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Difference between revisions of "2017 AIME II Problems/Problem 12"

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</asy>
 
</asy>
  
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==Solution==
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==Solution 1==
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Impose a coordinate system and let the center of <math>C_0</math> be <math>(0,0)</math> and <math>A_0</math> be <math>(1,0)</math>. Therefore <math>A_1=(1-r,r)</math>, <math>A_2=(1-r-r^2,r-r^2)</math>, <math>A_3=(1-r-r^2+r^3,r-r^2-r^3)</math>, <math>A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)</math>, and so on, where the signs alternate in groups of <math>2</math>. The limit of all these points is point <math>B</math>. Using the geometric series formula on <math>B</math> and reducing the expression, we get <math>B=(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1})</math>. The distance from <math>B</math> to the origin is <math>\sqrt{(\frac{1-r}{r^2+1})^2+(\frac{r-r^2}{r^2+1}))^2}=\frac{1-r}{\sqrt{r^2+1}}.</math> Let <math>r=\frac{11}{60}</math>, we find that the distance from the origin is <math>\frac{49}{61} \implies 49+61=\boxed{110}</math>.
+
Impose a coordinate system and let the center of <math>C_0</math> be <math>(0,0)</math> and <math>A_0</math> be <math>(1,0)</math>. Therefore <math>A_1=(1-r,r)</math>, <math>A_2=(1-r-r^2,r-r^2)</math>, <math>A_3=(1-r-r^2+r^3,r-r^2-r^3)</math>, <math>A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)</math>, and so on, where the signs alternate in groups of <math>2</math>. The limit of all these points is point <math>B</math>. Using the sum of infinite geometric series formula on <math>B</math> and reducing the expression: <math>(1-r)-(r^2-r^3)+(r^4-r^5)-...=\frac{1-r}{1+r^2}</math>, <math>(r-r^2)-(r^3-r^4)+(r^5-r^6)-...=\frac{r-r^2}{1+r^2}</math>. Thus, we get <math>B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)</math>. The distance from <math>B</math> to the origin is <math>\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.</math> Let <math>r=\frac{11}{60}</math>, and the distance from the origin is <math>\frac{49}{61}</math>.  <math>49+61=\boxed{110}</math>.
  
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=See Also=
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==Solution 2==
 +
Let the center of circle <math>C_i</math> be <math>O_i</math>. Note that <math>O_0BO_1</math> is a right triangle, with right angle at <math>B</math>. Also, <math>O_1B=\frac{11}{60}O_0B</math>, or <math>O_0B = \frac{60}{61}O_0O_1</math>. It is clear that <math>O_0O_1=1-r=\frac{49}{60}</math>, so <math>O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math>
 +
 
 +
-william122
 +
 
 +
==Solution 3==
 +
Note that there is an invariance, Consider the entire figure <math>\mathcal{F}</math>. Perform a <math>90^\circ</math> counterclockwise rotation, then scale by <math>r</math> with respect to <math>(1, 0)</math>. It is easy to see that the new figure <math>\mathcal{F}' \cup S^1 = \mathcal{F}</math>, so <math>B</math> is invariant.
 +
 
 +
Using the invariance, Let <math>B = (x,y)</math>. Then rotating and scaling, <math>B = (1-r(1+y), rx)</math>. Equating, we find <math>x = \frac{1-r}{r^2+1}, y = \frac{r-r^2}{r^2+1}</math>. The distance is thus <math>\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math>
 +
 
 +
-Isogonal
 +
 
 +
==Solution 4==
 +
Using the invariance again as in Solution 3, assume <math>B</math> is <math>d</math> away from the origin. The locus of possible points is a circle with radius <math>d</math>. Consider the following diagram.
 +
<asy>
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size(7cm);
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draw(circle((0,0), 49/61));
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draw((0,0)--(0.790110185, 0.144853534));
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draw((0,0)--(-0.144853534, 0.790110185));
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draw((-0.144853534, 0.790110185)--(1,0));
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draw((0,0)--(1,0));
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draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3));
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 +
label("$O$", (0,0), SW);
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label("$(1,0)$", (1,0), E);
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label("$B$", (0.790110185, 0.144853534), NE);
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label("$B'$", (-0.144853534, 0.790110185), N);
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label("$d$", (0.5 * 49/61, 0), S);
 +
 
 +
</asy>
 +
 
 +
Let the distance from <math>B</math> to <math>(1,0)</math> be <math>x</math>. As <math>B</math> is invariant, <math>x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}</math>. Then by Power of a Point, <math>x(BB' + x) = (1-d)(1+d) \implies xr(BB' + x) = r(1-d)(1+d) \implies x^2 = r(1-d^2) \implies d^2 = \left(1 + \frac{2r}{(1-r)^2}\right)</math>. Solving, <math>d = \frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math>
 +
 
