AIME 2014 II · 第 8 题

Solution 1

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

Solution 5 (Heron's Formula)

This solution focuses on the area of $\triangle DEC$. Because $EF$ is perpendicular to $AB$, it is an altitude of $\triangle DEC$. Therefore, we can express the area of $\triangle DEC$ as $\frac{1}{2}\cdot EF \cdot DC$. We can also express the area of $\triangle DEC$ using Heron's Formula. Let $r$ equal the radius of circle $E$. Then $DC$ = $CA - DA = 2-3r$. We also know that $CE = 2-r$ and $DE = 3r+r=4r$. The semi-perimeter of $\triangle DEC$ is $(DC+CE+DE)/2 = 2$.

Applying Heron's Formula, we get

$$\sqrt{2(2-4r)(r)(3r)} = \sqrt{6r^2(2-4r)}.$$ We set this equal to $\frac{1}{2}r(2-3r)$.

$$\sqrt{6r^2(2-4r)} = \frac{1}{2}r(2-3r).$$ This simplifies to the quadratic equation $0 = 9r^2+84r-44$. Remember that we are solving for $3r$, which we will set equal to $x$. Then we now have the equation $0 = x^2 +28x - 44$. Applying the quadratic formula, we get $x=-14 \pm 4\sqrt{15}$. We want the positive solution, so we take $4\sqrt{15}-14 = \sqrt{240}-14$. Our answer is therefore $240 + 14 = \boxed{254}$.

~lprado