Figure 8(B) shows a slice of ham on a piece of bread. Prove that it is possible to slice this open-faced sandwich so that each part has equal amounts of ham and bread. By Exercise 36, for all 0 ≤ θ ≤ π there is a line L(θ) of incline θ (which we assume is unique) that divides the ham into two equal pieces. Let B(θ) denote the amount of bread to the left of (or above) L(θ) minus the amount to the right (or below). Notice that L(π) and L(0) are the same line, but B(π) = −B(0) since left and right get interchanged as the angle moves from 0 to π. Assume that B is continuous and apply the IVT. (By a further extension of this argument, one can prove the full Ham Sandwich Theorem, which states that if you allow the knife to cut at a slant, then it is possible to cut a sandwich consisting of a slice of ham and two slices of bread so that all three layers are divided in half.)

Data From Exercise 36

Figure 8(A) shows a slice of ham. Prove that for any angle θ (0 ≤ θ ≤ π), it is possible to cut the slice in half with a cut of incline θ. The lines of inclination θ are given by the equations y = (tan θ)x + b, where b varies from −∞ to ∞. Each such line divides the slice into two pieces (one of which may be empty). Let A(b) be the amount of ham to the left of the line minus the amount to the right, and let A be the total area of the ham. Show that A(b) = −A if b is sufficiently large and A(b) = A if b is sufficiently negative. Then use the IVT. This works if θ ≠ 0 or π/2. If θ = 0, define A(b) as the amount of ham above the line y = b minus the amount below. How can you modify the argument to work when θ = π/2 (in which case tan θ = ∞)?

Transcribed Image Text:

y (5) L(6) 在 · L(0) = L(x) (B) A slice of ham on top of a slice of bread 一篇