The quadratic formula in a triangle

Using the usual notation of $a=BC$ , $b=CA=1$ , and $c=AB$ and $A$ , $B$ , and $C$ for the three angles, we have $1 + c = 2a\cos B$ . By cosine rule,

$\begin{aligned} 1 & = a^2 + c^2 - 2ac \cos B \\ & = a^2 + c^2 - c(1+c) \\ \implies c & = a^2 - 1 \\ \implies \cos B & = \frac {1+c}{2a} = \frac a2 \end{aligned}$

Area of $\triangle ABC$ is given:

$\begin{aligned} A_\triangle & = \frac {ac}2 \sin B = \frac {a(a^2-1)}2 \sin B \\ & = \cos B(4\cos^2 B-1)\sin B \\ & = \frac 12 \sin 2B(2\cos 2B+1) \\ & = \frac {\sin 4B+\sin 2B}2 \\ \frac {dA_\triangle}{dB} & = 2\cos 4B + \cos 2B & \small \blue{\text{Putting }\frac {dA_\triangle}{dB} = 0} \\ 4\cos^2 2B + \cos 2B - 2 & = 0 \\ \implies \cos 2B & = \frac {\pm\sqrt{33}-1}8 \end{aligned}$

Since $\dfrac {dA_\triangle}{dB} < 0$ when $\cos 2B = \dfrac {\sqrt{33}-1}8$ , $\implies A_\triangle$ is maximum when $\cos 2B = \dfrac {\sqrt{33}-1}8$ .

By sine rule, we have $\dfrac {\sin A}a = \dfrac {\sin B}b \implies \sin A = a\sin B = 2\cos B \sin B = \sin 2B \implies \cos A = \cos 2B = \dfrac {\sqrt{33}-1}8 \approx 0.593070331 \implies \lfloor 10000 \cos A \rfloor = \boxed{5930}$ .

Let $a = BC$ , $b = AC$ , and $c = AB$ . Then from the given information, $b = 1$ and $1 + c = 2a \cos B$ .

By the law of cosines, $\cos B = \frac{a^2 + c^2 - b^2}{2bc}$ . Substituting into the equation with $b = 1$ and simplifying gives $a = \sqrt{c + 1}$ .

Therefore, the three sides of the triangle are $1$ , $\sqrt{c + 1}$ , and $c$ . By Heron's formula, the area simplifies to $A = \frac{1}{4}c\sqrt{-c^2 + 2c + 3}$ which rearranges to $16A^2 = -c^4 + 2c^3 + 3c^2$ . Using implicit differentiation, the maximum area occurs when $-4c^3 + 6c^2 + 6c = -2c(c^2 - 3c -3) = 0$ , which by the quadratic equation is $c = \frac{1}{4}(3 + \sqrt{33})$ for $c > 0$ .

By the law of cosines, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ . Substituting $b = 1$ , $a = \sqrt{c + 1}$ , and $c = \frac{1}{4}(3 + \sqrt{33})$ gives $\cos A = \frac{1}{8}(\sqrt{33} - 1)$ , so that $\lfloor 10000 \cos \angle BAC \rfloor = \boxed{5930}$ .