Difference between revisions of "2015 AMC 12A Problems/Problem 25"
(→Solution) |
m (→Problem) |
||
(19 intermediate revisions by 6 users not shown) | |||
Line 40: | Line 40: | ||
}</asy> | }</asy> | ||
− | <math> \textbf{(A)} | + | <math> \textbf{(A) } \frac{286}{35} \qquad\textbf{(B) } \frac{583}{70} \qquad\textbf{(C) } \frac{715}{73} \qquad\textbf{(D) } \frac{143}{14} \qquad\textbf{(E) } \frac{1573}{146}</math> |
==Solution 1== | ==Solution 1== | ||
Line 47: | Line 47: | ||
We see that <math>a = 2\sqrt{xz}</math>, <math>b = 2\sqrt{yz}</math>, and <math>c = 2\sqrt{xy}</math>. Since <math>a + b = c</math>, we have that <math>2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}</math>. Divide this equation by <math>2\sqrt{xyz}</math>, and this equation becomes the well-known relation of Descartes's Circle Theorem <math>\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.</math> We can apply this relationship recursively with the circles in layers <math>L_2, L_3, \cdots, L_6</math>. | We see that <math>a = 2\sqrt{xz}</math>, <math>b = 2\sqrt{yz}</math>, and <math>c = 2\sqrt{xy}</math>. Since <math>a + b = c</math>, we have that <math>2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}</math>. Divide this equation by <math>2\sqrt{xyz}</math>, and this equation becomes the well-known relation of Descartes's Circle Theorem <math>\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.</math> We can apply this relationship recursively with the circles in layers <math>L_2, L_3, \cdots, L_6</math>. | ||
− | Here, let <math>S(n)</math> denote the sum of the reciprocals of the square roots of all circles in layer <math>n</math>. 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 <math>\textstyle\sum_{n=0}^{6}S(n)</math>. We already have that <math>S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}</math>. Then, <math>S(2) = 2S(1) + S(0) = 3S(0)</math>. Additionally, <math>S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)</math>, and <math>S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)</math>. Now, we notice that <math>S(n + 1) = 3S(n)</math> because <math>S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)</math>, which is a power of <math>3.</math> Hence, our desired sum is <math>(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)</math>. This simplifies to <math>365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{ (D)}</math>. | + | Here, let <math>S(n)</math> denote the sum of the reciprocals of the square roots of all circles in layer <math>n</math>. 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 <math>\textstyle\sum_{n=0}^{6}S(n)</math>. We already have that <math>S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}</math>. Then, <math>S(2) = 2S(1) + S(0) = 3S(0)</math>. Additionally, <math>S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)</math>, and <math>S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)</math>. Now, we notice that <math>S(n + 1) = 3S(n)</math> because <math>S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)</math>, which is a power of <math>3.</math> Hence, our desired sum is <math>(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)</math>. This simplifies to <math>365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{(D)}</math>. |
Note that the circles in this question are known as Ford circles. | Note that the circles in this question are known as Ford circles. | ||
− | ==Solution 2 ( | + | ==Solution 2 (Pythagorean Theorem)== |
− | Let the two circles from <math>L_0</math> be | + | Let the two circles from <math>L_0</math> be of radius <math>r_1</math> and <math>r_2</math>, with <math>r_1>r_2</math>. Let the circle of radius <math>r_1</math> be circle <math>A</math> and the circle of radius <math>r_2</math> be circle <math>B</math>. Now, let the circle of <math>L_1</math> have radius <math>r_3</math>. Let this circle be circle <math>C</math>. 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 <math>B</math> and the center point of circle <math>C</math>. Now, we have <math>3</math> right triangles with a line of common length (The two parallel lines). Using the pythagorean theorem, we get the formula <math>\sqrt{(r_2+r_1)^2-(r_2-r_1)^2}=\sqrt{(r_1+r_3)^2-(r_1-r_3)^2}+\sqrt{(r_2+r_3)^2-(r_2-r_3)^2}</math> Now we solve for <math>r_3</math>. Square both sides, use the identity <math>(a^2-b^2)=(a+b)(a-b)</math> and simplify: <math>(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}}</math> |
Now, let's change this into a function to clean things up: <math>f(x,y) = \frac{xy}{x+y+2\sqrt{xy}}=\frac{xy}{(\sqrt{x}+\sqrt{y})^2}</math> | Now, let's change this into a function to clean things up: <math>f(x,y) = \frac{xy}{x+y+2\sqrt{xy}}=\frac{xy}{(\sqrt{x}+\sqrt{y})^2}</math> | ||
