Difference between revisions of "2002 AIME II Problems/Problem 15"
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== Problem == | == Problem == | ||
+ | Circles <math>\mathcal{C}_{1}</math> and <math>\mathcal{C}_{2}</math> intersect at two points, one of which is <math>(9,6)</math>, and the product of the radii is <math>68</math>. The x-axis and the line <math>y = mx</math>, where <math>m > 0</math>, are tangent to both circles. It is given that <math>m</math> can be written in the form <math>a\sqrt {b}/c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, <math>b</math> is not divisible by the square of any prime, and <math>a</math> and <math>c</math> are relatively prime. Find <math>a + b + c</math>. | ||
+ | |||
+ | == Solution 1== | ||
+ | Let the smaller angle between the <math>x</math>-axis and the line <math>y=mx</math> be <math>\theta</math>. Note that the centers of the two circles lie on the angle bisector of the angle between the <math>x</math>-axis and the line <math>y=mx</math>. Also note that if <math>(x,y)</math> is on said angle bisector, we have that <math>\frac{y}{x}=\tan{\frac{\theta}{2}}</math>. Let <math>\tan{\frac{\theta}{2}}=m_1</math>, for convenience. Therefore if <math>(x,y)</math> is on the angle bisector, then <math>x=\frac{y}{m_1}</math>. Now let the centers of the two relevant circles be <math>(a/m_1 , a)</math> and <math>(b/m_1 , b)</math> for some positive reals <math>a</math> and <math>b</math>. These two circles are tangent to the <math>x</math>-axis, so the radii of the circles are <math>a</math> and <math>b</math> respectively. We know that the point <math>(9,6)</math> is a point on both circles, so we have that | ||
+ | |||
+ | <cmath>(9-\frac{a}{m_1})^2+(6-a)^2=a^2</cmath> | ||
+ | |||
+ | <cmath>(9-\frac{b}{m_1})^2+(6-b)^2=b^2</cmath> | ||
+ | |||
+ | Expanding these and manipulating terms gives | ||
+ | |||
+ | <cmath>\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0</cmath> | ||
+ | |||
+ | <cmath>\frac{1}{m_1^2}b^2-[(18/m_1)+12]b+117=0</cmath> | ||
+ | |||
+ | It follows that <math>a</math> and <math>b</math> are the roots of the quadratic | ||
+ | |||
+ | <cmath>\frac{1}{m_1^2}x^2-[(18/m_1)+12]x+117=0</cmath> | ||
+ | |||
+ | It follows from Vieta's Formulas that the product of the roots of this quadratic is <math>117m_1^2</math>, but we were also given that the product of the radii was 68. Therefore <math>68=117m_1^2</math>, or <math>m_1^2=\frac{68}{117}</math>. Note that the half-angle formula for tangents is | ||
+ | |||
+ | <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}</cmath> | ||
+ | |||
+ | Therefore | ||
+ | |||
+ | <cmath>\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}</cmath> | ||
+ | |||
+ | Solving for <math>\cos{\theta}</math> gives that <math>\cos{\theta}=\frac{49}{185}</math>. It then follows that <math>\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}</math>. | ||
+ | |||
+ | It then follows that <math>m=\tan{\theta}=\frac{12\sqrt{221}}{49}</math>. Therefore <math>a=12</math>, <math>b=221</math>, and <math>c=49</math>. The desired answer is then <math>12+221+49=\boxed{282}</math>. | ||
+ | |||
+ | == Solution 2 (Alcumus)== | ||
+ | Let <math>r_1</math> and <math>r_2</math> be the radii of the circles. Then the centers of the circles are of the form <math>(kr_1,r_1)</math> and <math>(kr_2,r_2)</math> for the same constant <math>k,</math> since the two centers are collinear with the origin. Since <math>(9,6)</math> lies on both circles, | ||
+ | <cmath>(kr_i - 9)^2 + (r_i - 6)^2 = r^2,</cmath>where <math>r_i</math> represents either radius. Expanding, we get | ||
