20061029, 12:28  #1 
Aug 2005
Brazil
552_{8} Posts 
Another easy one
Could anyone enlighten me with the solution:
Find all integer solutions of the equation and explain why your answer is correct. 
20061029, 15:29  #2  
Oct 2005
Fribourg, Switzerlan
2^{2}·3^{2}·7 Posts 
Quote:
therefore or Last fiddled with by victor on 20061029 at 15:29 Reason: thereforE< 

20061029, 16:16  #3 
Sep 2006
Brussels, Belgium
13×127 Posts 
the easy part is x=y and needs no explaining
This simplifies the problem to find the integer solutions of x^{2}+y^{2}+xy3x3y=0 or y^{2}+(x3)y+x^{2}3x=0 Wich has two solutions: y=(x+3+sqrt((x3)^{2}4(x^{2}3x)))/2 and y=(x+3+sqrt((x3)^{2}4(x^{2}3x)))/2 the determinant must be positive, thus (x3)^{2}4(x^{2}3x)=3x^{2}+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 20061029 at 16:17 
20061029, 18:23  #4 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} 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 20061029 at 18:23 
20061029, 19:21  #5  
Sep 2006
Brussels, Belgium
673_{16} Posts 
Quote:
(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)^{2}4(x^{2}3x)=)=3x^{2}+6x+9 and this is non negative for x larger or equal to 1 and less or equal to +3. As for: Quote:
Last fiddled with by S485122 on 20061029 at 19:29 

20061029, 20:00  #6  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
That is only because of the way that you interpret your loose usage of the language 
Quoting directly from your text, Quote:
Clearly, that is the ONLY place where you claim any solution(s). Further, you had stated, 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:
Last fiddled with by Wacky on 20061029 at 20:01 

20061029, 21:44  #7 
Sep 2006
Brussels, Belgium
13·127 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 x^{2}+y^{2}+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^{2}+y^{2}+xy3x3y=0 part (a double solution?) 
20061030, 12:32  #8 
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.

20061102, 16:01  #9 
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} 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 
20061102, 17:38  #10  
Aug 2005
Brazil
362_{10} Posts 
Quote:
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 20061102 at 17:44 

20061103, 03:29  #11  
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
TYPO.
Quote:
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 

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