Đến nội dung

Hình ảnh

Hanoi Open Mathematical Olympiad 2012 - Junior Section


  • Please log in to reply
Chủ đề này có 9 trả lời

#1
Zaraki

Zaraki

    PQT

  • Phó Quản lý Toán Cao cấp
  • 4273 Bài viết
Đề này của anh nguyenta_98 (anh ấy thi HMO 2012) đưa cho, xin post cho mọi người tham khảo.


Hanoi Mathematical Society

Hanoi Open Mathematical Olympiad 2012

Junior Section

Sunday, 11 April 2011

Important:
Answer all 15 questions.
Enter your answers on the answer sheet provided.
For the multiple choise questions, enter only the lettlers (A, B, C, D or E) corresponding to the correct answers in the answer sheet,
No caculators are allowed.

Q1. Assum that $a-b=-(a-b).$ Then:
$$(A) \; a=b; \qquad (B) \; a<b; \qquad (C) \; a>b \qquad (D) \; \text{ It is impossible to compare those of a and b.}$$
Q2. Let be given a parallegogram $ABCD$ with the area of $12 \ \text{cm}^2$. The line through $A$ and the midpoint $M$ of $BC$ mects $BD$ at $N.$ Compute the area of the quadrilateral $MNDC.$
$$(A) \; 4 \text{cm}^2; \qquad (B) \; 5 \text{cm}^2; \qquad (C ) \; 6 \text{cm}^2; \qquad (D) \; 7 \text{cm}^2; \qquad (E) \; \text{None of the above.}$$
Q3. For any possitive integer $a$, let $\left[ a\right]$ denote the smallest prime factor of $a.$ Which of the following numbers is equal to $\left[ 35 \right]$ ?
$$(A) \; \left[10 \right]; \qquad (B) \; \left[ 15 \right]; \qquad (C ) \; \left[45 \right]; \qquad (D) \; \left[ 55 \right]; \qquad (E) \; \left[75 \right].$$
Q4. A man travels from town $A$ to town $E$ through $B,C$ and $D$ with uniform speeds 3km/h, 2km/h, 6km/h and 3km/h on the horizontal, up slope, down slope and horizontal road, respectively. If the road between town $A$ and town $E$ can be classified as horizontal, up slope, down slope and horizontal and total length of each typr of road is the same, what is the average speed of his journey?
$$(A) \; 2 \text{km/h} \qquad (B) \; 2,5 \text{km/h} ; \qquad (C ) \; 3 \text{km/h} ; \qquad (D) \; 3,5 \text{km/h} ; \qquad (E) \; 4 \text{km/h}.$$
Q5. How many different 4-digit even integers can be form from the elements of the set $\{ 1,2,3,4,5 \}.$
$$(A) \; 4; \qquad (B) \; 5; \qquad (C ) \; 8; \qquad (D) \; 9; \qquad (E) \; \text{None of the above.}$$
Q6. At $3:00$ AM, the temperature was $13^o$ below zero. By none it has risen to $32^o$. What is the average hourly increase in temperature ?
Q7. Find all integers $n$ such that $60+2n-n^2$ is a perfect square.
Q8. Given a triangle $ABC$ and $2$ point $K \in AB, \; N \in BC$ such that $BK=2AK, \; CN=2BN$ and $Q$ is the common point of $AN$ and $CK$. Compute $\dfrac{ S_{ \triangle ABC}}{S_{\triangle BCQ}}.$
Q9. Evaluate the integer part of the number
$$H= \sqrt{1+2011^2+ \frac{2011^2}{2012^2}}+ \frac{2011}{2012}.$$
Q10. Solve the following equation $$\frac{1}{(x+29)^2}+ \frac{1}{(1+30)^2}= \frac{13}{36}.$$
Q11. Let be given a sequense $a_1=5, \; a_2=8$ and $a_{n+1}=a_n+3a_{n-1}, \qquad n=1,2,3,...$ Calculate the greatest common divisor of $a_{2011}$ and $a_{2012}$.
Q12. Find all positive integers $P$ such that the sum and product of all its divisors are $2P$ and $P^2$, respectively.
Q13. Determine the greatest value of the sum $M=11xy+3x+2012yz$, where $x,y,z$ are non negative integers satisfying condition $x+y+z=1000.$
Q14. Let be given a trinagle $ABC$ with $\angle A=90^o$ and the bisectrices of angles $B$ and $C$ meet at $I$. Suppose that $IH$ is perpendicular to $BC$ ($H$ belongs to $BC$). If $HB=5 \text{cm}, \; HC=8 \text{cm}$, compute the area of $\triangle ABC$.
Q15. Determine the greatest value of the sum $M=xy+yz+xyz$, where $x,y,z$ are real numbers satisfying the following condition $x^2+2y^2+5z^2=22.$






__________________


Bài viết đã được chỉnh sửa nội dung bởi Phạm Quang Toàn: 10-06-2012 - 10:03

Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.

