## Thursday, March 9, 2006

### It's The Real Deal. Topic: Algebra. Level: Olympiad.

Problem: (360 Problems for Mathematical Contests - 1.1.18) Find the real solutions to the system of equations

$\frac{1}{x}+\frac{1}{y} = 9$
$\left(\frac{1}{\sqrt[3]{x}}+\frac{1}{\sqrt[3]{y}}\right) \left(1+\frac{1}{\sqrt[3]{x}}\right) \left(1+\frac{1}{\sqrt[3]{y}}\right) = 18$.

Solution: To simplify things, make the substitution

$a = \frac{1}{\sqrt[3]{x}}$, $b = \frac{1}{\sqrt[3]{y}}$.

We now have the system

$a^3+b^3 = (a+b)(a^2-ab+b^2) = 9$
$(a+b)(1+a)(1+b) = (a+b)(1+a+b+ab) = 18$.

Letting $p = a+b$, $q = ab$, it becomes

$p(p^2-3q) = 9$ (1)
$p(1+p+q) = 18$ (2)

Adding three times (2) to (1), we have

$p(p^2+3p+3) = 63$,

or

$(p-3)(p^2+6p+21) = 0$ (3)

Since the discriminant of $p^2+6p+21$ is $6^2-4(21) < 0$, the only real solution is $p = 3$. Then from (1), we have

$3(9-3q) = 9 \Rightarrow q = 2$.

So $ab = 2 \Rightarrow a = \frac{2}{b}$. So

$p = \frac{2}{b}+b = 3 \Rightarrow b^2-3b+2 = (b-1)(b-2) = 0$.

Then $b = 1, 2$ giving the solutions

$(a,b) = (1,2); (2,1)$, which become

$(x,y) = \left(1, \frac{1}{8}\right); \left(\frac{1}{8},1\right)$. QED.

--------------------

Comment: The only really hard part about this problem was the ugliness of the original problem. Making the right substitutions reduces it to simple algebra and solving a cubic (with an easy root if you saw it).

--------------------

Practice Problem: Prove that the equation $a^2+4b^2+4ab+4a+8b+5 = 0$ has no solution in the reals.

1. Hey this is kind of random but what kinds of things do you learn at PROMYS? I might go. The problems on the quiz thingy aren't very hard.

2. Hey for that problem you can just view it as a quadratic in a (or b obviously) and the discrimant is less than 0.

3. Where'd you get that book? Did you order it directly from Mr. Andreescu?

4. Well...

At PROMYS, I basically learned all my Number Theory. It also helped build up my ability to write rigorous proofs (as in rigorous to the point of stating associativity, commutivity, etc.). It's a really good program and you'll meet some really cool people, so I'd definitely suggest applying.

As for the book, I just ordered it online, I think... it's a good book with problems of many types as well as solutions in case you can't figure something out.

5. I couldn't find it online, at least on amazon... where did you get it from, do you remember?

6. Ah, I remember now. I got the book as a USAMTS prize last year. I can't find it anywhere online, though!

7. Well thanks anyway. :)