Contest Problem from Berkeley math tournament – 2023 :
Maria and Skyler have a square shaped cookie with a side length of 1 inch.
They split the cookie by choosing two points on distinct sides of the cookie uniformly at random and cutting across the line segment formed by connecting the two points.
If Maria always gets the larger piece, what is the expected amount of extra cookie in Maria's piece compared to Skyler's in square inches.
Does this problem indicate that the choosing of sides and cutting across in the same cookie every time or a different square shaped cookie.
Best Answer
Let the Purple line indicate a "Cut" on the Green Cookie.
(A) Let the Cookie be cut on adjacent sides like this , where I am showing multiple random attempts :
(B) Let the Cookie be cut on opposite sides like this , where I am showing multiple random attempts :
We have 2 Cases for (B) & 4 Cases for (A) , via Basic Combinatorics.
We have to take "Weighted Average" to get total Expectation.
With Double Integration , we will get Area for Case (A) , where the 2 variables are between $0$ & $1$.
Case (B) is Symmetric , hence I assume that it will work out to Maria getting $3/4$
Let us try to derive that actual value.
When bottom Point is $x$ & top Point is $y$ & the Cut is made :
Area to the left is less than $0.5$ when Center is not included
Area to the right is less than $0.5$ when Center is included
The switch-over occurs when the Cut is through the Center.
Both $x$ $y$ vary between $0$ & $1$
We see that the larger Area is $[(1-x)+(1-y)]/2$ for $y$ between $y=0$ & $y=1-x$.
When switch-over occurs , the larger Area is $[x+y]/2$ for $y$ between $y=1-x$ & $y=1$.
Expected Area :
$$\frac{1}{2}\int_0^1 \int_0^{1-x} [(1-x)+(1-y)] dydx+\frac{1}{2}\int_0^1 \int_{1-x}^1 [x+y] dydx$$ $$\frac{1}{2}\int_0^1 \int_0^{1-x} [2-x-y] dydx+\frac{1}{2}\int_0^1 \int_{1-x}^1 [x+y] dydx$$ My assumption that it will be $3/4$ was wrong , Wolfram Online tool says that it is $2/3$.
Case (A) Double Integration for smaller piece :
$$\int_0^1\int_0^1 (xy/2) dx dy$$ $$\int_0^1 (x^2y/4)_0^1 dy$$ $$\int_0^1 (y/4) dy$$ $$(y^2/8)_0^1$$ $$1/8$$ Hence , larger piece is $7/8$
"Weighted Average" is $7/8 \times 4/6 + 2/3 \times 2/6 = 29/36$
Extra Amount = $29/36-7/36=11/18$