3
$\begingroup$

I am having issues to prove the back of this theorem:

Let $ABC$ be a triangle and fixed $D∈AB$. The Geometric Place of the $X$-points that form with $D$ and an arbitrary point $S∈AC$ an equilateral triangle $DSX$ is a straight segment.

Can someone save me?

$\endgroup$
1
$\begingroup$

enter image description here

Note that $a$ and $\beta$ are constants, only $s$ and $\delta$ are changed. Also note that according to law of sines:

$$a\sin\beta=s\sin\delta\tag{1}$$

Let us calculate $x$ and $y$ coordinates of point $X$:

$$x=BS+s\cos(180^\circ-(\delta+60^\circ))=a\cos\beta+s\cos\delta+s\cos(120^\circ-\delta)$$

$$x=a\cos\beta+s\cos\delta+s\cos120^\circ\cos\delta+s\sin120^\circ\sin\delta$$

$$x=(a\cos\beta+s\frac{\sqrt3}{2}\sin\delta)+\frac12 s\cos\delta$$

Now use (1) and you get:

$$x=(a\cos\beta+a\frac{\sqrt3}{2}\sin\beta)+\frac12 s\cos\delta\tag{2}$$

On the other side:

$$y=s\sin(180^\circ-(\delta+60^\circ))=s\sin(\delta+60^\circ)$$

$$y=s\sin60^\circ\cos\delta+s\cos60^\circ\sin\delta$$

Now use (1) and you get:

$$y=\frac12 a\sin\beta+\frac{\sqrt3}{2}s\cos\delta\tag{3}$$

Introduce variable $u=s\cos\delta$ and take a closer look at (2) and (3). As already mentioned, $a$ and $\beta$ are constants so we can write (2) and (3) in the following form:

$$x=c_1+c_2u\tag{4}$$

$$x=c_3+c_4u\tag{5}$$

...with $c_1,c_2,c_3,c_4$ being constant values and $u$ being a parameter. Because of that equations (4) and (5) are actully parametric equations of a straight line (you can eliminate $u$ and write $y$ as a linear function of $x$ directly if you want so, but that's really not necessary, except if you want to investigate the locus of point X further).

$\endgroup$
1
$\begingroup$

Most of the setup is irrelevant to the movement of $X$. $\Delta DSX$ being isosceles will suffice:

All we need is that $S,X$ are transformed by the same angle and the same scale of length relative to $D$. Then $\Delta DSS'$ and $\Delta DXX'$ are congruent. Here is a diagram. Details are left to you, but feel free to ask.

$\qquad\qquad\qquad$ enter image description here

$\endgroup$
0
$\begingroup$

It is well known that isometrie (i.e. rotation, reflection, translation, glide translation) takes line to a line, segment to a segment, circle to a circle.

Since $S$ goes to $X$ after a rotation around $D$ for $60^{\circ}$, we can aplay this fact a mentioned and we are done.

Notice that $D$ does not need to be on a segment $AB$ and this is still true.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.