BdMO Online Forum

Too much of the forum is dedicated to advanced problem solving, which I'm afraid is a level accessible to only a few in the country. Let's change that, shall we?


This thread is dedicated for a marathon on relatively easy problems. The rules are similar to the Geo Marathon or the Combi Marathon, just with a focus on easier problems.

We'll start by posting problems, and someone else will post the solution. If someone solves a problem, they should post a pretty solution, and they have the honor of posting the next problem. If the solver of the previous problem doesn't post a new problem along with his solution, anyone else can post a new problem. If no one solves a problem within 48 hours of posting, the problem poster should provide the solution or a link to the solution. Then we'll jump to the next problem.

The problem difficulty should be at most the hardest found in BDMO. The focus is on easy problems, so even problems of BDMO National #1 level are welcome. Ideally most of the problems will be somewhere between these two extremes, and help out all the people still looking to break at BDMO nationals.

Rules:
Posting this to junior since that is the level with the most potential for development and most in need of basic problems. But this thread is open to people from all age groups and skill levels, so get posting!

$\text{Problem} 1$

NASA has proposed populating Mars with 2,004 settlements. The only way to get
from one settlement to another will be by a connecting tunnel. A bored bureaucrat draws on a map of
Mars, randomly placing N tunnels connecting the settlements in such a way that no two settlements
have more than one tunnel connecting them. What is the smallest value of N that guarantees that,
no matter how the tunnels are drawn, it will be possible to travel between any two settlements?

It's the max number of edges in a disconnected graph + 1.
So suppoose we have a maximal disconnected graph. Then each connected component must be a clique.
So answer = K_2003 + isolated vertex.

$\text{Problem 2}$

Let $\triangle ABC$ be a triangle with $\angle A = 60^\circ$. The points $M, N, K$ lie on $BC, AC, AB$ respectively such that $BK = KM = MN = NC$. If $AN = 2AK$, find the values of $\angle B$ and $\angle C$.

Solution of Problem $\boxed{2}$:

$\frac{AK}{AN} = \frac{1}{2} = \cos 60$

Thus,$\triangle AKN$ is a $30-60-90$ triangle.

Let,$\angle B = b$.So,$\angle C = 120-b$

By,some angle chasing , we will get
$\angle KMN =180 - (\angle KMB + \angle NMC)$
$180 - (\angle KBM + \angle NCM) = 180 - (b+120-b)$
$=60$

This implies that,$\angle MKN = \angle MNK = 60$.

So,$\angle BKM = 180 - \angle MKN - \angle AKN$
$180-60-90= 30$

At last , $\angle B = (180-30)/2 = 75$ and
$\angle C = 120-75=45$

P3. Several stones are placed on an infinite (in both directions) strip of squares. As long as there are at least two stones on a single square, you may pick up two such stones, then move one to the preceding square and one to the following square. Is it possible to return to the starting configuration after a finite sequence of such moves?

Solution to problem #3:

Since the stones are finite, so let’s have a configuration with the 1st square having 2 stones & rest of the squares having 1 stone. If we apply the move to the 1st square, the preceding square will be the 1st square having any stone. No more stone can be put there. In the starting configuration, the 1st square had 2 stones. But after applying the move at that square, we can never have the 1st square having 2 stones. So it’s not always possible to return to the starting configuration after some finite moves.

I'm posting a new problem.

$\text {Problem 4}$

Let $ABC$ be an acute triangle with $D,E,F$ the feet of the altitudes lying on $BC, CA,AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP =AQ$.

(Source : IMO shortlist 2010/G1)

$\text{Solution 4}$

Case 1: $P$, on $\widehat{AB}$. that doesn't contain $C$.

Proof: $\angle ACB=\angle DHB=\angle DFB=\angle AFQ=a$
and,$\angle ACB=\angle QPA=a$
so, $AQPF$ is cyclc.
and,$\angle ACB=\angle AFE=\angle PFB=a$
we know that $AQPF$ is cyclic.
so, $\angle PFB=\angle AQP=a$
so, $\angle AQP=a=\angle APQ \Rightarrow AP=AQ$

Case 2: $P'$, on $\widehat {AB}$. that contains $C$.
Proof: $\angle ACB=\angle AP'B=b$
and, $\angle ACB=\angle BHD=\angle BFD=b$
so, $\angle AP'Q'=\angle Q'FB=b \Rightarrow AFQ'P'$ is cyclic.
and. $\angle ACD=\angle AHE=\angle AFP'=b$
so, $\angle APQ'=b=\angle AQP \Rightarrow AQ'=AP'$

$[DONE]$

$\text {Problem 5}$

Suppose there are $n$ points in a plane no three of which are collinear with the property that if we label these points as $A_1, A_2, . . . , A_n$ in any way whatsoever, the broken line $A_1A_2 . . . A_n$ does not intersect itself. Find the maximum value of $n$.