It is known that $a^{2005} + b^{2005}$ can be expressed as the polynomial of $a + b$ and $ab$. Find the coefficients' sum of this polynomial.

Click for solution If $a^{2005}+b^{2005}=P(a+b,ab)$ then the requested sum equals $P(1,1)$. Therefore, take $a=\frac{1+i\sqrt{3}}{2}$ and $b=\frac{1-i\sqrt{3}}{2}$. We obtain $P(1,1)=1.$

Given three points $P$, $A$, $B$ and a circle such that the lines $PA$ and $PB$ are tangent to the circle at the points $A$ and $B$, respectively. A line through the point $P$ intersects that circle at two points $C$ and $D$. Through the point $B$, draw a line parallel to $PA$; let this line intersect the lines $AC$ and $AD$ at the points $E$ and $F$, respectively. Prove that $BE = BF$.

Click for solution I have a little bit long proof using Euclidean Geometry. $pf$) Let $T \in circumcircle$ s.t $BT \parallel PD$, $S=AT \bot CD$. Here, $\angle PBA = \angle BTA = \angle TSD = \angle ASC$, or $ASBP$ is cyclic. And cuz $ABPO$ is cyclic, we get $ABPOS$ is cyclic. ($O$ is the circumcenter) therefore, $\angle OSP = \angle OAP = \frac{\pi}{2}$. or $DS = CS$. and cuz $CD \parallel BT$, $BC = DT$. or $\angle BAC = \angle TAD$......$(*)$. And, $\angle FEA = \angle CAP = \angle CDA$ or $\triangle AEF$ resembles $\triangle ADC$. cuz $AS$ is median of $\triangle ADC$, we get $AB$ is the median of $\triangle AEF$ from $(*)$. therefore, we get $BE=BF$. $Q.E.D.$

Set $S = \{1, 2, 3, ..., 2005\}$. If among any $n$ pairwise coprime numbers in $S$ there exists at least a prime number, find the minimum of $n$.

Click for solution The 14 prime numbers $< \sqrt{2005}$ are : $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ If we take 15, or more, numbers relatively prime they can't be all of the form $\prod^n p_i^{a_i}$ with $n=1$ : one must be prime (since it must be prime with all prime numbers $< \sqrt{2005}$).

Given is the positive integer $n > 2$. Real numbers $\mid x_i \mid \leq 1$ ($i = 1, 2, ..., n$) satisfying $\mid \sum_{i=1}^{n}x_i \mid > 1$. Prove that there exists positive integer $k$ such that $\mid \sum_{i=1}^{k}x_i - \sum_{i=k+1}^{n}x_i \mid \leq 1$.

Click for solution It seems like this can be done as follows: suppose WLOG that $\sum_{i=1}^n x_i > 0$. Then, let $S_k = \sum_{i=1}^k x_i - \sum_{i=k+1}^n x_i$; we have $S_n = \sum_{i=1}^n x_i > 1$ and $S_0 = -\sum_{i=1}^n x_i < -1$. Suppose that we never have $-1 \le S_k \le 1$. Then let $k_0$ be the maximal value of $k$ for which $S_{k_0} < 0$; we know $k_0$ exists as $k = 0$ certainly satisfies the given property; additionally, $k_0 \ne n$ as $k = n$ does NOT satisfy the property, so $S_{k_0 + 1}$ is defined. By the assumption, we have $S_{k_0} < -1$, and as $k_0$ was the maximal value, we must have $S_{k_0 + 1} \ge 0$ and thus by the assumption $S_{k_0 + 1} > 1$. Thus $2 < S_{k_0 + 1} - S_{k_0} =$ $\left(\sum_{i=1}^{k_0+1} x_i - \sum_{i=k_0+2}^n x_i\right) - \left(\sum_{i=1}^{k_0} x_i - \sum_{i=k_0 + 1}^n x_i\right) =$ $2x_{k_0+1} \le 2|x_{k_0+1}| \le 2$ which is sort of false. Thus there exists a $k$ with the desired property, i.e. that $|S_k| \le 1$.

Circles $C(O_1)$ and $C(O_2)$ intersect at points $A$, $B$. $CD$ passing through point $O_1$ intersects $C(O_1)$ at point $D$ and tangents $C(O_2)$ at point $C$. $AC$ tangents $C(O_1)$ at $A$. Draw $AE \bot CD$, and $AE$ intersects $C(O_1)$ at $E$. Draw $AF \bot DE$, and $AF$ intersects $DE$ at $F$. Prove that $BD$ bisects $AF$.

In isosceles right-angled triangle $ABC$, $CA = CB = 1$. $P$ is an arbitrary point on the sides of $ABC$. Find the maximum of $PA \cdot PB \cdot PC$.

If $a,b,c$ are positive reals such that $a+b+c=1$, prove that \[ 10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq 1 . \]

Click for solution We need to prove : $10(a^3+b^3+c^3)(a+b+c)^2-9(a^5+b^5+c^5) \geq (a+b+c)^5$ $\iff 15\sum_{sym}(a^4b-a^2b^2c) \geq 0$ : obviously true by Muirhead

For $n$ people, if it is known that (a) there exist two people knowing each other among any three people, and (b) there exist two people not knowing each other among any four people. Find the maximum of $n$. Here, we assume that if $A$ knows $B$, then $B$ knows $A$.