Geometry Problem: Number 1

ប្រធានលំហាត់៖
គេឲ្យត្រីកោណស្រួច $ABC$ មានបណ្តាកំពស់ $AA^{\prime},BB^{\prime},CC^{\prime}$ កាត់គ្នាត្រង់ $H$ ។
តាង $S_a,\ S_b,\ S_c$ តាមលំដាប់ ជាក្រឡាផ្ទៃរបស់ត្រីកោណ $AB^{\prime}C^{\prime},\ BC^{\prime}A^{\prime},\ CA^{\prime}B^{\prime}$។
ស្រាយបញ្ជាក់ថា $\dfrac{AH^2}{S_a}=\dfrac{BH^2}{S_b}=\dfrac{CH^2}{S_c}$

ដំណោះស្រាយ

យើងមាន
\[\dfrac{S_a}{S_{ABC}}=\dfrac{AB^{\prime}}{AC}\cdot\dfrac{AC^{\prime}}{AB}\ \Rightarrow\ \dfrac{AH^2}{S_a}=\dfrac{AH^2\cdot AB\cdot AC}{S_{ABC}\cdot AB^{\prime}\cdot AC^{\prime}}\quad\quad (1)\\ \dfrac{S_b}{S_{ABC}}=\dfrac{BC^{\prime}}{AB}\cdot \dfrac{BA^{\prime}}{BC}\ \Rightarrow\ \dfrac{BH^2}{S_b}=\dfrac{BH^2\cdot AB\cdot BC}{S_{ABC}\cdot BC^{\prime}\cdot BA^{\prime}}\quad\quad (2)\\ \dfrac{S_c}{S_{ABC}}=\dfrac{CA^{\prime}}{BC}\cdot \dfrac{CB^{\prime}}{AC}\ \Rightarrow\ \dfrac{CH^2}{S_c}=\dfrac{CH^2\cdot AC\cdot BC}{S_{ABC}\cdot CB^{\prime}\cdot CA^{\prime}}\quad\quad (3) \]
ដូចនោះ លំហាត់ក្លាយទៅជាការស្រាយបញ្ជាក់ថា
\[\dfrac{AH^2.AB.AC}{AB^{\prime}.AC^{\prime}}=\dfrac{BH^2.AB.BC}{BC^{\prime}.BA^{\prime}}=\dfrac{CH^2.AC.BC}{CB^{\prime}.CA^{\prime}}\]
យើងងាយនឹងស្រាយបានថា $\triangle AA^{\prime}C\sim\triangle BB^{\prime}C$ (ម.ម)
\[\Rightarrow\ \dfrac{BC}{AC}=\dfrac{B^{\prime}C^{\prime}}{A^{\prime}C}=\dfrac{BB^{\prime}}{AA^{\prime}}\ \Rightarrow\ \begin{cases} BC=\dfrac{BB^{\prime}}{AA^{\prime}}\cdot AC \\ B^{\prime}C=\dfrac{BB^{\prime}}{AA^{\prime}}\cdot A^{\prime}C \\ A^{\prime}C=\dfrac{AA^{\prime}}{BB^{\prime}}\cdot B^{\prime}C \\ AC=\dfrac{AA^{\prime}}{BB^{\prime}}. BC\end{cases}\]
ដូចនេះ យើងបាន $\dfrac{CH^2.AC.BC}{CB^{\prime}.CA^{\prime}}=\left(\dfrac{CH.AC}{A^{\prime}C}\right)^2=\left(\dfrac{CH\cdot BC}{B^{\prime}C}\right)^2 $
ស្រាយដូចគ្នាដែរ យើងបាន
\[\dfrac{BH^2. AB.BC}{BC^{\prime}.BA^{\prime}}=\left(\dfrac{BH.BC}{BC^{\prime}}\right)^2.\dfrac{AH^2.AC.AB}{AB^{\prime}.AC^{\prime}}=\left(\dfrac{AH\cdot AC}{AC^{\prime}}\right)^2\]
ដូចនេះ
$\dfrac{CH^2.AC.BC}{CB^{\prime}.CA^{\prime}}=\dfrac{BH^2.AB.BC}{BC^{\prime}.BA^{\prime}}\ \Leftrightarrow\ \left(\dfrac{CH.BC}{B^{\prime}C}\right)^2=\left(\dfrac{BH.BC}{BC^{\prime}}\right)^2$
$\Leftrightarrow\ \dfrac{CH}{B^{\prime}C}=\dfrac{BH}{BC^{\prime}}\ \Leftrightarrow\ \triangle C^{\prime}HB\sim\triangle B^{\prime}HC$ (ម.ម)
ហើយ
$\dfrac{CH^2.AC.BC}{CB^{\prime}.CA^{\prime}}=\dfrac{AH^2.AC.AB}{AB^{\prime}.AC^{\prime}}\ \Leftrightarrow\ \left(\dfrac{CH.AC}{A^{\prime}C}\right)^2=\left(\dfrac{AH.AC}{AC^{\prime}}\right)^2$
$\dfrac{CH}{A^{\prime}C}=\dfrac{AH}{AC^{\prime}}\ \Leftrightarrow\ \triangle C^{\prime}HA\sim\triangle A^{\prime}HC$ (ម.ម)
ដូចនេះ $\dfrac{AH^2.AB.AC}{AB^{\prime}.AC^{\prime}}=\dfrac{BH^2.AB.BC}{BC^{\prime}.BA^{\prime}}=\dfrac{CH^2.AC.BC}{CB^{\prime}.CA^{\prime}}$
យើងបានបញ្ហាត្រូវស្រាយបញ្ជាក់!
របៀបទី២៖
យើងងាយនឹងស្រាយបានថា
$\triangle CC^{\prime}B\sim \triangle AA^{\prime}B$ (ម.ម)$\  \Rightarrow \dfrac{BC^{\prime}}{BA^{\prime}}=\dfrac{BC}{BA}\ \Rightarrow\ \triangle A^{\prime}BC^{\prime}\sim\triangle ABC$ (ម.ម)
ដូចគ្នាដែរ យើងបាន $\triangle AB^{\prime}C^{\prime}\sim\triangle ABC,\ \triangle A^{\prime}B^{\prime}C\sim\triangle ABC$
យើងទាញបាន $\triangle A^{\prime}BC^{\prime}\sim\triangle AB^{\prime}C^{\prime}$ $\Rightarrow\dfrac{S_a}{S_b}=\dfrac{A{B^{\prime}}^2}{A^{\prime}B^2}$
ហើយដោយមាន $\triangle AHB^{\prime}\sim\triangle BHA^{\prime}$ (ម.ម) $\ \Rightarrow\ \dfrac{AB^{\prime}}{A^{\prime}B}=\dfrac{AH}{BH}$
ដូចនេះ $\dfrac{S_a}{S_b}=\dfrac{A{B^{\prime}}^2}{A^{\prime}B}^2=\dfrac{AH^2}{BH^2}\ \Rightarrow\ \dfrac{AH^2}{S_a}=\dfrac{BH^2}{S_b}$
ស្រាយបញ្ជាក់ដូចគ្នាដែរ យើងបាន $\dfrac{BH^2}{S_b}=\dfrac{CH^2}{S_c}$
យើងបាន បញ្ហាត្រូវបានស្រាយបញ្ជាក់រួចរាល់។