Sum of squares of distances from vertices of a triangle

 If $ABC$ is a triangle with centroid $M$, and $P$ is any point then show that

$$PA^2 + PB^2 + PC^2 = 3PM^2 + \frac{1}{3}(AB^2 + BC^2 + AC^2)$$

Scroll down for a surprising solution.

Let $A_1,B_1, C_1$ be the midpoints of $BC, AC$ and $AB$ respectively.

Triangle $A_1B_1C_1$ (referred to as medial triangle of $ABC$) is similar to $ABC$ 

Using Appolonius theorem we get

$$PA^2 + PB^2 = 2(PC_1^2 + AC_1^2) = 2PC_1^2 + 2\frac{AB^2}{4}$$

$$PB^2 + PC^2 = 2(PA_1^2 + BA_1^2) = 2PA_1^2 + 2\frac{BC^2}{4}$$

$$PC^2 + PA^2 = 2(PB_1^2 + CB_1^2) = 2PB_1^2 + 2\frac{AC^2}{4}$$

Adding we get

$$PA^2 + PB^2 + PC^2 = PA_1^2 + PB_1^2 + PC_1^2 + \frac{AB^2}{4} + \frac{BC^2}{4} + \frac{AC^2}{4}$$

Thus if $A_nB_nC_n$ is the medial triangle of $A_{n-1}B_{n-1}C_{n-1}$  ($ABC = A_0 B_0C_0$) we have that

$$PA_{n-1}^2 + PB_{n-1}^2 + PC_{n-1}^2 = PA_n^2 + PB_n^2 + PC_n^2 + \frac{A_{n-1}B_{n-1}^2}{4} + \frac{B_{n-1}C_{n-1}^2}{4} + \frac{A_{n-1}C_{n-1}^2}{4}$$

 

We can easily show that $A_n, B_n, C_n$ all converge to $M$ ($A_nM = 3A_{n+1}M$), and observing the above is a telescoping series and that sides of a medial triangle are half the original triangle leads us 

$$PA^2 + PB^2 + PC^2 = 3PM^2 + \frac{AB^2 + BC^2 + AC^2}{4}(1 + \frac{1}{4} + \frac{1}{4^2} + \dots)$$

$$= 3PM^2 + \frac{1}{3}(AB^2 + BC^2 + AC^2)$$

Allow opponents to make a mistake

 Long time no bridge hand!

Here is a puzzle.

You end up in 4S playing a team game (bidding and hands below, you are South).

IMPS  N/S	Partner ♠ xxx ♥ xxx ♦ Kxx ♣ Kxxx
	You ♠ QJT9xx ♥ Qx ♦ AQx ♣ AQ

W	N	E	S
			1S
P	2S	P	4S
P	P	P

LHO leads a low heart to RHO's Ace, who leads a heart back to LHO's King and a third heart which you ruff.

You have lost two hearts and have two spades to lose. Situation is hopeless. But, is there anything you can do? If you want to think about it, don't scroll further.

There are no legitimate chances and need a defensive error. If the spades divide Ax opposite Kx, you must hope to induce a first round duck and then crash the A and K together. How could you do that?

The opponents don't know you have a 6 card suit. Consider what might happen if you lead a diamond to dummy's K and then play a spade to the Q! (has to be the Q, no other card will do)

LHO looking at Ax of spades might very well duck, thinking you have KQT9x and have a guess in spades (they hope you go to dummy with CK next and play a spade to the K, setting up partner's J for the setting trick). If you indeed have KQT9x and they win the A from Ax, then you are forced to take the winning line of finessing their partner for the J.

It is a slim chance which is more likely to work against better opponents than not, but then what have you got to lose?