Showing posts with label Out Standing Test. Show all posts
Showing posts with label Out Standing Test. Show all posts
Tuesday, October 14, 2014
Monday, October 13, 2014
Sunday, October 12, 2014
Saturday, October 11, 2014
Friday, October 10, 2014
Function Equation: Problems Number 1
ប្រធានលំហាត់៖
រកគ្រប់បណ្តាអនុគមន៍ $f:\mathbb R^{+}\to\mathbb R^{+}$ ផ្ទៀងផ្ទាត់ៈ
$f(ab)f(bc)f(ca)f(a+b)f(b+c)f(c+a)=2014$ ចំពោះគ្រប់ $a,b,c>0$
ដំណោះស្រាយ
+ឲ្យ $a=b=c=t$ យើងបានៈ $\left[f(x^2)f(2t)\right]^3=2014$
ទាញបាន $f(t^2)f(2t)=\sqrt[3]{2014}$
+យក $a=b=t$ និង $c=1$ ទាញបានៈ $f^2(t)f^2(t^2)f(2t)f^2(t+1)=2014)$
នាំឲ្យយើងបាន $f(t)f(t+1)=\sqrt[3]{2014}$
+ជំនួស $t$ ដោយ $t+1$ ទាញបានៈ $f(t+1)f(t+2)=\sqrt[3]{2014}$
នាំឲ្យ $f(t)=f(t+2)$
+យើងយក $c=1$ យើងបានៈ $f(ab)f(a)f(b)f(a+b)f(a+1)f(b+1)=2014$
ទាញបាន $f(xy)f(x+y)=\sqrt[3]{2014}$
+ជំនួស $b=2$ និង $b=4$ តាមលំដាប់ យើងបានៈ
\[\begin{cases}f(2a)f(a+2)=\sqrt[3]{2014}\\f(4a)f(a+4)=\sqrt[3]{2014}\end{cases}\ \Rightarrow\ f(2a)=f(4a)\]
ដោយ $f(t)=f(t+2)$
ទាញបាន $f(a)f(a+2)=\sqrt[3]{2014}\ \Rightarrow\ f(a)=\sqrt[6]{2014}$
ដូចនេះ អនុគមន៍ដែលត្រូវរកគឺ $f(a)=\sqrt[6]{2014}$
រកគ្រប់បណ្តាអនុគមន៍ $f:\mathbb R^{+}\to\mathbb R^{+}$ ផ្ទៀងផ្ទាត់ៈ
$f(ab)f(bc)f(ca)f(a+b)f(b+c)f(c+a)=2014$ ចំពោះគ្រប់ $a,b,c>0$
ដំណោះស្រាយ
+ឲ្យ $a=b=c=t$ យើងបានៈ $\left[f(x^2)f(2t)\right]^3=2014$
ទាញបាន $f(t^2)f(2t)=\sqrt[3]{2014}$
+យក $a=b=t$ និង $c=1$ ទាញបានៈ $f^2(t)f^2(t^2)f(2t)f^2(t+1)=2014)$
នាំឲ្យយើងបាន $f(t)f(t+1)=\sqrt[3]{2014}$
+ជំនួស $t$ ដោយ $t+1$ ទាញបានៈ $f(t+1)f(t+2)=\sqrt[3]{2014}$
នាំឲ្យ $f(t)=f(t+2)$
+យើងយក $c=1$ យើងបានៈ $f(ab)f(a)f(b)f(a+b)f(a+1)f(b+1)=2014$
ទាញបាន $f(xy)f(x+y)=\sqrt[3]{2014}$
+ជំនួស $b=2$ និង $b=4$ តាមលំដាប់ យើងបានៈ
\[\begin{cases}f(2a)f(a+2)=\sqrt[3]{2014}\\f(4a)f(a+4)=\sqrt[3]{2014}\end{cases}\ \Rightarrow\ f(2a)=f(4a)\]
ដោយ $f(t)=f(t+2)$
ទាញបាន $f(a)f(a+2)=\sqrt[3]{2014}\ \Rightarrow\ f(a)=\sqrt[6]{2014}$
ដូចនេះ អនុគមន៍ដែលត្រូវរកគឺ $f(a)=\sqrt[6]{2014}$
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, October 7, 2014
46th Canada MO: Problem Number 1
ប្រធានលំហាត់៖
គេឲ្យ $ a_1,\ a_2,\ ...,\ a_n$ ជាបណ្តាចំនួនពិតវិជ្ជមាន ដែលមានផលគុណស្មើ 1។
បង្ហាញថាផលបូកៈ
\[\dfrac{a_1}{1+a_1}+\dfrac{a_2}{(1+a_1)(1+a_2)}+\dfrac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+...+\dfrac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}\]
គឺធំជាង រឺ ស្មើនឹង $\dfrac{2^n-1}{2^n}$ ។
ដំណោះស្រាយ
$\dfrac{a_m}{(1+a_1)(1+a_2)...(1+a_m)}$
$=\dfrac{1+a_m}{(1+a_1)(1+a_2)...(1+a_m)}-\dfrac{1}{(1+a_1)(1+a_2)...(1+a_m)}$
$=\dfrac{1}{(1+a_1)...(1+a_{m-1})}-\dfrac{1}{(1+a_1)...(1+a_m)}$
ដូចនេះ បើយើងយក $b_j=(1+a_1)(1+a_2)...(1+a_j)$, ដែល $b_0=0$, ពេលនោះយើងបានផលបូក
\[\sum_{j=1}^{n}\dfrac{a_j}{(1+a_1)...(1+a_j)}=\sum_{j=1}^{n}(\dfrac{1}{b_j-1}-\dfrac{1}{b_j})=1-\dfrac{1}{b_n}\]
យកចិត្តទុកដាក់ថា $b_n=(1+a_1)(1+a_2)...(1+a_n)\ge(2\sqrt{a_1})(2\sqrt{a_2})...(2\sqrt{a_n})=2^n$
