A lattice point in the coordinate plane with origin $O$ is called invisible if the segment $OA$ contains a lattice point other than $O,A$. Let $L$ be a positive integer. Show that there exists a square with side length $L$ and sides parallel to the coordinate axes, such that all points in the square are invisible.

A lattice point in the coordinate plane with origin $O$ is called invisible if the segment $OA$ contains a lattice point other than $O,A$. Let $L$ be a positive integer. Show that there exists a square with side length $L$ and sides parallel to the coordinate axes, such that all points in the square are invisible.

$M$ is an arbitrary point inside $\triangle ABC$. $AM$ intersects the circumcircle of the triangle again at $A_1$. Find the points $M$ that minimise $\frac{MB\cdot MC}{MA_1}$.

Let $B$ be a point on a circle $S_1$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$. Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$. Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$.

Click for solution I. for one, distinctly remember posting two or three of these on the forum, but what the heck! They're nice problems; we can solve them all over again. I remember giving an inversive proof, but I can't find it anymore, so here's another one: Let $M$ be the circumcenter of $BCD$. We have $\angle BMC=2\angle BDC$, and we want to show that $\angle BMC=\angle BAC$, so it's equivalent to showing that $\angle BAC=2\angle BDC$. Let $C'$ be the intersection of $DC$ with $S_1$. The tangent through $C'$ to $S_1$ is parallel to $AC$ because $C'$ is the image $C$ through the homothety centered at $D$ which turns $S_2$ into $S_1$. Now let $T$ be the intersection of the tangents to $S_1$ through $B,C'$. We now have $\angle BAC=\pi-\angle BTC'=2\angle TC'B=2\angle BDC'=2\angle BDC$, Q.E.D.

$10$ cities are connected by one-way air routes in a way so that each city can be reached from any other by several connected flights. Let $n$ be the smallest number of flights needed for a tourist to visit every city and return to the starting city. Clearly $n$ depends on the flight schedule. Find the largest $n$ and the corresponding flight schedule.

Let $p_1,p_2,\ldots,p_n$ be distinct primes greater than $3$. Show that $2^{p_1p_2\cdots p_n}+1$ has at least $4^n$ divisors.

Click for solution Lemma 1: If $(a,b)=1$ and $a,b$ are odd, then $(2^a+1,2^b+1)=3$. Proof: Let $d=(2^a+1,2^b+1)$. $3|d$ and $d|(2^{2a}-1,2^{2b}-1)=2^{(2a,2b)}-1=2^2-1=3$, so $d=3$. $\blacksquare$ Lemma 2: If $3\not |a$ ($a$ is odd), then $9\not |2^a+1$. Proof: If $a=3k+1$ we have $2^a+1=2(2^{3k}+1)-1=9t-1$, and if $a=3k+2$ we have $2^a+1=4(2^{3k}+1)-3$. $\blacksquare$ Let $N=p_1\ldots p_n,\ N'=\frac N{p_n}$. For $n=1$, the result is easy to prove: $2^p+1$ has the divisors $1, 3, \frac{2^p+1}3, 2^p+1$. We have $3\ne \frac{2^p+1}3$ because $p>3$. We can now use induction. Assume we have shown it for $n-1$. This means that $2^{N'}+1$ has at least $4^{n-1}$ divisors. We have $2^{N'}+1|2^N+1,\ \frac{2^{p_n}+1}3|2^N+1$, and we can apply the lemmata for $a=p_n, b=N'$ to get $\frac{2^{p_n}+1}3\big|\frac{2^N+1}{2^{N'}+1}$. On the other hand, we clearly have $\left(\frac{2^{p_n}+1}3\right)^2<\frac{2^N+1}{2^{N'}+1}$, which means that $\frac{2^N+1}{2^{N'}+1}$ has at least four divisors $1=d_1,d_2,d_3,d_4$. For each $d|2^{N'}+1$, each of $d_id$ is a divisor of $2^N+1$, and all $d_id$ are different because $(d_i,d)=1$, hence the result: $2^N+1$ has at least $4\cdot 4^{n-1}=4^n$ divisors.