Difference between revisions of "1972 USAMO Problems/Problem 1"

Revision as of 20:53, 31 March 2014

Problem

The symbols $(a,b,\ldots,g)$ and $[a,b,\ldots, g]$ denote the greatest common divisor and least common multiple, respectively, of the positive integers $a,b,\ldots, g$ . For example, $(3,6,18)=3$ and $[6,15]=30$ . Prove that

$\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$ .

Solutions

Solution 1

As all of the values in the given equation are positive, we can invert it to get an equivalent equation:

$\frac{[a,b][b,c][c,a]}{[a,b,c]^2}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}$ .

We will now prove that both sides are equal, and that they are integers.

Consider an arbitrary prime $p$ . Let $p^\alpha$ , $p^\beta$ , and $p^\gamma$ be the greatest powers of $p$ that divide $a$ , $b$ , and $c$ . WLOG let $\alpha \leq \beta \leq \gamma$ .

We can now, for each of the expressions in our equation, easily determine the largest power of $p$ that divides it. In this way we will find that the largest power of $p$ that divides the left hand side is $\beta+\gamma+\gamma-2\gamma = \beta$ , and the largest power of $p$ that divides the right hand side is $\alpha + \beta + \alpha - 2\alpha = \beta$ . $\blacksquare$

Solution 2

Let $p = (a, b, c)$ , $pq = (a, b)$ , $pr = (b, c)$ , and $ps = (c, a)$ . Then it follows that $q, r, s$ are pairwise coprime and $a = pqsa'$ , $b = pqrb'$ , and $c = prsc'$ , with $a', b', c'$ pairwise coprime as well. Then, we wish to show

\[\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} &= \frac{p^2}{(pq)(pr)(ps)},\] (Error compiling LaTeX. Unknown error_msg)

which can be checked fairly easily.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 3: / Line 3: @@
 <center><math>\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}</math>.</center>
-==Solution==
+==Solutions==
+===Solution 1===
 As all of the values in the given equation are positive, we can invert it to get an equivalent equation:
@@ Line 14: / Line 16: @@
 We can now, for each of the expressions in our equation, easily determine the largest power of <math>p</math> that divides it. In this way we will find that the largest power of <math>p</math> that divides the left hand side is <math>\beta+\gamma+\gamma-2\gamma = \beta</math>, and the largest power of <math>p</math> that divides the right hand side is <math>\alpha + \beta + \alpha - 2\alpha = \beta</math>. <math>\blacksquare</math>
+===Solution 2===
+Let <math>p = (a, b, c)</math>, <math>pq = (a, b)</math>, <math>pr = (b, c)</math>, and <math>ps = (c, a)</math>. Then it follows that <math>q, r, s</math> are pairwise coprime and <math>a = pqsa'</math>, <math>b = pqrb'</math>, and <math>c = prsc'</math>, with <math>a', b', c'</math> pairwise coprime as well. Then, we wish to show
+<cmath>\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} &= \frac{p^2}{(pq)(pr)(ps)},</cmath>
+which can be checked fairly easily.
 {{alternate solutions}}

1972 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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