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I have got one simple task to prove:

We have got a triangle $\triangle XYZ$.
Then we create points $A,B,C$ on $XY, YZ, ZX$ respectively, such that $XA = AB = BZ$ and $CZ = AY = AC$.
How to prove that $XY = \frac{XZ + YZ}{2}$?

Illustration of the situation

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Does the context of this homework give any indication about the mathematical tools you're expected to use? – MvG Oct 8 '12 at 12:55
I tried doing some distance geometry, writing down Cayley-Menger determinants for all combinations of 4 points as well as all sets of 3 collinear points. Using brute computer algebra (sage/Maxima), I found no solution yet. I would really appreciate an answer to my previous comment. – MvG Oct 9 '12 at 5:49
up vote 3 down vote accepted

I refer to the following figure:

enter image description here

The points $B$ and $C$ have to lie on the perpendicular bisector of $ZA$. The angles $\angle(AZX)$ and $\angle(AZY)$ will be fixed later; call them $\lambda-\delta$ and $\lambda+\delta$. Then the angles at $X$ and $Y$ will be $\lambda'-\epsilon$ and $\lambda'+\epsilon$ with $\lambda':={\pi\over2}-\lambda$ and some $\epsilon$. (The idea is that $|\delta|$, $|\epsilon|\ll 1$.)

The conditions $ZC=CA$ and $ZB=BA$ are already taken care of. The conditions $ZB=AX=:x$ and $ZC=AY=:y$ lead by means of the sine theorem to $$x={1\over\cos(\lambda+\delta)}={2\sin(\lambda-\delta)\over\sin(\lambda'-\epsilon)}\ ,\qquad y={1\over\cos(\lambda-\delta)}={2\sin(\lambda+\delta)\over\sin(\lambda'+\epsilon)}$$ or $$\sin(\lambda'-\epsilon)=2\sin(\lambda-\delta)\cos(\lambda+\delta)\ ,\qquad \sin(\lambda'+\epsilon)=2\sin(\lambda+\delta)\cos(\lambda-\delta)\ .\qquad(1)$$ Adding the two equations $(1)$ gives $2\sin\lambda'\cos\epsilon=2\sin(2\lambda)$ or $$\sin\lambda ={1\over2}\cos\epsilon\ .\qquad(2)$$ (Subtracting the two equations $(1)$ gives $2\cos\lambda'\sin\epsilon=2\sin2\delta$, so that together with $(2)$ we obtain $$\sin(2\delta)={1\over4}\sin(2\epsilon)\ .$$ This shows that there is in fact a one-parameter family of such configurations. The value $\epsilon=0$ corresponds to an equilateral triangle $XYZ$.)

We now compute $$\eqalign{XY=x+y&={2\sin(\lambda-\delta)\over\cos(\lambda+\epsilon)}+{2\sin(\lambda+\delta)\over\cos(\lambda-\epsilon)}\cr &={2\sin(\lambda-\delta)\cos(\lambda-\epsilon)+2\sin(\lambda+\delta)\cos(\lambda+\epsilon)\over \cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\cr &={\sin(2\lambda-\delta-\epsilon)+\sin(2\lambda+\delta+\epsilon)\over \cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\cr &={2\cos(\delta+\epsilon)\over \cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\sin(2\lambda)\ .\cr}$$ On the other hand, again by the sine theorem we have $$ZY=2{\cos(\delta+\epsilon)\over\cos(\lambda-\epsilon)}\ ,\qquad ZX=2{\cos(\delta+\epsilon)\over\cos(\lambda+\epsilon)}\ ,$$ so that $${ZX+ZY\over2}=\cos(\delta+\epsilon){\cos(\lambda+\epsilon)+\cos(\lambda-\epsilon)\over\cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}={2\cos(\delta+\epsilon)\over\cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\cos\lambda\cos\epsilon\ .$$ From $(2)$ follows $\ \cos\lambda\cos\epsilon=\sin(2\lambda)$, so that we indeed obtain $${ZX+ZY\over2}=XY\ .$$

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Choosing coordinates

Based on the variables $a$, $b$ and $c$ we define coordinates for your points like this, without loss of generality:

\begin{align*} X &= \begin{pmatrix}-1\\0\end{pmatrix} & Y &= \begin{pmatrix}1\\0\end{pmatrix} & Z &= \begin{pmatrix}b\\c\end{pmatrix} \\ A &= \begin{pmatrix}a\\0\end{pmatrix} & B &= Z + (1+a)\frac{Y-Z}{\lVert Y-Z\rVert} & C &= Z + (1-a)\frac{X-Z}{\lVert X-Z\rVert} \end{align*}

This choice of coordinates already ensures that $AX = BZ = (1+a)$ and $AY = CZ = (1-a)$.

