Can You Save Some Cold Pizza?

The Riddler for September 25, 2020. The Express asks how precise a golfer must be to make par when each swing closes a fixed fraction of the gap, and the Classic is a notoriously hard pizza-on-a-plate geometry problem.

Riddler Express

A golf hole is $400$ yards long; the cup is $4.25$ inches across. With every swing you hit the ball $X$ percent closer to the centre of the hole (so the remaining distance multiplies by $1 - X$). The ball drops once it is within the cup’s radius. What is the minimum $X$ for which you make par, sinking the ball in exactly four swings?

The Riddler, FiveThirtyEight, September 25, 2020(original post)

Solution

Work in inches. The hole is $400 \times 36 = 14{,}400$ inches away, and the ball falls once it is within the cup’s radius $4.25/2 = 2.125$ inches of centre. After four swings the remaining distance is $14{,}400\,(1 - X)^4$, so par needs $$14{,}400\,(1 - X)^4 \le 2.125 \quad\Longrightarrow\quad (1 - X)^4 = \frac{2.125}{14{,}400} = \frac{17}{115{,}200}.$$ Taking the fourth root, $$X = 1 - \left(\frac{17}{115{,}200}\right)^{1/4} \approx 1 - 0.11022 = \boxed{0.8898 \;(\approx 89\%)}.$$ You must erase almost $89\%$ of the gap on every swing. (A birdie, three swings, needs the cube root instead, about $95\%$, which shows how fast the bar rises as you demand fewer strokes.)

The computation

Encode the shrinking distance and find the threshold $X$ that lands the ball within the cup radius on the fourth swing.

dist = 400 * 36          # inches to the hole
radius = 4.25 / 2        # cup radius in inches
X = 1 - (radius / dist) ** (1 / 4)
print(round(100 * X, 2))                     # 88.98  (percent)

# check: four swings land just inside the cup
print(round(dist * (1 - X) ** 4, 4))         # 2.125

The minimum precision is $X \approx 88.98\%$, just under $89$.

Riddler Classic

You have a semicircular half of a $10$-inch pizza. The smallest circular plate holding the semicircle outright has diameter $10$ inches. But you may make a single straight cut and rearrange the two pieces (no overlap, nothing hanging off). What is the diameter of the smallest plate that then holds both pieces?

The Riddler, FiveThirtyEight, September 25, 2020(original post)

Deferred (open problem)

This one is, in the column’s own word, a “humdinger,” and it is not actually solved. A clean lower bound comes from the cheating arrangement where the two pieces may overlap: cut the semicircle into two quarter-circles and stack them, which fit a plate of diameter $5\sqrt2 \approx 7.07$ inches, so no honest answer can be smaller. For a genuine non-overlapping rearrangement the best results were found numerically by minimising the enclosing circle over three parameters (where on the arc the cut starts, where on the diameter it ends, and how the pieces are positioned). A cut perpendicular to the flat edge reaches a plate of about $8.79$ inches; allowing a non-perpendicular cut, the best anyone achieved was about $$\boxed{\approx 8.16 \text{ inches (best known; not proven optimal)}}.$$ The column flags this explicitly as an open problem, so there is no exact value or verifiable construction to compute here; the answer is recorded as the best-known $\approx 8.16$ inches, bounded below by $5\sqrt2 \approx 7.07$.