Difference between revisions of "2005 AIME II Problems/Problem 15"
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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>. | 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>. | + | 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>. |
+ | |||
+ | ==Solution 4== | ||
+ | [[Image:2005_AIME_II_-15.png||center|800px]] | ||
+ | First, rewrite the equations for the circles as <math>(x+5)^2+(y-12)^2=16^2</math> and <math>(x-5)^2+(y-12)^2=4^2</math>. | ||
+ | Then, choose a point <math>(a,b)</math> that is a distance of <math>x</math> from both circles. Use the distance formula between <math>(a,b)</math> and each of <math>A</math> and <math>C</math> (in the diagram above). The distances, as can be seen in the diagram above are <math>16-x</math> and <math>4+x</math>, respectively. | ||
+ | <cmath>(a-5)^2+(b-12)^2=(4+x)^2</cmath> | ||
+ | <cmath>(a+5)^2+(b-12)^2=(16-x)^2</cmath> | ||
+ | Subtracting the first equation from the second gives <cmath>20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2</cmath> | ||
+ | Substituting this into the first equation gives | ||
+ | <cmath>a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4</cmath> | ||
+ | <cmath>b^2-24b+69+\frac{3a^2}4=0</cmath> | ||
+ | Now, instead of converting this to the equation of an eclipse, solve for <math>b</math> and then divide by <math>a</math>. | ||
+ | <cmath>b=\frac{24\pm\sqrt{300-3a^2}}{2}</cmath> | ||
+ | We take the smaller root to minimize <math>\frac b a</math>. | ||
+ | <cmath>\frac b a=\frac{24-\sqrt{300-3a^2}}{2a}=\frac{24-\sqrt3\cdot\sqrt{100-a^2}}{2a}=\frac{12}a-\frac{\sqrt3}{2a}\sqrt{100-a^2}</cmath> | ||
+ | Now, let <math>10\cos\theta=a</math>. This way, <math>\sqrt{100-a^2}=10\sin\theta</math>. | ||
+ | Substitute this in. <math>\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta</math> | ||
+ | Then, take the derivative of this and set it to 0 to find the minimum value. | ||
+ | <math>\frac{6}{5}\sec\theta\tan\theta-\frac{\sqrt3}{2}\sec^2\theta=0\rightarrow\frac{6}{5}\sin\theta-frac{\sqrt3}{2}=0\rightarrow\sin\theta=\frac{5\sqrt3}{12}</math> | ||
+ | Then, use this value of <math>\sin\theta</math> to find the minimum of <math>\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta</math> to get <math>\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}</math> | ||
== See also == | == See also == |
Revision as of 18:27, 19 September 2017
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 .
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 .
Solution 4
First, rewrite the equations for the circles as and . Then, choose a point that is a distance of from both circles. Use the distance formula between and each of and (in the diagram above). The distances, as can be seen in the diagram above are and , respectively. Subtracting the first equation from the second gives Substituting this into the first equation gives Now, instead of converting this to the equation of an eclipse, solve for and then divide by . We take the smaller root to minimize . Now, let . This way, . Substitute this in. Then, take the derivative of this and set it to 0 to find the minimum value. Then, use this value of to find the minimum of to get
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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