Difference between revisions of "1972 USAMO Problems/Problem 1"
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<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> | <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: | As all of the values in the given equation are positive, we can invert it to get an equivalent equation: | ||
Line 12: | Line 14: | ||
WLOG let <math>\alpha \leq \beta \leq \gamma</math>. | WLOG let <math>\alpha \leq \beta \leq \gamma</math>. | ||
− | 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>, q. | + | 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. | ||
+ | |||
+ | ===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 <math>(a,b,c)^2</math>, and then multiplying both sides by <math>[a,b][b,c][c,a]</math> gives us <cmath>\left(\frac{[a,b,c]}{(a,b,c)}\right)^2 = \frac{[a,b][b,c][c,a]}{(a,b)(b,c)(c,a)}.</cmath> | ||
+ | Now, we look at the prime factorisations of <math>a,b,c</math>. Let | ||
+ | <cmath>a=2^{x_1}\cdot 3^{x_2}\cdot 5^{x_3}\cdots,</cmath> | ||
+ | <cmath>b=2^{y_1}\cdot 3^{y_2}\cdot 5^{y_3}\cdots,</cmath> | ||
+ | <cmath>c=2^{z_1}\cdot 3^{z_2}\cdot 5^{z_3}\cdots,</cmath> | ||
+ | where all of the <math>x_1, y_1, z_1</math> are nonnegative integers (they could be 0). | ||
+ | Thus, we have <cmath>[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</cmath> | ||
+ | and <cmath>(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.</cmath> | ||
+ | Now, we see that <math>\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.</math> Squaring the LHS will just double all the exponents on the RHS. | ||
+ | |||
+ | |||
+ | Also, we have <math>[a,b]=2^{\max(x_1,y_1)} \cdot 3^{\max(x_2,y_2)} \cdot 5^{\max(x_3,y_3)} \cdots</math>. We can also build similar equations for <math>[b,c],[c,a],(a,b),(b,c)(c,a)</math>, using 2 of <math>x,y,z</math> and either the minimum of the exponents or the maximum. | ||
+ | We see that when we multiply <math>[a,b]</math>, <math>[b,c]</math>, and <math>[a,c]</math> together, the exponents of the prime factorization will be <math>\max(x_i,y_i)+ \max(y_i,z_i)+ \max(z_i,x_1)</math>, where <math>i</math> is chosen for the <math>i</math>'th prime. | ||
+ | When we multiply <math>(a,b)</math>, <math>(b,c)</math>, and <math>(a,c)</math> together, the exponents of the prime factorization will be <math>\min(x_i,y_i)+ \min(y_i,z_i)+ \min(z_i,x_1)</math>, where <math>i</math> is chosen for the <math>i</math>'th prime. | ||
+ | |||
+ | Thus, when we divide the former by the latter, we have <math>\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))</math>. We wish to prove that this is equal to the exponents we got from <math>([a,b,c]/(a,b,c))^2</math>. | ||
+ | |||
+ | In other words, this problem is now down to proving that <cmath>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))</cmath> for any nonnegative integers <math>x_i, y_i, z_i</math>. WLOG, let <math>x_i \le y_i \le z_i</math>. Thus, computing maximums and minimums, this equation turns into <cmath>2(z_i - x_i) = y_i + z_i + z_i - x_i - y_i - x_i.</cmath> | ||
+ | |||
+ | Finally, canceling out the <math>y_i</math>'s and collecting like terms, we see the RHS is <math>2z_i - 2x_i</math>, which is equal to the LHS. Thus, this equation is true, and we are done. <math>\square</math> | ||
+ | |||
+ | ===Solution 4, Very Simple=== | ||
+ | Let | ||
+ | <cmath>a=dzxA</cmath><cmath>b=dxyB</cmath><cmath>c=dyzC</cmath>where <math>\gcd(A,B,C)=\gcd(x,y,z)=1</math>. Then, <math>\operatorname{lcm}(a,b,c)^2=(dxyzABC)^2</math> and <math>\operatorname{lcm}(a,b)=dxyzAB</math>. Using this cyclically, we have | ||
+ | <cmath>\text{LHS}=\frac{d^2x^2y^2z^2A^2B^2C^2}{(dxyzAB)(dxyzBC)(dxyzCA)}=\frac{1}{dxyz}.</cmath>Now, note that <math>\gcd(a,b,c)^2=d^2</math> and <math>\gcd(a,b)=dx</math>. Using this cyclically, we have | ||
+ | <cmath>\text{RHS}=\frac{d^2}{(dx)(dy)(dz)}=\frac{1}{dxyz}.</cmath> | ||
+ | -brainiacmaniac31 | ||
+ | |||
+ | |||
+ | {{alternate solutions}} | ||
− | ==See | + | ==See Also== |
{{USAMO box|year=1972|before=First Question|num-a=2}} | {{USAMO box|year=1972|before=First Question|num-a=2}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 12:28, 1 January 2021
Contents
Problem
The symbols and denote the greatest common divisor and least common multiple, respectively, of the positive integers . For example, and . Prove that
Solutions
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 . Let , , and be the greatest powers of that divide , , and . WLOG let .
We can now, for each of the expressions in our equation, easily determine the largest power of that divides it. In this way we will find that the largest power of that divides the left hand side is , and the largest power of that divides the right hand side is .
Solution 2
Let , , , and . Then it follows that are pairwise coprime and , , and , with pairwise coprime as well. Then, we wish to show 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 , and then multiplying both sides by gives us Now, we look at the prime factorisations of . Let where all of the are nonnegative integers (they could be 0). Thus, we have and Now, we see that Squaring the LHS will just double all the exponents on the RHS.
Also, we have . We can also build similar equations for , using 2 of and either the minimum of the exponents or the maximum.
We see that when we multiply , , and together, the exponents of the prime factorization will be , where is chosen for the 'th prime.
When we multiply , , and together, the exponents of the prime factorization will be , where is chosen for the 'th prime.
Thus, when we divide the former by the latter, we have . We wish to prove that this is equal to the exponents we got from .
In other words, this problem is now down to proving that for any nonnegative integers . WLOG, let . Thus, computing maximums and minimums, this equation turns into
Finally, canceling out the 's and collecting like terms, we see the RHS is , which is equal to the LHS. Thus, this equation is true, and we are done.
Solution 4, Very Simple
Let where . Then, and . Using this cyclically, we have Now, note that and . Using this cyclically, we have -brainiacmaniac31
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 (Problems • Resources) | ||
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.