Difference between revisions of "2019 AMC 10B Problems/Problem 18"
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− | + | ==Problem== | |
+ | |||
+ | Henry decides one morning to do a workout, and he walks <math>\tfrac{3}{4}</math> of the way from his home to his gym. The gym is <math>2</math> kilometers away from Henry's home. At that point, he changes his mind and walks <math>\tfrac{3}{4}</math> of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks <math>\tfrac{3}{4}</math> of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked <math>\tfrac{3}{4}</math> of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point <math>A</math> kilometers from home and a point <math>B</math> kilometers from home. What is <math>|A-B|</math>? | ||
+ | |||
+ | <math>\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 1 \frac{1}{5} \qquad \textbf{(D) } 1 \frac{1}{4} \qquad \textbf{(E) } 1 \frac{1}{2}</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | Let the two points that Henry walks in between be <math>P</math> and <math>Q</math>, with <math>P</math> being closer to home. As given in the problem statement, the distances of the points <math>P</math> and <math>Q</math> from his home are <math>A</math> and <math>B</math> respectively. By symmetry, the distance of point <math>Q</math> from the gym is the same as the distance from home to point <math>P</math>. | ||
+ | |||
+ | Thus, <math>A = 2 - B</math>. | ||
+ | |||
+ | In addition, when he walks from point <math>Q</math> to home, he walks <math>\frac{3}{4}</math> of the distance, ending at point <math>P</math>. Therefore, we know that <math>B - A = \frac{3}{4}B</math>. | ||
+ | |||
+ | By substituting, we get <math>B - (2-B) = \frac{3}{4}\cdot B</math> and we solve to get <math>B=\dfrac{8}{5}</math>, so <math>A=2-\dfrac{8}{5}=\dfrac{2}{5}</math>. | ||
+ | |||
+ | <math>|A-B|=\left|\dfrac{2}{5}-\dfrac{8}{5} \right|=\frac{6}{5}=\boxed{\textbf{(C) } 1 \frac{1}{5}}</math>. | ||
+ | |||
+ | ==Solution 2 (Not Rigorous)== | ||
+ | We assume that Henry is walking back and forth exactly between points <math>P</math> and <math>Q</math>, with <math>P</math> closer to Henry's home than <math>Q</math>. Denote Henry's home as a point <math>H</math> and the gym as a point <math>G</math>. Then <math>HP:PQ = 1:3</math> and <math>PQ:QG = 3:1</math>, so <math>HP:PQ:QG = 1:3:1</math>. Therefore, <math>|A-B| = PQ = \frac{3}{1+3+1} \cdot 2 = \frac{6}{5} = \boxed{\textbf{(C) } 1 \frac{1}{5}}</math>. | ||
+ | |||
+ | ==Solution 3 (not rigorous; similar to 2)== | ||
+ | Since Henry is very close to walking back and forth between two points, let us denote <math>A</math> closer to his house, and <math>B</math> closer to the gym. Then, let us denote the distance from <math>A</math> to <math>B</math> as <math>x</math>. If Henry was at <math>B</math> and walked <math>\frac{3}{4}</math> of the way, he would end up at <math>A</math>, vice versa. Thus we can say that the distance from <math>A</math> to the gym is <math>\frac{1}{4}</math> the distance from <math>B</math> to his house. That means it is <math>\frac{1}{3}x</math>. This also applies to the other side. Furthermore, we can say <math>\frac{1}{3}x</math> + <math>x</math> + <math>\frac{1}{3}x</math> = <math>2</math>. We solve for <math>x</math> and get <math>x=\frac{6}{5}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) } 1\frac{1}{5}}</math>. | ||
+ | |||
+ | ~aryam | ||
+ | |||
+ | ==Solution 4 == | ||
+ | |||
+ | Let <math>A</math> have a distance of <math>x</math> from the home. Then, the distance to the gym is <math>2-x</math>. This means point <math>B</math> and point <math>A</math> are <math>\frac{3}{4} \cdot (2-x)</math> away from one another. It also means that Point <math>B</math> is located at <math>\frac{3}{4} (2-x) + x.</math> So, the distance between the home and point <math>B</math> is also <math>\frac{3}{4} (2-x) + x.</math> | ||
+ | |||
+ | It follows that point <math>A</math> must be at a distance of <math>\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right)</math> from point <math>B</math>. However, we also said that this distance has length <math>\frac{3}{4} (2-x)</math>. So, we can set those equal, which gives the equation: <cmath>\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right) = \frac{3}{4} (2-x).</cmath> | ||
+ | |||
+ | Solving, we get <math>x = \frac{2}{5}</math>. This means <math>A</math> is at point <math>\frac{2}{5}</math> and <math>B</math> is at point <math>\frac{3}{4} \cdot \frac{8}{5} + \frac{2}{5} = \frac{8}{5}.</math> | ||
+ | |||
+ | So, <math>|A - B| = \frac{6}{5}=\boxed{\textbf{(C) } 1\frac{1}{5}}.</math> | ||
+ | |||
+ | == Solution 5 (Rigorous) == | ||
