Difference between revisions of "Mock AIME II 2012 Problems/Problem 12"

Latest revision as of 10:09, 8 July 2022

Problem

Let $\log_{a}b=5\log_{b}ac^4=3\log_{c}a^2b$ . Assume the value of $\log_ab$ has three real solutions $x,y,z$ . If $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, find $m+n$ .

Solution

Let $log_{a}b=15x$ . Then $log_{b}ac^4=3x$ and $\log_{c}a^2b=5x$ . From this, we have the system

$a^{15x}=b$ $b^{3x}=ac^4$ $c^{5x}=a^2b$

Substituting the first equation into the second, we obtain

$a^{45x^2}=ac^4\rightarrow a^{\frac{45x^2-1}{4}}=c$

Plugging this into the third equation yields $a^{225x^3-5x}=a^{60x+8}$ .

Thus, $225x^3-65x-8=0$ . Note that our three real roots multiply to $\frac{8}{225}$ . However, since $\log_{a}b=15x$ , we need to multiply by $15^3$ , so our $xyz$ is $\frac{8}{225}\cdot 15^3=8\cdot 15=120$

We need $xy+xz+yz$ . Using vieta’s and making sure we count for each factor of $15$ we divided off, we have $15^2\cdot\frac{-65}{225}$ .

Our answer is $-\frac{65}{8\cdot 15}=-\frac{13}{24}$ , thus $13+24=\boxed{037}$ .

Solution 2

Let $\log_a b = x$ and $\log_b c=y$ , where $x,y>0$ . Then, it is obvious that $log_c a = \frac{1}{xy}$ .

We first focus on the first equality: $\log_a b = 5 \log_b{ac^4}$ . This may be simplified using our logarithmic properties:

$x = 5(\log_b a + \log_b c^4)$

$x = 5(\frac{1}{x} + 4y)$

$x = \frac{5}{x} + 20y.$

Now, let's focus on the last expression: note that,

$3 \log_c{a^2b} = 3 (2 \log_c a + \log_c b) = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.$

We can equate all of these expressions:

$x = \frac{5}{x} + 20y = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.$

Multiplying all expressions by $xy$ gives us

$x^2y = 5y + 20xy^2 = 6+3x.$

Now, from our first equality we obtain

$x^2 y = 5y + 20xy^2.$

Since $y \ne 0$ , we may safely divide by $y$ :

$x^2 = 5+20xy$

$y = \frac{x^2-5}{20x}.$

From the first and last expressions we have:

$x^2y = 6+3x$

$y= \frac{6+3x}{x^2}.$

Equating our expressions for $y$ gives

$\frac{x^2-5}{20x}=\frac{6+3x}{x^2}$

$x^4 - 5x^2 = 120x+60x^2.$

Since $x \ne 0$ , we may safely divide by $x$ :

$x^3 - 5x = 120 + 60x$

$x^3 - 65x-120=0.$

By Vieta's formulas, we must have $r_1r_2+r_2r_3+r_1r_3 = -65$ and $r_1r_2r_3 = 120$ . Dividing the former by the latter gives

$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} = -\frac{65}{120} = -\frac{13}{24}$

and hence $m+n = 13+24 = \boxed{037}$ .

~FIREDRAGONMATH16

@@ Line 37: / Line 37: @@
 Now, let's focus on the last expression: note that,
-<cmath>3 \log_c{a^2b} = 3 (2 \log_c a + \log_c b) = 6 (\frac{1}{xy}) + \frac{1}{y}).</cmath>
+<cmath>3 \log_c{a^2b} = 3 (2 \log_c a + \log_c b) = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.</cmath>
 We can equate all of these expressions:
-<cmath>x = \frac{5}{x} + 20y = 6 (\frac{1}{xy}) + \frac{1}{y}).</cmath>
+<cmath>x = \frac{5}{x} + 20y = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.</cmath>
 Multiplying all expressions by <math>xy</math> gives us
@@ Line 80: / Line 80: @@
 and hence <math>m+n = 13+24 = \boxed{037}</math>.
+~FIREDRAGONMATH16

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Difference between revisions of "Mock AIME II 2012 Problems/Problem 12"

Latest revision as of 10:09, 8 July 2022

Problem

Solution

Solution 2