Difference between revisions of "2024 AMC 8 Problems/Problem 6"

(Video Solution 1(easy to digest) by Power Solve)
Line 1: Line 1:
 +
==Solution 1 (Analysis)==
 +
 +
<math>R</math> skips around the boundary of the rink, while <math>P</math> goes around the whole boundary. Hence, the length of path <math>P</math> is less than the length of path <math>R</math>. Now, using the fact that the hypotenuse of a right triangle is greater than both of its legs, it is clear that the path described in <math>S</math> is longer than <math>P</math>. Finally, each V-shaped zag path from path <math>Q</math> is longer than a diagonal in path <math>S</math>, so the length of path <math>Q</math> is greater than that of <math>S</math>. Collectively, we obtain the answer <math>\boxed{\textbf{(D)}~R, P, S, Q}</math>.
 +
 
==Video Solution 1(easy to digest) by Power Solve==
 
==Video Solution 1(easy to digest) by Power Solve==
 
https://yo
 
https://yo

Revision as of 18:10, 25 January 2024

Solution 1 (Analysis)

$R$ skips around the boundary of the rink, while $P$ goes around the whole boundary. Hence, the length of path $P$ is less than the length of path $R$. Now, using the fact that the hypotenuse of a right triangle is greater than both of its legs, it is clear that the path described in $S$ is longer than $P$. Finally, each V-shaped zag path from path $Q$ is longer than a diagonal in path $S$, so the length of path $Q$ is greater than that of $S$. Collectively, we obtain the answer $\boxed{\textbf{(D)}~R, P, S, Q}$.

Video Solution 1(easy to digest) by Power Solve

https://yo