Difference between revisions of "2019 AMC 10B Problems/Problem 18"
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Sevenoptimus (talk | contribs) m (Fixed references to points A and B (whereas in the problem statement, these are distances)) |
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==Solution 1== | ==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> | + | 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 substuting, we get <math>B - A = \frac{3}{4}(2 - A)</math>. Adding these equations now gives <math>2(B - A) = \frac{3}{4}(2 + B - A)</math>. Multiplying by <math>4</math>, we get <math>8(B - A) = 6 + 3(B - A)</math>, so <math>B - A = \frac{6}{5} = \boxed{\textbf{(C) } 1\frac{1}{5}}</math>. |
− | ==Solution 2 ( | + | ==Solution 2 (not rigorous)== |
− | We assume that Henry is walking back and forth exactly between points <math> | + | 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>. |
==Video Solution== | ==Video Solution== | ||
For those who want a video solution: https://youtu.be/45kdBy3htOg | For those who want a video solution: https://youtu.be/45kdBy3htOg | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2019|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:44, 18 February 2019
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 substuting, we get
. Adding these equations now gives
. Multiplying by
, we 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,
.
Video Solution
For those who want a video solution: https://youtu.be/45kdBy3htOg
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.