Difference between revisions of "2014 AIME II Problems/Problem 8"
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Consider a reflection of circle <math>E</math> over diameter <math>\overline{AB}</math>. By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii <math>r</math>, <math>r</math>, and <math>3r</math>, and the big circle has radius <math>2</math>. | Consider a reflection of circle <math>E</math> over diameter <math>\overline{AB}</math>. By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii <math>r</math>, <math>r</math>, and <math>3r</math>, and the big circle has radius <math>2</math>. | ||
| − | Descartes' Circle Theorem gives <math>(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12)^2 = 2((\frac{1}{r})^2+(\frac{1}{r})^2+(\frac{1}{3r})^2+(-\frac12)^2)</math> | + | Descartes' Circle Theorem gives <math>\left(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12\right)^2 = 2\left(\left(\frac{1}{r}\right)^2+\left(\frac{1}{r}\right)^2+\left(\frac{1}{3r}\right)^2+\left(-\frac12\right)^2\right)</math> |
Note that the big circle has curvature <math>-\frac12</math> because it is internally tangent. | Note that the big circle has curvature <math>-\frac12</math> because it is internally tangent. | ||
Revision as of 19:39, 8 March 2021
Contents
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 
.
- Notice that C, E and the point of tangency to circle C for circle E will be collinear because C and E intersect the tangent line at a right angle, implying they must be on the same line.
Solution 2
Consider a reflection of circle over diameter . By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii , , and , and the big circle has radius 
.
Descartes' Circle Theorem gives 
Note that the big circle has curvature 
because it is internally tangent.
Solving gives for a final answer of .
See also
| 2014 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. 
