Difference between revisions of "2005 AIME II Problems/Problem 15"
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Let <math> w_1 </math> and <math> w_2 </math> denote the [[circle]]s <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positive value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math> | Let <math> w_1 </math> and <math> w_2 </math> denote the [[circle]]s <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positive value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math> | ||
− | == Solution == | + | == Solution 1 == |
Rewrite the given equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>. | Rewrite the given equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>. | ||
Line 43: | Line 43: | ||
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>\boxed{169}</math>. | Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>\boxed{169}</math>. | ||
+ | == Solution 2 == | ||
+ | As above, we rewrite the equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>. Let <math>F_1=(-5,12)</math> and <math>F_2=(5,12)</math>. If a circle with center <math>C=(a,b)</math> and radius <math>r</math> is externally tangent to <math>w_2</math> and internally tangent to <math>w_1</math>, then <math>CF_1=16-r</math> and <math>CF_2=4+r</math>. Therefore, <math>CF_1+CF_2=20</math>. In particular, the locus of points <math>C</math> that can be centers of circles must be an ellipse with foci <math>F_1</math> and <math>F_2</math> and major axis <math>20</math>. | ||
+ | |||
+ | Clearly, the minimum value of the slope <math>a</math> will occur when the line <math>y=ax</math> is tangent to this ellipse. Suppose that this point of tangency is denoted by <math>T</math>, and the line <math>y=ax</math> is denoted by <math>\ell</math>. Then we reflect the ellipse over <math>\ell</math> to a new ellipse with foci <math>F_1'</math> and <math>F_2'</math> as shown below. | ||
+ | <center><asy> | ||
+ | size(220); | ||
+ | pair F1 = (-5, 12), F2 = (5, 12),C=(0,12); | ||
+ | draw(circle(F1,16)); | ||
+ | draw(circle(F2,4)); | ||
+ | draw(ellipse(C,10,5*sqrt(3))); | ||
+ | xaxis("$x$",Arrows); | ||
+ | yaxis("$y$",Arrows); | ||
+ | dot(F1^^F2^^C); | ||
+ | |||
+ | real l(real x) {return sqrt(69)*x/10;} | ||
+ | path g=graph(l,-7,14); | ||
+ | draw(g); | ||
+ | draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3))); | ||
+ | pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10))); | ||
+ | dot(T); | ||
+ | pair F1P=reflect((0,0),(10,l(10)))*F1; | ||
+ | pair F2P=reflect((0,0),(10,l(10)))*F2; | ||
+ | dot(F1P^^F2P); | ||
+ | dot((0,0)); | ||
+ | label("$F_1$",F1,N,fontsize(9)); | ||
+ | label("$F_2$",F2,N,fontsize(9)); | ||
+ | label("$F_1'$",F1P,SE,fontsize(9)); | ||
+ | label("$F_2'$",F2P,SE,fontsize(9)); | ||
+ | label("$O$",(0,0),NW,fontsize(9)); | ||
+ | label("$\ell$",(13,l(13)),SE,fontsize(9)); | ||
+ | label("$T$",T,NW,fontsize(9)); | ||
+ | draw((0,0)--F1--F2--F2P--F1P--cycle); | ||
+ | draw(F1--F2P^^F2--F1P); | ||
+ | </asy></center> | ||
+ | By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that <math>F_1</math>, <math>T</math>, and <math>F_2'</math> are collinear, and similarly, <math>F_2</math>, <math>T</math> and <math>F_1'</math> are collinear. Therefore, <math>OF_1F_2F_2'F_1'</math> is a pentagon with <math>OF_1=OF_2=OF_1'=OF_2'=13</math>, <math>F_1F_2=F_1'F_2'=10</math>, and <math>F_1F_2'=F_1'F_2=20</math>. Note that <math>\ell</math> bisects <math>\angle F_1'OF_1</math>. We can bisect this angle by bisecting <math>\angle F_1'OF_2</math> and <math>F_2OF_1</math> separately. | ||
+ | |||
+ | We proceed using complex numbers. Triangle <math>F_2OF_1'</math> is isosceles with side lengths <math>13,13,20</math>. The height of this from the base of <math>20</math> is <math>\sqrt{69}</math>. Therefore, the complex number <math>\sqrt{69}+10i</math> represents the bisection of \angle <math>F_1'OF_2</math>. | ||
+ | |||
+ | Similarly, using the 5-12-13 triangles, we easily see that <math>12+5i</math> represents the bisection of the angle <math>F_2OF_1</math>. Therefore, we can add these two angles together by multiplying the complex numbers, finding | ||
+ | <cmath>\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.</cmath> | ||
+ | Now the point <math>F_1</math> is given by the complex number <math>-5+12i</math>. Therefore, to find a point on line <math>\ell</math>, we simply subtract <math>\frac{1}{2}\angle F_1'OF_1</math>, which is the same as multiplying <math>-5+12i</math> by the conjugate of <math>(\sqrt{69}+10i)(12+5i)</math>. We find | ||
+ | <cmath>(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).</cmath> | ||
+ | In particular, note that the tangent of the argument of this complex number is <math>\sqrt{69}/10</math>, which must be the slope of the tangent line. Hence <math>a^2=69/100</math>, and the answer is <math>\boxed{169}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2005|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2005|n=II|num-b=14|after=Last Question}} |
Revision as of 15:55, 1 August 2015
Contents
Problem
Let and denote the circles and respectively. Let be the smallest positive value of for which the line contains the center of a circle that is externally tangent to and internally tangent to Given that where and are relatively prime integers, find
Solution 1
Rewrite the given equations as and .
Let have center and radius . Now, if two circles with radii and are externally tangent, then the distance between their centers is , and if they are internally tangent, it is . So we have
Solving for in both equations and setting them equal, then simplifying, yields
Squaring again and canceling yields
So the locus of points that can be the center of the circle with the desired properties is an ellipse.
Since the center lies on the line , we substitute for and expand:
We want the value of that makes the line tangent to the ellipse, which will mean that for that choice of there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is , so .
Solving yields , so the answer is .
Solution 2
As above, we rewrite the equations as and . Let and . If a circle with center and radius is externally tangent to and internally tangent to , then and . Therefore, . In particular, the locus of points that can be centers of circles must be an ellipse with foci and and major axis .
Clearly, the minimum value of the slope will occur when the line is tangent to this ellipse. Suppose that this point of tangency is denoted by , and the line is denoted by . Then we reflect the ellipse over to a new ellipse with foci and as shown below.
By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that , , and are collinear, and similarly, , and are collinear. Therefore, is a pentagon with , , and . Note that bisects . We can bisect this angle by bisecting and separately.
We proceed using complex numbers. Triangle is isosceles with side lengths . The height of this from the base of is . Therefore, the complex number represents the bisection of \angle .
Similarly, using the 5-12-13 triangles, we easily see that represents the bisection of the angle . Therefore, we can add these two angles together by multiplying the complex numbers, finding Now the point is given by the complex number . Therefore, to find a point on line , we simply subtract , which is the same as multiplying by the conjugate of . We find In particular, note that the tangent of the argument of this complex number is , which must be the slope of the tangent line. Hence , and the answer is .
See also
2005 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 | ||
All AIME Problems and Solutions |
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