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.
![[asy] size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d; pair A = (-5, 12), B = (5, 12), C = (0, 0); D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red); D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow(4)); void bluecirc (real x) { pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue); D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype("4 4")); } bluecirc(-9.2); bluecirc(-4); bluecirc(3); [/asy]](http://latex.artofproblemsolving.com/3/9/5/3952f7611dcec95c9d68289eb69eb74de86b62bc.png)
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.
![[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]](http://latex.artofproblemsolving.com/3/9/f/39f7436c9b561798a9afaf9bddc9c97fd6de020d.png)
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.