 +
-Isogonal
 +
 
 +
==Solution 5 (complex)==
 +
Let <math>A_0</math> be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to <math>r</math>. Now accounting for rotation by <math>\frac{\pi}{2}</math> radians, we see that the common ratio is <math>ri</math>. Thus since our first term is <math>A_1=-r+ri</math>, the total sum (by geometric series formula) is <math>\frac{-r+ri}{1-ri}=\frac{-781+538i}{3721}</math>. We need the distance from <math>C_0=-1</math> so our distance is <math>|B-C_0|=\sqrt{\left(\frac{-781}{3721}-(-1)\right)^2+\left(\frac{538i}{3721}\right)^2}=\sqrt{\frac{2401}{3721}}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math>
 +
 
 +
-chrisdiamond10
 +
 
 +
By the way, [https://www.geogebra.org/calculator/wrtgmqmq here's] a Geogebra Diagram. ~r00tsOfUnity
 +
 
 +
==Video Solution by mop 2024==
 +
https://youtube.com/watch?v=5keaS1CZlLo
 +
 
 +
~r00tsOfUnity
 +
 
 +
==See Also==
 
{{AIME box|year=2017|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2017|n=II|num-b=11|num-a=13}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:35, 23 January 2024

Problem

Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("$A_0$",(125,0),E); dot((25,100)); label("$A_1$",(25,100),SE); dot((-55,20)); label("$A_2$",(-55,20),E); [/asy]

Solution 1

Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the sum of infinite geometric series formula on $B$ and reducing the expression: $(1-r)-(r^2-r^3)+(r^4-r^5)-...=\frac{1-r}{1+r^2}$, $(r-r^2)-(r^3-r^4)+(r^5-r^6)-...=\frac{r-r^2}{1+r^2}$. Thus, we get $B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)$. The distance from $B$ to the origin is $\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.$ Let $r=\frac{11}{60}$, and the distance from the origin is $\frac{49}{61}$. $49+61=\boxed{110}$.

Solution 2

Let the center of circle $C_i$ be $O_i$. Note that $O_0BO_1$ is a right triangle, with right angle at $B$. Also, $O_1B=\frac{11}{60}O_0B$, or $O_0B = \frac{60}{61}O_0O_1$. It is clear that $O_0O_1=1-r=\frac{49}{60}$, so $O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}$. Our answer is $49+61=\boxed{110}$

-william122

Solution 3

Note that there is an invariance, Consider the entire figure $\mathcal{F}$. Perform a $90^\circ$ counterclockwise rotation, then scale by $r$ with respect to $(1, 0)$. It is easy to see that the new figure $\mathcal{F}' \cup S^1 = \mathcal{F}$, so $B$ is invariant.

Using the invariance, Let $B = (x,y)$. Then rotating and scaling, $B = (1-r(1+y), rx)$. Equating, we find $x = \frac{1-r}{r^2+1}, y = \frac{r-r^2}{r^2+1}$. The distance is thus $\frac{49}{61}$. Our answer is $49+61=\boxed{110}$

-Isogonal

Solution 4

Using the invariance again as in Solution 3, assume $B$ is $d$ away from the origin. The locus of possible points is a circle with radius $d$. Consider the following diagram. [asy] size(7cm); draw(circle((0,0), 49/61)); draw((0,0)--(0.790110185, 0.144853534)); draw((0,0)--(-0.144853534, 0.790110185)); draw((-0.144853534, 0.790110185)--(1,0)); draw((0,0)--(1,0)); draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3)); label("$O$", (0,0), SW); label("$(1,0)$", (1,0), E); label("$B$", (0.790110185, 0.144853534), NE); label("$B'$", (-0.144853534, 0.790110185), N); label("$d$", (0.5 * 49/61, 0), S); [/asy]

Let the distance from $B$ to $(1,0)$ be $x$. As $B$ is invariant, $x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}$. Then by Power of a Point, $x(BB' + x) = (1-d)(1+d) \implies xr(BB' + x) = r(1-d)(1+d) \implies x^2 = r(1-d^2) \implies d^2 = \left(1 + \frac{2r}{(1-r)^2}\right)$. Solving, $d = \frac{49}{61}$. Our answer is $49+61=\boxed{110}$

-Isogonal

Solution 5 (complex)

Let $A_0$ be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to $r$. Now accounting for rotation by $\frac{\pi}{2}$ radians, we see that the common ratio is $ri$. Thus since our first term is $A_1=-r+ri$, the total sum (by geometric series formula) is $\frac{-r+ri}{1-ri}=\frac{-781+538i}{3721}$. We need the distance from $C_0=-1$ so our distance is $|B-C_0|=\sqrt{\left(\frac{-781}{3721}-(-1)\right)^2+\left(\frac{538i}{3721}\right)^2}=\sqrt{\frac{2401}{3721}}=\frac{49}{61}$. Our answer is $49+61=\boxed{110}$

-chrisdiamond10

By the way, here's a Geogebra Diagram. ~r00tsOfUnity

Video Solution by mop 2024

https://youtube.com/watch?v=5keaS1CZlLo

~r00tsOfUnity

See Also

2017 AIME II (Problems • Answer Key • Resources)
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Problem 11 		Followed by
Problem 13
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