Boxing at the 1952 Summer Olympics
Final results for the boxing competition at the 1952 Summer Olympics. The events were held at Messuhalli. From this edition of the Olympic Games, the bronze medal match was abolished.[1]
Boxing at the 1952 Summer Olympics | |
---|---|
Flyweight | men |
Bantamweight | men |
Featherweight | men |
Lightweight | men |
Light welterweight | men |
Welterweight | men |
Light middleweight | men |
Middleweight | men |
Light heavyweight | men |
Heavyweight | men |
Medal summary
Medal table
Rank | Nation | Gold | Silver | Bronze | Total |
---|---|---|---|---|---|
1 | 5 | 0 | 0 | 5 | |
2 | 1 | 1 | 1 | 3 | |
3 | 1 | 1 | 0 | 2 | |
4 | 1 | 0 | 4 | 5 | |
5 | 1 | 0 | 0 | 1 | |
1 | 0 | 0 | 1 | ||
7 | 0 | 2 | 4 | 6 | |
8 | 0 | 1 | 3 | 4 | |
9 | 0 | 1 | 1 | 2 | |
0 | 1 | 1 | 2 | ||
0 | 1 | 1 | 2 | ||
0 | 1 | 1 | 2 | ||
13 | 0 | 1 | 0 | 1 | |
14 | 0 | 0 | 1 | 1 | |
0 | 0 | 1 | 1 | ||
0 | 0 | 1 | 1 | ||
0 | 0 | 1 | 1 | ||
Totals (17 nations) | 10 | 10 | 20 | 40 |
gollark: If you multiply the `(x-1)` by `(ax^3+bx^2+cx+d)` it should expand out into having an x^4 term.
gollark: I'm probably explaining this badly, hmmm.
gollark: Then set the x^4/x^3/x^2/x^1 coefficients and constant terms on each side to be equal and work out a/b/c/d.
gollark: Set it equal to `(x-1)(ax^3+bx^2+cx+d)` (the thing you know it's divisible by times the generalized cubic thingy), and expand that out/simplify.
gollark: It would be annoying and inconsistent if it was 0. It's 1.
References
- "Boxing at the 1952 Helsinki Summer Games". Sports Reference. Archived from the original on 17 April 2020. Retrieved 7 November 2018.
External links
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