Austrian Mathematical Olympiad 2012
#1
Đã gửi 23-05-2012 - 23:02
Problem 1:
Prove that the inequality \[ a + a^3 - a^4 - a^6 < 1\] holds for all real numbers $a$.
Problem 2:
Determine all integer solutions $(x, y)$ of the equation \[(x - 1)x(x + 1) + (y - 1)y(y + 1) = 24 - 9xy\mbox{.}\]
Problem 3:
In an arithmetic sequence, the difference of consecutive terms in constant. We consider sequences of integers in which the difference of consecutive terms equals the sum of the differences of all preceding consecutive terms. Which of these sequences with $a_0 = 2012$ and $1\leqslant d = a_1-a_0 \leqslant 43$ contain square numbers?
Problem 4:
In a triangle $ABC$, let $H_a$, $H_b$ and $H_c$ denote the base points of the altitudes on the sides $BC$, $CA$ and $AB$, respectively. Determine for which triangles $ABC$ two of the lengths $H_aH_b$, $H_bH_c$ and $H_aH_c$ are equal.
Time: 4.5 hours.
#2
Đã gửi 23-05-2012 - 23:03
Problem 1:
Determine all functions $f: \mathbb{Z}\to\mathbb{Z}$ satisfying the following property: For each pair of integers $m$ and $n$ (not necessarily distinct), $\mathrm{gcd}(m, n)$ divides $f(m) + f(n)$.
Note: If $n\in\mathbb{Z}$, $\mathrm{gcd}(m, n)=\mathrm{gcd}(|m|, |n|)$ and $\mathrm{gcd}(n, 0)=n$.
Problem 2:
Determine all solutions $(n, k)$ of the equation $n!+An = n^k$ with $n, k \in\mathbb{N}$ for $A = 7$ and for $A = 2012$.
Problem 3:
Consider a stripe of $n$ fieds, numbered from left to right with the integers $1$ to $n$ in ascending order. Each of the fields is colored with one of the colors $1$, $2$ or $3$. Even-numbered fields can be colored with any color. Odd-numbered fields are only allowed to be colored with the odd colors $1$ and $3$. How many such colorings are there such that any two neighboring fields have different colors?
Problem 4:
Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the intersection of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$. Show that $\gamma$ is the second-largest angle in the triangle $ABC$.
Time: 4.5 hours.
AoPS
#3
Đã gửi 24-05-2012 - 07:23
$\fbox{1}.$ Chứng minh rằng bất đẳng thức \[ a + a^3 - a^4 - a^6 < 1\] đúng với mọi số thực $a$.Problem 1:
Prove that the inequality \[ a + a^3 - a^4 - a^6 < 1\] holds for all real numbers $a$.
Giải phương trình nghiệm nguyên \[(x - 1)x(x + 1) + (y - 1)y(y + 1) = 24 - 9xy\mbox{.}\]Problem 2:
Determine all integer solutions $(x, y)$ of the equation \[(x - 1)x(x + 1) + (y - 1)y(y + 1) = 24 - 9xy\mbox{.}\]
$\fbox{1}.$ Xác định mọi hàm số $f: \mathbb{Z}\to\mathbb{Z}$ thỏa mãn: Với mỗi cặp số nguyên $m$ và $n$ (không nhất thiết phải khác nhau) $\mathrm{gcd}(m, n)$ là ước của $f(m) + f(n)$Problem 1:
Determine all functions $f: \mathbb{Z}\to\mathbb{Z}$ satisfying the following property: For each pair of integers $m$ and $n$ (not necessarily distinct), $\mathrm{gcd}(m, n)$ divides $f(m) + f(n)$.
Note: If $n\in\mathbb{Z}$, $\mathrm{gcd}(m, n)=\mathrm{gcd}(|m|, |n|)$ and $\mathrm{gcd}(n, 0)=n$.
Chú ý: Nếu $n\in\mathbb{Z}$, $\mathrm{gcd}(m, n)=\mathrm{gcd}(|m|, |n|)$ và $\mathrm{gcd}(n, 0)=n$.
Tìm nghiệm $(n, k)$ của phương trình $n!+An = n^k$ với $n, k \in\mathbb{N}$ nếu $A = 7$ và nếu $A = 2012$.Problem 2:
Determine all solutions $(n, k)$ of the equation $n!+An = n^k$ with $n, k \in\mathbb{N}$ for $A = 7$ and for $A = 2012$.
Problem 4:
Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the intersection of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$. Show that $\gamma$ is the second-largest angle in the triangle $ABC$.
Cho $ABC$ là tam giác không cân. Lấy $U$ là tâm đường tròn ngoại tiếp của tam giác này và $I$ là tâm đường tròn nội tiếp. Giả sử rằng các giao điểm của phân giác góc $\gamma = \angle ACB$ với đường tròn ngoại tiếp $ABC$ nằm trên đường vuông góc với phân giác của $UI$. Hãy chỉ ra rằng $\gamma$ là góc lớn thứ hai trong tam giác $ABC$.
Bài viết đã được chỉnh sửa nội dung bởi Phạm Quang Toàn: 24-05-2012 - 07:30
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
Đã gửi 21-08-2012 - 22:19
$a^6-a^4-a^3-a+1=(a^3-\dfrac{1}{2})^2+(a^2-\dfrac{1}{2})^2+(a-\dfrac{1}{2})^2+\dfrac{1}{4} >0$$\fbox{1}.$ Chứng minh rằng bất đẳng thức \[ a + a^3 - a^4 - a^6 < 1\] đúng với mọi số thực $a$.
A1K39PBC
#5
Đã gửi 21-08-2012 - 22:35
$(x-1)x(x+1)+(y-1)y(y+1)=24-9xy$Giải phương trình nghiệm nguyên \[(x - 1)x(x + 1) + (y - 1)y(y + 1) = 24 - 9xy\mbox{.}\]
$\Rightarrow x^3+y^3-x-y+9xy-24=0$
$\Rightarrow (x+y-3)(x^2-4xy+y^2+3x+3y+8)=0$
$\Rightarrow (x+y-3)(4x^2-4xy+4y^2+12x+12y+32)=0$
$\Rightarrow (x+y-3)[(2x-y)^2+3(y-3)^2-4]=0$
$\begin{bmatrix}
x+y=3\\
(2x-y)^2+3(y-3)^2=4
\end{bmatrix}$
Ta có: $4=1^2+3.1^2 \Rightarrow \left\{\begin{matrix}
(2x-y)^2=1\\
(y-3)^2=1
\end{matrix}\right.$
Do đó pt có nghiệm $(x; y)$ là $(n; 3-n), (-2,-2),(-3,-4),(-3,-2),(-4;-4),(-2,-3),(-4,-3)$
A1K39PBC
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