Let $a \in \mathbb{R}$ be a parameter. (1) Prove that the curves of $y = x^2 + (a + 2)x - 2a + 1$ pass through a fixed point; also, the vertices of these parabolas all lie on the curve of a certain parabola. (2) If the function $x^2 + (a + 2)x - 2a + 1 = 0$ has two distinct real roots, find the value range of the larger root.

Click for solution (1) It is clear that all parabolas pass through $(2,9)$, and the vertices lie on the parabola $y = -x^2+4x+5$. (2) The larger root is $ \frac{-a-2+\sqrt{a^2+12a}}{2}$, so $a \in (-\infty , -12)$ or $(0,\infty )$. Now, as $a$ gets smaller, the root gets bigger. On the other hand, the root is negative in $(0,\frac{1}{2})$, and the function is concave, so the lower bound is $-1$.

Circle $C$ (with center $O$) does not have common point with line $l$. Draw $OP$ perpendicular to $l$, $P \in l$. Let $Q$ be a point on $l$ ($Q$ is different from $P$), $QA$ and $QB$ are tangent to circle $C$, and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$. $PM$, $PN$ are perpendicular to $QB$, $QA$, respectively, $M \in QB$, $N \in QA$. Prove that segment $KP$ is bisected by line $MN$.

Click for solution draw PX perpendicular to AB ( X is a point on AB) We know that O, A, P, Q, B are cyclic. Then apply Simson's theorem from point P to the triangle ABQ. K, N, M are on the same line. angle QOP = angle QAP = angle NXP = angle KPX (because OQ are parallel to XP) So KP is bisected by line NM

Let $n$ be positive integer, set $M = \{ 1, 2, \ldots, 2n \}$. Find the minimum positive integer $k$ such that for any subset $A$ (with $k$ elements) of set $M$, there exist four pairwise distinct elements in $A$ whose sum is $4n + 1$.

Click for solution $k=n+3$ will do. to start with note that the set $\{n-1,n,n+1,n+2,\ldots,2n\}$ is a set of $n+2$ elements and has a minimum sum (with $4$ distinct elements) of $n-1+n+n+1+n+2=4n+2$, so $k\geq n+3$.also for a set of size atleast $n+2$ we can form pairs of distinct elements that sum to $2n$ and pigeonhole $\Rightarrow$ there exists a pair of distinct elements with sum $2n$.once we remove such a pair from the $n+3$-set and we are still left with an $n+1$ set from which another pair,this time with sum $2n+1$ (again pigeonhole) can be extracted and so we are through.

Find all positive integer solutions $(a, b, c)$ to the function $a^{2} + b^{2} + c^{2} = 2005$, where $a \leq b \leq c$.

