Difference between revisions of "2005 AIME II Problems/Problem 8"
(→Solution) |
(→Solution) |
||
Line 10: | Line 10: | ||
pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-arccos(3/7)); | pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-arccos(3/7)); | ||
− | path cir1 = CR(C1,4), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H); | + | 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); | pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t); | ||
Line 34: | Line 34: | ||
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>, | 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> | + | <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}</math>. By the [[Pythagorean Theorem]] on <math>\triangle ATO_3</math>, we find that | It follows that <math>HO_1 = \frac{28}{3}</math>, and that <math>O_3T = \frac{58}{7}</math>. By the [[Pythagorean Theorem]] on <math>\triangle ATO_3</math>, we find that |
Revision as of 01:49, 20 December 2010
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
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-arccos(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); draw((-14,0)--draw(MP("H",H,W))--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(C_3,T,H)); (Error making remote request. Unknown error_msg)
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 .
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 |