Prove that there exist infinitely many positive integers which can't be represented as sum of less than $10$ odd positive integers' perfect squares.

Given a solid $R$ contained in a semi cylinder with the hight $1$ which has a semicircle with radius $1$ as the base. The cross section at the hight $x\ (0\leq x\leq 1)$ is the form combined with two right-angled triangles as attached figure as below. Answer the following questions. (1) Find the cross-sectional area $S(x)$ at the hight $x$. (2) Find the volume of $R$. If necessary, when you integrate, set $x=\sin t.$

Let $a_1, a_2, ..., a_m$ be positive integers, none of which is equal to $10$, such that $a_1 + a_2 + ...+ a_m = 10m$. Prove that $(a_1a_2a_3 \cdot ...\cdot a_m)^{1/m} \le 3\sqrt{11}$.

Last year, Master Cheung is famous for multi-rotation. This year, he comes to DAMO to make noodles for sweeping monk. One day, software engineer Xiao Li talks with Master Cheung about his job. Xiao Li mainly researches and designs the algorithm to adjust the paramter of different kinds of products. These paramters can normally be obtainly by minimising loss function $f$ on $\mathbb{R}^n$. In the recent project of Xiao Li, this loss function is obtained by other topics. For safety consideration and technique reasons, this topic makes Xiao Li difficult to find the interal details of the function. They only provide a port to calculate the value of $f(\text x)$ for any $\text x\in\mathbb{R}^n$. Therefore, Xiao Li must only use the value of the function to minimise $f$. Also, every times calculating the value of $f$ will use a lot of calculating resources. It is good to know that the dimension $n$ is not very high (around $10$). Also, colleague who provides the function tells Xiao Li to assume $f$ is smooth first. This problem reminds Master Cheung of his antique radio. If you want to hear a programme from the radio, you need to turn the knob of the radio carefully. At the same time, you need to pay attention to the quality of the radio received, until the quality is the best. In this process, no one knows the relationship between the angle of turning the knob and the quality of the radio received. Master Cheung and Xiao Li realizes that minimising $f$ is same as adjusting the machine with multiple knobs: Assume every weight of $\text x$ is controlled by a knob. $f(\text x)$ is a certain performance of the machine. We only need to adjust every knobs again and again and observes the value of $f$ in the same time. Maybe there is hope to find the best $\text x$. As a result, two people suggest an iteration algorithm (named Automated Forward/Backward Tuning, $\text{AFBT}$, to minimise $f$. In $k$-th iteration, the algorithm adjusts the individual weight of $\text{x}_k$ to $2n$ points $\{\text x_k\pm t_k\text e^i:i=1,...,n\}$, where $t_k$ is the step size; then, make $y_k$ be the smallest one among the value of the function of thosse points. Then check if $\text y_k$ sufficiently makes $f$ decrease; then, take $\text x_{k+1}=\text y_k$, then make the step size doubled. Otherwise, make $\text x_{k+1}=\text x_k$ and makes the step size decrease in half. In the algorithm, $\text e^i$ is the $i$-th coordinate vector in $\mathbb{R}^n$. The weight of $i$-th is $1$. Others are $0$; $\mathbf{1}(\cdot)$ is indicator function. If $f(\text x_k)-f(\text y_k)$ is at least the square of $t_k$, then take the value of $\mathbf{1}(f(\text k)-f(y_k)\ge t^2_k)$ as $1$. Otherwise, take it as $0$. $\text{AFBT}$ algorithm Input $\text{x}_0\in \mathbb{R}^n$, $t_0>0$. For $k=0, 1, 2, ...$, perform the following loop: 1: #Calculate loss function. 2: $s_k:=\mathbb{1}[f(\text{x}_k)-f(\text{y}_k)\ge t^2_k]$ #Is it sufficiently decreasing? Yes: $s_k=1$; No: $s_k=0$. 3: $\text{x}_{k+1}:=(1-s_k)\text{x}_k+s_k\text{y}_k$ #Update the point of iteration. 4: $t_{k+1}:=2^{2S_k-1}t_k$ #Update step size. $s_k=1$: Step size doubles; $s_k=0$: Step size decreases by half. Now, we made assumption to the loss function $f:\mathbb{R}^n\to \mathbb{R}$. Assumption 1. Let $f$ be a convex function. For any $\text{x}, \text{y}\in \mathbb{R}^n$ and $\alpha \in [0, 1]$, we have $f((1-\alpha)\text{x}+\text{y})\le (1-\alpha)f(\text{x})+\alpha f(\text{y})$. Assumption 2. $f$ is differentiable on $\mathbb{R}^n$ and $\nabla f$ is L-Lipschitz continuous on $\mathbb{R}^n$. Assumption 3. The level set of $f$ is bounded. For any $\lambda\in\mathbb{R}$, set $\{\text x\in \mathbb{R}^n:f(\text x)\le \lambda\}$ is all bounded. Based on assumption 1 and 2, we can prove that $\left\langle \nabla f(\text x),\text y-\text x \right\rangle \le f(\text y)-f(\text x)\le \left\langle \nabla f(\text x),\text y-\text x\right\rangle+\frac{L}{2}||\text x-\text y||^2$ You can refer to any convex analysis textbook for more properties of convex function. Prove that under the assumption 1-3, for $AFBT$, $\lim_{k \to \infty}f(\text{x}_k)=f^*$

