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September 3rd, 2018, 10:30 PM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 310 Thanks: 1  How to solve this circular track problem?
A, B, C are running clockwise around a circular track. When A completes the first round, the distance between B and C is 2/3 the distance between A and B. When A completes the 2nd round, C reaches the same position where B was at the time A finished the Ist round. If A is the fastest and C is the slowest among them. What is the ratio of speeds of A, B and C? I tried: Round 1: A is at X, B is at Y, and C is at (2/3)(XY) Round 2: A is at 2X, B is at 2Y, and C is at Y. From here, how to proceed? 
September 4th, 2018, 03:19 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,541 Thanks: 1750 
After round 2, C's total distance run has doubled to be Y, so C runs half as fast as B. Hence Y/2 = (2/3)(X  Y), and so Y/X = 4/7. Hence the desired ratio is 1:4/7:2/7, which is 7:4:2. Note that although A has returned to the start position after round 1, the distance between A and B must then be X  Y rather than Y. 
September 4th, 2018, 07:01 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,462 Thanks: 106 
I get inconsistent conditions: After first lap for A: DB1DC1=2/3 DB1 > DC1=1/3 DB1 After second lap for A, ie, after same period of time has elapsed DC2=2DC1=DB1  > DC1=1/2 DB1 EDIT: So my assumptions were wrong. Either B, or B and C must have completed a lap. I'll assume skipjack's proof is correct. . Last edited by zylo; September 4th, 2018 at 07:15 AM. 
September 4th, 2018, 10:40 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,541 Thanks: 1750 
The problem states that C is the slowest runner, so B must complete more than a lap overall. In round 1, B covered 4/7 of a lap, reaching a point Y that was 3/7 of a lap from the starting point, so the distance that the problem refers to as "the distance between A and B" is 3/7 of a lap, not 4/7 of a lap. 
September 4th, 2018, 01:17 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,462 Thanks: 106 
The problem is not solvable. Assume they all start together and A has run a lap ccw. If you draw a circle, A is then at the top of the circle, and ccw from A is C and then B. First condition: Let b, c be respective distances of B, C from A after one lap of A. bc=2/3 b, c=1/3b In the time A runs a second lap, C moves a distance c again to 2/3 b, and so can't reach B. After the third lap of A, C reaches the initial position of b. Last edited by skipjack; September 4th, 2018 at 02:07 PM. 
September 4th, 2018, 02:07 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,541 Thanks: 1750 
That makes essentially the same mistake you made previously.

September 5th, 2018, 03:33 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,462 Thanks: 106  
September 5th, 2018, 09:09 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,541 Thanks: 1750 
As explained in post #4.

September 5th, 2018, 09:18 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,129 Thanks: 914 
Assume track is straight road = 28 Assume Skip's ratio: 7:4:2. Make up simple scenario, like speeds A=28, B=16, C=8 (LCM(7,4,2): A...............28................>(28)28>A(56) B..........16.........>(16)16B(32) C...8...>(8)8>C(16) "When A completes the first round, the distance between B and C is 2/3 the distance between A and B." B  C = 8 = 2/3(12) "When A completes the 2nd round, C reaches the same position where B was at the time A finished the 1st round." B  C = 2/3(A  B) 8 = 2/3(12) 8 = 8 Soooooo: Skip is correct! Last edited by Denis; September 5th, 2018 at 09:34 AM. 
September 5th, 2018, 11:59 AM  #10  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,462 Thanks: 106 
Reference previous post. After first lap A is at beginning, B is at 16 and C is at 8. Distance between B and C is 8 and distance between A and B is 16. Quote:
Soooooooo. skip AND Denis are wrong.  

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