 

Grothendieck, Récoltes et Semailles (“Crops and Seeds”). 


#2
perfectstrong

perfectstrong

    $LOVE(x)|_{x =\alpha}^\Omega=+\infty$

  • Quản lý Toán Ứng dụng
  • 4991 Bài viết
Trả lời bằng tiếng việt vậy :P Không biết làm toán tiếng anh :D
Q11:
\[gt \Rightarrow {a_{n + 1}} - {a_n} \vdots 3,\forall n\]
Gọi $d = \left( {{a_{2011}};{a_{2012}}} \right) \Rightarrow d|3 \Rightarrow d \in \left\{ {1;3} \right\}$
Mà dễ thấy \[{a_n} \equiv 2\left( {\bmod 3} \right) \Rightarrow d = 1\]

Q12:
Dễ thấy $P=1$ không thỏa nên $P\geq 2$
Gọi các ước nguyên dương của $P$ là $1 = {d_1} < {d_2} < ... < {d_k} = P$
\[gt \Rightarrow {d_1}{d_2}...{d_k} = {P^2}\]
Mà ta lại có:
\[\begin{array}{l}
{d_1}{d_k} = P;{d_2}{d_{k - 1}} = P;...;{d_{k - 1}}{d_2} = P;{d_k}{d_1} = P \\
\Rightarrow {\left( {{d_1}{d_2}...{d_k}} \right)^2} = {P^k} \Leftrightarrow {P^4} = {P^k} \Leftrightarrow k = 4 \\
\end{array}\]
\[{d_1} + {d_2} + {d_3} + {d_4} = 2P \Leftrightarrow {d_2} + {d_3} = P - 1\]
Theo định lý Viete, $d_2;d_3$ là nghiệm của pt:
\[{t^2} - \left( {P - 1} \right)t + P = 0\]
Để pt có nghiệm nguyên thì $\Delta$ là số chính phương.
\[\begin{array}{l}
\Leftrightarrow {\Delta _t} = {\left( {P - 1} \right)^2} - 4P = {k^2} \Leftrightarrow {P^2} - 6P + 1 = {k^2} \Leftrightarrow {\left( {P - 3} \right)^2} - {k^2} = 8 \\
\Leftrightarrow \left( {P - 3 - k} \right)\left( {P - 3 + k} \right) = 8 \\
\end{array}\]
Do $P-3+k \geq 0; P-3+k>P-3-k$ nên ta có các TH sau:
\[\begin{array}{l}
\left\{ \begin{array}{l}
P - 3 + k = 4 \\
P - 3 - k = 2 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
P = 6 \\
k = 1 \\
\end{array} \right.:True \\
\left\{ \begin{array}{l}
P - 3 + k = 8 \\
P - 3 - k = 1 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
P = \frac{{15}}{2} \\
k = \frac{7}{2} \\
\end{array} \right.:False \\
\end{array}\]
Vậy $P=6$

Bài viết đã được chỉnh sửa nội dung bởi perfectstrong: 13-03-2012 - 22:13

Luôn yêu để sống, luôn sống để học toán, luôn học toán để yêu!!! :D
$$\text{LOVE}\left( x \right)|_{x = \alpha}^\Omega = + \infty $$
I'm still there everywhere.

#3
Zaraki

Zaraki

    PQT

  • Phó Quản lý Toán Cao cấp
  • 4273 Bài viết
Q15.
Phân tích. Gỉa sử $M$ đạt max khi $x=a,y=b,z=c$. Ta có $\begin{cases} xb=ya \\ yc=zb \\ za=xc \end{cases}$. Khi đó $$\begin{array}{l} xyab \le \frac{x^2b^2+y^2a^2}{2} \\ yzbc \le \frac{y^2c^2+z^2b^2}{2} \\ zxac \le \frac{z^2a^2+x^2c^2}{2} \end{array} \Rightarrow \begin{array}{l} xyabc \le \frac{(x^2b^2+y^2a^2)c}{2} \\ yzabc \le \frac{a(y^2c^2+z^2b^2)}{2} \\ zxabc \le \frac{b(z^2a^2+x^2c^2)}{2} \end{array}$$
Cộng ba vế BĐT ta được $$abc(xy+yz+zx) \le \frac{x^2(c^2b+b^2c)+y^2(a^2c+c^2a)+z^2(a^2b+b^2a)}{2}$$
Ta thấy để vận dụng được giả thiết thiết $x^2+2y^2+5z^2=22$ thì $$\begin{cases} \frac{bc(c+b)}{2}=1 \\ \frac{ac(c+a)}{2}=2 \\ \frac{ab(a+b)}{2}=5 \\ a^2+2b^2+5c^2=22 \end{cases}$$
Nghĩ được đến đây thì không biết giải cái hệ thế nào :(

Bài viết đã được chỉnh sửa nội dung bởi Phạm Quang Toàn: 02-02-2013 - 22:17

Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.

