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

(Solution 3)
Line 48: Line 48:
{{alternate solutions}}
{{alternate solutions}}
===Solution 3===
==See Also==
==See Also==

Revision as of 01:27, 19 February 2019


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



Solution 1

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


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 \[\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)},\] which can be checked fairly easily.

Solution 3 (very long, but detailed)

Note: This solution is more of a "bashy" solution, and is easier to think of than the first 2 solutions.

Dividing both sides of the equation by $(a,b,c)^2$, and then multiplying both sides by $[a,b][b,c][c,a]$ gives us \[\left(\frac{[a,b,c]}{(a,b,c)}\right)^2 = \frac{[a,b][b,c][c,a]}{(a,b)(b,c)(c,a)}.\] Now, we look at the prime factorisations of $a,b,c$. Let \[a=2^{x_1}\cdot 3^{x_2}\cdot 5^{x_3}\cdots,\] \[b=2^{y_1}\cdot 3^{y_2}\cdot 5^{y_3}\cdots,\] \[c=2^{z_1}\cdot 3^{z_2}\cdot 5^{z_3}\cdots,\] where all of the $x_1, y_1, z_1$ are nonnegative integers (they could be 0). Thus, we have \[[a,b,c]=2^{\max(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3)} \cdots\] and \[(a,b,c)=2^{\min(x_1,y_1,z_1)} \cdot 3^{\min(x_2,y_2,z_2)} \cdot 5^{\min(x_3,y_3,z_3)} \cdots.\] Now, we see that $\left(\frac{[a,b,c]}{(a,b,c)}\right) = 2^{\max(x_1,y_1,z_1) - \min(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2) - \min(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3) - \min(x_3,y_3,z_3)} \cdots.$ Squaring the LHS will just double all the exponents on the RHS.

Also, we have $[a,b]=2^{\max(x_1,y_1)} \cdot 3^{\max(x_2,y_2)} \cdot 5^{\max(x_3,y_3)} \cdots$. We can also build similar equations for $[b,c],[c,a],(a,b),(b,c)(c,a)$, using 2 of $x,y,z$ and either the minimum of the exponents or the maximum. We see that when we multiply $[a,b]$, $[b,c]$, and $[a,c]$ together, the exponents of the prime factorization will be $\max(x_i,y_i)+ \max(y_i,z_i)+ \max(z_i,x_1)$, where $i$ is chosen for the $i$'th prime. When we multiply $(a,b)$, $(b,c)$, and $(a,c)$ together, the exponents of the prime factorization will be $\min(x_i,y_i)+ \min(y_i,z_i)+ \min(z_i,x_1)$, where $i$ is chosen for the $i$'th prime.

Thus, when we divide the former by the latter, we have $\max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))$. We wish to prove that this is equal to the exponents we got from $([a,b,c]/(a,b,c))^2$.

In other words, this problem is now down to proving that \[2(\max(x_i,y_i,z_i) - \min(x_i,y_i,z_i)) = \max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))\] for any nonnegative integers $x_i, y_i, z_i$. WLOG, let $x_i \le y_i \le z_i$. Thus, computing maximums and minimums, this equation turns into \[2(z_i - x_i) = y_i + z_i + z_i - x_i - y_i - x_i.\]

Finally, canceling out the $y_i$'s and collecting like terms, we see the RHS is $2z_i - 2x_i$, which is equal to the LHS. Thus, this equation is true, and we are done. $\square$

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

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS