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Difference between revisions of "2015 AMC 12A Problems/Problem 18"

(Created page with "The problem asks us to find the sum of every integer value of <math>a</math> such that the roots of <math>x^2 - ax + 2a = 0</math> are both integers. The quadratic formula gives...")
 
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There are a total of four possible values for <math>a</math>: <math>-1, 0, 8,</math> and <math>9</math>. Hence, the sum of all of the possible values of <math>a</math> is '''16 (C)'''.
 
There are a total of four possible values for <math>a</math>: <math>-1, 0, 8,</math> and <math>9</math>. Hence, the sum of all of the possible values of <math>a</math> is '''16 (C)'''.
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'''[[Solution 2]]'''
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Let <math>m</math> and <math>n</math> be the roots of <math>x^2-ax+2a</math>
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By Vieta's Formulas, <math>n + m = a</math> and <math>mn = 2a</math>
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Substituting gets us <math>n + m = \frac{mn}{2}</math>
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<math>2n - mn + 2m = 0</math>
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Using Simon's Favorite Factoring Trick:
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<math>n(2-m) + 2m = 0</math>
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<math>-n(2-m) - 2m = 0</math>
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<math>-n(2-m) - 2m + 4 = 4</math>
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<math>(2-n)(2-m) = 4</math>
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This means that the values for <math>(m,n)</math> are <math>(0,0),(4,4),(3,6),(1,-2)</math> giving us <math>a</math> values of <math>-1, 0, 8,</math> and <math>9</math>. Adding these up gets '''16 (C)'''

Revision as of 23:39, 4 February 2015

The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.

The quadratic formula gives the roots of the quadratic equation: $x = \frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg)

As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$, for some nonnegative integer $k$.

$a^2 - 8a = k^2$

$a(a - 8) = k^2$

$((a - 4) + 4)((a - 4) - 4) = k^2$

$(a - 4)^2 - 4^2 = k^2$

$(a - 4)^2 = k^2 + 4^2$

From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k, 4, |a - 4|)$ must be a Pythagorean triple unless $k = 0$.

In the case $k = 0$, the equation simplifies to $|a - 4| = 4$. From this equation, we have $a = 0, 8$. For both $a = 0$ and $a = 8$, $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg) yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")

If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k, 4, |a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3, 4, 5)$ triple. Here, $k = 3$ and $|a - 4| = 5$. Hence, $a = -1, 9$. Again, $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg) yields two integers for both $a = -1$ and $a = 9$, so these two values also satisfy the original constraints.

There are a total of four possible values for $a$: $-1, 0, 8,$ and $9$. Hence, the sum of all of the possible values of $a$ is 16 (C).

Solution 2

Let $m$ and $n$ be the roots of $x^2-ax+2a$

By Vieta's Formulas, $n + m = a$ and $mn = 2a$

Substituting gets us $n + m = \frac{mn}{2}$

$2n - mn + 2m = 0$

Using Simon's Favorite Factoring Trick:

$n(2-m) + 2m = 0$

$-n(2-m) - 2m = 0$

$-n(2-m) - 2m + 4 = 4$

$(2-n)(2-m) = 4$

This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1, 0, 8,$ and $9$. Adding these up gets 16 (C)

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