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考题25 2015 AMC数学竞赛 12A

2018-08-22 重点归纳

AMC12是针对高中学生的数学测验，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 25

A collection of circles in the upper half-plane, all tangent to the $x$ -axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k \ge 1$ , the circles in $\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$ -axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\bigcup_{j=0}^{6}L_j$ , and for every circle $C$ denote by $r(C)$ its radius. What is $\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?$

$AMC数学竞赛$

$AMC数学竞赛12A$

Solution 1

Let us start with the two circles in $L_0$ and the circle in $L_1$ . Let the larger circle in $L_0$ be named circle $X$ with radius $x$ and the smaller be named circle $Y$ with radius $y$ . Also let the single circle in $L_1$ be named circle $Z$ with radius $z$ . Draw radii $x$ , $y$ , and $z$ perpendicular to the x-axis. Drop altitudes $a$ and $b$ from the center of $Z$ to these radii $x$ and $y$ , respectively, and drop altitude $c$ from the center of $Y$ to radius $x$ perpendicular to the x-axis. Connect the centers of circles $x$ , $y$ , and $z$ with their radii, and utilize the Pythagorean Theorem. We attain the following equations. $(x - z)^2 + a^2 = (x + z)^2 \implies a^2 = 4xz$ $(y - z)^2 + b^2 = (y + z)^2 \implies b^2 = 4yz$ $(x - y)^2 + c^2 = (x + y)^2 \implies c^2 = 4xy$

We see that $a = 2\sqrt{xz}$ , $b = 2\sqrt{yz}$ , and $c = 2\sqrt{xy}$ . Since $a + b = c$ , we have that $2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}$ . Divide this equation by $2\sqrt{xyz}$ , and this equation becomes the well-known relation of Descartes's Circle Theorem $\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.$ We can apply this relationship recursively with the circles in layers $L_2, L_3, \cdots, L_6$ .

Here, let $S(n)$ denote the sum of the reciprocals of the square roots of all circles in layer $n$ . The notation in the problem asks us to find the sum of the reciprocals of the square roots of the radii in each circle in this collection, which is $\textstyle\sum_{n=0}^{6}S(n)$ . We already have that $S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}$ . Then, $S(2) = 2S(1) + S(0) = 3S(0)$ . Additionally, $S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)$ , and $S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)$ . Now, we notice that $S(n + 1) = 3S(n)$ because $S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)$ , which is a power of $3.$ Hence, our desired sum is $(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)$ . This simplifies to $365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{ (D)}$ .

Note that the circles in this question are known as Ford circles.

Solution 2 (Pythagorean Theorem)

Let the two circles from $L_0$ be of radius $r_1$ and $r_2$ , with $r_1>r_2$ . Let the circle of radius $r_1$ be circle $A$ and the circle of radius $r_2$ be circle $B$ . Now, let the circle of $L_1$ have radius $r_3$ . Let this circle be circle $C$ . Draw the radii of the three circles down to the common tangential line and connect the radii. Draw two lines parallel to the common tangential line of the two layers intersecting the center point of circle $B$ and the center point of circle $C$ . Now, we have $3$ right triangles with a line of common length (The two parallel lines). Using the pythagorean theorem, we get the formula $amc数学竞赛试题及答案$ Now we solve for $r_3$ . Square both sides, use the identity $(a^2-b^2)=(a+b)(a-b)$ and simplify: $(2r_2)(2r_1) = (2r_1)(2r_3)+2\sqrt{16r_1r_3r_2r_3}+(2r_2)(2r_3)=4(r_1r_3+r_2r_3+2r_3\sqrt{r_1r_2})=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \\ 4r_2r_1=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \implies r_3=\frac{r_2r_1}{r_1+r_2+2\sqrt{r_1r_2}}$

Now, let's change this into a function to clean things up: $f(x,y) = \frac{xy}{x+y+2\sqrt{xy}}=\frac{xy}{(\sqrt{x}+\sqrt{y})^2}$ Let's begin to rewrite the sum we want to find in terms of the radii of the circles, call this $g(x)$ : $g(x) = \frac{1}{\sqrt{r}}=\frac{1}{\sqrt{\frac{xy}{(\sqrt{x}+\sqrt{y})^2}}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}} = \frac{\sqrt{y}}{y}+\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}$ Using this, we can find the sum of some layers: $L_0$ , $\frac{1}{70}+\frac{1}{73}$ , $L_0$ and $L_1$ : $\frac{1}{70}+\frac{1}{73}+\frac{1}{70}+\frac{1}{73} = 2(\frac{1}{70}+\frac{1}{73})$ This is interesting, we have that the sum of Layer 0 and Layer 1 is equal to twice of Layer 0. If we continue and find the sum of layers 0, 1 and 2, we see it is equal to $5(L_0)$ . This is getting very interesting, there must be some pattern. First of all, we should observe that finding $g(x)$ of a circle is equivalent to adding up those of the 2 larger circles to construct the smaller one. Second, upon further observation, we can draw out the layers. When we're finding the next layer, we can split the current layers across the center, so that each half includes the center circle $L_1$ . Now, if we were to find $g(x)$ , we notice we are doubling the current sum and including the center circle twice. So, the recursive sum would be $a_n=3a_{n-1}-1$ . So, applying this new formula, we get $\sum_{C \in S}\frac{1}{\sqrt{r}} = (3(3(3(3(3(3-1)-1)-1)-1)-1)-1)(\frac{1}{70}+\frac{1}{73})=365\cdot(\frac{1}{70}+\frac{1}{73})=365\cdot\frac{143}{70\cdot73}=\boxed{\frac{143}{14}}$