Let's begin to rewrite the sum we want to find in terms of the radii of the circles, call this <math>g(x)</math>: <math>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}}</math> Using this, we can find the sum of some layers: <math>L_0</math>, <math>\frac{1}{70}+\frac{1}{73}</math>, <math>L_0</math> and <math>L_1</math>: <math>\frac{1}{70}+\frac{1}{73}+\frac{1}{70}+\frac{1}{73} = 2(\frac{1}{70}+\frac{1}{73})</math> 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 <math>5(L_0)</math>. This is getting very interesting, there must be some pattern. First of all, we should observe that finding <math>g(x)</math> 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 <math>L_1</math>. Now, if we were to find <math>g(x)</math>, we notice we are doubling the current sum and including the center circle twice. So, the recursive sum would be <math>a_n=3a_{n-1}-1</math>. So, applying this new formula, we get <math>\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}}</math> | Let's begin to rewrite the sum we want to find in terms of the radii of the circles, call this <math>g(x)</math>: <math>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}}</math> Using this, we can find the sum of some layers: <math>L_0</math>, <math>\frac{1}{70}+\frac{1}{73}</math>, <math>L_0</math> and <math>L_1</math>: <math>\frac{1}{70}+\frac{1}{73}+\frac{1}{70}+\frac{1}{73} = 2(\frac{1}{70}+\frac{1}{73})</math> 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 <math>5(L_0)</math>. This is getting very interesting, there must be some pattern. First of all, we should observe that finding <math>g(x)</math> 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 <math>L_1</math>. Now, if we were to find <math>g(x)</math>, we notice we are doubling the current sum and including the center circle twice. So, the recursive sum would be <math>a_n=3a_{n-1}-1</math>. So, applying this new formula, we get <math>\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}}</math> | ||
+ | |||
+ | === Video Solution by Richard Rusczyk === | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc12a/402 | ||
+ | |||
+ | ~ dolfin 7 | ||
+ | |||
+ | == See Also == | ||
+ | {{AMC12 box|year=2015|ab=A|num-b=24|after=Last Problem}} |
Latest revision as of 14:24, 23 November 2023
Contents
Problem
A collection of circles in the upper half-plane, all tangent to the -axis, is constructed in layers as follows. Layer
consists of two circles of radii
and
that are externally tangent. For
, the circles in
are ordered according to their points of tangency with the
-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
consists of the
circles constructed in this way. Let
, and for every circle
denote by
its radius. What is
Solution 1
Let us start with the two circles in and the circle in
. Let the larger circle in
be named circle
with radius
and the smaller be named circle
with radius
. Also let the single circle in
be named circle
with radius
. Draw radii
,
, and
perpendicular to the x-axis. Drop altitudes
and
from the center of
to these radii
and
, respectively, and drop altitude
from the center of
to radius
perpendicular to the x-axis. Connect the centers of circles
,
, and
with their radii, and utilize the Pythagorean Theorem. We attain the following equations.
We see that ,
, and
. Since
, we have that
. Divide this equation by
, and this equation becomes the well-known relation of Descartes's Circle Theorem
We can apply this relationship recursively with the circles in layers
.
Here, let denote the sum of the reciprocals of the square roots of all circles in layer
. 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
. We already have that
. Then,
. Additionally,
, and
. Now, we notice that
because
, which is a power of
Hence, our desired sum is
. This simplifies to
.
Note that the circles in this question are known as Ford circles.
Solution 2 (Pythagorean Theorem)
Let the two circles from be of radius
and
, with
. Let the circle of radius
be circle
and the circle of radius
be circle
. Now, let the circle of
have radius
. Let this circle be circle
. 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
and the center point of circle
. Now, we have
right triangles with a line of common length (The two parallel lines). Using the pythagorean theorem, we get the formula
Now we solve for
. Square both sides, use the identity
and simplify:
Now, let's change this into a function to clean things up:
Let's begin to rewrite the sum we want to find in terms of the radii of the circles, call this
:
Using this, we can find the sum of some layers:
,
,
and
:
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
. This is getting very interesting, there must be some pattern. First of all, we should observe that finding
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
. Now, if we were to find
, we notice we are doubling the current sum and including the center circle twice. So, the recursive sum would be
. So, applying this new formula, we get
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc12a/402
~ dolfin 7
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |