Another easy one

[fetofs]

Could anyone enlighten me with the solution:

Find all integer solutions of the equation and explain why your answer is correct.

[victor]

Originally Posted by fetofs

Could anyone enlighten me with the solution:

Find all integer solutions of the equation and explain why your answer is correct.

therefore

or

[S485122]

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 :-(

[Wacky]

But don't leave out the x=y part. As noted, the relationship holds for any integer k and x = y = k.

[S485122]

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:

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 :

Originally Posted by Jacob Visser

the easy part is x=y and needs no explaining

[Wacky]

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,

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,

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

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.

[S485122]

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?)

[fetofs]

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.

[mfgoode]

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

[fetofs]

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.

[mfgoode]

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