mersenneforum.org (https://www.mersenneforum.org/index.php)
-   Puzzles (https://www.mersenneforum.org/forumdisplay.php?f=18)
-   -   Another easy one (https://www.mersenneforum.org/showthread.php?t=6525)

 fetofs 2006-10-29 12:28

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.

 victor 2006-10-29 15:29

[QUOTE=fetofs;90221]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.[/QUOTE]

$$\large{x^3-y^3=3(x^2-y^2)}$$
$$\large{(x-y)(x^2+x\cdot y+y^2)=3(x+y)(x-y)}$$
$$\large{x^2+xy+y^2=3(x+y)}$$
therefore
$$\large{x=\frac{sqrt{3-y}sqrt{3(y+1)}-y+3}{2}}$$
or
$$\large{x=\frac{-(sqrt{3-y}sqrt{3(y+1)}+y-3)}{2}}$$
:smile:

 S485122 2006-10-29 16:16

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]+xy-3x-3y=0 or y[sup]2[/sup]+(x-3)y+x[sup]2[/sup]-3x=0

Wich has two solutions:
y=(-x+3+sqrt((x-3)[sup]2[/sup]-4(x[sup]2[/sup]-3x)))/2
and
y=(-x+3+sqrt((x-3)[sup]2[/sup]-4(x[sup]2[/sup]-3x)))/2

the determinant must be positive, thus (x-3)[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 :-(

 Wacky 2006-10-29 18:23

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

 S485122 2006-10-29 19:21

[QUOTE=Jacob Visser;90232]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)[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((x-3)2-4(x2-3x)))/2
and
y=(-x+3-sqrt((x-3)2-4(x2-3x)))/2

(x-3)[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]

 Wacky 2006-10-29 20:00

[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 --
[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+xy-3x-3y=0 or y2+(x-3)y+x2-3x=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.

 S485122 2006-10-29 21:44

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]+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 x[sup]2[/sup]+y[sup]2[/sup]+xy-3x-3y=0 part (a double solution?)

 fetofs 2006-10-30 12:32

1 Attachment(s)
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 2006-11-02 16:01

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:

 fetofs 2006-11-02 17:38

[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.

 mfgoode 2006-11-03 03:29

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 (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 :coffee:

 All times are UTC. The time now is 15:22.