Difference between revisions of "2014 AIME II Problems/Problem 8"

Latest revision as of 20:31, 9 January 2024

Problem

Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$ .

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$", (5.812051070003994,-1.276266740352400), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

Using the diagram above, let the radius of $D$ be $3r$ , and the radius of $E$ be $r$ . Then, $EF=r$ , and $CE=2-r$ , so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$ . Also, $CD=CA-AD=2-3r$ , so $DF=DC+CF=2-3r+\sqrt{4-4r}.$ Noting that $DE=4r$ , we can now use the Pythagorean theorem in $\triangle DEF$ to get $(2-3r+\sqrt{4-4r})^2+r^2=16r^2.$

Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$ .

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 $E$ over diameter $\overline{AB}$ . 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 $r$ , $r$ , and $3r$ , and the big circle has radius $2$ .

Descartes' Circle Theorem gives $\left(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12\right)^2 = 2\left(\left(\frac{1}{r}\right)^2+\left(\frac{1}{r}\right)^2+\left(\frac{1}{3r}\right)^2+\left(-\frac12\right)^2\right)$

Note that the big circle has curvature $-\frac12$ because it is internally tangent. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$ .

Solution 3

We use the notation of Solution 1 for triangle $\triangle DEC$ $\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC = \frac {\sqrt{15}}{4}.$ We use Cosine Law for $\triangle DEC$ and get: $(4r)^2 +(2 – 3r)^2 – 2 \cdot 4r \cdot (2 – 3r) \cdot \frac {\sqrt{15}}{4} = (2 – r)^2$ . $(24 + 6 \sqrt{15} ) r^2 = (8 + 4 \sqrt {15})r \implies 3r = 4 \sqrt{15} – 14 \implies \boxed{\textbf{254}}.$ vladimir.shelomovskii@gmail.com, vvsss

Solution 4

This problem can be very easily solved using Descartes' Circle Theorem. It states that if we have 4 circles that are all tangent with each other, $(k_1 + k_2 + k_3 + k_4)^{2} = 2(k_1^{2} + k_2^{2} + k_3^{2} + k_4^{2})$ , where $k_i$ is the curvature of circle $i$ , meaning $k_i = \dfrac{1}{r}$ . When three of the circles are internally tangent to the fourth one, the fourth circle has a negative curvature. Suppose we reflect Circle $E$ over $\overline{AB}$ . Now, we have our four circles to apply that theorem. First, lets scale our image down such that Circle $C$ has radius $1$ , for ease of computation. Let the radius of Circle $D$ be $r$ , so Circle $E$ has radius $\dfrac{r}{3}$ . Then, we have that $(-1 + \dfrac{1}{r} + \dfrac{3}{r} + \dfrac{3}{r})^{2} = 2(1 + \dfrac{1}{r^{2}} + \dfrac{9}{r^{2}} + \dfrac{9}{r^{2}})$ . This simplifies to $\dfrac{49}{r^{2}} - \dfrac{14}{r} + 1 = \dfrac{2r^{2} + 38}{r^{2}}$ . Multiplying both sides by $r^{2}$ , we get that $49 - 14r + r^{2} = 2r^{2} + 38$ , or $r^2 + 14r - 11 = 0$ . We get $r = -7 \pm 2\sqrt{15}$ , but we want the positive solution, which is $r = 2\sqrt{15} - 7$ . We have to rescale back up, so we get $r = 4\sqrt{15} - 14 = \sqrt{240} - 14$ , so we get that our answer is $240 + 14 = \boxed{254}$ . ~Puck_0

@@ Line 51: / Line 51: @@
 ==Solution 3==
 We use the notation of Solution 1 for triangle <math>\triangle DEC</math>
-<math></math>\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC =  \frac {\sqrt{15}}{4}.<math> </math>
+<cmath>\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC =  \frac {\sqrt{15}}{4}.</cmath>
-We use Cosine Law and get:
+We use Cosine Law for <math>\triangle DEC</math> and get:
 <cmath>(4r)^2 +(2 – 3r)^2 – 2 \cdot 4r \cdot (2 – 3r) \cdot  \frac {\sqrt{15}}{4} = (2 – r)^2 </cmath>.
 <cmath>(24 + 6 \sqrt{15} ) r^2 = (8 + 4 \sqrt {15})r \implies 3r = 4 \sqrt{15} – 14 \implies \boxed{\textbf{254}}.</cmath>
 '''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution 4==
+This problem can be very easily solved using Descartes' Circle Theorem. It states that if we have 4 circles that are all tangent with each other, <math>(k_1 + k_2 + k_3 + k_4)^{2} = 2(k_1^{2} + k_2^{2} + k_3^{2} + k_4^{2})</math>, where <math>k_i</math> is the curvature of circle <math>i</math>, meaning <math>k_i = \dfrac{1}{r}</math>. When three of the circles are internally tangent to the fourth one, the fourth circle has a negative curvature. Suppose we reflect Circle <math>E</math> over <math>\overline{AB}</math>. Now, we have our four circles to apply that theorem. First, lets scale our image down such that Circle <math>C</math> has radius <math>1</math>, for ease of computation. Let the radius of Circle <math>D</math> be <math>r</math>, so Circle <math>E</math> has radius <math>\dfrac{r}{3}</math>. Then, we have that <math>(-1 + \dfrac{1}{r} + \dfrac{3}{r} + \dfrac{3}{r})^{2} = 2(1 + \dfrac{1}{r^{2}} + \dfrac{9}{r^{2}} + \dfrac{9}{r^{2}})</math>. This simplifies to <math>\dfrac{49}{r^{2}} - \dfrac{14}{r} + 1 = \dfrac{2r^{2} + 38}{r^{2}}</math>. Multiplying both sides by <math>r^{2}</math>, we get that <math>49 - 14r + r^{2} = 2r^{2} + 38</math>, or <math>r^2 + 14r - 11 = 0</math>. We get <math>r = -7 \pm 2\sqrt{15}</math>, but we want the positive solution, which is <math>r = 2\sqrt{15} - 7</math>. We have to rescale back up, so we get <math>r = 4\sqrt{15} - 14 = \sqrt{240} - 14</math>, so we get that our answer is <math>240 + 14 = \boxed{254}</math>.
+~Puck_0
 == 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

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