Showing posts with label Geometry. Show all posts
Showing posts with label Geometry. Show all posts
Monday, October 13, 2014
Sunday, October 12, 2014
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។
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
\[\dfrac{S_a}{S_{ABC}}=\dfrac{AB^{\prime}}{AC}\cdot\dfrac{AC^{\prime}}{AB}\ \Rightarrow\ \dfrac{AH^2}{S_a}=\dfrac{AH^2\cdot AB\cdot AC}{S_{ABC}\cdot AB^{\prime}\cdot AC^{\prime}}\quad\quad (1)\\ \dfrac{S_b}{S_{ABC}}=\dfrac{BC^{\prime}}{AB}\cdot \dfrac{BA^{\prime}}{BC}\ \Rightarrow\ \dfrac{BH^2}{S_b}=\dfrac{BH^2\cdot AB\cdot BC}{S_{ABC}\cdot BC^{\prime}\cdot BA^{\prime}}\quad\quad (2)\\ \dfrac{S_c}{S_{ABC}}=\dfrac{CA^{\prime}}{BC}\cdot \dfrac{CB^{\prime}}{AC}\ \Rightarrow\ \dfrac{CH^2}{S_c}=\dfrac{CH^2\cdot AC\cdot BC}{S_{ABC}\cdot CB^{\prime}\cdot CA^{\prime}}\quad\quad (3) \]
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។
Tuesday, April 29, 2014
Free Maths Book: Geometry-4th-edition _ Schaums'outline
This book include geometry in plan theorem and problems with full solution good for all high school student prepare for outstanding student test and use as fundamental for geometry studying for APMO and IMO test.
Hope this book will be good for all of you who like mathematics. Let enjoy download it for free.
About this book:
+Tittle: Geometry-4th-edition _ Schaums'outline
+Language: English
+Language: English
+Size: 3.85MB
+File Type: PDF
+Link Download: Dropbox File
Download More Book Here:
http://highschoolcam.blogspot.com/p/english-maths-books.html
Download More Book Here:
http://highschoolcam.blogspot.com/p/english-maths-books.html
Sunday, December 8, 2013
Maths Exercise #20: Geometry Again ^_^
Now let focus on geometry problems. Let taste out this one, do you have another solution? If have let send to me in my mail box so that i can share it to all other readers.
Maths Exercise #18: Choose From AMMC 2000
Let enjoyt AMMC 2000 exercise was solve by Vichet Sang in Math Dynasty Group
And here is the full solution from brother Mengsay Loem and Somaisatchya Un by following
Wednesday, November 6, 2013
Maths Exercise #10: Geometry Again ^_^
Enjoy inequality in geometry, this exercise i choose from geometry book for grade 8 in VN book.
Saturday, November 2, 2013
Maths Exercise #6: Inequality in geometry
This exercise is for grade 8, 9, 10 students in high school. Let try to do it now.
Monday, October 28, 2013
Maths Exericse #1: Geometry Grade 8, 9, 10
គេឲ្យ M ជាចំនុចស្ថិតនៅក្នុងត្រីកោណ ABC ។
ស្រាយបញ្ជាក់ថា៖ MB+MC < AB+AC ។
តាមនោះទាញថា MA+MB+MC < AB+AC+BC ។
ស្រាយបញ្ជាក់ថា៖ MB+MC < AB+AC ។
តាមនោះទាញថា MA+MB+MC < AB+AC+BC ។
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