+ | <cmath>k^2 r^2 - (18k + 12) r + 117 = 0.</cmath>We are told the product of the circles is 68, so by Vieta's formulas, <math>\frac{117}{k^2} = 68.</math> Hence, <math>k^2 = \frac{117}{68},</math> and <math>k = \sqrt{\frac{117}{68}}.</math> | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.25 cm); | ||
+ | |||
+ | pair[] O; | ||
+ | real[] r; | ||
+ | pair P; | ||
+ | |||
+ | r[1] = 4.096; | ||
+ | r[2] = 16.6; | ||
+ | O[1] = (r[1]/(2/3*sqrt(17/13)),r[1]); | ||
+ | O[2] = (r[2]/(2/3*sqrt(17/13)),r[2]); | ||
+ | P = reflect(O[1],O[2])*(9,6); | ||
+ | |||
+ | draw(Circle(O[1],r[1])); | ||
+ | //draw(Circle(O[2],r[2])); | ||
+ | draw(arc(O[2],r[2],130,300)); | ||
+ | draw((0,0)--(8,12*sqrt(221)/49*8)); | ||
+ | draw((0,0)--(30,0)); | ||
+ | draw((0,0)--O[1]--(O[1].x,0)); | ||
+ | draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); | ||
+ | |||
+ | label("$y = mx$", (8,12*sqrt(221)/49*8), N); | ||
+ | |||
+ | dot("$(9,6)$", (9,6), NE); | ||
+ | dot("$(kr,r)$", O[1], N); | ||
+ | dot(P,red); | ||
+ | </asy> | ||
+ | |||
+ | Since the circle is tangent to the line <math>y = mx,</math> the distance from the center <math>(kr,r)</math> to the line is <math>r.</math> We can write <math>y = mx</math> as <math>y - mx = 0,</math> so from the distance formula, | ||
+ | <cmath>\frac{|r - krm|}{\sqrt{1 + m^2}} = r.</cmath>Squaring both sides, we get | ||
+ | <cmath>\frac{(r - krm)^2}{1 + m^2} = r^2,</cmath>so <math>(r - krm)^2 = r^2 (1 + m^2).</math> Since <math>r \neq 0,</math> we can divide both sides by r, to get | ||
+ | <cmath>(1 - km)^2 = 1 + m^2.</cmath>Then <math>1 - 2km + k^2 m^2 = 1 + m^2,</math> so <math>m^2 (1 - k^2) + 2km = 0.</math> Since <math>m \neq 0,</math> | ||
+ | <cmath>m(1 - k^2) + 2k = 0.</cmath>Hence, | ||
+ | <cmath>m = \frac{2k}{k^2 - 1} = \frac{2 \sqrt{\frac{117}{68}}}{\frac{117}{68} - 1} = \frac{12 \sqrt{221}}{49}.</cmath> | ||
+ | Our answer is thus <math>12 + 221 + 49 = \boxed{282}</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | Let the centers of <math>C_1</math> and <math>C_2</math> be <math>A</math> and <math>B</math>, respectively, and let the point <math>(9, 6)</math> be <math>P</math>. | ||
+ | |||
+ | Because both <math>C_1</math> and <math>C_2</math> are tangent to the x-axis, and both of them pass through <math>P</math>, both <math>A</math> and <math>B</math> must be equidistant from <math>P</math> and the x-axis. Therefore, they must both be on the parabola with <math>P</math> as the focus and the x-axis as the directrix. Since the coordinates of <math>P</math> is <math>(9, 6)</math>, we see that this parabola is the graph of the function <cmath>y=\frac{1}{12}(x-9)^2+3=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}.</cmath> | ||
+ | |||
+ | Let <math>AB</math> be <math>y=kx</math>. Because <math>C_1</math> and <math>C_2</math> are both tangent to the x-axis, the y-coordinates of <math>A</math> and <math>B</math> are <math>r_1</math> and <math>r_2</math>, respectively, so the x-coordinates of <math>A</math> and <math>B</math> are <math>\frac{r_1}{k}</math> and <math>\frac{r_2}{k}</math>. But since <math>A</math> and <math>B</math> are also on the graph of the function <math>y=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}</math>, the x-coordinates of <math>A</math> and <math>B</math> are also the roots of the equation <math>kx=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}</math>, and by Vieta's Formulas, their product is <math>\frac{\frac{39}{4}}{\frac{1}{12}}=117</math>. So we have <math>\frac{r_1}{k}\cdot \frac{r_2}{k}=117</math>. | ||
+ | |||
