2025 AMC 12A Problems/Problem 12: Difference between revisions
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== Problem == | |||
== Solution 1 == | |||
We will need to determine the sum of the reciprocals of the roots. To find the sum of the reciprocals of the roots <imath>p,q</imath> of the quadratic <imath>ax^2+bx+c</imath>, we use Vieta's formulas. Recall that <imath>p+q = -b/a</imath> and <imath>pq = c/a</imath>. Therefore, <cmath>\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-b/a}{c/a} = \frac{-b}{c},</cmath> | |||
which doesn't depend on <imath>a</imath>. | |||
The sum of the reciprocals of the roots of the quadratic <imath>x^2-4x-3</imath> is <imath>\frac{-(-4)}{-3} = -4/3.</imath> The same is true for every quadratic in the form <imath>ax^2-4x-3</imath>. The sum of all the reciprocals of the roots of <imath>ax^2+bx+c</imath> is <imath>2025 \cdot \left(-\frac{4}{3}\right).</imath> | |||
Because we have <imath>2025</imath> quadratics, there are <imath>2 \cdot 2025 = 4050</imath> total roots. Our answer is <cmath>\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} = \boxed{-\frac{3}{2}}.</cmath> | |||
~lprado | |||
Revision as of 14:36, 6 November 2025
Problem
Solution 1
We will need to determine the sum of the reciprocals of the roots. To find the sum of the reciprocals of the roots 
of the quadratic 
, we use Vieta's formulas. Recall that 
and 
. Therefore, ![\[\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-b/a}{c/a} = \frac{-b}{c},\]](http://latex-new.aopstest.com/e/f/6/ef6e050c17f57e240479ac3a1290e33129dafa38.png)
which doesn't depend on 
.
The sum of the reciprocals of the roots of the quadratic 
is 
The same is true for every quadratic in the form 
. The sum of all the reciprocals of the roots of is 
Because we have 
quadratics, there are 
total roots. Our answer is ![\[\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} = \boxed{-\frac{3}{2}}.\]](http://latex-new.aopstest.com/b/5/d/b5d88b2eeaea06eb0abee83bf056178d6076d0a3.png)
~lprado