ដែល សមភាពកើតមានលុះត្រាតែគ្រប់ $a_i$ ស្មើនឹង $1$ ។ ដូចនេះ,
\[1-\dfrac{1}{b_n}\ge 1-\dfrac{1}{2^n}=\dfrac{2^n-1}{2^n}\]
ដើម្បីដឹងថាតម្លៃតូចបំផុតនេះ អាចកើតមានរឺទេ, គឺយើងជំនួនគ្រប់ $a_i=1$ ចូល យើងបានៈ
\[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^n}=\dfrac{2^{n-1}+2^{n-2}+...+1}{2^n}=\dfrac{2^n-1}{2^n}\]
បញ្ហាត្រូវបានស្រាយបញ្ជាក់។
Saturday, May 3, 2014
Physics Exercise #23: VNPhO 30-4-2004, Grade 10 P4
Enjoy new exercise today all physics lovers!!
You can see full solution by the link below!
https://www.dropbox.com/s/q0nnut778krhvwx/P4_2004%20Answer.pdfFriday, May 2, 2014
Free Maths Book: IMO 2011 Short List With Full Solution
52nd International Mathematical Olympiad Problem Shortlist with solutions is a good document for all high school student who need to prepare for IMO
So let enjoy download this book for fee now by the link below.
About this book:
+Tittle: IMO 2011 Short List With Full Solution
+Language: English
+Language: English
+Size: 700KB
+File Type: PDF
+Link Download: Dropbox File
Need More Book?? Let check the link below ៖
Need More Book?? Let check the link below ៖
- Khmer Maths Books៖ http://highschoolcam.blogspot.com/p/khmer-maths-books.html
- English Maths Books៖ http://highschoolcam.blogspot.com/p/english-maths-books.html
- Vietnam Maths Books៖ http://highschoolcam.blogspot.com/p/vietnam-maths-books.html
Wednesday, April 30, 2014
Free Maths Book: Mathematics Problems Collection For Outstanding Students Competition
គណិតវិទ្យាជ្រើសរើសសម្រាប់ត្រៀមប្រឡងសិស្ស ពូកែជាសៀវភៅដែលរៀបរៀងដោយអតីតជ័យលាភី ទូទាំងប្រទេសឆ្នាំ២០១១ ចំណាត់ថ្នាក់លេខ៥។
សៀវភៅនេះ រួមមាន៧៤លំហាត់ ត្រូវបានជ្រើសរើស ដោយសម្រិតសម្រាំង សម្រាប់ត្រៀមប្រឡងសិស្ស ពូកែថ្នាក់ខេត្ត និងថ្នាក់ជាតិ។
សូមអរគុណចំពោះប្អូនប្រុស សុខ ពិសី ជាអ្នករៀប រៀងសៀវភៅនេះ និងបានផ្ញើរមកកាន់ខ្ញុំផ្ទាល់ដើម្បីធ្វើ ការចែកចាយបន្ត ដល់ប្អូនៗជំនាន់ក្រោយ។
About this book:
+Tittle: គណិតវិទ្យាជ្រើសរើសត្រៀមសិស្សពូកែ
+Language: Khmer
+Language: Khmer
+Size: 1.72MB
+File Type: PDF
+Link Download: Dropbox File
ទាញយកសៀវភៅជាច្រើនទៀតនៅទីនេះ ៖
ទាញយកសៀវភៅជាច្រើនទៀតនៅទីនេះ ៖
- សៀវភៅគណិតជាភាសាខ្មែរ៖ http://highschoolcam.blogspot.com/p/khmer-maths-books.html
- សៀវភៅគណិតជាភាសាអង់គ្លេស៖ http://highschoolcam.blogspot.com/p/english-maths-books.html
- សៀវភៅគណិតជាភាសាវៀតណាម៖ http://highschoolcam.blogspot.com/p/vietnam-maths-books.html
Sunday, April 27, 2014
Free Test Paper Collection: Fermat Test (Grade 11) 1997-2014
Congratulation to brother Mengsay Loem who got medal in Fermat Test for Grade 11 this year
- Fermat is one of the good test for grade 11 students to test your maths skill, and this year we just have Fermat for the first year in our country that should thank to teacher.Vin Visnuka for his great job. Hope this test will be continue every year and we can improve our education system step by step by this kind of test. Who joint this test next year??
