Difference between revisions of "2011 AIME I Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | If the vertex is at <math>\left(\frac{1}{4}, -\frac{9}{8}\right)</math>, the equation of the parabola can be expressed in the form < | + | If the vertex is at <math>\left(\frac{1}{4}, -\frac{9}{8}\right)</math>, the equation of the parabola can be expressed in the form <cmath>y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}.</cmath> |
− | Expanding, we find that < | + | Expanding, we find that <cmath>y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8},</cmath> and <cmath>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.</cmath> From the problem, we know that the parabola can be expressed in the form <math>y=ax^2+bx+c</math>, where <math>a+b+c</math> is an integer. From the above equation, we can conclude that <math>a=a</math>, <math>-\frac{a}{2}=b</math>, and <math>\frac{a}{16}-\frac{9}{8}=c</math>. Adding up all of these gives us <cmath>\frac{9a-18}{16}=a+b+c.</cmath> We know that <math>a+b+c</math> is an integer, so <math>9a-18</math> must be divisible by <math>16</math>. Let <math>9a=z</math>. If <math>{z-18}\equiv {0} \pmod{16}</math>, then <math>{z}\equiv {2} \pmod{16}</math>. Therefore, if <math>9a=2</math>, <math>a=\frac{2}{9}</math>. Adding up gives us <math>2+9=\boxed{011}</math> |
+ | |||
+ | ==Solution 2== | ||
+ | Complete the square. Since <math>a>0</math>, the parabola must be facing upwards. <math>a+b+c=\text{integer}</math> means that <math>f(1)</math> must be an integer. The function can be recasted into <math>a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}</math> because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than <math>-\frac{9}{8}</math> is <math>-1</math>. So the <math>y</math>-coordinate must change by <math>\frac{1}{8}</math> and the <math>x</math>-coordinate must change by <math>1-\frac{1}{4}=\frac{3}{4}</math>. Thus, <math>a\left(\frac{3}{4}\right)^2=\frac{1}{8}\implies \frac{9a}{16}=\frac{1}{8}\implies a=\frac{2}{9}</math>. So <math>2+9=\boxed{011}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | To do this, we can use the formula for the minimum (or maximum) value of the <math>x</math> coordinate at a vertex of a parabola, <math>-\frac{b}{2a}</math> and equate this to <math>\frac{1}{4}</math>. Solving, we get <math>-\frac{a}{2}=b</math>. Enter <math>x=\frac{1}{4}</math> to get <math>-\frac{9}{8}=\frac{a}{16}+\frac{b}{4}+c=-\frac{a}{16}+c</math> so <math>c=\frac{a-18}{16}</math>. This means that <math>\frac{9a-18}{16}\in \mathbb{Z}</math> so the minimum of <math>a>0</math> is when the fraction equals -1, so <math>a=\frac{2}{9}</math>. Therefore, <math>p+q=2+9=\boxed{011}</math>. | ||
+ | -Gideontz | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Write this as <math>a\left( x- \frac 14 \right)^2 - \frac 98</math>. Since <math>a+b+c</math> is equal to the value of this expression when you plug <math>x=1</math> in, we just need <math>\frac{9a}{16}- \frac 98</math> to be an integer. Since <math>a>0</math>, we also have <math>\frac{9a}{16}>0</math> which means <math>\frac{9a}{16}- \frac 98 > -\frac{9}{8}</math>. The least possible value of <math>a</math> is when this is equal to <math>-1</math>, or <math>a=\frac 29</math>, which gives answer <math>11</math>. | ||
+ | |||
+ | -bobthegod78, krwang, Simplest14 | ||
+ | |||
+ | ==Solution 5 (You don't remember conic section formulae)== | ||
+ | |||
+ | Take the derivative to get that the vertex is at <math>2ax+b=0</math> and note that this implies <math>\frac{1}{2} \cdot a = -b</math> and proceed with any of the solutions above. | ||
+ | |||
+ | ~Dhillonr25 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=vkniYGN45F4 | ||
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=5|num-a=7}} | {{AIME box|year=2011|n=I|num-b=5|num-a=7}} | ||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | |||
* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:18, 11 December 2023
Contents
Problem
Suppose that a parabola has vertex and equation
, where
and
is an integer. The minimum possible value of
can be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution
If the vertex is at , the equation of the parabola can be expressed in the form
Expanding, we find that
and
From the problem, we know that the parabola can be expressed in the form
, where
is an integer. From the above equation, we can conclude that
,
, and
. Adding up all of these gives us
We know that
is an integer, so
must be divisible by
. Let
. If
, then
. Therefore, if
,
. Adding up gives us
Solution 2
Complete the square. Since , the parabola must be facing upwards.
means that
must be an integer. The function can be recasted into
because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than
is
. So the
-coordinate must change by
and the
-coordinate must change by
. Thus,
. So
.
Solution 3
To do this, we can use the formula for the minimum (or maximum) value of the coordinate at a vertex of a parabola,
and equate this to
. Solving, we get
. Enter
to get
so
. This means that
so the minimum of
is when the fraction equals -1, so
. Therefore,
.
-Gideontz
Solution 4
Write this as . Since
is equal to the value of this expression when you plug
in, we just need
to be an integer. Since
, we also have
which means
. The least possible value of
is when this is equal to
, or
, which gives answer
.
-bobthegod78, krwang, Simplest14
Solution 5 (You don't remember conic section formulae)
Take the derivative to get that the vertex is at and note that this implies
and proceed with any of the solutions above.
~Dhillonr25
Video Solution
https://www.youtube.com/watch?v=vkniYGN45F4
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.