Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AIME II Problems/Problem 6"

(→‎Solution)
(→‎Solution)
Line 35: Line 35:
 
These triples give <math>a</math> values of <math>13</math>, <math>15</math>, and <math>15</math>, respectively, and <math>c</math> values of <math>64</math>, <math>216</math>, and <math>224</math>, respectively.   
 
These triples give <math>a</math> values of <math>13</math>, <math>15</math>, and <math>15</math>, respectively, and <math>c</math> values of <math>64</math>, <math>216</math>, and <math>224</math>, respectively.   
  
−
We know that Jon still found two possible values of <math>c</math> when Steve told him the <math>a</math> value, so the <math>a</math> value must be <math>15</math>.  Thus, the two <math>c</math> values are <math>216</math> and <math>224</math>, which sum to <math>\boxed{\text{240}}</math>.
+
We know that Jon still found two possible values of <math>c</math> when Steve told him the <math>a</math> value, so the <math>a</math> value must be <math>15</math>.  Thus, the two <math>c</math> values are <math>216</math> and <math>224</math>, which sum to <math>\boxed{\text{440}}</math>.
  
 
~BealsConjecture~
 
~BealsConjecture~

Revision as of 23:57, 26 March 2015

Problem

Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?"

After some calculations, Jon says, "There is more than one such polynomial."

Steve says, "You're right. Here is the value of $a$." He writes down a positive integer and asks, "Can you tell me the value of $c$?"

Jon says, "There are still two possible values of $c$."

Find the sum of the two possible values of $c$.

Solution

We call the three roots (some may be equal to one another) $x_1$, $x_2$, and $x_3$. Using Vieta's formulas, we get $x_1+x_2+x_3 = a$, $x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}$, and $x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}$.

Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \cdot x_1 \cdot x_2+2 \cdot x_1 \cdot x_3+2 \cdot x_2 \cdot x_3 = a^2$.

We can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \cdot \frac{a^2-81}{2}$.

Simplifying the right side:

$a^2-2 \cdot \frac{a^2-81}{2}$

$a^2-a^2+81$

$81$

So, we know $x_1^2+x_2^2+x_3^2 = 81$.

We can then list out all the triples of positive integers whose squares sum to $81$:

We get $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$.

These triples give $a$ values of $13$, $15$, and $15$, respectively, and $c$ values of $64$, $216$, and $224$, respectively.

We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$. Thus, the two $c$ values are $216$ and $224$, which sum to $\boxed{\text{440}}$.

~BealsConjecture~

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_6&oldid=69584"