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

m (→‎Solution 2 (Algebra+ Brute Force))
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<cmath>16>a(5)</cmath>
 
<cmath>16>a(5)</cmath>
 
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.
 
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.
 +
 +
==Solution 3 (Calculus, not as good)==
 +
Since each of the roots is positive, the local maximum of the function must occur at a positive value of <math>x</math>. Taking <math>\frac{d}{dx}</math> of the polynomial yields <math>6x^2-4ax+a^2-81</math>, which is equal to <math>0</math> at the local maximum. Since this is a quadratic in <math>a</math>, we can find an expression for <math>a</math> in terms of <math>x</math>. The quadratic formula gives <math>a=\frac{4x\pm\sqrt{324-8x^2}}{2}</math>, which simplifies to <math>a=2x\pm\sqrt{81-2x^2}</math>. We know that <math>a</math> is a positive integer, and testing small positive integer values of <math>x</math> yields <math>a=15</math> or <math>a=1</math> when <math>x=4</math>, and <math>a=15</math> or <math>a=9</math> when <math>x=6</math>. Because there are two possible values of <math>c</math> for <math>a</math>, <math>a</math> must equal <math>15</math>.
 +
 +
Now our polynomial equals <math>2x^3-30x^2+144x-c</math>. Because one root is less than the <math>x</math> value at the local maximum (picture the graph of a cubic equation), it suffuces to synthetically divide by small integer values of <math>x</math> to see if the resulting quadratic also has integer roots. Dividing by <math>x=3</math> leaves a quotient of <math>2x^2-24x+72=2(x-6)^2</math>, and dividing by <math>x=4</math> leaves a quotient of <math>2x^2-22x+56=2(x-4)(x-7)</math>. Thus, c=<math>2\cdot 3\cdot 6\cdot 6=216</math>, or <math>c=2\cdot 4\cdot 4\cdot 7=224</math>. Our answer is <math>216+224=\boxed{440}</math> ~bad_at_mathcounts
  
 
==See also==
 
==See also==

Revision as of 12:56, 21 October 2021

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.

Solution 3 (Calculus, not as good)

Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic formula gives $a=\frac{4x\pm\sqrt{324-8x^2}}{2}$, which simplifies to $a=2x\pm\sqrt{81-2x^2}$. We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$, and $a=15$ or $a=9$ when $x=6$. Because there are two possible values of $c$ for $a$, $a$ must equal $15$.

Now our polynomial equals $2x^3-30x^2+144x-c$. Because one root is less than the $x$ value at the local maximum (picture the graph of a cubic equation), it suffuces to synthetically divide by small integer values of $x$ to see if the resulting quadratic also has integer roots. Dividing by $x=3$ leaves a quotient of $2x^2-24x+72=2(x-6)^2$, and dividing by $x=4$ leaves a quotient of $2x^2-22x+56=2(x-4)(x-7)$. Thus, c=$2\cdot 3\cdot 6\cdot 6=216$, or $c=2\cdot 4\cdot 4\cdot 7=224$. Our answer is $216+224=\boxed{440}$ ~bad_at_mathcounts

See also

2015 AIME II (Problems • Answer Key • Resources)
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Problem 7
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