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Thursday, October 9, 2014

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។

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