Area probability for two random points

Problem

On the sides $AB$ and $AC$ of an equilateral triangle $ABC$, points $P$ and $Q$ are chosen independently and uniformly at random. The question asks which interval contains the probability that the area of $△APQ$ is less than half the area of $△ABC$. There are five choices, and each choice is an interval.

The full problem and the choices are on the AoPS problem page.

Where it goes wrong

• Randomness plus symmetry can look like a probability of $\frac{1}{2}$. The area ratio is $xy$, not $x$ or $\frac{x+y}{2}$. When both points are midpoints the ratio is only $\frac{1}{4}$, so the cases below one half are the majority.
• Once the probability is that of $xy<\frac{1}{2}$, it is easy to treat the boundary $xy=\frac{1}{2}$ as a straight line. It is a hyperbola, and the area of the region then comes out wrong.
• It is also easy to think the problem needs $\int \frac{1}{x}dx=\ln x$. The question only asks which interval the probability falls in, and a squeeze is enough.

Idea

Let $x=\frac{AP}{AB}$ and $y=\frac{AQ}{AC}$. They are independent and uniform on $[0,1]$. The two triangles share $\angle A$:

The probability is the area of the unit square where $xy<\frac{1}{2}$. The complement $xy\ge \frac{1}{2}$ lies above the curve $y=\frac{1}{2x}$, and only inside ${\left[\frac{1}{2},1\right]}^2$:

(The integral of $\frac{1}{x}$ and the natural logarithm $\ln$ are beyond Taiwan’s high-school curriculum. They are used here only to name the exact value.)

An integral is not required. The complement contains the triangle with vertices $\left(\frac{1}{2},1\right)$, $(1,1)$, and $\left(1,\frac{1}{2}\right)$. The curve bends upward, so the chord lies above the curve, and the complement has area greater than $\frac{1}{8}$. The complement is also contained in the square of side $\frac{1}{2}$, so its area is less than $\frac{1}{4}$. The probability therefore lies between $\frac{3}{4}$ and $\frac{7}{8}$, which is choice (D).

What the figure shows

• Drag $P$ and $Q$ on the left, or the point on the right, or move x = AP/AB and y = AQ/AC. Gold $△APQ$ stays in step with the point in the square. Once that point is above and to the right of the gold curve $xy=\frac{1}{2}$, the area ratio is no longer less than one half.
• Twenty thousand random pairs $(x,y)$ are drawn. Blue points sit under the curve and red points above it. The polyline below is the running proportion. It settles near $0.8466$ and stops inside the gold band $\left(\frac{3}{4},\frac{7}{8}\right]$.
• Press Show the bounds. The red region is contained in the dashed square and contains the gold triangle, so the probability is trapped between $\frac{3}{4}$ and $\frac{7}{8}$ with no integral.