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Here is the problem:

We start with a triangle ABC with area 1. We choose a point (F) on side AB, then someone else chooses a point (G) on side BC. We then choose the last point (H) on side CA. Our goal is to mximize the area of the triangle FGH. What is the maximum area we can get no matter where the other person puts his point?

I am stuck with this problem. I think that the best place for point F is the midpoint of AB, but I'm not sure how to prove it or what to do next. I appreciate any help you can give me.

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No, you can't prove that coz. it does not hold. See this enter image description here

The diagram shows triangle ABC, mid points of sides D,E,F and some random points G,H,I.

$ar(\Delta GHI)>ar(\Delta DEF)$

You can see this , the area will maximize when the points on the sides will coincide with original triangle.

Well it is the case for minimum area.

Now we see that max. area comes when the points tends to the original triangle.

So if we consider max area from our side and that the others try to minimise it, then the'll probably place points at mid points and then you may solve where your point must lie.

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I thought about the midpoint because the problem is not about the maximal possible area, but about the maximum area you can get when the second point is chosen not by you. It asks what is the area you can always get, even if the person who chooses the second point tries to minimize the area of FGH. – suomynona Apr 25 '13 at 16:46
So, it's not a team effort. let me see again.But still i'll go with limit $F\to A$ @suomynona See the edit – ABC Apr 25 '13 at 16:49

The area of triangle FGH = [FG] x height/2. So once F and G are given, you need to maximize the height. If FG is equilateral to BC, it doesn't matter where you put H. Otherwize, you either put it on exactly B or C, whichever gives you the most height.

Let now $f = [BF]/[BA]$ and g = [BG]/[BC]. With $G_1$ at g=f, you got the equilateral case, with the area: $[FG_1A]=[FG_1C]=[FG_1H] = ([FG_1]h_H)/2= ((f[AC])((1-f)h_B))/2 = [ABC]f(1-f)$ for any point H between A and C, $h_H$ being the distance between the equilaterals $FG_1$ and $AC$, and $h_B$ being the height of B above [AC].

With $G_2$ at g < f, the area of $FG_2C$ is trivially larger than $FG_1C$, just draw it.

With $G_3$ at g > f, the area of $FG_3A$ is larger than $FG_1A$, because the height of $G_3$ over $FA$ is larger then of $G_1$.

So for any f, the best strategy of your opponent is to make g=f. Then the area is $[ABC]f(1-f)$. This is a nice parabola with maximum at f=1/2 with area $[ABC]/4$.


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This is true, but it asks what exactly is the area of FGH that you can always get when the point G is somewhere on BC and you choose the other two... – suomynona Apr 25 '13 at 16:48

The triangle $A=(0,1)$, $B=(0,0)$, $C=(1,0)$ has area ${1\over2}$ only; but we shall compensate for this in a moment.

Let $F=(0,f)$, $G=(g,0)$, $H=(1-h,h)$ be the three chosen points. Then the quantity $Q$ in question is given by $$Q:=2\>\bigl|\triangle(FGH)\bigr|=\vec{GH}\wedge\vec{GF}=(1-g)f+h(g-f)\ .$$

In order to maximize $Q$ after both $f$ and $g$ have been chosen we should choose $h:=0$ if $g<f$, and $h:=1$ if $g>f$, and $h\in[0,1]$ arbitrary if $g=f$. In this way we would realize $$Q=\cases{(1-g) f\quad&$(g\leq f)$\cr g(1-f)&$(g\geq f)$\cr}\ .\tag{1}$$ Now we go one step back. After we have chosen $f$ our adversary will try to minimize the $Q$ in $(1)$ by proper choice of $g$. It is easy to see that he will then choose $g=f$, in which case $Q=f(1-f)$. This implies that our choice of $f$ guarantees a $$Q\geq f(1-f)\ ,$$ whatever the choice of $g$. Therefore we should choose $f:={1\over2}$, which will guarantee $Q\geq{1\over4}$ with proper choice of $h$.

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