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

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Find the sum of the two possible values of <math>c</math>.
 
Find the sum of the two possible values of <math>c</math>.
  
==Solution==
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==Solution 1 (Algebra)==
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We call the three roots (some may be equal to one another) <math>x_1</math>, <math>x_2</math>, and <math>x_3</math>.  Using Vieta's formulas, we get <math>x_1+x_2+x_3 = a</math>, <math>x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}</math>, and <math>x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}</math>.
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Squaring our first equation we get <math>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</math>.
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We can then subtract twice our second equation to get <math>x_1^2+x_2^2+x_3^2 = a^2-2 \cdot \frac{a^2-81}{2}</math>.
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Simplifying the right side:
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<cmath> \begin{align*}
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a^2-2 \cdot \frac{a^2-81}{2} &= a^2-a^2+81\\
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&= 81.\\
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\end{align*} </cmath>
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So, we know <math>x_1^2+x_2^2+x_3^2 = 81</math>.
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We can then list out all the triples of positive integers whose squares sum to <math>81</math>:
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We get <math>(1, 4, 8)</math>, <math>(3, 6, 6)</math>, and <math>(4, 4, 7)</math>. 
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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. 
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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>.
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~BealsConjecture~
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==Solution 2 (Algebra+ Brute Force)==
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First things first. Vietas gives us the following:
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 +
<cmath>\begin{align}
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x_1+x_2+x_3 = a\\
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x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}\\
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x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}
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\end{align}</cmath>
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From <math>(2)</math>, <math>a</math> must have odd parity, meaning <math>a^2-81</math> must be a multiple of <math>4</math>, which implies that both sides of <math>(2)</math> are even. Then, from <math>(1)</math>, we see that an odd number of <math>x_1</math>, <math>x_2</math>, and <math>x_3</math> must be odd, because we have already deduced that <math>a</math> is odd. In order for both sides of <math>(2)</math> to be even, there must only be one odd number and two even numbers.
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Now, the theoretical maximum value of the left side of <math>(2)</math> is <math>3 \cdot \frac{a}{3}^2=\frac{a^2}{3}</math>. That means that the maximum bound of <math>a</math> is where
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<cmath>\frac{a^2}{3}> \frac{a^2-81}{2} \indent (4)</cmath>
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which simplifies to <cmath>\sqrt{243}>a</cmath> meaning
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<cmath>16>a(5)</cmath>
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So now we have that <math>9<a</math> from <math>(2)</math>, <math>a<16</math> from <math>(5)</math>, and <math>a</math> is odd from <math>(2)</math>. This means that <math>a</math> could equal <math>11</math>, <math>13</math>, or <math>15</math>. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of  <math>(1, 4, 8)</math>, <math>(3, 6, 6)</math>, and <math>(4, 4, 7)</math>, of which the last two return equal <math>a</math> values. Then, <math>2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}</math> AWD.
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==See also==
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{{AIME box|year=2015|n=II|num-b=5|num-a=7}}
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 16:18, 18 October 2020

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 1 (Algebra)

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:


\begin{align*} a^2-2 \cdot \frac{a^2-81}{2} &= a^2-a^2+81\\ &= 81.\\ \end{align*}

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~

Solution 2 (Algebra+ Brute Force)

First things first. Vietas gives us the following:

\begin{align} 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}\\ x_1 \cdot x_2 \cdot x_3 = \frac{c}{2} \end{align}

From $(2)$, $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$, which implies that both sides of $(2)$ are even. Then, from $(1)$, we see that an odd number of $x_1$, $x_2$, and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of $(2)$ to be even, there must only be one odd number and two even numbers.

Now, the theoretical maximum value of the left side of $(2)$ is $3 \cdot \frac{a}{3}^2=\frac{a^2}{3}$. That means that the maximum bound of $a$ is where \[\frac{a^2}{3}> \frac{a^2-81}{2} \indent (4)\] which simplifies to \[\sqrt{243}>a\] meaning \[16>a(5)\] So now we have that $9<a$ from $(2)$, $a<16$ from $(5)$, and $a$ is odd from $(2)$. This means that $a$ could equal $11$, $13$, or $15$. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$, of which the last two return equal $a$ values. Then, $2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}$ AWD.

See also

2015 AIME II (ProblemsAnswer KeyResources)
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

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