Click for solution I do not know the official solution, but here's my solution anyway. We consider $\mod 8$. Since all squares are $\equiv 0, 1$ or $4 \pmod{8}$ and $2005 \equiv 5 \pmod{8}$ it follows that $a^2, b^2, c^2$ are $\equiv 0, 1, 4 \pmod{8}$ in some order. WLOG, we ignore the order of $a,b,c$ (we drop the assumption of $a \leq b \leq c$) and assume $a^2 \equiv 0 \pmod{8}, b^2 \equiv 1 \pmod{8}, c^2 \equiv 4 \pmod{8}$. Let $a=4a_1, b=2b_1+1, c=4c_1+2$. We then have $a^2+b^2+c^2=16{a_1}^2+4{b_1}^2+4b_1+16{c_1}^2+16c_1+5 = 2005$ $\Leftrightarrow 4{a_1}^2+{b_1}^2+b_1 + 4{c_1}^2+4c_1 = 500$. Taking $\mod 4$, we have ${b_1}^2 + b_1 \equiv 0 \pmod{4} \Rightarrow b_1 \equiv 0$ or $-1 \pmod{4}$. Case 1: $b_1 \equiv 0 \pmod{4}$. Let $b_1 = 4b_2$ We have $4{a_1}^2+16{b_2}^2+4b_2 + 4{c_1}^2+4c_1 = 500 \Leftrightarrow {a_1}^2+4{b_2}^2+b_2 + {c_1}^2+c_1 = 125$. Since $b_2 \in \mathbb{Z}^+$ and $4{b_2}^2 > 125 \ \forall \ b_2 > 5$, we have $1 \leq b_2 \leq 5$. When $b_2 = 5$, we have ${a_1}^2 +{c_1}^2+c_1=20$. Note that ${c_1}^2+c_1 \equiv 0 \pmod{2}$, thus ${a_1}^2 = 4$ or $16$. We then have ${c_1}^2+c_1 = 16$ or $4$ respectively, which yield no solutions. When $b_2=4$, we have ${a_1}^2 +{c_1}^2+c_1=57$. Note that ${c_1}^2+c_1 \equiv 0 \pmod{2}$, thus ${a_1}^2 = 1, 9, 25$ or $49$. We then have ${c_1}^2+c_1 = 56, 48, 32$ or $8$ respectively, which yield the only possible solution $c_1 = 7 \Rightarrow (a,b,c) = (4,33,30)$. When $b_2=3$, we have ${a_1}^2 +{c_1}^2+c_1=86$. Similarly, ${a_1}^2$ can only be $4, 16, 36$, or $64$, which means ${c_1}^2+c_1 = 82, 70, 50$ or $22$ respectively, which yield no solutions. When $b_2=2$, we have ${a_1}^2 +{c_1}^2+c_1=107$. Similarly, ${a_1}^2$ can only be $1, 9, 25, 49$ or $81$, which means ${c_1}^2+c_1 = 106, 98, 82, 58$ or $26$ respectively, which yield no solutions. When $b_2=1$, we have ${a_1}^2 +{c_1}^2+c_1=120$. Similarly, ${a_1}^2$ can only be $4, 16, 36, 64$ or $100$, which means ${c_1}^2+c_1 = 116, 104, 84, 56$ or $20$ respectively, which yield the possible solutions $c_1 = 7, 4 \Rightarrow (a,b,c) = (32, 9, 30), (40, 9, 18)$. Case 2: $b_1 \equiv -1 \pmod{4}$. Let $b_1 = 4b_2 -1$ We have $4{a_1}^2+16{b_2}^2-4b_2 + 4{c_1}^2+4c_1 = 500 \Leftrightarrow {a_1}^2+4{b_2}^2-b_2 + {c_1}^2+c_1 = 125$. Since $b_2 \in \mathbb{Z}^+$ and $4{b_2}^2 -b_2 > 125 \forall b_2 > 5$, we have $1 \leq b_2 \leq 5$ When $b_2 = 5$, we have ${a_1}^2 +{c_1}^2+c_1=30$. Similarly, by considering parity, we must have ${a_1}^2 = 4$ or $16$, which means ${c_1}^2+c_1 = 26$ or $14$ respectively, which yield no solutions. When $b_2=4$, we have ${a_1}^2 +{c_1}^2+c_1=65$. Similarly, we have ${a_1}^2 = 1, 9, 25$ or $49$, meaning ${c_1}^2+c_1 = 64, 56, 40$ or $16$ respectively, which yield the only possible solution $c_1 = 7 \Rightarrow (a,b,c) = (12, 31, 30)$. When $b_2=3$, we have ${a_1}^2 +{c_1}^2+c_1=92$. Similarly, we have ${a_1}^2 = 4, 16, 36$ or $64$, meaning ${c_1}^2+c_1 = 88, 76, 56$ or $28$ respectively, which yield the only possible solution $c_1 = 7 \Rightarrow (a,b,c) = (24, 23, 30)$. When $b_2=2$, we have ${a_1}^2 +{c_1}^2+c_1=111$. Similarly, we have ${a_1}^2 = 1, 9, 25, 49$ or $81$, meaning ${c_1}^2+c_1 = 110, 102, 86, 62$ or $30$ respectively, which yield the possible solutions $c_1 = 5, 10 \Rightarrow (a,b,c) = (4, 15, 22), (36, 15, 42)$. When $b_2=1$, we have ${a_1}^2 +{c_1}^2+c_1=122$. Similarly, ${a_1}^2$ can only be $1, 9, 25, 49, 81$ or $121$, which means ${c_1}^2+c_1 = 121, 113, 97, 73, 41$ or $1$ respectively, which yield no solutions. Therefore, by considering the above $10$ cases, the only seven possible solutions are $(4,33,30), (32,9,30), (40,9,18), (12,31,30), (24,23,30), (4,15,22), (36,15,42)$. As a side note, we can actually make use of what Singular has mentioned regarding the solvability of $a^2+b^2 =k$ to reduce the number of cases, but this does not necessarily make the solution shorter. Using the "Dirichlet Class-Number Theorem", or rather, finding the appropriate class numbers, to compute $\frac{r_3(2005)}{2^3 \cdot 3!} - \frac{r_2(2005)}{2^2 \cdot 2!} = 7$ is but an unnecessary complication. And besides, I haven't seen any olympiad problems that ask "Find the number of solutions..." instead of the usual "Find all solutions to...", so I doubt this theorem will be used that often.

