Given a circle $(O,R)$ and a point $P$ in its interior. For any two points $A,B$ on its perimeter, we define $f(A,B)$ to be the point $C$ such that $APBC$ is a parallelogram.
Determine the image of $f$ when $A,B$ vary over all possible pairs of points on the perimeter. That is, determine the set $S = \{f(A,B) | AO = BO = R\}$
Solution Let $O'$ be a point on the extension of $PO$ such that $OO' = 2PO$. We shall show that $S$ is a circle (including the interior) centered at $O'$ with radius $2R$.

Outline of proof:

Let $T$ to be the circle we claim. Note that $T$ is a 2x dilation of the original circle with center $P$. We show that the boundary of $S$ is boundary of $T$ as well, by letting $A=B$ in the opration

Next we show that any point in $T$ is an image of a specific application of $f$. Let $X' \in T$, then $X'$ is an image of a 2x dilation. Suppose $X$ is the original point. We claim that there exists $A,B$ such that $PX$ is a median of $PAB$. Indeed, if $OAB$ is a triangle such that $OX$ is its median, then $PX$ is median of $PAB$.

Then it is routine to show that $f(A,B) = X'$.

Note: this problem can be extended to any point $P$, and any convex shape $V$ in $R^n$. The sum of vectors $PA + PB$ where $A,B$ are any two points on its boundary is equivalent to the 2x dilation of $V$ with center $P$. However, $V$ must be bounded.

Challenge: prove this generalization. (Hint: use the connectedness of $S^{n-1}$ in $R^n$).

Challenge: give a counter example of this result of $V$ is unbounded

Game making expression positive definite

On the board there are six numbers $A,B,C,D,E,F$, all initially zero. Mary and Nancy are playing a game as follows. On Mary's turn, she increases one of $A,B,C$ by 1, and on Nancy's turn, she increases one of $D,E,F$ by 1. They take turns alternatingly until each has gone 2019 times. Each pair of turns is called a "set" (for example if Mary moves first, then 1 set consists of Mary's move followed by Nancy's move).

Winning condition: Mary wins if at any point at the end of a set the following inequality is true for all $x,y$ real numbers: $$Ax^2 + By^2 + C \geq Dxy + Ex + Fy$$ Determine who has the winning strategy if:

1. Mary goes first

2. Nancy goes first

Solution

If Mary goes first Nancy has a winning strategy.

Replace $x$ with $x/z$ and $y$ with $y/z$ to make the inequality homogeneous: $$Ax^2 + By^2 + Cz^2 \geq Dxy + Exz + Fyz$$ Note that the following form is always true if $u,v,w \geq 0$ $$u(x-y)^2 + v(x-z)^2 + w(y-z)^2 \geq 0$$ So if at any point after a set the inequality can be reduced to that form, then Mary wins. We claim the following two things:

1. That Nancy can always avoid that form, and

2. That if the form is not achieved then there exists a $x,y,z$ to make the inequality false

First to prove 1, note the following rearrangement: $$u(x-y)^2 + v(x-z)^2 + w(y-z)^2 \geq 0$$ $$(u+v) x^2 + (u+w)y^2 + (v+w)z^2 \geq 2u xy + 2v xz + 2w yz$$ Therefore if $D = (A+B-C), E = (A+C-B), F = (B+C-A)$ then that form is achieved. This is the unique solution to achieve that form, and that solution may be negative. If the solution to that form contains negative number then no matter what Nancy does, that form is not achieved. But even if it is a triple of non-negative integers, Nancy has 3 choices of moves to make, so she still has at least 2 moves that won't result in that form.

Now to prove 2, we show that we can find $x,y,z$ that violates the inequality.

First, if $A,B,C$ do not form a triangle, then $A>B+C$ (or its permutation). Note that if we assume (WLOG) that $A>B+C$ then $A+B>C,A+C>B$. Furthermore WLOG we may assume that $B \geq C$. $$Ax^2 + By^2 + Cz^2 \geq Dxy + Exz + Fyz$$ $$\iff (A+B-C)(x-y)^2 + (A+C-B)(y-z)^2 + (B+C-A)(x-z)^2 \geq -(2A+2B-2C-2D)xy - (2A+2C-2B-2E)yz - (2B+2C-2A-2F)xz$$ In that case we can choose the following variables. Let $S,R$ be large numbers. Let $x= 1/R, y = R + 1/R, z = SR + 1/R$. Then $$ (A+B-C)R^2 + (A+C-B)(S-1)^2R^2 \geq (A-B-C)S^2R^2 + ...$$ The "..." part in RHS consists of terms $SR^2$ or lower. As we let $S,R$ become large, the dominant terms are $(S-1)^2R^2$ versus $S^2R^2$. For large enough $S$, $(A+B-C)+(A+B-C)(S-1)^2 < (A-B-C)S^2$. At that point we just fix $S$ but let $R$ become larger. Since the coefficient of $R^2$ in the RHS is larger, then the RHS grows faster, invalidating the inequality.

Now if $A,B,C$ form a triangle, $$\iff (A+B-C)(x-y)^2 + (A+C-B)(y-z)^2 + (B+C-A)(x-z)^2 \geq (2A+2B-2C+2D)xy + (2A+2C-2B+2E)yz + (2B+2C-2A+2F)xz$$ Note that the sum of coefficients in RHS is zero, so the terms are not all positive. We shall divide into two cases: one positive two negatives or vice versa

Case 1: if the RHS is of the form $Uxy + Vxz - (U+V) yz$ with $U,V \geq 0$. Then $$RHS = Uy(x-z) + Vz(x-y)$$ Then we set $x = R+1/R, y = R, z= R$. The terms in LHS will be zero or $(1/R^2)$ whereas the terms in RHS will be $(U + V)$. By setting $R$ large enough we can make LHS < RHS.

Case 2: if the RHS is of the form $(U+V)xy - Uyz - Vxz$ with $U,V \geq 0$ Then $$RHS = Uy(x-z) + Vx(y-z)$$ Then we set $x = R+1/R, y = R+1/R, z= R$. The terms in LHS will be zero or $(1/R^2)$ whereas the terms in RHS will be $(U + V)(1+1/R)$. Again, by setting $R$ large enough we can make LHS < RHS.

Challenge Can you generalize this to third power and four variables? In other words, can the game be extended into the following inequality? $$Ax^3 + By^3 + Cz^3 + Dw^3 \geq Exyz + Fxyw + Gxzw + H yzw$$ Answer: yes. We make use of the following property: $f(a,b,c) = a^3 + b^3 + c^3 - 3abc = (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)/2 \geq 0$ If the inequality can be expressed as a positive sum of $f(x,y,z), f(x,y,w), f(x,z,w), f(y,zw)$ then the inequality holds. If not, then there are two cases:

1. One or more of the $f$ form occurs on the RHS, in which case we choose $x,y,z,w$ to maximize the growth of the ones in the RHS. For example, $x=y=1/R, z = R+1/R, w = SR + 1/R$ for large $S,R$.

2. The $f$ forms all occur on the LHS, so there are terms of $xyz$ etc on the RHS whose coefficients all sum to zero. We can pick $x,y,z,w$ whose differences are small but themselves are large, such as $x=R, Y = R+1/R, z = R+2/R, w = R+3/R$. That way, the values of $f$ will be small but values of $xyz$ will be big. By judiciously permuting those values depending on which coefficients are negative, we can make LHS to be < RHS.