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Suppose we have a triangle with vertices $A,B,C$ lying in the plane with angles $\alpha , \beta , \gamma$ respectively. I am trying to show that a successive rotation about each vertex by $\theta, \phi, \psi$ is the identity map if and only if $\theta = 2 \alpha, \phi = 2 \beta, \psi = 2 \gamma$.

If we write this out in terms of a map on arbitrary point $x \in \mathbb{R}^2$, I see that we must have $\theta + \phi + \psi = 2k \pi$ and $R_{\theta}(a) + R_{\theta + \phi}(b) + c = 0$ as an iff condition, where $R_{q}$ is rotation (clockwise) by $q$. However I do not see how to deduce the result from here. I can relate the angles and sides together by rotating edges of the triangle however using this directly seems to create a mess. Is there a neat solution?

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    $\begingroup$ Hint: If the successive rotations are the identity, then rotating around $A$ and then $B$ must be the identity on $C$. $\endgroup$ – Michael Burr Feb 8 '18 at 22:00
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Observe that if successive rotations are the identity, then, rotating around $A$ and then around $B$ must fix $C$. This follows from the fact that rotation around $C$ fixes $C$. Moreover, the distance between the image of $C$ and $B$ must remain the same as the distance between $B$ and $C$.

More precisely, Let $d(A,C)$ be the distance between $A$ and $C$. Then the distance between $R_{\theta}(C)$ and $A$ must be the same as the distance between $A$ and $C$ since this is a rotation about $A$. Therefore, $R_{\theta}(C)$ lies on the circle of radius $d(A,C)$ about $A$. By a similar argument, $R_{\theta}(C)$ lies on the circle of radius $d(B,C)$ about $C$. Since these two circles intersect in at most two points, we can conclude that $\theta=2\alpha$ and $\phi=2\beta$ since the line between $A$ and $B$ must be a line of symmetry.

By considering the same argument with $A$ playing the role of $C$ finishes the argument.

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  • $\begingroup$ Thanks. But why must the line joining $A,B$ be a line of symmetry ? $\endgroup$ – Evgeny T Feb 9 '18 at 14:48
  • $\begingroup$ Because you have two circles centered at $A$ and $B$ intersecting. Draw two circles that intersect and then draw the line through their centers. $\endgroup$ – Michael Burr Feb 9 '18 at 15:27
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Unlike my other text, this is only a partial answer the "if" part only, using geometrical properties that, I think, shed some light on the issue.

Let

$$\tag{1}R:=R_{B,2\beta} \circ R_{C,2\gamma} \circ R_{A,2\alpha}$$

We can establish that $R=Id$ by 2 methods. The first one (A) is straightforward ; the second one (B) is less direct, but helps to understand how points can be found is the same place after undergoing these three rotations.

A) Denoting by $S_{XY}$ the symmetry with respect to line $XY$, we have (taking care to place symmetries in the right order):

$$\tag{1}\begin{cases}R_{A,2\alpha}&=&S_{AC} \circ S_{AB}\\R_{B,2\beta}&=&S_{AB} \circ S_{BC}\\R_{C,2\gamma}&=&S_{BC} \circ S_{AC}\end{cases}$$

(See Appendix).

Plugging expressions (2) in (1) gives immediately $R=Id$.

B) Another proof stems from the following fact. Consider the orthocenter $H$ (intersection of altitudes) of triangle $ABC$ (see Fig. 1). A property of $H$ is that its reflected points $H_A, H_B, H_C$ with respect to sides $BC,AC,AB$ (in this resp. order) belong to the circumcircle (https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml). $$H_C \xrightarrow{R_{A,2\alpha}}H_B \xrightarrow{R_{C,2\gamma}}H_A \xrightarrow{R_{B,2\beta}}H_C$$ enter image description here

Fig. 1.