+ | We are also given that <math>r_1r_2=68</math>, so <math>k^2=\frac{r_1r_2}{117}=\frac{68}{117}</math>, which means that <math>k=\sqrt{\frac{68}{117}}</math>. Note that the line <math>AB</math> is the angle bisector of the angle between the line <math>y=mx</math> and the x-axis. Therefore, we apply the double-angle formula for tangents and get <cmath>m=\frac{2k}{1-k^2}=\frac{2\sqrt{\frac{68}{117}}}{1-\frac{68}{117}}=\frac{12\sqrt{221}}{49}.</cmath> Thus, the answer is <math>12+221+49=\boxed{282}</math>. | ||
+ | |||
+ | == Sidenote == | ||
+ | The two circles are centered at <cmath>\left (\frac{1}{13}\left (117 + 4 \sqrt{221} \pm 2 \sqrt{13(18 \sqrt{221} - 49)} \right), \frac{2}{39} \left(68 + 9 \sqrt {221} \pm 2 \sqrt{17(18 \sqrt{221} - 49)} \right) \right)</cmath> | ||
− | |||
− | |||
== See also == | == See also == | ||
− | + | {{AIME box|year=2002|n=II|num-b=14|after=Last Question}} | |
− | + | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:46, 23 June 2024
Problem
Circles and
intersect at two points, one of which is
, and the product of the radii is
. The x-axis and the line
, where
, are tangent to both circles. It is given that
can be written in the form
, where
,
, and
are positive integers,
is not divisible by the square of any prime, and
and
are relatively prime. Find
.
Solution 1
Let the smaller angle between the -axis and the line
be
. Note that the centers of the two circles lie on the angle bisector of the angle between the
-axis and the line
. Also note that if
is on said angle bisector, we have that
. Let
, for convenience. Therefore if
is on the angle bisector, then
. Now let the centers of the two relevant circles be
and
for some positive reals
and
. These two circles are tangent to the
-axis, so the radii of the circles are
and
respectively. We know that the point
is a point on both circles, so we have that
Expanding these and manipulating terms gives
It follows that and
are the roots of the quadratic
It follows from Vieta's Formulas that the product of the roots of this quadratic is , but we were also given that the product of the radii was 68. Therefore
, or
. Note that the half-angle formula for tangents is
Therefore
Solving for gives that
. It then follows that
.
It then follows that . Therefore
,
, and
. The desired answer is then
.
Solution 2 (Alcumus)
Let and
be the radii of the circles. Then the centers of the circles are of the form
and
for the same constant
since the two centers are collinear with the origin. Since
lies on both circles,
where
represents either radius. Expanding, we get
We are told the product of the circles is 68, so by Vieta's formulas,
Hence,
and
Since the circle is tangent to the line the distance from the center
to the line is
We can write
as
so from the distance formula,
Squaring both sides, we get
so
Since
we can divide both sides by r, to get
Then
so
Since
Hence,
Our answer is thus
Solution 3
Let the centers of and
be
and
, respectively, and let the point
be
.
Because both and
are tangent to the x-axis, and both of them pass through
, both
and
must be equidistant from
and the x-axis. Therefore, they must both be on the parabola with
as the focus and the x-axis as the directrix. Since the coordinates of
is
, we see that this parabola is the graph of the function
Let be
. Because
and
are both tangent to the x-axis, the y-coordinates of
and
are
and
, respectively, so the x-coordinates of
and
are
and
. But since
and
are also on the graph of the function
, the x-coordinates of
and
are also the roots of the equation
, and by Vieta's Formulas, their product is
. So we have
.
We are also given that , so
, which means that
. Note that the line
is the angle bisector of the angle between the line
and the x-axis. Therefore, we apply the double-angle formula for tangents and get
Thus, the answer is
.
Sidenote
The two circles are centered at
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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