Line $l$ tangents unit circle $S$ in point $P$. Point $A$ and circle $S$ are on the same side of $l$, and the distance from $A$ to $l$ is $h$ ($h > 2$). Two tangents of circle $S$ are drawn from $A$, and intersect line $l$ at points $B$ and $C$ respectively. Find the value of $PB \cdot PC$.

Click for solution Consider the circle $(S)$ of radius r = 1 and the line $l$ directed. Let $k$ be a directed line parallel to the line $l$ at a distance $h$ and A a point on this line outside of the circle $(S)$. Let A be a point on the line $k$ outside of the circle $(S)$ and H the foot of a normal from the point A to the line $l$. Let $b, c$ be the 2 directed tangents from the point A to the circle $(S)$ intersecting the line $l$ at points B, C and let $\beta = \widehat{(b, k)} = \widehat{(b, l)} = \measuredangle ABH$ $\gamma = \widehat{(c, k)} = \widehat{(c, l)} = 180^\circ - \measuredangle ACH$ be the directed angles formed by the directed tangents $b, c$ with the directed lines $k, l$. (A directed tangent must have the same direction as the directed circle at their tangency point. As opposed to 2 undirected lines, which form 2 different complementary angles, 2 directed lines form a unique angle.) Then $\frac{r}{PB} = \frac{PS}{PB} = \tan \widehat{SBP} = \tan \left(90^\circ - \frac{\widehat{ABH}}{2}\right) = \tan \left(90^\circ - \frac{\beta}{2}\right) = \frac{1}{\tan \frac \beta 2}$ ${\frac{r}{PC} = \frac{PS}{PC} = \tan \widehat{SCP}} = \tan \frac{\widehat{ACH}}{2} = \tan \left(90^\circ - \frac{\gamma}{2}\right) = \frac{1}{\tan \frac \gamma 2}$ The power of the directed line $k$ to the directed circle $(S)$ is equal to $\tan \frac{\widehat{(b, k)}}{2} \cdot \tan \frac{\widehat{(c, k)}}{2} = \tan \frac{\beta}{2} \cdot \tan \frac{ \gamma}{2} = \frac{PB \cdot PC}{r^2}$ and it does not depend on the position of the point A on the directed line $k$. Obviously, this is a dual concept of the power of a point to an undirected circle, which does not depend on the angle of the circle secant from this point. I guess it is not considered a pure geometry, because it contains trigonometric functions . To calculate the power of the directed line $k$ to the directed circle $(S)$, we can move the point A into the foot $A_0$ of a normal from the tangency point P to the line $k$, because then $\gamma_0 = \beta_0$ and $\tan \frac{\beta}{2} \cdot \tan \frac{ \gamma}{2} = \tan \frac{\beta_0}{2} \cdot \tan \frac{ \gamma_0}{2} = \tan^2 \frac{\beta_0}{2}$ Let $U_0, V_0$ be the tangency points of the directed tangents from the point $A_0$ to the directed circle $(S)$ intersecting the directed line $l$ at points $B_0, C_0$. $\frac{r}{h - r} = \frac{U_0S}{A_0S} = \cos \widehat{A_0SU_0} = \cos \widehat{A_0BP} = \cos (180^\circ - \beta_0) = -\cos \beta_0$ $\cos \beta_0 = \cos^2 \frac{\beta_0}{2} - \sin^2 \frac{\beta_0}{2} = \cos^2 \frac{\beta_0}{2} \left(1 - \tan^2 \frac{\beta_0}{2}\right) = \frac{1 - \tan^2 \frac{\beta_0}{2}}{1 + \tan^2 \frac{\beta_0}{2}} = \frac{1 - x}{1 + x}$ Solving for x, we get $x = \tan^2 \frac{\beta_0}{2} = \frac{1 - \cos \beta_0}{1 + \cos \beta_0} = \frac{1 + \frac{r}{h - r}}{1 - \frac{r}{h - r}} = \frac{h - r + r}{h - r - r} = \frac{h}{h - 2r}$ $PB \cdot PC = r^2 \tan^2 \frac{\beta_0}{2} = \frac{r^2h}{h - 2r}$ and for a unit circle r = 1, $PB \cdot PC = \frac{h}{h - 2}$