Some computer algebra

Now you have to ensure that $AB$ and $AC$ play along as well, which leads to two equations that my computer algebra system initially writes as

\begin{alignat*}{2} \lVert A-B\rVert &\,=\,& \sqrt{{\left| \frac{{\left(a + 1\right)} c}{\sqrt{c^{2} + {\left| b - 1 \right|}^{2}}} - c \right|}^{2} + {\left| \frac{{\left(b - 1\right)} {\left(a + 1\right)}}{\sqrt{c^{2} + {\left| b - 1 \right|}^{2}}} + a - b \right|}^{2}} &= a + 1 \\ \lVert A-C\rVert &\,=\,& \sqrt{{\left| -\frac{{\left(a - 1\right)} c}{\sqrt{c^{2} + {\left| b + 1 \right|}^{2}}} - c \right|}^{2} + {\left| -\frac{{\left(b + 1\right)} {\left(a - 1\right)}}{\sqrt{c^{2} + {\left| b + 1 \right|}^{2}}} + a - b \right|}^{2}} &= -a + 1 \end{alignat*}

Now you square the equations, multiply by the common denominator, put the remaining root on one side and the rest on the other, square again, and eventually solve for $a$ and $c$ depending on $b$. There will be multiple solutions, most of them complex, and one degenerate with $c=0$ which I'll discuss below. The remaining solutions only differ in signs, so we can write them as one using $\pm$.

The expression for $a$ is quite complex, but the one for $b$ is simply

$$c = \pm\frac12\sqrt{12-3b^2}$$

which you can write as

$$3b^2 + 4c^2 - 12 = 0\;.\tag{1}$$

It's an ellipse

Our aim is to prove the equation from your question,


As Arthur pointed out in his comment, this condition can be interpreted as a condition that $Z$ lies on a certain ellipse with $X$ and $Y$ as its focal points. There are many such ellipses, so you still need one more point on the ellipse to specify which one you actually want. In our coordinate setup, $(2,0)^T$ is an example of a point which satisfies your equation and therefore has to lie on the ellipse.

Equation $(1)$ does indeed describe an ellipse. I don't feel like computing its focal points from the equation just now. Instead, observe that from the form of this equation (i.e. only pure squared monomials and constant term, no bare $x$, $y$ or mixed $xy$), the ellipse has its center of symmetry at the origin, and its axes coincide with your coordinate axes. By simply inserting coordinates into the equation, you can check that the following points lie on this ellipse:

$$ \begin{pmatrix}\pm2\\0\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}0\\\pm\sqrt3\end{pmatrix} $$

Both of these pairs of symmetric points satisfy $(2)$: the first can be easily checked as all points are on a line, whereas the second corresponds to the equilateral case. For an ellipse whose axes coincide with the coordinate axes, the above points already define both radii, and therefore uniquey define the ellipse.

To conclude, we know that all points satisfying $(1)$ lie on an ellipse, we know that $(2)$ describes an ellipse, and we know that these two ellipses are the same. As all non-degenerate configurations satisfying the initial length equalities will lead to $(1)$, we now have proven $(2)$ to hold.

Degenerate situation

The equations stated above also have one solution

$$a=b\quad c=0\;.$$

This means that you can also satisfy your length equalities by choosing $A=Z$, so that your whole configuration collapses to a single line. In that case, $(2)$ does not necessarily hold, as the following example demonstrates:

\begin{align*} X &= \begin{pmatrix}-2\\0\end{pmatrix} & C &= \begin{pmatrix}-1\\0\end{pmatrix} & A=Z &= \begin{pmatrix}0\\0\end{pmatrix} & Y &= \begin{pmatrix}1\\0\end{pmatrix} & B &= \begin{pmatrix}2\\0\end{pmatrix} \end{align*}

This leads to

\begin{align*} XA=AB=BZ &= 2 \\ CZ=AY=AC &= 1 \end{align*}


$$ XY = 3 \neq \frac32 = \frac{2+1}2 = \frac{XZ + YZ}{2}\;. $$

So you should require that your triangle is non-degenerate.

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enter image description here

From the above figure, it can be prove that $\triangle ABC$ and $\triangle BCZ$ are congruent. Or $\triangle ABC = \triangle BCZ$

I am stuck at this point.

Note: The above relation as asked in the question, $XY=\frac{XZ+YZ}{2}$ is true for equilateral triangles and can be proved as well.

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About the note: the relation as asked in the question is true if and only if $Z$ lies on the unique ellipse with $X$ and $Y$ as focal points, allowing for the equilateral triangle. This by the definition of ellipse and the rewriting of the relation as $XZ + YZ = 2XY$. – Arthur Oct 8 '12 at 10:52
And what will happen if this $\triangle XYZ$ isn't equilateral? – Edger Bachmann Oct 8 '12 at 19:20
Maybe a help: diagram now shows that line BC is perpendicular bisector of segment AZ. – coffeemath Oct 9 '12 at 0:02

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