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Câu hỏi: Cho tam giác ABC vuông tại A. Gọi I là tâm đường tròn nội tiếp tam giác ABC. Đặt $IA=x;IB=y;IC=z$, biết rằng $\dfrac{1}{{{x}^{2}}}=\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}+\dfrac{\sqrt{a}}{yz}$. Giá trị của a bằng:
A. 2.
B. 1.
C. 3.
D. 5.
A. 2.
B. 1.
C. 3.
D. 5.
Gọi r là bán kính đường tròn nội tiếp tam giác ABC.
Ta có: $IA=\dfrac{r}{\sin \dfrac{A}{2}};IB=\dfrac{r}{\sin \dfrac{B}{2}};IC=\dfrac{r}{\sin \dfrac{C}{2}}$.
Vế trái $\dfrac{1}{{{x}^{2}}}=\dfrac{{{\sin }^{2}}\dfrac{A}{2}}{{{r}^{2}}}=\dfrac{{{\sin }^{2}}45{}^\circ }{{{r}^{2}}}=\dfrac{1}{2{{r}^{2}}}$.
Vế phải:
$\begin{aligned}
& \dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}+\dfrac{\sqrt{a}}{yz}=\dfrac{{{\sin }^{2}}\dfrac{B}{2}}{{{r}^{2}}}+\dfrac{{{\sin }^{2}}\dfrac{C}{2}}{{{r}^{2}}}+\dfrac{\sqrt{a}.\sin \dfrac{B}{2}.\sin \dfrac{C}{2}}{{{r}^{2}}} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{{{r}^{2}}}\left[ \dfrac{1-\cos B+1-\cos C}{2}+\dfrac{\sqrt{a}.\left( \cos \dfrac{B-C}{2}-\cos \dfrac{B+C}{2} \right)}{2} \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-2\cos \dfrac{B+C}{2}\cos \dfrac{B-C}{2}+\sqrt{a}\cos \dfrac{B-C}{2}-\sqrt{a}\cos \dfrac{B+C}{2} \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-2\cos 45{}^\circ .\cos \dfrac{B-C}{2}+\sqrt{a}\cos \dfrac{B-C}{2}-\sqrt{a}\cos 45{}^\circ \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-2.\dfrac{\sqrt{2}}{2}.\cos \dfrac{B-C}{2}+\sqrt{a}\cos \dfrac{B-C}{2}-\sqrt{a}.\dfrac{\sqrt{2}}{2} \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-\dfrac{\sqrt{2a}}{2}-\left( \sqrt{2}-\sqrt{a} \right)\cos \dfrac{B-C}{2} \right] \\
\end{aligned}$
Do $\dfrac{1}{{{x}^{2}}}=\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}+\dfrac{\sqrt{a}}{yz}\Rightarrow \dfrac{1}{2{{r}^{2}}}=\dfrac{1}{2{{r}^{2}}}\left[ 2-\dfrac{\sqrt{2a}}{2}+\left( \sqrt{a}-\sqrt{2} \right)\cos \dfrac{B-C}{2} \right]\Rightarrow \left\{ \begin{aligned}
& 2-\dfrac{\sqrt{2a}}{2}=1 \\
& \sqrt{a}-\sqrt{2}=0 \\
\end{aligned} \right.\Rightarrow a=2$.
Ta có: $IA=\dfrac{r}{\sin \dfrac{A}{2}};IB=\dfrac{r}{\sin \dfrac{B}{2}};IC=\dfrac{r}{\sin \dfrac{C}{2}}$.
Vế trái $\dfrac{1}{{{x}^{2}}}=\dfrac{{{\sin }^{2}}\dfrac{A}{2}}{{{r}^{2}}}=\dfrac{{{\sin }^{2}}45{}^\circ }{{{r}^{2}}}=\dfrac{1}{2{{r}^{2}}}$.
Vế phải:
$\begin{aligned}
& \dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}+\dfrac{\sqrt{a}}{yz}=\dfrac{{{\sin }^{2}}\dfrac{B}{2}}{{{r}^{2}}}+\dfrac{{{\sin }^{2}}\dfrac{C}{2}}{{{r}^{2}}}+\dfrac{\sqrt{a}.\sin \dfrac{B}{2}.\sin \dfrac{C}{2}}{{{r}^{2}}} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{{{r}^{2}}}\left[ \dfrac{1-\cos B+1-\cos C}{2}+\dfrac{\sqrt{a}.\left( \cos \dfrac{B-C}{2}-\cos \dfrac{B+C}{2} \right)}{2} \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-2\cos \dfrac{B+C}{2}\cos \dfrac{B-C}{2}+\sqrt{a}\cos \dfrac{B-C}{2}-\sqrt{a}\cos \dfrac{B+C}{2} \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-2\cos 45{}^\circ .\cos \dfrac{B-C}{2}+\sqrt{a}\cos \dfrac{B-C}{2}-\sqrt{a}\cos 45{}^\circ \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-2.\dfrac{\sqrt{2}}{2}.\cos \dfrac{B-C}{2}+\sqrt{a}\cos \dfrac{B-C}{2}-\sqrt{a}.\dfrac{\sqrt{2}}{2} \right] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2{{r}^{2}}}\left[ 2-\dfrac{\sqrt{2a}}{2}-\left( \sqrt{2}-\sqrt{a} \right)\cos \dfrac{B-C}{2} \right] \\
\end{aligned}$
Do $\dfrac{1}{{{x}^{2}}}=\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}+\dfrac{\sqrt{a}}{yz}\Rightarrow \dfrac{1}{2{{r}^{2}}}=\dfrac{1}{2{{r}^{2}}}\left[ 2-\dfrac{\sqrt{2a}}{2}+\left( \sqrt{a}-\sqrt{2} \right)\cos \dfrac{B-C}{2} \right]\Rightarrow \left\{ \begin{aligned}
& 2-\dfrac{\sqrt{2a}}{2}=1 \\
& \sqrt{a}-\sqrt{2}=0 \\
\end{aligned} \right.\Rightarrow a=2$.
Đáp án A.