Difference between revisions of "2014 AIME II Problems/Problem 8"
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| + | ==Problem== | ||
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| + | Circle <math>C</math> with radius 2 has diameter <math>\overline{AB}</math>. Circle D is internally tangent to circle <math>C</math> at <math>A</math>. Circle <math>E</math> is internally tangent to circle <math>C</math>, externally tangent to circle <math>D</math>, and tangent to <math>\overline{AB}</math>. The radius of circle <math>D</math> is three times the radius of circle <math>E</math>, and can be written in the form <math>\sqrt{m}-n</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m+n</math>. | ||
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==Solution 1== | ==Solution 1== | ||
<center> | <center> | ||
Revision as of 22:52, 8 April 2014
Problem
Circle 
with radius 2 has diameter 
. Circle D is internally tangent to circle at 
. Circle 
is internally tangent to circle , externally tangent to circle 
, and tangent to . The radius of circle is three times the radius of circle , and can be written in the form 
, where 
and 
are positive integers. Find 
.
Solution 1
![[asy] import graph; size(7.99cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079; draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000)); draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670)); draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416)); draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432)); draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666)); draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227)); draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002)); dot((7.780000000000009,-1.320000000000002),dotstyle); label("$C$", (7.707377218471464,-1.576266740352400), NE * labelscalefactor); dot((7.271934046987930,-1.319179731427737),dotstyle); label("$D$", (7.303064016111554,-1.276266740352400), NE * labelscalefactor); dot((9.198812158392690,-0.8249788848962227),dotstyle); label("$E$", (9.225301294671791,-0.7792624249832147), NE * labelscalefactor); dot((9.198009254448635,-1.322289365031666),dotstyle); label("$F$", (9.225301294671791,-1.276266740352400), NE * labelscalefactor); dot((9.779997393419690,-1.323228980404432),dotstyle); label("$B$", (9.810012253929656,-1.276266740352400), NE * labelscalefactor); dot((5.780002606580324,-1.316771019595571),dotstyle); label("$A$", (5.812051070003994,-1.276266740352400), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/7/4/2/74210cde32b228b8c40db58832312c9112770eca.png)
Using the diagram above, let the radius of be 
, and the radius of be 
. Then, 
, and 
, so the Pythagorean theorem in 
gives 
. Also, 
, so ![\[DF=DC+CF=2-3r+\sqrt{4-4r}.\]](http://latex.artofproblemsolving.com/a/1/0/a100f0ee8e2f480eefd5981f7ab6440b683155ed.png)
Noting that 
, we can now use the Pythagorean theorem in 
to get
![\[(2-3r+\sqrt{4-4r})^2+r^2=16r^2.\]](http://latex.artofproblemsolving.com/b/b/f/bbff5121ff5a612b18bd1fe289947dcdc6f65322.png)
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives 
for a final answer of 
.
