Handling a 5-1 trump split

This is a first round hand from yesterday's open KO at the Bothell memorial day sectional.

Partner opens a 15-17 NT and you end up in 6C. LHO leads a high diamond spot (likely a doubleton/singleton).

This is what you see:

IMPS None	North ♠ AQ82 ♥ T2 ♦ KJ92 ♣ AJ7
	South ♠ K3 ♥ AKJ97 ♦ AT ♣ KT54

W	N	E	S
	1NT	P	2D1
P	2H	P	3C
P	4C2	P	4NT
P	5H3	P	6C4
P	P	P

1  transfer
2: ?
3: 2 Keycards, no Q
4: 6NT is better perhaps.

Anyway, LHO leads a high diamond spot, and RHO follows low.

You win the DT, and play a club to the J which wins! Now you cash the CA, RHO shows out, throwing a diamond.

How will you play?




If LHO is exactly 3=3=2=5 then you can make.

Cash AK heart, ruff a heart. Unblock DA, cash three rounds of spades (throwing a heart).  You are left with Hx, CKT in hand while LHO has to follow suit all along and is now left with CQ9x.

Now play the DK and  throw the last heart. LHO has to ruff this and lead into your KT.

So did we win IMPS on this? (or at least not lose big to 6NT)

No, we were actually missing the CT, so 6C was down one. LHO was indeed 3=3=2=5, so if we had the CT we would have made.

The other table was in 3NT.

Perhaps there are additional chances you have catered to? If so, please comment.

Odd points even triangles [Solution]

The problem was

You are given a set $S$ of $2005$ points (no three collinear) in the 2D plane. For each point $P$ in $S$, you count the number of triangles (formed by points in $S$) within which $P$ lies ($P$ must be strictly inside the triangle).

Show that this number is even, irrespective of the point $P$.


A possible solution


Given a point $P$ form a graph $G$ of the triangles that contain it.

In this graph, two triangles are adjacent if they share a side.

The claim is that each triangle has degree exactly $2001$ in $G$.

To show that,  consider a triangle $T = \triangle{ABC}$ which contains $P$ and another point $P'$.

The claim is that exactly one of the triangles $P'AB$, $P'AC$, $P'BC$ contains $P$ (haven't tried proving this rigorously, hence this is only a possible solution)

There are $2001$ such points $P'$ and thus degree of $T$ is $2001$.

Since the degree of each vertex in $G$ is 2001, the number of vertices must be even ($\sum d_i = 2|E| \implies 2001|V| = 2|E| \implies |V|$ is even).