MHB How Many Races to Find the Top 3 Horses from 25 with Only 5 Tracks?

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To determine the top three fastest horses out of 25 using only five tracks, a strategic approach is required. Initially, conduct five races with five horses each, ranking the horses within each race. Next, take the winners of these races and race them against each other to identify the fastest horse. The second and third fastest horses can then be deduced from the horses that finished closely behind the top contenders in the previous races. This method ensures that the minimum number of races needed to accurately identify the top three horses is 7.
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you have 25 horses and you have to pick fastest 3 out of the 25. In each race
only 5 horses can run at the same time as there are only 5 tracks. what is the
minimax number of races to ensure the 3 horses can be chosen without using a stopwatch ?
(suppose the speeds of all horses are different)
 
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Albert said:
you have 25 horses and you have to pick fastest 3 out of the 25. In each race
only 5 horses can run at the same time as there are only 5 tracks. what is the
minimax number of races to ensure the 3 horses can be chosen without using a stopwatch ?
(suppose the speeds of all horses are different)
my solution:
after each race two horses will be eliminated from competition ,
so when race 8 is finished ,there are only 9 horses remained marked with $A_1,A_2,A_3,A_4,A_5,A_6,A_7,A_8,A_9$
we arrange race 9:$A_1,A_2,A_3,A_4,A_5$
race 10: $A_1,A_2,A_3,A_6,A_7$ $(A_4,A_5)$ out from race 9
race 11:$A_1,A_2,A_3,A_8,A_9$ $(A_6,A_7)$ out from race 10
after race 11 the top 3 can be produced
$25-11\times 2=3$
 
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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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