# Putnam, Probabilities, and Math (in Contests)

## [P]MittNovember 10 2006 3:39 PM EST

This is a contest for all you math-letes.

You have a square board. You throw a dart at the board. Assuming that you have an equal probability of hitting anywhere on that board, and you do not miss, find the probability that you hit closer to the center than you do to the edge.

The prize will be 120k and will go up by 10k each day (starting tomorrow)

There is no entry fee.

A percentage in the form of a decimal, rounding to the 9th decimal place (so I know that you have the correct answer, not guessing or checking)
(example: 70% would be written as 0.700000000)

of (A + B^(1/2))/(C) - may need more than just 3 numbers in the answer to solve) - make sure your parenthesis are in the right spot!

Have fun and don't stress out too much :)

Note: the answer is NOT 25%. It's a square, think about it.

0.333333333

## TezmacNovember 10 2006 3:47 PM EST

1/9, or .111111111

## AdminG BeeeNovember 10 2006 3:54 PM EST

Am I drunk or sober when throwing it ?

.200000000

## TezmacNovember 10 2006 3:57 PM EST

Aw man, I assumed that the person would be able sober! If the person has had 5 drinks, it drops to 1/14, 8 drinks drops it to 1/20, and being Scottish (which implies you've had more than 8 24-7) then it drops to 1/93. :O)

.500000000

## [P]MittNovember 10 2006 4:10 PM EST

deifeln - 1/3
Tezmac - 1/9, 1/93 (=P)
ser!alk!ller - 1/5
It is not a simple answer, that's why I put in the extra 10k per day.

In its simplest form, it is

a√(b)-c
d

G Beee: . Assuming that you have an equal probability of hitting anywhere on that board

Sobriety does not affect your dart-throwing ability! you can drink 20 beers and throw as if you were sober!

## AdminG BeeeNovember 10 2006 4:13 PM EST

If I drink 20 beers I'll miss the wall, never mind the dart board !

## Caedmon[Revenge of the Forgers]November 10 2006 4:15 PM EST

(4 sqrt(2)-5 )/3, or just under 22%.

## [P]MittNovember 10 2006 4:25 PM EST

oh my, that was fast =)

Mitt [t] (Contest Mitt) 71.202.240.77 Caedmon [t] (Caedmon) \$120000 4:24 PM EST

Do you mind telling everyone how you got it?

## Caedmon[Revenge of the Forgers]November 10 2006 4:38 PM EST

Sure. For sake of concreteness, take a square with corners (-1,-1), (-1,1),(1,1) and (1,-1) -- so area is 4. Consider just the eighth above y=x and to the right of the y-axis. Within this slice, points are closer to the center if

sqrt(x^2+y^2)<1-y

since the closest edge is the top. Solving for y gives y<(1-x^2)/2.

So, the area in question lies between y=x and y=(1-x^2)/2. Integrating the function (1-x^2)/2-x from x=0 to x=[sqrt(2)-1]/2 (the intersection of the two) gives [4sqrt(2)-5]/6. Eight symmetric pieces gives a total area of 4[4 sqrt(2)-5]/3, and dividing by the square's area of 4 finishes the problem.

I don't recall if this was a Putnam problem or not, but I really enjoy my bronze medal from '93. ;)
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