x^3-y^3=3(x^2-y^2)}$
$\large{(x-y)(x^2+x\cdot y+y^2)=3(x+y)(x-y)}$
$\large{x^2+xy+y^2=3(x+y)}$
therefore
$\large{x=\frac{sqrt{3-y}sqrt{3(y+1)}-y+3}{2}}$
or
$\large{x=\frac{-(sqrt{3-y}sqrt{3(y+1)}+y-3)}{2}}$

Last fiddled with by victor on 2006-10-29 at 15:29 Reason: thereforE<-

 2006-10-29, 16:16 #3 S485122     Sep 2006 Brussels, Belgium 31378 Posts the easy part is x=y and needs no explaining This simplifies the problem to find the integer solutions of x2+y2+xy-3x-3y=0 or y2+(x-3)y+x2-3x=0 Wich has two solutions: y=(-x+3+sqrt((x-3)2-4(x2-3x)))/2 and y=(-x+3+sqrt((x-3)2-4(x2-3x)))/2 the determinant must be positive, thus (x-3)2-4(x2-3x)=-3x2+2x+3 >= 0 Which implies that x is bounded by -1 and 3 The integer solutions are {-1,2}, {0,3}, {2,-1} and {3,0} and of course the solution [0,0] I must learn to use tex :-( Last fiddled with by S485122 on 2006-10-29 at 16:17
 2006-10-29, 18:23 #4 Wacky     Jun 2003 The Texas Hill Country 32·112 Posts But don't leave out the x=y part. As noted, the relationship holds for any integer k and x = y = k. Last fiddled with by Wacky on 2006-10-29 at 18:23
2006-10-29, 19:21   #5
S485122

Sep 2006
Brussels, Belgium

110010111112 Posts

Quote:
 Originally Posted by Jacob Visser y=(-x+3+sqrt((x-3)2-4(x2-3x)))/2 and y=(-x+3+sqrt((x-3)2-4(x2-3x)))/2 the determinant must be positive, thus (x-3)2-4(x2-3x)=-3x2+2x+3 >= 0
Miscutting and pasting and completely wrong calculation on my part :-(
(I had the values already and just needed to justify them ;-)

It should be
y=(-x+3+sqrt((x-3)2-4(x2-3x)))/2
and
y=(-x+3-sqrt((x-3)2-4(x2-3x)))/2

(x-3)2-4(x2-3x)=)=-3x2+6x+9
and this is non negative for x larger or equal to -1 and less or equal to +3.

As for:
Quote:
 Originally Posted by Wacky But don't leave out the x=y part. As noted, the relationship holds for any integer k and x = y = k.
It was covered in my first sentence :
Quote:
 Originally Posted by Jacob Visser the easy part is x=y and needs no explaining

Last fiddled with by S485122 on 2006-10-29 at 19:29

2006-10-29, 20:00   #6
Wacky

Jun 2003
The Texas Hill Country

32×112 Posts

Quote:
 Originally Posted by Jacob Visser It was covered in my first sentence :
That is only because of the way that you interpret your loose usage of the language --
Quoting directly from your text,
Quote:
 The integer solutions are {-1,2}, {0,3}, {2,-1} and {3,0} and of course the solution [0,0]
.
Clearly, that is the ONLY place where you claim any solution(s).

Further, you had stated,
Quote:
 This simplifies the problem to find the integer solutions of x2+y2+xy-3x-3y=0 or y2+(x-3)y+x2-3x=0
.

No where do you state that it is the "OR" rather than the "AND" of the two conditions.

So, I do not think it to be a totally unreasonable reading of your statements to be that there is only one solution
Quote:
 the solution [0,0]
Please understand that my complaint is only with the method that you chose to present the "solutions". Because of multiple possible interpretations of the words, your answer lacks clarity.

Last fiddled with by Wacky on 2006-10-29 at 20:01

 2006-10-29, 21:44 #7 S485122     Sep 2006 Brussels, Belgium 7·233 Posts Sorry, as you will have understood English is not my mother tongue. Especially not for mathematics. And I must admit that restarting to do computations after 33 years does not go very smoothly. I indeed used "or" where I meant "which can also be written as". My presentation of the solutions was indeed sloppy, I should have repeated the first set of solutions (any integer x with y=x), plus the solutions of the x2+y2+xy-3x-3y=0 part. Finally I mentioned the {0,0} pair since it is not only a solution of the x=y part but also of the x2+y2+xy-3x-3y=0 part (a double solution?)
 2006-10-30, 12:32 #8 fetofs     Aug 2005 Brazil 2·181 Posts Hi again! These problems are from BMO (last weekend), and they only publish the solutions 2 years after the competition, and I'm really curious. So, if you could solve the other problem I couldn't do, I'd appreciate it (the picture is attached). Note that angle B (ABC) = 70º, AM=BM, AN=CN, AR=HR, H is the orthocenter, and it's asking for angle MNR. Attached Thumbnails
 2006-11-02, 16:01 #9 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22×33×19 Posts An easy one. Well fetofs I have come up with a very elegant derivation of angle MNR Which is required. In triangle ABC drop a perpendicular from C to AB and call it P. Drop a perpendicular from A to BC intersecting CP at H (the orthocentre) In triangle PBC, angle PBC = 70* (given) Ang. BPC = 90* (construction) Therefore Ang. BCP = 20* Now line MN ( M and N being midpoints) is parallel (//) to base BC in Triangle ABC. In triangle AHC , R and N are midpoints. Therefore RH is // to HC Therefore Ang. MNR = Ang. BCP = 20* Because of being angle between //’s. Q.E.D. Mally
2006-11-02, 17:38   #10
fetofs

Aug 2005
Brazil

2×181 Posts

Quote:
 Originally Posted by mfgoode In triangle AHC , R and N are midpoints. Therefore RH is // to HC
A typo, perhaps? RH and HC should make an X, unless I miss something.

P.S: A clearer drawing could've helped me see some things. The triangle AMN is similar to triangle ABC, in the sense that is only scaled down. AM/2 = AB, MN/2 = BC and AN = AC/2, therefore their angles are equal.

Last fiddled with by fetofs on 2006-11-02 at 17:44

2006-11-03, 03:29   #11
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
TYPO.

Quote:
 Originally Posted by fetofs A typo, perhaps? RH and HC should make an X, unless I miss something. P.S: A clearer drawing could've helped me see some things. The triangle AMN is similar to triangle ABC, in the sense that is only scaled down. AM/2 = AB, MN/2 = BC and AN = AC/2, therefore their angles are equal.
:smile

Yes you are right fetofs. It was meant to be RN // HC. Hence Triangle ARN is similar to triangle AHC.

Agreed: A good figure (36-24-36 inclusive!) is half the solution to the problem.

Your proportions are entirely wrong; AB = 2 AM, BC = 2 MN and AC = 2 AN
This similarity does not get you very far.

Given details are never superfluous and you must make use of the given orthocentre. Hence I have used the triangle AHC also which is important for the proof.

Im sorry I dont know how to make diagrams on the pc. Otherwise I would have given one for clarity.

Regards, and feel free to send some more problems from BMO (whatever that is)

Mally

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