 

Grothendieck, Récoltes et Semailles (“Crops and Seeds”). 


#4
Joker9999

Joker9999

    Thiếu úy

  • Thành viên
  • 659 Bài viết

Q15.

Cái này là làm nháp thôi em, đj thi tìm xong $a,b,c$ thì viết thẳng chứ viết lại cái kia là chi.:D Hệ đj thi thường cho đẹp, nghiệm nguyên, mò là ra k cần giải chi tiết đâu
p/s: $a=c=1,b=2$

<span style="font-family: trebuchet ms" ,="" helvetica,="" sans-serif'="">Nỗ lực chưa đủ để thành công.


.if i sad, i do Inequality to become happy. when i happy, i do Inequality to keep happy.

#5
Zaraki

Zaraki

    PQT

  • Phó Quản lý Toán Cao cấp
  • 4273 Bài viết

Cái này là làm nháp thôi em, đj thi tìm xong $a,b,c$ thì viết thẳng chứ viết lại cái kia là chi. :D Hệ đj thi thường cho đẹp, nghiệm nguyên, mò là ra k cần giải chi tiết đâu
p/s: $a=c=1,b=2$

Vâng, cái khoản đó là em nháp mà :icon6:

Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.

 

Grothendieck, Récoltes et Semailles (“Crops and Seeds”). 


#6
duaconcuachua98

duaconcuachua98

    Sĩ quan

  • Thành viên
  • 461 Bài viết
Không biết dịch TA có đúng không nữa? :lol:
Q7:
We have equation $-n^{2}+2n+60=k^{2}(k\in \mathbb{Z}^{+})\Leftrightarrow -(n-1)^{2}+61=k^{2}\Leftrightarrow (n-1)^{2}+k^{2}=61$
Because $n$ and $k$ is two integer so $61$ is the total of two perfect square.
We have the following cases $61=25+36$
$\bullet$ If $(n-1)^{2}=25\Leftrightarrow n=6$ (satisfies)
$\bullet$ If $(n-1)^{2}=36\Leftrightarrow n=7$ (satisfies)
So $n=7;n=6$ are the values need to find.

Bài viết đã được chỉnh sửa nội dung bởi duaconcuachua98: 04-02-2013 - 12:10


#7
duaconcuachua98

duaconcuachua98

    Sĩ quan

  • Thành viên
  • 461 Bài viết
Q14:
aaaaaaaaaaaaaaaaaaaaaa.JPG

We have $2HB=AB+BC-AC=10$; $2HC=AC+BC-AB=16$
$\Rightarrow AC-AB=3$
We have $AB^{2}+AC^{2}=169$ so we have system of equation
$\left\{\begin{matrix} AC-AB=3 & \\ AB^{2}+AC^{2}=169 & \end{matrix}\right.$
Solve the system of equation we have $AB=\frac{\sqrt{329}-3}{2};AC=\frac{\sqrt{329}+3}{2}$
So $S_{\Delta ABC}=\frac{AB.AC}{2}=\sqrt{5}(cm^{2})$

Bài viết đã được chỉnh sửa nội dung bởi duaconcuachua98: 03-02-2013 - 22:39


#8
vutuanhien

vutuanhien

    Thiếu úy

  • ĐHV Toán Cao cấp
  • 690 Bài viết
Bài 15: Ta có $x^2+2y^2+5z^2-2(xy+yz+zx)=(x-y-z)^2+(2z-y)^2\geq 0\Rightarrow xy+yz+zx\leq 11$.

"The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months—when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!" - Grothendieck


#9
Primary

Primary

    Sĩ quan

  • Thành viên
  • 316 Bài viết
Q 9:
$H=\frac{2011}{2012}+\sqrt{2012^2-2.2011+\frac{2011^2}{2012^2}}=\frac{2011}{2012}+2012-\frac{2011}{2012}=2012$

#10
phuongmai3199

phuongmai3199

    Lính mới

  • Thành viên
  • 5 Bài viết
Q13;

Đề này của anh nguyenta_98 (anh ấy thi HMO 2012) đưa cho, xin post cho mọi người tham khảo.


Hanoi Mathematical Society

Hanoi Open Mathematical Olympiad 2012

Junior Section

Sunday, 11 April 2011

Important:
Answer all 15 questions.
Enter your answers on the answer sheet provided.
For the multiple choise questions, enter only the lettlers (A, B, C, D or E) corresponding to the correct answers in the answer sheet,
No caculators are allowed.

