Let side $BC$ of $\bigtriangleup ABC$ be the diameter of a semicircle which cuts $AB$ and $AC$ at $D$ and $E$ respectively. $F$ and $G$ are the feet of the perpendiculars from $D$ and $E$ to $BC$ respectively. $DG$ and $EF$ intersect at $M$. Prove that $AM \perp BC$.

Click for solution Orl, you could remember this one! [url=http://groups.yahoo.com/group/Hyacinthos/message/9531?expand=1]Hyacinthos message #9531[/url] wrote: From: "orl_ml" (Orlando Doehring) Subject: AM is perpendicular to BC Let side BC of triangle ABC be the diameter of a semicircle which cuts AB and AC at D and E respectively. F and G are the feet of the perpendiculars from D and E to BC respectively. DG and EF intersect at M. Prove that AM is perpendicular to BC. [url=http://groups.yahoo.com/group/Hyacinthos/message/9550?expand=1]Hyacinthos message #9550[/url] wrote: From: Darij Grinberg Subject: Re: AM is perpendicular to BC Dear Orlando, In Hyacinthos message #9531, you wrote: >> Let side BC of triangle ABC be the diameter >> of a semicircle which cuts AB and AC at D and >> E respectively. F and G are the feet of the >> perpendiculars from D and E to BC respectively. >> DG and EF intersect at M. Prove that AM is >> perpendicular to BC. Well, D and E are just the feet of the altitudes from C and B, respectively. Let S be the foot of the altitude from A, and let H be the orthocenter of triangle ABC. Let the line DG meet the line AS at M', and let the line EF meet the line AS at M". We will show that the points M' and M" coincide. Since the point M' lies on the lines AS and DG, the Menelaos theorem applied to triangle ABS yields AM' SG BD --- * -- * -- = -1, M'S GB DA and therefore AM' DA GB AD GB --- = - -- * -- = -- * --. M'S BD SG BD SG Since EG || HS, we have GB / SG = EB / HE, so that AM' AD EB AD DH EB AE --- = -- * -- = -- * -- * -- * --. M'S BD HE DH BD AE HE The triangles BDH and BEA are similar, so DH AE DH EB -- = --, and -- * -- = 1, BD EB BD AE so that our equation above simplifies to AM' AD AE AD AE --- = -- * -- = - -- * --. M'S DH HE DH EH Now, this expression is symmetric in B and C; hence, by analogy, AM" AD AE --- = - -- * --, M"S DH EH and the points M' and M" coincide. Thus, the lines DG, EF and AS meet at one point M' = M" = M, and AM is perpendicular to BC. There should be a simpler proof. Thanks for this and all the other problems, many of them are still too difficult or non-standard for me. Friendly, Darij Grinberg [url=http://groups.yahoo.com/group/Hyacinthos/message/9553?expand=1]Hyacinthos message #9553[/url] wrote: From: Vladimir Dubrovsky Subject: Re: [EMHL] Re: AM is perpendicular to BC Dear Darij, here is a shorter solution to Hyacinthos message #9531 -- > >> Let side BC of triangle ABC be the diameter > >> of a semicircle which cuts AB and AC at D and > >> E respectively. F and G are the feet of the > >> perpendiculars from D and E to BC respectively. > >> DG and EF intersect at M. Prove that AM is > >> perpendicular to BC. Let D', E' be reflections of D, E in BC, N the intersection of DE' and D'E. Obviously, N lies on BC and MN is parallel to DD' and EE', or perpendicular to BC (because M and N divide DG and DE', resp., in the same ratio DF:EG=DD':EE'). So it remains to show that, in your notation, N=S, the foot of the altitude from A, or, equivalently, that angle NEC = angle ABC.. But angle NEC = angle D'EC = angle DE'C = angle DBC. QED Best, Vladimir Dubrovsky Darij

$S$ is the set of functions $f:\mathbb{N} \to \mathbb{R}$ that satisfy the following conditions: I. $f(1) = 2$ II. $f(n+1) \geq f(n) \geq \frac{n}{n + 1} f(2n)$ for $n = 1, 2, \ldots$ Find the smallest $M \in \mathbb{N}$ such that for any $f \in S$ and any $n \in \mathbb{N}, f(n) < M$.

Click for solution That's a serious misuse of limit language- something that's eventually true doesn't have to be immediately true. $f(2)\le\frac21f(1)=4$ $f(4)\le\frac32f(2)\le6$ $f(3)\le f(4)\le6$ $f(8)\le\frac54f(4)\le7.5$ $f(5)\le f(6)\le f(7)\le f(8)\le7$ $f(16)\le\frac98f(8)\le7+\frac78$ $f(n)\le7$ for all $n$ and $f$. The example $f(1)=2,f(2)=4,f(3)=f(4)=6,f(n)=7$ for $n\ge5$ shows that this estimate is the best possible.

Let $ M = \lbrace 2, 3, 4, \ldots\, 1000 \rbrace$. Find the smallest $ n \in \mathbb{N}$ such that any $ n$-element subset of $ M$ contains 3 pairwise disjoint 4-element subsets $ S, T, U$ such that I. For any 2 elements in $ S$, the larger number is a multiple of the smaller number. The same applies for $ T$ and $ U$. II. For any $ s \in S$ and $ t \in T$, $ (s,t) = 1$. III. For any $ s \in S$ and $ u \in U$, $ (s,u) > 1$.

3 countries $A, B, C$ participate in a competition where each country has 9 representatives. The rules are as follows: every round of competition is between 1 competitor each from 2 countries. The winner plays in the next round, while the loser is knocked out. The remaining country will then send a representative to take on the winner of the previous round. The competition begins with $A$ and $B$ sending a competitor each. If all competitors from one country have been knocked out, the competition continues between the remaining 2 countries until another country is knocked out. The remaining team is the champion. I. At least how many games does the champion team win? II. If the champion team won 11 matches, at least how many matches were played?

Let $\alpha_1, \alpha_2, \dots, \alpha_n$, and $\beta_1, \beta_2, \ldots, \beta_n$, where $n \geq 4$, be 2 sets of real numbers such that \[\sum_{i=1}^{n} \alpha_i^2 < 1 \qquad \text{and} \qquad \sum_{i=1}^{n} \beta_i^2 < 1.\]Define \begin{align*} A^2 &= 1 - \sum_{i=1}^{n} \alpha_i^2,\\ B^2 &= 1 - \sum_{i=1}^{n} \beta_i^2,\\ W &= \frac{1}{2} (1 - \sum_{i=1}^{n} \alpha_i \beta_i)^2. \end{align*}Find all real numbers $\lambda$ such that the polynomial \[x^n + \lambda (x^{n-1} + \cdots + x^3 + Wx^2 + ABx + 1) = 0,\]only has real roots.

Does there exist non-zero complex numbers $a, b, c$ and natural number $h$ such that if integers $k, l, m$ satisfy $|k| + |l| + |m| \geq 1996$, then $|ka + lb + mc| > \frac {1}{h}$ is true?