Given regular $1987$ -gon on plane with vertices $A_1, A_2,..., A_{1987}$. Find locus of points M of the plane sych that $$\left|\overrightarrow{MA_1}+\overrightarrow{MA_2}+...+\overrightarrow{MA_{1987}}\right| \le 1987$$.

Let $a, b$ be positive real numbers such that there exist infinite number of natural numbers $k$ such that $\lfloor a^k \rfloor + \lfloor b^k \rfloor = \lfloor a \rfloor ^k + \lfloor b \rfloor ^k$ . Prove that $\lfloor a^{2014} \rfloor + \lfloor b^{2014} \rfloor = \lfloor a \rfloor ^{2014} + \lfloor b \rfloor ^{2014}$

A portion of a corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles? [asy]/* AMC8 2002 #23 Problem */ fill((0,2)--(1,3)--(2,3)--(2,4)--(3,5)--(4,4)--(4,3)--(5,3)--(6,2)--(5,1)--(4,1)--(4,0)--(2,0)--(2,1)--(1,1)--cycle, mediumgrey); fill((7,1)--(6,2)--(7,3)--(8,3)--(8,4)--(9,5)--(10,4)--(7,0)--cycle, mediumgrey); fill((3,5)--(2,6)--(2,7)--(1,7)--(0,8)--(1,9)--(2,9)--(2,10)--(3,11)--(4,10)--(4,9)--(5,9)--(6,8)--(5,7)--(4,7)--(4,6)--cycle, mediumgrey); fill((6,8)--(7,9)--(8,9)--(8,10)--(9,11)--(10,10)--(10,9)--(11,9)--(11,7)--(10,7)--(10,6)--(9,5)--(8,6)--(8,7)--(7,7)--cycle, mediumgrey); draw((0,0)--(0,11)--(11,11)); for ( int x = 1; x < 11; ++x ) { draw((x,11)--(x,0), linetype("4 4")); } for ( int y = 1; y < 11; ++y ) { draw((0,y)--(11,y), linetype("4 4")); } clip((0,0)--(0,11)--(11,11)--(11,5)--(4,1)--cycle);[/asy] $ \textbf{(A)}\ \frac13\qquad\textbf{(B)}\ \frac49\qquad\textbf{(C)}\ \frac12\qquad\textbf{(D)}\ \frac59\qquad\textbf{(E)}\ \frac58$

Each square of the chessboard was painted in one of eight colors in such a way that the number of squares colored by all the colors are equal. Is it always possible to put $8$ rooks not threatening each other on multi-colored cells?

Prove that for any $n$ ($n \geq 2$) pairwise distinct fractions in the interval $(0,1)$, the sum of their denominators is no less than $\frac{1}{3} n^{\frac{3}{2}}$.

Define a sequence $(a_n)$ by $a_0 =-4 , a_1 =-7$ and $a_{n+2}= 5a_{n+1} -6a_n$ for $n\geq 0.$ Prove that there are infinitely many positive integers $n$ such that $a_n$ is composite.

Let be a positive real number $ r, $ a natural number $ n, $ and a function $ f:\mathbb{R}\longrightarrow\mathbb{R} $ satisfying $ f(rxy)=(f(x)f(y))^n, $ for any real numbers $ x,y. $ [b]a)[/b] Give three distinct examples of what $ f $ could be if $ n=1. $ [b]b)[/b] For a fixed $ n\ge 2, $ find all possibilities of what $ f $ could be. [i]Bogdan Blaga[/i]

Find all functions $f : \mathbb R\to\mathbb R $ such that \[f(x+yf(x^2))=f(x)+xf(xy)\] for all real numbers $x$ and $y$.

