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September 3rd, 2018, 10:30 PM   #1
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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)(X-Y)

Round 2: A is at 2X, B is at 2Y, and C is at Y.

From here, how to proceed?
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September 4th, 2018, 03:19 AM   #2
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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.
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September 4th, 2018, 07:01 AM   #3
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I get inconsistent conditions:

After first lap for A:

DB1-DC1=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.
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September 4th, 2018, 10:40 AM   #4
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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.
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September 4th, 2018, 01:17 PM   #5
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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.
b-c=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.
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September 4th, 2018, 02:07 PM   #6
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That makes essentially the same mistake you made previously.
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September 5th, 2018, 03:33 AM   #7
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Quote:
Originally Posted by skipjack View Post
That makes essentially the same mistake you made previously.
Which is?
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September 5th, 2018, 09:09 AM   #8
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As explained in post #4.
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September 5th, 2018, 09:18 AM   #9
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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)--------16--------B(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.
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September 5th, 2018, 11:59 AM   #10
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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:
Originally Posted by Ganesh Ujwal View Post
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.....
8 is not 2/3 of 16.

Soooooooo. skip AND Denis are wrong.
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