This will be my first question :-)

Let $\mathcal{D}_1$ and $\mathcal{D}_2$ two concurrent lines, and $F$ a point in the plane, and $H$ and $G$ its images by the symmetries of axis $\mathcal{D}_1$ and $\mathcal{D}_2$.

1) ( I solved this one, but I'm attaching it if it is somehow related ) : Find the locus of the point $F$ which satisfies $HG=\lambda$ ( where $\lambda > 0$ ).

Easy : If we call $E$ the intersection of the two lines and $\alpha$ the angle between them, we find that it is the circle of center $E$ and radius $\frac{\lambda}{2\sin \alpha}$.

2) Same question but this time we fix $FG+FH$ and not $HG$, in other words, the equation becomes $FG+FH=\lambda$.

Here I don't have any idea. First I conjectured that it might be a circle of center $E$ too, but later I found out that it is impossible...

Any hints please ? I'm not good in geometry, in fact, I'm doing this kind of exercises this summer in order to level up a little :-)

Thank you !

Here's a construction I did with GeoLabo :

enter image description here


The post by Jyrki Lahtonen directly addressed the orthogonal case. It can be generalized in a straightforward way. We do so, in order to address the belief by the OP that in the general case the locus is hyperbolic.

Let $\alpha$ be the angle between the two lines. We assume that $0< \alpha \le \pi/2$. We work in the usual coordinate system. Let the lines meet at the origin. In order to work out the geometry, we can place the lines wherever we wish.

For the sake of familiarity, and more importantly, symmetry, let the equations of the two lines be $y=mx=(\tan\beta)x$ and $y=-mx=-(\tan\beta)x$, where $\beta=\alpha/2$.

As in Jyrki Lahtonen's solution, the problem is the same as the problem of finding the locus of points $F$ such that the sum of the perpendicular distances from $F$ to the two lines is $\lambda/2$.

Recall that if a line has equation $ax+by+c=0$, then the perpendicular distance from $(x_0,y_0)$ to the line is equal to $$\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}.$$

Thus the distance from $(x_0,y_0)$ to $y=mx$ and $y=-mx$ are, respectively, $$\frac{|y_0-mx_0|}{\sqrt{1+m^2}}\qquad\text{and}\qquad \frac{|y_0+mx_0|}{\sqrt{1+m^2}}.$$ Thus the locus of $F$ is given by the equation $$|y-mx| +|y+mx|=\frac{\lambda\sqrt{1+m^2}}{2}.$$

This is a rectangle with sides parallel to the axes. The corner in the first quadrant has coordinates $(x,y)$ where $$x=\frac{\lambda\sqrt{1+m^2}}{4m}\qquad \text{and}\qquad y=\frac{\lambda\sqrt{1+m^2}}{4}$$ and the remaining corners are obtained by symmetry. Things look a lot better if we use the fact $m=\tan\beta$. That yields $$x\sin\beta=y\cos\beta=\lambda/4.$$

The simplicity of the answer shows that there must be a very simple synthetic argument. Now that we know the answer, producing a "geometric" argument should be easy. One advantage of the geometric approach is that the analytical work that went into it will disappear, so the proof will appear magical. Sometime, when I am in a geometric mood?

  • $\begingroup$ Now I understand, I completely ignored the rectangle thing that I just continued squaring until getting a hyperbola-like equation ( which in fact is not, I reverified this morning ). Thank you both user6312 and Jyrki Lahtonen ! $\endgroup$
    – BS.
    Jul 13 '11 at 13:23

Well done with part 1. Here's a hint for part two: $|FG|+|FH|$ is twice the sum of the distances from $F$ to the two lines. So if the two lines happened to be orthogonal, then the answer (in the obvious coordinate system) would be the set of solutions of $$|x|+|y|=\frac{\lambda}2.$$

The solution in the special case above is a square. In the general case we get a rectangle as user6312 showed anaytically. Perhaps a bit more `geometric' way of constructing it goes as follows. There are two points on $D_1$ at a distance $\lambda/2$ from $D_2$ and two points on $D_2$ at a distance $\lambda/2$ from $D_1$. A reflection with respect to the bisector of (any) angle between $D_1$ and $D_2$ is a symmetry of the picture, so these 4 points are the vertices of a rectangle: the sides are parallel to the two angle bisectors. Let $A$ and $B$ be two adjacent vertices of this rectangle. Say that $A$ is at a distance 0 from $D_1$ and at a distance $\lambda/2$ from $D_2$ - for $B$ the two distances are interchanged. If $P$ is any point on the line segment $AB$, then $\vec{EP}=t\vec{EA}+(1-t)\vec{EB}$ for some $t\in[0,1]$. It is then clear that $P$ is at a distance $t(\lambda/2)$ from $D_1$, and at a distance $(1-t)(\lambda/2)$ from $D_2$. Therefore $P$ has the desired property. IOW in the general case I used a basis other than the usual orthonormal $\{\vec{i},\vec{j}\}$ that is good enough for the simpler case.

  • $\begingroup$ Thank you for the hint ! Unless I'm mistaken, if the lines aren't orthogonal we get a conic section ( an hyperbola to be precise ). $\endgroup$
    – BS.
    Jul 12 '11 at 22:05
  • $\begingroup$ @Bouazza: No, we get a rectangle as shown by user6312. I added a spoiler. It uses vectors so is not really purely geometric. That solution really depends on the fact that if you multiply a vector by a scalar, the length of the projection is scaled accordingly. A better solution may still be coming... Somebody with a suitable drawing tool might be able to describe this idea in a way that makes it more obvious. $\endgroup$ Jul 13 '11 at 7:21

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