Transformations & Symmetry

Solution. pick two arbitrary points $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. Their images are $P^{\prime }=(x_1+3,y_1+4)$ and $Q^{\prime }=(x_2+3,y_2+4)$. Step 2 — the distance between the images is

Solution. recognize it. This is rotation by $90^{\circ }$ counterclockwise about the origin: the standard rotation formula $(x,y)\mapsto (x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta )$ with $\theta =90^{\circ }$ gives $\cos 90^{\circ }=0$, $\sin 90^{\circ }=1$, so $(x,y)\mapsto (-y,x)$. Step 2 — check it is orientation-preserving. The matrix is $\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$, whose determinant is $0\cdot 0-(-1)\cdot 1=1>0$. A positive determinant means handedness is kept, so this is a proper (orientation-preserving) isometry — a rotation, as claimed, not a reflection. Sanity check: test the point $(1,0)$. It maps to $(0,1)$, which is indeed $(1,0)$ turned a quarter-turn counterclockwise. And both are distance $1$ from the origin. $✓$□

Solution. align the first side. Since $AB=DE$, there is an isometry $g$ (a translation followed by a rotation) sending $A\mapsto D$ and $B\mapsto E$. Isometries acting on the plane are exactly determined once we say where two points go, plus a choice of side, so such a $g$ exists. Step 2 — locate the third vertex. Under $g$, vertex $C$ lands at a point $C^{\prime }$ with $D{C}^{\prime }=CA=FD$ and $E{C}^{\prime }=BC=EF$. So $C^{\prime }$ is a point at distance $FD$ from $D$ and distance $EF$ from $E$. Step 3 — resolve the two candidates. Two circles (one centered at $D$ of radius $FD$, one centered at $E$ of radius $EF$) meet in at most two points, symmetric across line $DE$. Point $F$ is one of them. If $C^{\prime }=F$ we are done; if $C^{\prime }$ is the mirror point, compose with the reflection over line $DE$, which fixes $D$ and $E$ and swaps the two candidates, sending $C^{\prime }$ to $F$. Conclusion: an isometry (direct, or direct-then-reflected) maps $△ABC$ exactly onto $△DEF$. That is precisely what "congruent" means. This is why SSS is a valid congruence test — it is a guarantee that the required rigid motion exists, not a computation of area.□

Solution. move each endpoint: $A^{\prime }=(0+3,0-1)=(3,-1)$ and $B^{\prime }=(1+3,4-1)=(4,3)$. Step 2 — original length: $AB=\sqrt{(1-0{)}^2+(4-0{)}^2}=\sqrt{1+16}=\sqrt{17}$. Step 3 — image length: $A^{\prime }{B}^{\prime }=\sqrt{(4-3{)}^2+(3-(-1){)}^2}=\sqrt{1+16}=\sqrt{17}$. Sanity check: $AB=A^{\prime }{B}^{\prime }=\sqrt{17}$, so the segment moved rigidly. There are no fixed points here since $v\ne 0$. $✓$□

Solution. apply: $(2+(-1),5+3)=(1,8)$. Step 2 — undo by translating by the inverse vector $⟨1,-3⟩$: $(1+1,8-3)=(2,5)$. Sanity check: we returned to the original point $(2,5)$, confirming $T_{-v}$ is the inverse of $T_v$. $✓$□

Solution. choose the translation vector as $v=D-A=⟨5-0,2-0⟩=⟨5,2⟩$. Step 2 — apply $T_v$ to every vertex. Then $A\mapsto (5,2)=D$, and $B,C$ move by the same $⟨5,2⟩$, so the whole triangle keeps its exact shape and size (translation is an isometry). Step 3 — after this slide, side $AB$ points in the same direction as before but now starts at $D$. A rotation about $D$ (and, if needed, a reflection across line $DE$) then swings $AB$ onto $DE$; because $AB=DE$, endpoint $B$ lands exactly on $E$, and the equal remaining sides force $C$ onto $F$. Takeaway: the translation does the "get one vertex in place" part of superposition; rotation/reflection finish the alignment. This is exactly the machinery behind SSS.□

Solution. plug into the formula with $\theta =90^{\circ }$, so $\cos \theta =0$ and $\sin \theta =1$:

Solution. use the arbitrary-center formula with $\theta =180^{\circ }$, so $\cos \theta =-1$, $\sin \theta =0$. The $\sin$ terms vanish and each coordinate becomes $h+(x-h)(-1)=2h-x$. Thus $(x,y)\mapsto (2\cdot 2-x,2\cdot 3-y)=(4-x,6-y)$. Step 2 — apply to $(5,7)$: $(4-5,6-7)=(-1,-1)$. Sanity check: the midpoint of $(5,7)$ and $(-1,-1)$ is $(\frac{5+(-1)}{2},\frac{7+(-1)}{2})=(2,3)=C$. The center is the midpoint of every point and its image, exactly as central symmetry requires. $✓$□

