Difference between revisions of "2005 AIME II Problems/Problem 15"
Mathgeek2006 (talk | contribs) |
Mathgeek2006 (talk | contribs) |
||
Line 86: | Line 86: | ||
<cmath>(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).</cmath> | <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>. | 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>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | We use the same reflection as in Solution 2. As <math>OF_1'=OF_2=13</math>, we know that <math>\triangle OF_1'F_2</math> is isosceles. Hence <math>\angle F_2F_1'O=\angle F_1'F_2O</math>. But by symmetry, we also know that <math>\angle OF_1T=\angle F_2F_1'O</math>. Hence <math>\angle OF_1T=\angle F_1'F_2O</math>. In particular, as <math>\angle OF_1T=\angle OF_2T</math>, this implies that <math>O, F_1, F_2</math>, and <math>T</math> are concyclic. | ||
+ | |||
+ | Let <math>X</math> be the intersection of <math>F_2F_1'</math> with the <math>x</math>-axis. As <math>F_1F_2</math> is parallel to the <math>x</math>-axis, we know that <cmath>\angle TXO=180-\angle F_1F_2T.\tag{1}</cmath> But <cmath>180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}</cmath> By the fact that <math>OF_1F_2T</math> is cyclic, <cmath>\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}</cmath> Therefore, combining (1), (2), and (3), we find that | ||
+ | <cmath>\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}</cmath> | ||
+ | |||
+ | By symmetry, we also know that | ||
+ | <cmath>\angle F_1TO=\angle OTF_1'.\tag{5}</cmath> | ||
+ | Therefore, (4) and (5) show by AA similarity that <math>\triangle F_1OT\sim \triangle OXT</math>. Therefore, <math>\angle XOT=\angle OF_1T</math>. | ||
+ | |||
+ | Now as <math>OF_1=OF_2'=13</math>, we know that <math>\triangle OF_1F_2'</math> is isosceles, and as <math>F_1F_2'=20</math>, we can drop an altitude to <math>F_1F_2'</math> to easily find that <math>\tan \angle OF_1T=\sqrt{69}/10</math>. Therefore, <math>\tan\angle XOT</math>, which is the desired slope, must also be <math>\sqrt{69}/10</math>. As before, we conclude that 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 16:42, 1 August 2015
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 .
Solution 3
We use the same reflection as in Solution 2. As , we know that is isosceles. Hence . But by symmetry, we also know that . Hence . In particular, as , this implies that , and are concyclic.
Let be the intersection of with the -axis. As is parallel to the -axis, we know that But By the fact that is cyclic, Therefore, combining (1), (2), and (3), we find that
By symmetry, we also know that Therefore, (4) and (5) show by AA similarity that . Therefore, .
Now as , we know that is isosceles, and as , we can drop an altitude to to easily find that . Therefore, , which is the desired slope, must also be . As before, we conclude that 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.