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

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-Isogonal
 
-Isogonal
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==Solution 5==
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We place this on the complex plane with <math>C_0=0</math>. Notice that <math>B</math> is the center of <math>C_k</math> as <math>k</math> approaches <math>\infty</math>. We have <math>C_1=C_0+1-\frac{11}{60}=C_0+\frac{49}{60}</math>. Similarly, <math>C_2=C_1+\frac{11}{60}\cdot\frac{49}{60}i</math>, <math>C_3=C_2+\left(\frac{11}{60}\right)^2\cdot\frac{49}{60}i^2</math>, and so on. Therefore, <math>B=\frac{49}{60}+\frac{49}{60}\cdot\frac{11}{60}i+\frac{49}{60}\cdot\left(\frac{11}{60}i\right)^2+\ldots</math>
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It follows that <math>B=\frac{\frac{49}{60}}{1-\frac{11}{60}i}=\frac{49}{60-11i}</math>. We seek <math>|B|</math>, which is <math>\frac{49}{|60-11i|}=\frac{49}{61}</math>, and the answer is <math>49+61=\boxed{110}</math>.
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-TheUltimate123
  
 
=See Also=
 
=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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:07, 4 March 2018

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 geometric series formula on $B$ and reducing the expression, 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

We place this on the complex plane with $C_0=0$. Notice that $B$ is the center of $C_k$ as $k$ approaches $\infty$. We have $C_1=C_0+1-\frac{11}{60}=C_0+\frac{49}{60}$. Similarly, $C_2=C_1+\frac{11}{60}\cdot\frac{49}{60}i$, $C_3=C_2+\left(\frac{11}{60}\right)^2\cdot\frac{49}{60}i^2$, and so on. Therefore, $B=\frac{49}{60}+\frac{49}{60}\cdot\frac{11}{60}i+\frac{49}{60}\cdot\left(\frac{11}{60}i\right)^2+\ldots$

It follows that $B=\frac{\frac{49}{60}}{1-\frac{11}{60}i}=\frac{49}{60-11i}$. We seek $|B|$, which is $\frac{49}{|60-11i|}=\frac{49}{61}$, and the answer is $49+61=\boxed{110}$.

-TheUltimate123

See Also

2017 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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