- Because this is the new test for ours country and ours next generation so it may be difficult for them to prepare for this test if they don't have any of previous test paper to learn, so now i have collect all the test paper with its full solution for you all next generation to prepare for this test.
- Hope you will be a successful generation and can be improve your maths skill more than your previous generation too. Try harder and become strong from now on!!
- You can download this test collection by this https://www.dropbox.com/s/5p7nvq5zwddmzsv/Fermat%20%28Grade%2011%29%201997-2014.pdf
- You also can go to the official link to know more about this test and other test similar to this test by this link: http://www.cemc.uwaterloo.ca/contests/pcf.html
Saturday, April 26, 2014
Free Maths Test Collection: Pascal Test For Grade 9 From 1997-2014
- What is Pascal test? who can and should joint this test?
- Pascal is one of the good test for grade 9 students to test your maths skill, and this year we just have Pascal for the first year in our country that should thank to teacher.Vin Visnuka for his great job. Hope this test will be continue every year and we can improve our education system step by step by this kind of test. Who joint this test next year??
- Because this is the new test for ours country and ours next generation so it may be difficult for them to prepare for this test if they don't have any of previous test paper to learn, so now i have collect all the test paper with its full solution for you all next generation to prepare for this test.
- Hope you will be a successful generation and can be improve your maths skill more than your previous generation too. Try harder and become strong from now on!!
- You can download this test collection by this https://www.dropbox.com/s/ckviwiuy3l50o50/Pascal%20Grade%209_1997-2014.pdf
- You also can go to the official link to know more about this test and other test similar to this test by this link: http://www.cemc.uwaterloo.ca/contests/pcf.html
Friday, April 25, 2014
Free Maths Test Collection: APMO 1989-2014 Full Test Paper With Full Some Full Solutions
ដើម្បីរួមចំនែកក្នុងការស្វែងរកឯកសារត្រៀមប្រឡង គណិវិទ្យាអូឡាំព្យាដអាស៊ីប៉ាស៊ីភិក នៅឆ្នាំក្រោយ របស់ប្អូនៗ ខ្ញុំបានប្រមូលផ្តុំរាល់វិញ្ញាសាប្រឡង និងអត្រាកំនែនៅបណ្តាឆ្នាំមុនៗ ក្នុងការប្រឡងនេះ ចងក្រងទៅជាសៀវភៅមួយក្បាលនេះ។
សង្ឃឹមថាឯកសារមួយនេះ នឹងចូលរួមចំនែកយ៉ាងសកម្ម ក្នុងការពង្រឹង និងពង្រឺកបន្ថែម ទៀតលើមុខវិជ្ជាគណិតវិទ្យារបស់ប្អូនៗ ជាពិសេសសម្រាប់អ្នកដែលត្រៀមខ្លួនក្នុងការប្រកួត ប្រជែងលក្ខណៈអន្តរជាតិមួយនេះ។
ប្អូនៗ និងមិត្តអ្នកសិក្សាអាចចូលទៅទាញយកឯកសារមួយនេះ តាមរយៈតំនភ្ជាប់ខាងក្រោមៈ
https://www.dropbox.com/s/ar4obgynjn7juov/APMO%201989-2014.pdf
Thursday, April 17, 2014
Free Maths Book: International Mathematics Competition (IMC) Collection 1999-2013
IMC is a test that ours country should joint. It has a lot of good problems in strange type that we should let ours Khmer students have fun with this test.
Here is the book of IMC collection by me ^_^
About this book:
+Tittle: International Mathematics Competition (IMC) Collection 1999-2013
+Size: 4.72MB
+File Type: PDF
+Link Download: Dropbox File
For more info about this test go to the link below: http://www.imc-official.org/en_US/
For more info about this test go to the link below: http://www.imc-official.org/en_US/
Saturday, April 12, 2014
Free Physics Book: Physics for scientists and engineers 5th edition (Instructor's Solution Manual)
Physics for scientists and engineers 5th edition (Instructor's Solution Manual)
Big thank to brother Leav Menghuy for this book.
Hope this book will be good for all of you who like mathematics. Let enjoy download it for free.
About this book:
+Tittle: Physics for scientists and engineers 5th edition (Instructor's Solution Manual)
+Size: 20.0MB
+File Type: PDF
+Link Download: Dropbox File
Friday, April 11, 2014
Test Paper: Cambodia Physics Olympiad Nation Round 2014 For Grade 12
Sorry for this document that it not so clear, please what for new clear version one.
I will update it when i got a new one. Big thank to brother Ham Senghong for this test paper!
Thursday, April 10, 2014
Maths Exercise #37: Grade 9 Nation Round 2014, P5,P6
Maths Exercises #37: Grade 9 Nation Round 2014, 1st Day, P2,P3,P4
Let me know if you have different solution with me, or you can send and comment in the comment box below. Let enjoy with my solution ^_^
For problem #2, thank to sis.Seavhouy that remind me with the solution !
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