Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 6"
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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>. | 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]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]] |
− | *[[Mock AIME 1 2006-2007/Problem 7 | Next Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 7 | Next Problem]] |
*[[Mock AIME 1 2006-2007]] | *[[Mock AIME 1 2006-2007]] |
Latest revision as of 15:52, 3 April 2012
Problem
Let and
be two parabolas in the Cartesian plane. Let
be the common tangent line of
and
that has a rational slope. If
is written in the form
for positive integers
where
, find
.
Solution
From the condition that is tangent to
we have that the system of equations
and
has exactly one solution, so
has exactly one solution. A quadratic equation with only one solution must have discriminant equal to zero, so we must have
or equivalently
. Applying the same process to
, we have that
has a unique root so
or equivalently
. We multiply the first of these equations through by
and the second through by
and subtract in order to eliminate
and get
. We know that the slope of
,
, is a rational number, so we divide this equation through by
and let
to get
. Since we're searching for a rational root, we can use the Rational Root Theorem to search all possibilities and find that
is a solution. (The other two roots are the roots of the quadratic equation
, both of which are irrational.) Thus
. Now we go back to one of our first equations, say
, to get
. (We can reject the alternate possibility
because that would give
and our "line" would not exist.) Then
and since the greatest common divisor of the three numbers is 1,
and
.