China Team Selection Test

Given a convex quadrilateral $ABCD$ with side lengths $AB = a$, $BC = b$, $CD = c$, $DA = d$, and diagonals $AC = e$, $BD = f$, where $\max \{a, b, c, d, e, f\} = 1$, we aim to find the maximum value of $abcd$.

We claim that the maximum value of $abcd$ is $2 - \sqrt{3}$.

To show that this value is attainable, consider an equilateral triangle $\triangle ABC$ with side length 1. Let $D$ be the unique point such that $BD = 1$, $DA = DC$, and $ABCD$ is a convex quadrilateral. In this configuration, we have:
$$abcd = 1 \cdot 1 \cdot 2 \cos 15^\circ \cdot 2 \cos 15^\circ = 2 - \sqrt{3}.$$

To prove that this is the optimal value, we redefine "convex" to permit angles of the quadrilateral to be $180^\circ$. We call a convex quadrilateral satisfying the conditions of the problem a "tapir" if it has the maximum possible area. We show that all tapirs have area $\leq 2 - \sqrt{3}$, and we already know that all tapirs have area $\geq 2 - \sqrt{3}$.

### Lemma 1
No tasty quadrilateral has three collinear vertices.

Proof: Suppose $A, B, C$ were collinear. Then, we have:
$$AD \cdot DC \cdot CB \cdot BA \leq 1 \cdot 1 \cdot \frac{1}{4} (AB + BC)^2 \leq 1 \cdot 1 \cdot \frac{1}{4} \cdot 1 = \frac{1}{4} < 2 - \sqrt{3},$$
which contradicts the fact that $ABCD$ was a tapir.
$\blacksquare$

### Lemma 2
For every tapir $ABCD$, we have that $(A, C)$ and $(B, D)$ are both tasty.

Proof: Start with an arbitrary tapir $ABCD$. Suppose $(A, C)$ was not tasty. If $(A, D)$ is also not tasty, then rotating $A$ away from $D$ about $B$ increases $\angle ABD$ and $\angle ABC$. This process preserves the lengths of $AB, BC, CD$, while increasing the length of $AD$. Since $ABCD$ was a tapir, this process must break some condition of the problem. If $A, B, C$ are collinear, it contradicts Lemma 1. Therefore, $(A, D)$ must be tasty. By similar reasoning, $(A, B), (C, B), (C, D)$ are all tasty, implying $ABCD$ is a rhombus of side length 1, contradicting $AC, BD \leq 1$.
$\blacksquare$

### Lemma 3
All tapirs have at least one side of length 1.

Proof: Assume the contrary. Let $\theta_1, \theta_2$ denote $\angle BDA, \angle CDB$ respectively. By Lemma 1, $\theta_1, \theta_2 > 0$. By Lemma 2, $BD = 1$. Rotating $B$ about $C$ decreases $\theta_2$, preserving $c, d$. Consider $a^2 b^2 = (d^2 + 1 - 2d \cos \theta_1) (c^2 + 1 - 2c \cos \theta_2)$ as a function of $\theta_1$. The derivative must be zero, implying:
$$2a^2 c \sin \theta_2 = 2b^2 d \sin \theta_1,$$
yielding:
$$\frac{c}{d} \cdot \frac{\sin \theta_2}{\sin \theta_1} = \frac{b^2}{a^2}.$$
By the Sine Law in $\triangle CDA$, $E = BD \cap AC$ satisfies $\frac{CE}{EA} = \frac{b^2}{a^2}$, making $ABCD$ a cyclic harmonic quadrilateral. Since $AC = BD = 1$, $ABCD$ is an isosceles trapezoid. Let $EA = EB = x, EC = ED = 1-x$ and $\angle BEC = \theta$. Then:
$$abcd = 4 \cos^2 \left(\frac{\theta}{2}\right) x (1-x) \cdot \left(x^2 + (1-x)^2 - 2x(1-x) \cos \theta\right).$$
Noting $4 \cos^2 \left(\frac{\theta}{2}\right) = 2 \cos \theta + 2$, we rewrite:
$$[(2 \cos \theta + 2) x (1-x)] \cdot [1 - (2 \cos \theta + 2) x (1-x)].$$
Letting $t = (2 \cos \theta + 2) x (1-x)$, the above is $t(1-t) \leq \frac{1}{4} < 2 - \sqrt{3}$, contradicting $ABCD$ being a tapir.
$\blacksquare$

By Lemmas 1, 2, and 3, all tapirs satisfying $CA = AB = BD = 1$ have $abcd \leq 2 - \sqrt{3}$. Let $P$ be the point such that $\triangle APB$ is equilateral, and $P, C, D$ are on the same side of $AB$. The conditions imply $\angle DBA, \angle CAB \leq 60^\circ$, giving $CD \leq 1$.

#### Case 1: $P \in \{C, D\}$
Suppose $P = C$. Let $\angle DBA = 2\theta$ for $0 \leq \theta \leq 30^\circ$. Then:
$$abcd = 2 \sin \theta \cdot 2 \sin (30^\circ - \theta) = 2(\cos (2\theta - 30^\circ) - \cos 30^\circ).$$
Maximizing at $\theta = 15^\circ$, we get $abcd = 2 - \sqrt{3}$.

#### Case 2: $P \notin \{C, D\}$
Let $\angle CAB = 2\alpha, \angle DBA = 2\beta$ with $0 \leq \alpha, \beta \leq 30^\circ$. Then $AD, BC = 2 \sin \beta, 2 \sin \alpha$. By Pythagorean Theorem:
$$c = \sqrt{(\cos 2\alpha + \cos 2\beta - 1)^2 + (\sin 2\beta - \sin 2\alpha)^2}.$$
Considering $bcd$ as a function of $\alpha$, its derivative must be zero:
$$2 \cos \alpha \cdot c + 2 \sin \alpha \cdot \frac{\partial c}{\partial \alpha} = 0.$$
Thus:
$$4 \cos \alpha \cdot c^2 + 2(\cos 2\alpha + \cos 2\beta - 1)(-2 \sin 2\alpha) + 2(\sin 2\beta - \sin 2\alpha)(-2 \cos 2\alpha) = 0.$$
Analogously:
$$4 \cos \beta \cdot c^2 + 2(2 \cos 2\alpha + \cos 2\beta - 1)(-2 \sin 2\beta) + 2(\sin 2\beta - \sin 2\alpha)(-2 \cos 2\beta) = 0.$$
If $\alpha > \beta$, the LHS of the first equation is less than the second, contradicting equal RHS's. Thus, $\alpha = \beta$. Then:
$$abcd = 2 \sin \alpha \cdot 2 \sin \alpha \cdot (2 \cos 2\alpha - 1) = 4 \sin^2 \alpha \cdot (1 - 4 \sin^2 \alpha).$$
Letting $\gamma = 4 \sin^2 \alpha$, we get $abcd = \gamma (1 - \gamma) \leq \frac{1}{4} < 2 - \sqrt{3}$, contradicting $ABCD$ being a tapir.

Thus, all tapirs have $abcd = 2 - \sqrt{3}$, and all tapirs are the same up to rotation and relabeling of vertices.

The answer is: $\boxed{2 - \sqrt{3}}$.