Each map is bijective, the displayed formulas establish identity, closure, and inverses, and function composition is associative.

No. The identity translates by zero, which is excluded, and an inverse would translate by a negative distance, also excluded. Closure under composition alone does not suffice.

Translation cancels in a difference; a unit multiplier does not change its modulus.

Identity gives \(A\sim A\). If \(g(A)=B\), then \(g^{-1}(B)=A\), so symmetry follows from inverses. If \(g(A)=B\) and \(h(B)=C\), closure supplies \(hg\in G\) and \(hg(A)=C\). Thus transitivity holds.

The origin is a one-point orbit. A nonzero point stays on the circle with its modulus, and every point on that circle is reached by choosing the difference of arguments as the rotation angle.

Both have corresponding side lengths \(2,1,\sqrt5\), but signed areas \(1\) and \(-1\). Direct motions preserve signed area, so the requested map does not exist. Reflection in the horizontal axis does match the labels in the full isometry group. Reordering the labels would be a different problem.

The displacement is a positive real multiple of the opposite side, so it has the same direction. Taking moduli gives half its length. The triangle is nondegenerate, so the compared displacements are nonzero.

Let \(m=(q+r)/2\). Then

This is the point two-thirds of the way from the vertex to the opposite midpoint. Repeating with the other vertices gives the same point, establishing concurrence and the ratio.

These are rotations about the center of the disk of radius three centered at five.

Identity also transports to identity, and all the maps are bijections. This checks the remaining requirements.

No. For vertices \(0,1,i\), the sum is \(2\). Translate by \(10\); one vertex alone is then ten units from the origin. The new sum is greater than two, providing a counterexample.

Approach. Think of a relation defined by a class of invertible transformations.

Similarity of figures is an example. The identity similarity proves reflexivity; the inverse of a similarity proves symmetry; and composing similarities proves transitivity. Another example is equal length of segments. “Strictly parallel distinct lines” would not work without allowing a line to be parallel to itself, since reflexivity would fail.

Approach. Match the three equivalence properties with identity, inverse, and composition.

If \(G\) acts on figures, the identity gives \(I(A)=A\), proving reflexivity. If \(T(A)=B\) with \(T\in G\), then \(T^{-1}\in G\) and \(T^{-1}(B)=A\), proving symmetry. If also \(U(B)=C\), closure gives \(UT\in G\) and \(UT(A)=C\), proving transitivity.

Approach. Compute the inverse and the composition explicitly.

The identity uses \(\theta=0,b=0\). Every map is bijective with inverse \(w\mapsto e^{-i\theta}w-e^{-i\theta}b\), again in \(E\). If \(T(z)=e^{i\theta}z+b\) and \(U(z)=e^{i\phi}z+c\), then \(T(U(z))=e^{i(\theta+\phi)}z+(e^{i\theta}c+b)\), which is in \(E\). Function composition is associative. Thus all group conditions hold.

Approach. Choose a common center for all rotations.

Use rotations about the origin: \(R=\{z\mapsto e^{i\theta}z:\theta\in\mathbb R\}\). The identity has angle zero, the inverse has angle \(-\theta\), and compositions add angles. Every such rotation lies in \(E\) by taking \(b=0\). Thus \((\mathbb C,R)\) is a geometry.

Approach. Write down the identity, inverse, and composition.

For \(R_\theta(z)=e^{i\theta}z\), \(R_0=I\), \(R_\theta^{-1}=R_{-\theta}\), and \(R_\theta R_\phi=R_{\theta+\phi}\). Each is a bijection of the nonempty set \(\mathbb C\); hence the defining group axioms hold.

Approach. A distinguished center is allowed in rotational geometry.

Take the family consisting only of the unit circle \(C=\{|z|=1\}\). Each origin-centered rotation sends \(C\) to itself because \(|e^{i\theta}z|=|z|\). Translation by \(2\) sends it to \(|z-2|=1\), a different circle outside the one-element family. Therefore the family is rotationally invariant but not Euclidean invariant.

Approach. Use a point’s distance from the fixed center.

On single-point figures, use \(f(\{z\})=|z|\). Rotations preserve it because \(|e^{i\theta}z|=|z|\). Translation by \(1\) sends the origin, where \(f=0\), to \(1\), where \(f=1\). This one counterexample disproves invariance under all rigid motions.

Approach. Subtract the two transformed points before taking the modulus.

For \(T(z)=e^{i\theta}z+b\),

This uses \(|e^{i\theta}|=1\). The same translation cancels from both arguments; transforming only one argument would not give invariance.

Approach. Distinguish a vector’s direction from its length.

A translation \(T(z)=z+b\) gives \(v(Tz_1,Tz_2)=z_1+b-z_2-b=v(z_1,z_2)\). Under rotation \(R_\theta\), the vector becomes \(e^{i\theta}v(z_1,z_2)\). For example, \(v(1,0)=1\) but \(v(i,0)=i\) after a quarter-turn, so invariance under all rotations fails.

Approach. Translations cancel from both differences; a line direction is defined modulo π.

The nonzero difference \(v=z_1-z_2\) has nonzero conjugate, so the quotient is defined. Translating both points by \(b\) leaves the numerator \(v\) unchanged and also leaves its conjugate unchanged. Thus the ratio remains \(v/\bar v\). In polar form, if \(v=re^{i\theta}\), then \(v/\bar v=e^{2i\theta}\); reversing direction adds \(\pi\) to the angle and leaves the direction number unchanged.

Approach. Write the quotient in terms of the ratio of the two nonzero vectors.