Q1. Assum that $a-b=-(a-b).$ Then:
$$(A) \; a=b; \qquad (B) \; a<b; \qquad © \; a>b \qquad (D) \; \text{ It is impossible to compare those of a and b.}$$
Q2. Let be given a parallegogram $ABCD$ with the area of $12 \ \text{cm}^2$. The line through $A$ and the midpoint $M$ of $BC$ mects $BD$ at $N.$ Compute the area of the quadrilateral $MNDC.$
$$(A) \; 4 \text{cm}^2; \qquad (B) \; 5 \text{cm}^2; \qquad (C ) \; 6 \text{cm}^2; \qquad (D) \; 7 \text{cm}^2; \qquad (E) \; \text{None of the above.}$$
Q3. For any possitive integer $a$, let $\left[ a\right]$ denote the smallest prime factor of $a.$ Which of the following numbers is equal to $\left[ 35 \right]$ ?
$$(A) \; \left[10 \right]; \qquad (B) \; \left[ 15 \right]; \qquad (C ) \; \left[45 \right]; \qquad (D) \; \left[ 55 \right]; \qquad (E) \; \left[75 \right].$$
Q4. A man travels from town $A$ to town $E$ through $B,C$ and $D$ with uniform speeds 3km/h, 2km/h, 6km/h and 3km/h on the horizontal, up slope, down slope and horizontal road, respectively. If the road between town $A$ and town $E$ can be classified as horizontal, up slope, down slope and horizontal and total length of each typr of road is the same, what is the average speed of his journey?
$$(A) \; 2 \text{km/h} \qquad (B) \; 2,5 \text{km/h} ; \qquad (C ) \; 3 \text{km/h} ; \qquad (D) \; 3,5 \text{km/h} ; \qquad (E) \; 4 \text{km/h}.$$
Q5. How many different 4-digit even integers can be form from the elements of the set $\{ 1,2,3,4,5 \}.$
$$(A) \; 4; \qquad (B) \; 5; \qquad (C ) \; 8; \qquad (D) \; 9; \qquad (E) \; \text{None of the above.}$$
Q6. At $3:00$ AM, the temperature was $13^o$ below zero. By none it has risen to $32^o$. What is the average hourly increase in temperature ?
Q7. Find all integers $n$ such that $60+2n-n^2$ is a perfect square.
Q8. Given a triangle $ABC$ and $2$ point $K \in AB, \; N \in BC$ such that $BK=2AK, \; CN=2BN$ and $Q$ is the common point of $AN$ and $CK$. Compute $\dfrac{ S_{ \triangle ABC}}{S_{\triangle BCQ}}.$
Q9. Evaluate the integer part of the number
$$H= \sqrt{1+2011^2+ \frac{2011^2}{2012^2}}+ \frac{2011}{2012}.$$
Q10. Solve the following equation $$\frac{1}{(x+29)^2}+ \frac{1}{(1+30)^2}= \frac{13}{36}.$$
Q11. Let be given a sequense $a_1=5, \; a_2=8$ and $a_{n+1}=a_n+3a_{n-1}, \qquad n=1,2,3,...$ Calculate the greatest common divisor of $a_{2011}$ and $a_{2012}$.
Q12. Find all positive integers $P$ such that the sum and product of all its divisors are $2P$ and $P^2$, respectively.
Q13. Determine the greatest value of the sum $M=11xy+3x+2012yz$, where $x,y,z$ are non negative integers satisfying condition $x+y+z=1000.$
Q14. Let be given a trinagle $ABC$ with $\angle A=90^o$ and the bisectrices of angles $B$ and $C$ meet at $I$. Suppose that $IH$ is perpendicular to $BC$ ($H$ belongs to $BC$). If $HB=5 \text{cm}, \; HC=8 \text{cm}$, compute the area of $\triangle ABC$.
Q15. Determine the greatest value of the sum $M=xy+yz+xyz$, where $x,y,z$ are real numbers satisfying the following condition $x^2+2y^2+5z^2=22.$







__________________

Q13 là 3xz chứ bạn. lúc ấy sẽ có:

Q13:
x=1000-y-z. Thay vào ta có:
M= 11(1000-y-z)y + 3z(1000-y-z) + 2012yz
=11000y -11y2 -11yz + 3000z - 3zy - 3z2 + 2012yz
= -11(y2-1000y+5002) - 3(z2-1000z+5002) + 11.5002 + 3.5002 + 1998yz
file:///C:\DOCUME~1\Pc\LOCALS~1\Temp\msohtml1\01\clip_image002.gif 11.5002 + 3.5002 + 1998yz
file:///C:\DOCUME~1\Pc\LOCALS~1\Temp\msohtml1\01\clip_image002.gif11.5002 + 3.5002 + 1998.5002 = 2012.5002.
Vậy Max M=2012.5002 khi x = 0; y = z = 500
Proud to be a LT-er and a member of A3!!!




2 người đang xem chủ đề

0 thành viên, 2 khách, 0 thành viên ẩn danh