Let $P(A)$ be the arithmetic-means of all elements of set $A = \{ a_1, a_2, \ldots, a_n \}$, namely $P(A) = \frac{1}{n} \sum^{n}_{i=1}a_i$. We denote $B$ "balanced subset" of $A$, if $B$ is a non-empty subset of $A$ and $P(B) = P(A)$. Let set $M = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$. Find the number of all "balanced subset" of $M$.

Click for solution I'd look at the numbers not as they are written down now, but at their difference with 5. So: 4, 3, 2, 1, 0, 1, 2, 3, 4 At both the left and the right part, the sum of the differences must be the same. D is the total difference at one side: If D = 0: One possibility at both sides: (empty) 1*1=1 If D = 1: One possibility at both sides: 4, 6 1*1=1 If D = 2: One possibility at both sides: 3, 7 1*1=1 If D = 3: Two possibilities at both sides: 3+4 / 2, 6+7 / 8 2*2=4 If D = 4: Two possibilities at both sides: 2+4 / 1, 6+8 / 9 2*2=4 If D = 5: Two possibilities at both sides: 1+4 / 2+3, 6+9 / 7+8 2*2=4 If D = 6: Two possibilities at both sides: 1+3 / 2+3+4, 7+9 / 6+7+8 2*2=4 If D = 7: Two possibilities at both sides: 1+2 / 1+3+4, 8+9 / 6+7+9 2*2=4 If D = 8: One possibility: no 3, 7 1*1=1 If D=9: One possibility: no 4,6 1*1=1 If D=10: all numbers: 1 Counting gives us 26. Multiplied by 2 for including or excluding 5 gives 52, minus the empty subset gives us 51 as well!

(1) Find the possible number of roots for the equation $|x + 1| + |x + 2| + |x + 3| = a$, where $x \in R$ and $a$ is parameter. (2) Let $\{ a_1, a_2, \ldots, a_n \}$ be an arithmetic progression, $n \in \mathbb{N}$, and satisfy the condition \[ \sum^{n}_{i=1}|a_i| = \sum^{n}_{i=1}|a_{i} + 1| = \sum^{n}_{i=1}|a_{i} - 2| = 507. \] Find the maximum value of $n$.

Click for solution First, observe that if $a_{1}<a_{2}<\ldots<a_{n}$ are real numbers and $f:\mathbb{R}\rightarrow\mathbb{R}$ is defined as \[ f\left( x\right) =\left| x-a_{1}\right| +\left| x-a_{2}\right| +\ldots+\left| x-a_{n}\right| , \] then $f$ is first decreasing, and then increasing. More precisely, if $n$ is odd, say $n=2k+1,$ then $f$ is strictly decreasing on the interval $(-\infty,a_{k}]$ and strictly increasing on $[a_{k},+\infty)$, while if $n$ is even, say $n=2k,$ then $f$ is strictly decreasing on the interval $(-\infty,a_{k}],$ constant on $[a_{k},a_{k+1}]$ and strictly increasing on $[a_{k+1},+\infty).$ Now, let WLOG $a_{1}\leq\ldots\leq a_{n}$ be our arithmetic sequence and $f$ defined as above. We see that $f(0)=f\left( -1\right) =f\left( 2\right)$, hence three values of $f$ are equal. This is possible only if $n$ is even (say $n=2k$) and $0,-1,2\in\lbrack a_{k},a_{k+1}].$ We deduce that the common difference $d$ is at least $3,$ and $a_{k}\leq-1,a_{k+1}\geq2.$ A short computation shows that in this case \[ 507=\sum\left| a_{i}\right| \geq3k^{2}, \] hence $k\leq13,$ that is, $n\leq26.$ Equality is obtained when $a_{1}=-37,a_{2}=-34,\ldots,a_{26}=38.$

Let $0 < \alpha, \beta, \gamma < \frac{\pi}{2}$ and $\sin^{3} \alpha + \sin^{3} \beta + \sin^3 \gamma = 1$. Prove that \[ \tan^{2} \alpha + \tan^{2} \beta + \tan^{2} \gamma \geq \frac{3 \sqrt{3}}{2} . \]