Thus $R$ has $H_C$ as a fixed point ; as the composition of rotations can be either a rotation or a translation, we can thus rule out the case of a translation: $R$ is a rotation. Knowing that the composition of rotations with angles $u$, $v$, $w$, when it is not a translation, is a rotation with angle $u+v+w$ whatever the centers of rotation. Here, as $2\alpha+2\beta+2\gamma=2\pi$, $R$ is a rotation with zero angle, thus the identity transform.

Remark 1 : This "angle doubling" that is so often met can be given a natural understanding by reference to the inscribed angle property (https://en.wikipedia.org/wiki/Inscribed_angle). Thus, the central angles $2 \alpha, 2 \beta, 2 \gamma$ are a partition of the central $2\pi$ angle.

Appendix : The composition of two symmetries with respect to two non parallel lines $\ell_1$ and $\ell_2$ making an angle $\alpha$ amounts to a rotation with respect to the intersection of the two lines with angle $2 \alpha $ (https://math.stackexchange.com/q/2235338).

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Here is a complete answer : Instead of adding a big "Edit" to the previous partial answer I gave, I provide a separate text.

This problem can be treated WLOG inside the unit circle, using the complex number interpretation of points in $\mathbb{R}^2$ with

$$\begin{cases}A=e^{ia}\\B=e^{i0}=1\\C=e^{ic}\end{cases}.$$

(Take for example $a=\tfrac{4 \pi}{3}, c=\tfrac{\pi}{3}$, so that triangle $ABC$ has a direct orientation with $\alpha=\tfrac{\pi}{4}$, etc. due to central angle property.

I give to the "unknown" rotation angles the names $u,v,w$ (instead of $\theta,\phi,\psi$ only because it's easier to type the former ones...) and

$$\begin{cases}U:=e^{iu}\\V:=e^{iv}\\W:=e^{iw}\end{cases}$$

Another remark : I work with anticlockwise ( = trigonometric) orientation, which is the most common convention.

Let us recall the complex representation of a rotation $r$ with center $c$ and angle $a$ (Composition of Rotation and Translation in the Complex Plane -- Finding Angle of Rotation and Point) :

$$r(z)=e^{ia}(z-c)+c$$

Let us now compose the three unknown rotations and express that we obtain the identity transform:

$$W(V(U(Z-A)+A-B)+B)-C)+C)=$$

$$=\underbrace{(UVW)}_{=1}Z+\underbrace{[C+(B-C)W+(A-B)VW-AUVW]}_{=0}$$

The second relationship can be written under the form :

$$(C-A)+(B-C)W+(A-B)VW=0$$

i.e.,

$$\tag{1}(C-A)+(B-C)e^{iw}+(A-B)e^{-iu}=0$$

(explanation for the presence of $e^{-iu}$: $VW=e^{i(v+w)}=\underbrace{e^{i(u+v+w)}}_{=1}e^{-iu}$).

If we substract, to (1), the identity $(C-A)+(B-C)+(A-B)=0$, we get:

$$(B-C)(1-e^{iw})+(A-B)(1-e^{-iu})=0$$

$$\tag{2}\iff \ \dfrac{C-B}{A-B}=\dfrac{1-e^{-iu}}{1-e^{iw}}$$

$$\tag{3}\iff \ \dfrac{1-e^{ic}}{1-e^{ia}}=\dfrac{1-e^{-iu}}{1-e^{iw}}$$

First of all, we notice that (3) is verified for $u=-c=2\alpha$ and $w=a=2\gamma$ (we use the inscribed angle property).

But (3) expresses much more than that: it expresses that triangles with vertices

$$(e^{ic},1,e^{ia}) \ \ \text{and} \ \ (e^{-iu},1,e^{iw})$$

are similar (https://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbersGeometry.shtml).

Besides, they have the same circumscribed circle, i.e., the unit circle. In such a case, similar triangles (with same orientation) are images one of the other through a rotation. As the triangles have a common point $1$, these triangles must be identical.

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  • $\begingroup$ I have almost completely modified my answer. $\endgroup$ – Jean Marie Feb 12 '18 at 15:05

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