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected? $ \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$

Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) [i]Proposed by Sergey Berlov, Russia[/i]

Find a triple $(l, m, n)$ of positive integers $(1<l<m<n)$, such that $\sum_{k=1}^{l}k, \sum_{k=l+1}^{m}k, \sum_{k=m+1}^{n}k$ form a geometric sequence in order.

Find all ordered pairs $(a, b)$ of positive integers for which \[\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}.\]

[b]a)[/b] Find all $ 2\times 2 $ complex matrices $ A $ which have the property that there are two complex numbers $ \alpha ,\gamma $ with $ \alpha \neq \text{tr} (A) $ or $ \gamma\neq \det (A) $ such that $ A^2-\alpha A+\gamma I=0. $ [b]b)[/b] Consider $ B\not\in\{ 0,I\} $ as a matrix having the property mentioned at [b]a).[/b] Solve in the complex numbers the system $ xB-yI-B^2=xB^2-yI-B^4=0. $ [i]Adrian Troie[/i]

Some time ago, on a radio program, a baker announced a special promotion in the purchase of two stuffed cakes. Each cake could contain up to five fillings of which had in the pastry. On the show, a lady said there were $1,048,576$ different possibilities to choose the two stuffed cakes. How many different fillings did the pastry chef have?

A plane $p$ passes through a vertex of a cube so that the three edges at the vertex make equal angles with $p$. Find the cosine of this angle. Find the positions of the feet of the perpendiculars from the vertices of the cube onto $p$. There are 28 lines through two vertices of the cube and 20 planes through three vertices of the cube. Find some relationship between these lines and planes and the plane $p$.

A circle $S$ bisects a circle $S'$ if it cuts $S'$ at opposite ends of a diameter. $S_A$, $S_B$,$S_C$ are circles with distinct centers $A, B, C$ (respectively). Show that $A, B, C$ are collinear iff there is no unique circle $S$ which bisects each of $S_A$, $S_B$,$S_C$ . Show that if there is more than one circle $S$ which bisects each of $S_A$, $S_B$,$S_C$ , then all such circles pass through two fixed points. Find these points. [b]Original Statement:[/b] A circle $S$ is said to cut a circle $\Sigma$ [b]diametrically[/b] if and only if their common chord is a diameter of $\Sigma.$ Let $S_A, S_B, S_C$ be three circles with distinct centres $A,B,C$ respectively. Prove that $A,B,C$ are collinear if and only if there is no unique circle $S$ which cuts each of $S_A, S_B, S_C$ diametrically. Prove further that if there exists more than one circle $S$ which cuts each $S_A, S_B, S_C$ diametrically, then all such circles $S$ pass through two fixed points. Locate these points in relation to the circles $S_A, S_B, S_C.$

The circles $C_1$ and $C_2$ intersect at the distinct points $M$ and $N$. Points $A$ and $B$ belong respectively to the circles $C_1$ and $C_2$ so that the chords $[MA]$ and $[MB]$ are tangent at point $M$ to the circles $C_2$ and $C_1$, respectively. To prove it that the angles $\angle MNA$ and $\angle MNB$ are equal.

The polynomial $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ has real coefficients and satisfies $ \left|f(x)\right|\le 1$ for all $ x\in [0, 1]$. Find the maximal value of $ |a|\plus{}|b|\plus{}|c|$.

An infinite arithmetic progression is given. The products of the pairs of its members are considered. Prove that two of these numbers differ by no more than 1. [i]Proposed by A. Kuznetsov[/i]

Given positive integers $a_1,a_2,\ldots, a_n$ with $a_1<a_2<\cdots<a_n)$, and a positive real $k$ with $k\geq 1$. Prove that \[\sum_{i=1}^{n}a_i^{2k+1}\geq \left(\sum_{i=1}^{n}a_i^k\right)^2.\]

In triangle $ABC$ $\angle B > 90^\circ$. Tangents to this circle in points $A$ and $B$ meet at point $P$, and the line passing through $B$ perpendicular to $BC$ meets the line $AC$ at point $K$. Prove that $PA = PK$. [i](Proposed by Danylo Khilko)[/i]