+ | Let A be the point closer to Henry’s home, and B be the point closer to the gym. Define <math>(a_n)</math> to be the position of Henry after <math>2n</math> walks. Similarly, define <math>(b_n)</math> to be the position of Henry after <math>2n - 1</math> walks. Thus, <math>a_1 = \frac{1}{4} \cdot (\frac{3}{4} \cdot 2) = \frac{3}{8}</math> and <math>b_1 = \frac{3}{4} \cdot 2 = \frac{3}{2}</math>. We can also deduce that <cmath>a_n = \frac{1}{4} ( \frac{3}{4} (2 - a_{n-1}) + a_{n-1} ) = \frac{1}{16} a_{n-1} + \frac{3}{8}</cmath> (<math>2 - a_{n-1}</math> is Henry's distance to the gym, so we take <math>\frac{3}{4}</math> of that and add it to our original position. Then, we take <math>\frac{1}{4}</math> of that to obtain Henry's distance from home). Similarly, we can deduce that <cmath>b_n = \frac{3}{4} (2 - \frac{1}{4} b_{n-1}) + \frac{1}{4} b_{n-1} = \frac{1}{16} b_{n-1} + \frac{3}{2}</cmath> Now, we follow the standard procedure to convert this arithmetico geometric recursion into a closed form. Let <math>a_n - k = \frac{1}{16} (a_{n-1} -k)</math> for some constant <math>k</math>. Then, <math>a_n = \frac{1}{16} a_{n-1} + \frac{15}{16} k</math>. So, <math>\frac{1}{16} a_{n-1} + \frac{15}{16} k = \frac{1}{16} a_{n-1} + \frac{3}{8} \Rightarrow k = \frac{3}{8} \cdot \frac{16}{15} = \frac{2}{5}</math>. This means that <cmath>a_n - \frac{2}{5} = \frac{1}{16} (a_{n-1} - \frac{2}{5}) \Rightarrow a_n - \frac{2}{5} = (\frac{1}{16})^{n-1} (a_1 - \frac{2}{5}) = (\frac{1}{16})^{n-1} (\frac{3}{8} - \frac{2}{5}) = (\frac{1}{16})^{n-1} \cdot -\frac{1}{40} \Rightarrow a_n = \frac{2}{5} - \frac{1}{16^{n-1} \cdot 40}</cmath> Now, calculating <cmath>\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{2}{5} - \frac{1}{16^{n-1} \cdot 40} = \frac{2}{5} - \lim_{n \to \infty} \frac{1}{16^{n-1} \cdot 40} = \frac{2}{5} - 0 = \frac{2}{5} </cmath> Thus, <math>A = \frac{2}{5}</math>. Taking a similar process for <math>B</math>, we derive that <math>b_n = \frac{8}{5} - \frac{1}{16^{n-1} \cdot 10}</math>, so <math>B = \lim_{n \to \infty} \frac{8}{5} - \frac{1}{16^{n-1} \cdot 10} = \frac{8}{5}</math>. Finally, <math>|A-B| = |\frac{2}{5} - \frac{8}{5}| = \boxed{\frac{6}{5}}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:CrazyVideoGamez CrazyVideoGamez] | ||
+ | |||
+ | == Note == | ||
+ | This problem can be estimated by drawing out a diagram. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/4WttvHavnkM?t=55 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution by On The Spot STEM== | ||
+ | https://youtu.be/45kdBy3htOg | ||
+ | |||
+ | == Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/U5PjjZ-5MIQ | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2019|ab=B|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:28, 15 July 2024
Contents
[hide]Problem
Henry decides one morning to do a workout, and he walks of the way from his home to his gym. The gym is
kilometers away from Henry's home. At that point, he changes his mind and walks
of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks
of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked
of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point
kilometers from home and a point
kilometers from home. What is
?
Solution 1
Let the two points that Henry walks in between be and
, with
being closer to home. As given in the problem statement, the distances of the points
and
from his home are
and
respectively. By symmetry, the distance of point
from the gym is the same as the distance from home to point
.
Thus, .
In addition, when he walks from point to home, he walks
of the distance, ending at point
. Therefore, we know that
.
By substituting, we get and we solve to get
, so
.
.
Solution 2 (Not Rigorous)
We assume that Henry is walking back and forth exactly between points and
, with
closer to Henry's home than
. Denote Henry's home as a point
and the gym as a point
. Then
and
, so
. Therefore,
.
Solution 3 (not rigorous; similar to 2)
Since Henry is very close to walking back and forth between two points, let us denote closer to his house, and
closer to the gym. Then, let us denote the distance from
to
as
. If Henry was at
and walked
of the way, he would end up at
, vice versa. Thus we can say that the distance from
to the gym is
the distance from
to his house. That means it is
. This also applies to the other side. Furthermore, we can say
+
+
=
. We solve for
and get
. Therefore, the answer is
.
~aryam
Solution 4
Let have a distance of
from the home. Then, the distance to the gym is
. This means point
and point
are
away from one another. It also means that Point
is located at
So, the distance between the home and point
is also
It follows that point must be at a distance of
from point
. However, we also said that this distance has length
. So, we can set those equal, which gives the equation:
Solving, we get . This means
is at point
and
is at point
So,
Solution 5 (Rigorous)
Let A be the point closer to Henry’s home, and B be the point closer to the gym. Define to be the position of Henry after
walks. Similarly, define
to be the position of Henry after
walks. Thus,
and
. We can also deduce that
(
is Henry's distance to the gym, so we take
of that and add it to our original position. Then, we take
of that to obtain Henry's distance from home). Similarly, we can deduce that
Now, we follow the standard procedure to convert this arithmetico geometric recursion into a closed form. Let
for some constant
. Then,
. So,
. This means that
Now, calculating
Thus,
. Taking a similar process for
, we derive that
, so
. Finally,
.
Note
This problem can be estimated by drawing out a diagram.
Video Solution by OmegaLearn
https://youtu.be/4WttvHavnkM?t=55
~ pi_is_3.14
Video Solution by On The Spot STEM
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.