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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 6"

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==Problem==
 
==Problem==
  
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Let <math>P_{1}: y=x^{2}+\frac{101}{100}</math> and <math>P_{2}: x=y^{2}+\frac{45}{4}</math> be two parabolas in the cartesian plane. Let <math>\mathcal{L}</math> be the common tangent of <math>P_{1}</math> and <math>P_{2}</math> that has a rational slope. If <math>\mathcal{L}</math> is written in the form <math>ax+by=c</math> for positive integers <math>a,b,c</math> where <math>\gcd(a,b,c)=1</math>. Find <math>a+b+c</math>.
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Let <math>P_{1}: y=x^{2}+\frac{101}{100}</math> and <math>P_{2}: x=y^{2}+\frac{45}{4}</math> be two [[parabola]]s in the [[Cartesian plane]]. Let <math>\mathcal{L}</math> be the common [[tangent line]] of <math>P_{1}</math> and <math>P_{2}</math> that has a [[rational number | rational]] [[slope]]. If <math>\mathcal{L}</math> is written in the form <math>ax+by=c</math> for [[positive integer]]s <math>a,b,c</math> where <math>\gcd(a,b,c)=1</math>, find <math>a+b+c</math>.
  
 
==Solution==
 
==Solution==
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{{solution}}
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From the condition that <math>\mathcal L</math> is tangent to <math>P_1</math> we have that the system of equations <math>ax + by = c</math> and <math> {y = x^2 + \frac{101}{100}}</math> has exactly one solution, so <math>ax + b(x^2 + \frac{101}{100}) = c</math> has exactly one solution.  A [[quadratic equation]] with only one solution must have [[discriminant]] equal to [[zero (constant) | zero]], so we must have <math>a^2 - 4\cdot b \cdot (\frac{101}{100}b - c) = 0</math> or equivalently <math>25a^2 -101b^2 + 100bc = 0</math>.  Applying the same process to <math>P_2</math>, we have that <math>a(y^2 + \frac{45}4) + by = c</math> has a unique [[root]] so <math>b^2 - 4\cdot a \cdot (\frac{45}4a - c) = 0</math> or equivalently <math>b^2 - 45a^2 + 4ac = 0</math>.  We multiply the first of these equations through by <math>a</math> and the second through by <math>25b</math> and subtract in order to eliminate <math>c</math> and get <math>25a^3 + 1125 a^2b - 101ab^2 - 25b^3 = 0</math>.  We know that the slope of <math>\mathcal L</math>, <math>-\frac b a</math>, is a rational number, so we divide this equation through by <math>-a^3</math> and let <math>\frac b a = q</math> to get <math>25q^3 +101q^2 - 1125q - 25 = 0</math>.  Since we're searching for a rational root, we can use the [[Rational Root Theorem]] to search all possibilities and find that <math>q = 5</math> is a solution.  (The other two roots are the roots of the quadratic equation <math>25q^2 + 226q +5 = 0</math>, both of which are [[irrational number | irrational]].)  Thus <math>b = 5a</math>.  Now we go back to one of our first equations, say <math>b^2 - 45a^2 + 4ac = 0</math>, to get <math>25a^2 - 45a^2 + 4ac = 0 \Longrightarrow c = 5a</math>.  (We can reject the alternate possibility <math>a = 0</math> because that would give <math>a = b = 0</math> and our "[[line]]" would not exist.)  Then <math>a : b : c = 1 : 5 : 5</math> and since the [[greatest common divisor]] of the three numbers is 1, <math>a = 1, b = 5, c = 5</math> and <math>a + b + c = 011</math>.
  
 
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*[[Mock AIME 1 2006-2007/Problem 5 | Previous Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]]
  
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*[[Mock AIME 1 2006-2007/Problem 7 | Next Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 7 | Next Problem]]
  
 
*[[Mock AIME 1 2006-2007]]
 
*[[Mock AIME 1 2006-2007]]
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 14:52, 3 April 2012

Problem

Let $P_{1}: y=x^{2}+\frac{101}{100}$ and $P_{2}: x=y^{2}+\frac{45}{4}$ be two parabolas in the Cartesian plane. Let $\mathcal{L}$ be the common tangent line of $P_{1}$ and $P_{2}$ that has a rational slope. If $\mathcal{L}$ is written in the form $ax+by=c$ for positive integers $a,b,c$ where $\gcd(a,b,c)=1$, find $a+b+c$.

Solution

From the condition that $\mathcal L$ is tangent to $P_1$ we have that the system of equations $ax + by = c$ and ${y = x^2 + \frac{101}{100}}$ has exactly one solution, so $ax + b(x^2 + \frac{101}{100}) = c$ has exactly one solution. A quadratic equation with only one solution must have discriminant equal to zero, so we must have $a^2 - 4\cdot b \cdot (\frac{101}{100}b - c) = 0$ or equivalently $25a^2 -101b^2 + 100bc = 0$. Applying the same process to $P_2$, we have that $a(y^2 + \frac{45}4) + by = c$ has a unique root so $b^2 - 4\cdot a \cdot (\frac{45}4a - c) = 0$ or equivalently $b^2 - 45a^2 + 4ac = 0$. We multiply the first of these equations through by $a$ and the second through by $25b$ and subtract in order to eliminate $c$ and get $25a^3 + 1125 a^2b - 101ab^2 - 25b^3 = 0$. We know that the slope of $\mathcal L$, $-\frac b a$, is a rational number, so we divide this equation through by $-a^3$ and let $\frac b a = q$ to get $25q^3 +101q^2 - 1125q - 25 = 0$. Since we're searching for a rational root, we can use the Rational Root Theorem to search all possibilities and find that $q = 5$ is a solution. (The other two roots are the roots of the quadratic equation $25q^2 + 226q +5 = 0$, both of which are irrational.) Thus $b = 5a$. Now we go back to one of our first equations, say $b^2 - 45a^2 + 4ac = 0$, to get $25a^2 - 45a^2 + 4ac = 0 \Longrightarrow c = 5a$. (We can reject the alternate possibility $a = 0$ because that would give $a = b = 0$ and our "line" would not exist.) Then $a : b : c = 1 : 5 : 5$ and since the greatest common divisor of the three numbers is 1, $a = 1, b = 5, c = 5$ and $a + b + c = 011$.

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