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

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==Problem==
 
==Problem==
4 random points are chosen on a sphere. What is the probability that the tetrahedron with vertices of the 4 points contains the center of the sphere?
 
  
(A) 1/2 (B) 1/4 (C) 3/8 (D) 1/8 (E) 3/10
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Sergai skated around an ice rink, gliding along different paths. The gray lines in the figures below show four of the paths labeled P, Q, R, and S. What is the sorted order of the four paths from shortest to longest?
(Source: Putnam)
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lmao
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[[Image:2024_AMC_8-Problem_6.png|500px]]
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<math>\textbf{(A)}\ P,Q,R,S \qquad \textbf{(B)}\ P,R,S,Q \qquad \textbf{(C)}\ Q,S,P,R \qquad \textbf{(D)}\ R,P,S,Q \qquad \textbf{(E)}\ R,S,P,Q</math>
  
 
==Solution 1==
 
==Solution 1==
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You can measure the lengths of the paths until you find a couple of guaranteed true inferred statements as such:
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Q is greater than S,
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P is greater than R,
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and R and P are the smallest two, therefore the order is R, P, S, Q.
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Thus we get the answer (D) R, P, S, Q
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- U-King
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==Solution 2 (Intuitive)==
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Obviously Path Q is the longest path, followed by Path S.
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So, it is down to Paths P and R.
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Notice that curved lines are always longer than the straight ones that meet their endpoints, therefore Path P is longer than Path R.
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Thus, the order from shortest to longest is <math>\boxed{\textbf{(D) } \text{R, P, S, Q}}</math>.
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~MrThinker
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==Video Solution 1 by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=V-xN8Njd_Lc
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~NiuniuMaths
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==Video Solution by Math-X (First fully understand the problem!!!)==
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https://youtu.be/BaE00H2SHQM?si=ZedvqIYTDG3D20Rp&t=1301
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~Math-X
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==Video Solution by Power Solve (easy to digest!)==
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https://www.youtube.com/watch?v=16YYti_pDUg
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==Video Solution by Interstigation==
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https://youtu.be/ktzijuZtDas&t=386
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==See Also==
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{{AMC8 box|year=2024|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 09:41, 19 February 2024

Problem

Sergai skated around an ice rink, gliding along different paths. The gray lines in the figures below show four of the paths labeled P, Q, R, and S. What is the sorted order of the four paths from shortest to longest?

2024 AMC 8-Problem 6.png

$\textbf{(A)}\ P,Q,R,S \qquad \textbf{(B)}\ P,R,S,Q \qquad \textbf{(C)}\ Q,S,P,R \qquad \textbf{(D)}\ R,P,S,Q \qquad \textbf{(E)}\ R,S,P,Q$

Solution 1

You can measure the lengths of the paths until you find a couple of guaranteed true inferred statements as such: Q is greater than S, P is greater than R, and R and P are the smallest two, therefore the order is R, P, S, Q. Thus we get the answer (D) R, P, S, Q

- U-King

Solution 2 (Intuitive)

Obviously Path Q is the longest path, followed by Path S.

So, it is down to Paths P and R.

Notice that curved lines are always longer than the straight ones that meet their endpoints, therefore Path P is longer than Path R.

Thus, the order from shortest to longest is $\boxed{\textbf{(D) } \text{R, P, S, Q}}$.

~MrThinker

Video Solution 1 by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths


Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=ZedvqIYTDG3D20Rp&t=1301

~Math-X

Video Solution by Power Solve (easy to digest!)

https://www.youtube.com/watch?v=16YYti_pDUg

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=386

See Also

2024 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 AJHSME/AMC 8 Problems and Solutions

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