Difference between revisions of "2011 AIME I Problems/Problem 6"

Revision as of 03:37, 29 March 2011

Problem

Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ .

Solution

If the vertex is at $\left(\frac{1}{4}, -\frac{9}{8}\right)$ , the equation of the parabola can be expressed in the form $y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}$ . Expanding, we find that $y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8}$ , and $y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}$ . From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$ , where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$ , $-\frac{a}{2}=b$ , and $\frac{a}{16}-\frac{9}{8}=c$ . Adding up all of these gives us $\frac{9a-18}{16}=a+b+c$ . We know that $a+b+c$ is an integer, so 9a-18 must be divisible by 16. Let $9a=z$ . If ${z-18}\equiv {0} \pmod{16}$ , then ${z}\equiv {2} \pmod{16}$ . Therefore, if $9a=2$ , $a=\frac{2}{9}$ . Adding up gives us $2+9=\boxed{011}$

@@ Line 7: / Line 7: @@
 == See also ==
-{{AIME box|year=2011|n=I|before=Problem 5|num-a=7}}
+{{AIME box|year=2011|n=I|num-b=5|num-a=7}}
 * [[AIME Problems and Solutions]]
 * [[American Invitational Mathematics Examination]]
 * [[Mathematics competition resources]]

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Difference between revisions of "2011 AIME I Problems/Problem 6"

Revision as of 03:37, 29 March 2011

Problem

Solution

See also