Difference between revisions of "2017 AIME II Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | 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=\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>. | + | 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>. |
==Solution 2== | ==Solution 2== | ||
Line 23: | Line 23: | ||
-william122 | -william122 | ||
− | =See Also= | + | ==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> | ||
+ | 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 <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 02:35, 23 January 2024
Contents
Problem
Circle has radius
, and the point
is a point on the circle. Circle
has radius
and is internally tangent to
at point
. Point
lies on circle
so that
is located
counterclockwise from
on
. Circle
has radius
and is internally tangent to
at point
. In this way a sequence of circles
and a sequence of points on the circles
are constructed, where circle
has radius
and is internally tangent to circle
at point
, and point
lies on
counterclockwise from point
, as shown in the figure below. There is one point
inside all of these circles. When
, the distance from the center
to
is
, where
and
are relatively prime positive integers. Find
.
Solution 1
Impose a coordinate system and let the center of be
and
be
. Therefore
,
,
,
, and so on, where the signs alternate in groups of
. The limit of all these points is point
. Using the sum of infinite geometric series formula on
and reducing the expression:
,
. Thus, we get
. The distance from
to the origin is
Let
, and the distance from the origin is
.
.
Solution 2
Let the center of circle be
. Note that
is a right triangle, with right angle at
. Also,
, or
. It is clear that
, so
. Our answer is
-william122
Solution 3
Note that there is an invariance, Consider the entire figure . Perform a
counterclockwise rotation, then scale by
with respect to
. It is easy to see that the new figure
, so
is invariant.
Using the invariance, Let . Then rotating and scaling,
. Equating, we find
. The distance is thus
. Our answer is
-Isogonal
Solution 4
Using the invariance again as in Solution 3, assume is
away from the origin. The locus of possible points is a circle with radius
. Consider the following diagram.
Let the distance from to
be
. As
is invariant,
. Then by Power of a Point,
. Solving,
. Our answer is
-Isogonal
Solution 5 (complex)
Let be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to
. Now accounting for rotation by
radians, we see that the common ratio is
. Thus since our first term is
, the total sum (by geometric series formula) is
. We need the distance from
so our distance is
. Our answer is
-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) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.