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?

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


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

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).


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


(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


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) :


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



The second relationship can be written under the form :




(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:


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

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

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.

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

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.