Maximize area of a triangle

Question

Given a triangle $\Delta ABC$ in the plane and points $X,Y$ on line segments $BC$ and $CA$, respectively, so that the lines $AB$ and $XY$ are parallel, find the locations of $X$ and $Y$ so that the area of a triangle $\Delta XYZ$ with $Z$ on line segment $AB$ becomes maximal.

(Remark: Obviously, the area of $\Delta XYZ$ is independent of the location of $Z$ as the perpendicular height stays the same.)

After experimenting a bit with GeoGebra, I'm assuming that $X$ and $Y$ have to be the centers of the segments $BC$ and $CA$. However, I am having some trouble proving this.

Any hints are appreciated!

Answer 1

$ABC$ is given, so we can give names to relevant quantities of that triangle. I will call $a=AC,b=BC,c=AB,\hat B=A\hat BC$, where $A\hat BC$ is the angle in the vertex $B$. Let $x=BX$. You can easily verify that $AY=\frac abx$. We need to work out $XY$ and the height. Let us call $t=XY$ and $h$ the height. My drawing skills are plain c**p, but hopefully this will help visualize:

It is easy to see that $BC:CH=x:h$. But $CH=BC\sin\hat B$. So $h=x\sin\hat B$. We can work out that sine, but it is a constant so no point in doing it. I did work it out, but I'm too lazy to copy it here. It involves the cosine law ($a^2=b^2+c^2-2bc\cos\hat B$) and $\sin^t=\sqrt{1-\cos^2t}$ when the sine is positive as in this case. It is easy to see that $t:(b-x)=c:b$, or $t=(b-x)\frac cb$. So:

$$A_{XYZ}=\frac{th}{2}=\frac12(b-x)\frac cb\cdot x\sin\hat B\frac{c}{2b}\sin\hat B(bx-x^2).$$

If you differentiate that, you get:

$$\frac{d}{dx}A_{XYZ}=\frac{c}{2b}\sin\hat B(b-2x),$$

hence the maximum is in $x=\frac b2$, so it turns out you were correct :).

Answer 2

We may assume $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, $Y=(0,y)$, $X=(1-y,y)$. It follows that $$\bigl|\triangle(X,Y,Z)\bigr|={1\over2}(1-y)y\ ,$$ and this is maximal when $y={1\over2}$. It follows that $Y$ has to be the midpoint of $AC$, and $X$ has to be the midpoint of $BC$, in order to make $\bigl|\triangle(X,Y,Z)\bigr|$ maximal. When $X$, $Y$ are chosen in this way then $$\bigl|\triangle(X,Y,Z)\bigr|={1\over4}\bigl|\triangle(A,B,C)\bigr|\ .$$