Difference between revisions of "1995 AIME Problems/Problem 4"
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− | == Solution == | + | == Solution 1 == |
− | We label the points as following: the centers of the circles of radii <math>3,6,9</math> are <math>O_3,O_6,O_9</math> respectively, and the endpoints of the chord are <math>P,Q</math>. Let <math>A_3,A_6,A_9</math> be the feet of the [[perpendicular]]s from <math>O_3,O_6,O_9</math> to <math>\overline{PQ}</math> (so <math>A_3,A_6</math> are the points of [[tangent (geometry)|tangency]]). Then we note that <math>\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}</math>, and <math>O_6O_9 : O_9O_3 = 3:6 = 1:2</math>. Thus, <math>O_9A_9 = \frac{ | + | We label the points as following: the centers of the circles of radii <math>3,6,9</math> are <math>O_3,O_6,O_9</math> respectively, and the endpoints of the chord are <math>P,Q</math>. Let <math>A_3,A_6,A_9</math> be the feet of the [[perpendicular]]s from <math>O_3,O_6,O_9</math> to <math>\overline{PQ}</math> (so <math>A_3,A_6</math> are the points of [[tangent (geometry)|tangency]]). Then we note that <math>\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}</math>, and <math>O_6O_9 : O_9O_3 = 3:6 = 1:2</math>. Thus, <math>O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5</math> (consider similar triangles). Applying the [[Pythagorean Theorem]] to <math>\triangle O_9A_9P</math>, we find that |
<cmath>PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}</cmath> | <cmath>PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}</cmath> | ||
<center><asy> | <center><asy> | ||
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D(A--MP("A_9",G,N)); D(B--MP("A_3",F,N)); D(C--MP("A_6",D,N)); D(A--P); D(rightanglemark(A,G,P,12)); | D(A--MP("A_9",G,N)); D(B--MP("A_3",F,N)); D(C--MP("A_6",D,N)); D(A--P); D(rightanglemark(A,G,P,12)); | ||
</asy></center> | </asy></center> | ||
+ | |||
+ | == Solution 2 (Analytic Geometry) == | ||
+ | <center><asy> | ||
+ | pointpen = black; pathpen = black + linewidth(0.7); size(150); | ||
+ | pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); | ||
+ | D(CR(D(MP("E",A)),9)); D(CR(D(MP("F",B)),3)); D(CR(D(MP("D",C)),6)); D((-9,0)--(9,0)); D(MP("",P,NW)--MP("",Q,NE)); | ||
+ | D(A--MP("B",G,N)); D(B--MP("C",F,N)); D(C--MP("A",D,N)); D(rightanglemark(A,G,P,12)); D(rightanglemark(C,D,P,12)); D(rightanglemark(B,F,P,12)); | ||
+ | </asy></center> | ||
+ | |||
+ | Let <math>A</math> be defined as the origin of a coordinate plane with the <math>y</math>-axis running across the chord and <math>C(6\sqrt{2},0)</math> by the [[Pythagorean Theorem]]. Then we have <math>D(0,-6)</math> and <math>F(6\sqrt{2},-3)</math>, and since <math>\frac{DE}{DF}=\frac{1}{3}</math>, the point <math>E</math> is one-third of the way from <math>D</math> to <math>F</math>, so point <math>E</math> has coordinates <math>(2\sqrt{2},-5)</math>. <math>E</math> is the center of the circle with radius <math>9</math>, so the equation of this circle is <math>(x-2\sqrt{2})^2+(y+5)^2=81</math>. Since the chord's equation is <math>y=0</math>, we must find all values of <math>x</math> satisfying the equation of the circle such that <math>y=0</math>. We find that <math>x-2\sqrt{2}=\pm\sqrt{56}</math>, so the chord has length <math>|\sqrt{56}+2\sqrt{2}-(-\sqrt{56}+2\sqrt{2})|=2\sqrt{56}</math> and the answer is <math>(2\sqrt{56})^2=\boxed{224}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:30, 5 May 2024
Problem
Circles of radius and
are externally tangent to each other and are internally tangent to a circle of radius
. The circle of radius
has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.
![[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(A,9)); D(CR(B,3)); D(CR(C,6)); D(P--Q); [/asy]](http://latex.artofproblemsolving.com/1/b/4/1b49c3112557e935af0792920ce28eea2617eab6.png)
Solution 1
We label the points as following: the centers of the circles of radii are
respectively, and the endpoints of the chord are
. Let
be the feet of the perpendiculars from
to
(so
are the points of tangency). Then we note that
, and
. Thus,
(consider similar triangles). Applying the Pythagorean Theorem to
, we find that
![[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP("O_9",A)),9)); D(CR(D(MP("O_3",B)),3)); D(CR(D(MP("O_6",C)),6)); D(MP("P",P,NW)--MP("Q",Q,NE)); D((-9,0)--(9,0)); D(A--MP("A_9",G,N)); D(B--MP("A_3",F,N)); D(C--MP("A_6",D,N)); D(A--P); D(rightanglemark(A,G,P,12)); [/asy]](http://latex.artofproblemsolving.com/4/8/b/48be03ee754e516e7858ae5ff4b144b4d32e4cf3.png)
Solution 2 (Analytic Geometry)
![[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP("E",A)),9)); D(CR(D(MP("F",B)),3)); D(CR(D(MP("D",C)),6)); D((-9,0)--(9,0)); D(MP("",P,NW)--MP("",Q,NE)); D(A--MP("B",G,N)); D(B--MP("C",F,N)); D(C--MP("A",D,N)); D(rightanglemark(A,G,P,12)); D(rightanglemark(C,D,P,12)); D(rightanglemark(B,F,P,12)); [/asy]](http://latex.artofproblemsolving.com/a/2/b/a2ba3edd7981fa26cb04db42091fc4f1dd9ad3a2.png)
Let be defined as the origin of a coordinate plane with the
-axis running across the chord and
by the Pythagorean Theorem. Then we have
and
, and since
, the point
is one-third of the way from
to
, so point
has coordinates
.
is the center of the circle with radius
, so the equation of this circle is
. Since the chord's equation is
, we must find all values of
satisfying the equation of the circle such that
. We find that
, so the chord has length
and the answer is
.
~eevee9406
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.