Question

Catetele unui triunghi dreptunghic satisfac relația:
$\sqrt{b^2 - 2 \sqrt{14}b+30 } + \sqrt{c^2 - 10 \sqrt{2}c + 51} \le 5$.
Calculați aria triunghiului.

Answer 1

$\sqrt{b^2-2\sqrt{14}b+30}+\sqrt{c^2-10\sqrt2c+51}\leq 5 \\ \\b^2-2\sqrt{14}b +30 = b^2-2\cdot b\cdot \sqrt{14}+\sqrt{14}^2+16 =\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~x^2-2\cdot x\cdot y~~~ +~~~y^2 \\ \\ =(b-\sqrt{14})^2+16 \\ \\ c^2-10\sqrt{2}c +51 = c^2-2\cdot c\cdot 5\sqrt2+(5\sqrt{2})^2+1 =\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~x^2-2\cdot x\cdot y~~~ +~~~y^2 \\ \\ =(c-2\sqrt{5})^2+1$

$\Rightarrow \underset{\geq 4}{\underbrace{\sqrt{\underset{\geq 0}{\underbrace{(b-\sqrt{14})^2}}+16}}} + \underset{\geq 1}{\underbrace{\sqrt{\underset{\geq 0}{\underbrace{(c-2\sqrt5)^2}}+1}}} \leq 5 \\ \\ \Rightarrow \sqrt{(b-\sqrt{14})^2+16}+\sqrt{(c-2\sqrt5)^2+1}\geq 5$

$\Rightarrow 5\leq \sqrt{(b-\sqrt{14})^2+16}+\sqrt{(c-2\sqrt5)^2+1} \leq 5 \\ \\ \Rightarrow \sqrt{(b-\sqrt{14})^2+16}+\sqrt{(c-2\sqrt5)^2+1} = 5$

$\Rightarrow \left\{ \begin{array}{ll} b-\sqrt{14}=0 \\ c-2\sqrt5 = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} b = \sqrt{14}\\ c = 2\sqrt{5} \end{array} \right| \Rightarrow A = \dfrac{b\cdot c}{2}\Rightarrow \\ \\ \\ \Rightarrow A = \dfrac{\sqrt{14}\cdot 2\sqrt5}{2} \Rightarrow \boxed{A = \sqrt{70}}$