Another easy one
Could anyone enlighten me with the solution:
Find all integer solutions of the equation [tex]x^3y^3=3(x^2y^2)[/tex] and explain why your answer is correct. 
[QUOTE=fetofs;90221]Could anyone enlighten me with the solution:
Find all integer solutions of the equation [tex]x^3y^3=3(x^2y^2)[/tex] and explain why your answer is correct.[/QUOTE] [tex]\large{x^3y^3=3(x^2y^2)}[/tex] [tex]\large{(xy)(x^2+x\cdot y+y^2)=3(x+y)(xy)}[/tex] [tex]\large{x^2+xy+y^2=3(x+y)}[/tex] therefore [tex]\large{x=\frac{sqrt{3y}sqrt{3(y+1)}y+3}{2}}[/tex] or [tex]\large{x=\frac{(sqrt{3y}sqrt{3(y+1)}+y3)}{2}}[/tex] :smile: 
the easy part is x=y and needs no explaining
This simplifies the problem to find the integer solutions of x[sup]2[/sup]+y[sup]2[/sup]+xy3x3y=0 or y[sup]2[/sup]+(x3)y+x[sup]2[/sup]3x=0 Wich has two solutions: y=(x+3+sqrt((x3)[sup]2[/sup]4(x[sup]2[/sup]3x)))/2 and y=(x+3+sqrt((x3)[sup]2[/sup]4(x[sup]2[/sup]3x)))/2 the determinant must be positive, thus (x3)[sup]2[/sup]4(x[sup]2[/sup]3x)=3x[sup]2[/sup]+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 :( 
But don't leave out the x=y part. As noted, the relationship holds for any integer k and x = y = k.

[QUOTE=Jacob Visser;90232]y=(x+3+sqrt((x3)24(x23x)))/2
and y=(x+3+sqrt((x3)24(x23x)))/2 the determinant must be positive, thus (x3)[sup]2[/sup]4(x[sup]2[/sup]3x)=3x[sup]2[/sup]+2x+3 >= 0[/QUOTE] 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((x3)24(x23x)))/2 and y=(x+3sqrt((x3)24(x23x)))/2 (x3)[sup]2[/sup]4(x[sup]2[/sup]3x)=)=3x[sup]2[/sup]+6x+9 and this is non negative for x larger or equal to 1 and less or equal to +3. As for: [QUOTE=Wacky;90244]But don't leave out the x=y part. As noted, the relationship holds for any integer k and x = y = k.[/QUOTE] It was covered in my first sentence : [QUOTE=Jacob Visser;90232]the easy part is x=y and needs no explaining[/QUOTE] 
[QUOTE=Jacob Visser;90249]It was covered in my first sentence :[/QUOTE]
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] [/QUOTE]. 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+xy3x3y=0 or y2+(x3)y+x23x=0[/QUOTE]. 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] [/QUOTE] 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. 
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 x[sup]2[/sup]+y[sup]2[/sup]+xy3x3y=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 x[sup]2[/sup]+y[sup]2[/sup]+xy3x3y=0 part (a double solution?) 
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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.

An easy one.
:smile:
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 :coffee: 
[QUOTE=mfgoode;90515]
In triangle AHC , R and N are midpoints. Therefore RH is // to HC [/QUOTE] 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. 
TYPO.
[QUOTE=fetofs;90520]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.[/QUOTE] :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 (362436 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 :coffee: 
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