Let \(v=z_1-z_2,w=z_3-z_4\). Under \(T(z)=e^{i\theta}z+b\), these differences become \(e^{i\theta}v,e^{i\theta}w\). Each direction number is therefore multiplied by \(e^{2i\theta}\), which cancels in their quotient. Further,

This equals \(1\) exactly when \(v/w=\overline{v/w}\), that is, when \(v/w\) is a nonzero real number. This is precisely linear dependence over the reals, or equal unoriented line directions. It includes collinear segments as having parallel directions.

Approach. Compute the two diagonal midpoints after placing one vertex at the origin.

Rigid motions preserve midpoints, since \(T((z+w)/2)=(Tz+Tw)/2\), and preserve rectangles. Place one vertex at zero and one side on the real axis. The four vertices are \(0,a,a+ib,ib\) with real nonzero \(a,b\). The diagonal from \(0\) to \(a+ib\) has midpoint \((a+ib)/2\); the diagonal from \(a\) to \(ib\) has exactly the same midpoint. Both diagonal segments therefore pass through that point and are bisected there. Apply the inverse rigid motion to return to the original rectangle.

Approach. Test the average of the three vertex coordinates.

For noncollinear vertices \(p,q,r\), let \(g=(p+q+r)/3\). The opposite midpoint to \(p\) is \(m_p=(q+r)/2\), and

Thus \(g\) lies on that median and is two-thirds of its length from \(p\). The identical calculation with \(q\) and \(r\) proves it lies on all three medians. Two different medians cannot be the same line: otherwise two vertices and their opposite midpoints force all three vertices to be collinear. Hence their common point is unique. The ratio statement follows because the parameter \(2/3\) is real and between zero and one.

Approach. Compute the height squared; factor the resulting difference of squares.

Use a rigid motion to place a side of length \(c>0\) on the real axis from \(0\) to \(c\). Let the third vertex be \(x+iy\), with distances \(b\) from zero and \(a\) from \(c\). Then \(x^2+y^2=b^2\) and \((x-c)^2+y^2=a^2\). Subtraction gives \(2cx=b^2+c^2-a^2\). Since \(A=c|y|/2\),

These four factors are \(2(s-b),2(s-c),2(s-a),2s\); hence \(A^2=s(s-a)(s-b)(s-c)\). Triangle inequalities make each factor positive, and area is nonnegative, so taking the nonnegative square root proves the formula.

Approach. Consider the orbit of one figure.

Choose a figure \(A\) in the family \(E\). Its orbit \(\mathcal O(A)=\{T(A):T\in G\}\) is nonempty, lies in \(E\) by invariance, and is itself invariant: applying \(U\in G\) gives \(UT(A)\) and composition remains in the group. Minimality forces \(\mathcal O(A)=E\). Every \(B\in E\) therefore equals \(T(A)\) for some \(T\in G\), which is exactly congruence. Conversely, an orbit has no proper nonempty invariant subfamily, because any of its members can be moved to any other by a group transformation.

Approach. Describe exactly what a permitted motion can change about a line.

• Euclidean. All straight lines: rotate the original direction to any target direction, then translate to the target line.
• Translational. All lines \(2x+y=c\) for real \(c\). Translation preserves direction; every offset is possible.
• Rotational. All lines whose Euclidean distance from the origin is \(1/\sqrt5\). Rotation preserves this distance and can rotate the unit normal to any chosen direction. Writing lines as \(n\cdot(x,y)=1/\sqrt5\) with unit normal \(n\) describes the entire orbit.

Each listed family is an orbit and therefore minimal by Exercise 8.

Approach. Represent a circle by its center and radius.

• Euclidean. For each \(r>0\), all circles of radius \(r\). Motions preserve radius and translations move the center arbitrarily.
• Translational. Again, for each fixed positive radius, all centers are possible. A circle has no additional orientation to preserve.
• Rotational. Fix both the positive radius \(r\) and the distance \(\rho\ge0\) of its center from zero. The centers run along \(|c|=\rho\); if \(\rho=0\), the orbit consists of just the one origin-centered circle.

These conditions are necessary because the indicated quantities are preserved, and sufficient because the permitted motions move any admissible center to any other. Hence each family is exactly one orbit.

Approach. For rotations, track the edge direction relative to the direction from zero to the center.

Represent a square by center \(c\), side length \(l>0\), and edge direction \(\theta\) modulo \(\pi/2\).

• Euclidean. Fix only \(l\). A rotation adjusts edge direction and a translation adjusts center, giving every square of that size.
• Translational. Fix \(l\) and \(\theta\pmod{\pi/2}\); allow any center. Translation cannot turn the square.
• Rotational. If \(c=0\), fix only \(l\); all orientations form one orbit. If \(c\ne0\), fix \(l\), \(|c|\), and \(\theta-\arg c\pmod{\pi/2}\). A rotation adds the same angle to the center argument and edge direction. Conversely, for two squares with these same quantities, rotate the first center to the second; the fixed relative direction ensures the edges also agree modulo the square’s quarter-turn symmetry.

Necessity and sufficiency show that these are precisely the orbits, hence the minimal invariant families.

For translations, equal length is necessary but insufficient: displacements 1 and i have equal modulus but different directions. Equality of the displacement vectors is necessary and sufficient, since translating the first initial endpoint to the second then also matches the final endpoint. For direct Euclidean motions, equal length is sufficient: the quotient of the target displacement by the source displacement has modulus one, so rotate by that unit multiplier and then translate the initial endpoint into place.

An isometry preserves a circle's diameter, the maximum distance between its points, which equals twice its radius. Thus radii must agree. With equal radii, translation by the difference of centers matches the circles. For centers a,b and radii R,S>0, the similarity \(T(z)=b+(S/R)(z-a)\) maps \(|z-a|=R\) onto \(|T(z)-b|=S\). Its inverse exists because S/R is positive. The larger group therefore removes radius as an orbit distinction.