Difference between revisions of "2005 AIME II Problems/Problem 8"
Firebolt360 (talk | contribs) (→Solution) |
|||
(13 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | [[Circle]]s <math> C_1 </math> and <math> C_2 </math> are externally [[tangent]], and they are both internally tangent to circle <math> C_3. </math> The radii of <math> C_1 </math> and <math> C_2 </math> are 4 and 10, respectively, and the [[center]]s of the three circles are all [[collinear]]. A [[chord]] of <math> C_3 </math> is also a common external tangent of <math> C_1 </math> and <math> C_2. </math> Given that the length of the chord is <math> \frac{m\sqrt{n}}p </math> where <math> m,n, </math> and <math> p </math> are positive integers, <math> m </math> and <math> p </math> are [[relatively prime]], and <math> n </math> is not divisible by the square of any [[prime]], find <math> m+n+p. </math> | ||
− | + | == Solution == | |
+ | <center> | ||
+ | <asy> | ||
+ | pointpen = black; | ||
+ | pathpen = black + linewidth(0.7); | ||
+ | size(200); | ||
+ | |||
+ | pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); | ||
+ | path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H); | ||
+ | pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t); | ||
+ | |||
+ | draw(cir1); draw(cir2); draw(cir3); | ||
+ | draw((14,0)--(-14,0)); | ||
+ | draw(A--B); | ||
+ | MP("H",H,W); | ||
+ | draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4")); | ||
+ | |||
+ | draw(MP("O_1",C1)); | ||
+ | draw(MP("O_2",C2)); | ||
+ | draw(MP("O_3",C3)); | ||
− | == | + | draw(MP("T",T,N)); |
+ | draw(MP("A",A,NW)); | ||
+ | draw(MP("B",B,NE)); | ||
+ | draw(C1--MP("T_1",T1,N)); | ||
+ | |||
+ | draw(C2--MP("T_2",T2,N)); | ||
+ | draw(C3--T); | ||
+ | draw(rightanglemark(C3,T,H)); | ||
+ | </asy></center> | ||
+ | |||
+ | Let <math>O_1, O_2, O_3</math> be the centers and <math>r_1 = 4, r_2 = 10,r_3 = 14</math> the radii of the circles <math>C_1, C_2, C_3</math>. Let <math>T_1, T_2</math> be the points of tangency from the common external tangent of <math>C_1, C_2</math>, respectively, and let the extension of <math>\overline{T_1T_2}</math> intersect the extension of <math>\overline{O_1O_2}</math> at a point <math>H</math>. Let the endpoints of the chord/tangent be <math>A,B</math>, and the foot of the perpendicular from <math>O_3</math> to <math>\overline{AB}</math> be <math>T</math>. From the similar [[right triangle]]s <math>\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T </math>, | ||
+ | |||
+ | <cmath> \frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}. </cmath> | ||
+ | |||
+ | It follows that <math>HO_1 = \frac{28}{3}</math>, and that <math>O_3T = \frac{58}{7}\dagger</math>. By the [[Pythagorean Theorem]] on <math>\triangle ATO_3</math>, we find that | ||
+ | |||
+ | <cmath>AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}</cmath> | ||
+ | |||
+ | and the answer is <math>m+n+p=\boxed{405}</math>. | ||
− | {{ | + | <math>\dagger</math> Alternatively, drop an altitude from <math>O_1</math> to <math>O_3T</math> at <math>O_3'</math>, and to <math>O_2T_2</math> at <math>O_2'</math>. Then, <math>O_2O_2'=10-4=6</math>, and <math>O_1O_2=14</math>. But <math>O_1O_3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}</math>. From rectangles, <math>O_3'T=O_1T_1=4</math> so <math>O_3T=4+\frac{30}{7}=\frac{58}{7}</math>. |
− | == | + | ==Solution 2== |
− | + | Call our desired length <math>x</math>. Note for any <math>X</math> on <math>\overline{AB}</math> and <math>Y</math> on <math>\overline{O_1O_2}</math> such that <math>\overline{XY}\perp\overline{AB}</math> that the function <math>f</math> such that <math>f(\overline{O_1Y})=\overline{XY}</math> is linear. Since <math>(0,4)</math> and <math>(14,10)</math>, we can quickly interpolate that <math>f(10)=\overline{O_3T}=\frac{58}{7}</math>. Then, extend <math>\overline{O_3T}</math> until it reaches the circle on both sides; call them <math>P,Q</math>. By Power of a Point, | |
+ | <math>\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}</math>. | ||
+ | Since <math>\overline{AT}=\overline{TB}=\frac{1}{2}x</math>, | ||
+ | <cmath>(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2</cmath> | ||
+ | <cmath>\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2</cmath> | ||
+ | After solving for <math>x</math>, we get <math>x=\frac{8\sqrt{390}}{7}</math>, so our answer is <math>8+390+7=\boxed{405}</math> | ||
+ | == See also == | ||
+ | {{AIME box|year=2005|n=II|num-b=7|num-a=9}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:03, 28 December 2021
Contents
[hide]Problem
Circles and
are externally tangent, and they are both internally tangent to circle
The radii of
and
are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of
is also a common external tangent of
and
Given that the length of the chord is
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime, find
Solution
![[asy] pointpen = black; pathpen = black + linewidth(0.7); size(200); pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H); pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t); draw(cir1); draw(cir2); draw(cir3); draw((14,0)--(-14,0)); draw(A--B); MP("H",H,W); draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4")); draw(MP("O_1",C1)); draw(MP("O_2",C2)); draw(MP("O_3",C3)); draw(MP("T",T,N)); draw(MP("A",A,NW)); draw(MP("B",B,NE)); draw(C1--MP("T_1",T1,N)); draw(C2--MP("T_2",T2,N)); draw(C3--T); draw(rightanglemark(C3,T,H)); [/asy]](http://latex.artofproblemsolving.com/5/0/5/505c60ef19e1317234ada04bf987e05e5689cd50.png)
Let be the centers and
the radii of the circles
. Let
be the points of tangency from the common external tangent of
, respectively, and let the extension of
intersect the extension of
at a point
. Let the endpoints of the chord/tangent be
, and the foot of the perpendicular from
to
be
. From the similar right triangles
,
It follows that , and that
. By the Pythagorean Theorem on
, we find that
and the answer is .
Alternatively, drop an altitude from
to
at
, and to
at
. Then,
, and
. But
is similar to
so
. From rectangles,
so
.
Solution 2
Call our desired length . Note for any
on
and
on
such that
that the function
such that
is linear. Since
and
, we can quickly interpolate that
. Then, extend
until it reaches the circle on both sides; call them
. By Power of a Point,
.
Since
,
After solving for
, we get
, so our answer is
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.