Solution. the center of rotation is the centroid (which for an equilateral triangle coincides with the circumcenter). Set $n=3$, so the symmetry angles are multiples of $\frac{360^{\circ }}{3}=120^{\circ }$. Step 2 — list $k\cdot 120^{\circ }$ for $k=0,1,2$: that is $0^{\circ }$ (the identity), $120^{\circ }$, and $240^{\circ }$. So the rotation group has order $3$. Step 3 — check closure: $120^{\circ }+240^{\circ }=360^{\circ }\equiv 0^{\circ }$, and $240^{\circ }+240^{\circ }=480^{\circ }\equiv 120^{\circ }$. Every composition lands back in the set, confirming it is a group (the cyclic group $C_3$). Sanity check: a $120^{\circ }$ turn sends each vertex to the next vertex, so the triangle looks identical afterward. $✓$□

Solution. apply $(x,y)\mapsto (y,x)$: the image is $(4,3)$. Step 2 — check the midpoint lies on $y=x$. Midpoint of $(3,4)$ and $(4,3)$ is $(\frac{3+4}{2},\frac{4+3}{2})=(3.5,3.5)$, and $3.5=3.5$, so it is on the line. $✓$ Step 3 — check the segment is perpendicular to the line. The segment from $(3,4)$ to $(4,3)$ has slope $\frac{3-4}{4-3}=-1$; the line $y=x$ has slope $1$; and $(-1)(1)=-1$, so they are perpendicular. $✓$ Conclusion: the mirror line is the perpendicular bisector of the segment joining a point to its image — the defining property of a reflection. (For a slanted line like $y=2x$, use the matrix formula above with $\tan \varphi =2$ instead.)□

Solution. reflect over the $x$ -axis: $(x,y)\mapsto (x,-y)$. Step 2 — reflect that over the $y$ -axis: $(x,-y)\mapsto (-x,-y)$. Step 3 — the net map is $(x,y)\mapsto (-x,-y)$, which is rotation by $180^{\circ }$ about the origin. Sanity check via the angle rule: the two mirrors (the axes) intersect at $90^{\circ }$, and $2\times 90^{\circ }=180^{\circ }$, matching the rotation we found. Two orientation reversals compose to an orientation-preserving map, consistent with getting a rotation rather than a reflection. $✓$□

Solution. rotations: $k\cdot \frac{360^{\circ }}{6}=k\cdot 60^{\circ }$ for $k=0,\dots ,5$, giving $6$ rotations (including the identity). Step 2 — reflections: a regular hexagon has $6$ lines of symmetry — $3$ through pairs of opposite vertices and $3$ through midpoints of opposite sides. That is $6$ reflections. Step 3 — total: $6+6=12$ symmetries, which is the dihedral group $D_6$ of order $2n=2\cdot 6=12$. $✓$ Note on parity: for a regular $n$ -gon, when $n$ is even the mirror lines split into two kinds (vertex-to-vertex and edge-to-edge); when $n$ is odd every mirror line runs from a vertex to the midpoint of the opposite side.□

Solution. rotations about the center: $0^{\circ }$ (identity), $120^{\circ }$, $240^{\circ }$ — three elements. Step 2 — reflections over the three altitudes (each through a vertex and the midpoint of the opposite side) — three elements. Step 3 — total $3+3=6$ elements. This is $D_3$, of order $2n=2\cdot 3=6$, and it equals $3!=6$ because every permutation of the three vertices is realizable by a symmetry. Group check: composing a $120^{\circ }$ rotation with a reflection gives another reflection (still in the set); the identity is present; each reflection is its own inverse and the two nontrivial rotations are inverses of each other. Closed, with identity and inverses — a genuine group. $✓$□

Solution. translations: shifting by any vector $⟨a,b⟩$ with $a,b$ integers maps the lattice onto itself. These already give infinitely many symmetries, so the group is infinite and discrete (a wallpaper group). Step 2 — rotations: a $90^{\circ }$ rotation about any lattice point permutes the lattice onto itself (e.g. about the origin, $(1,0)\mapsto (0,1)\mapsto (-1,0)\mapsto (0,-1)$, all lattice points). So the point group includes order-$4$ rotations — allowed by the crystallographic restriction. Step 3 — reflections: mirroring over the horizontal, vertical, and both diagonal lines through lattice points preserves the lattice. Conclusion: translations $+$ $4$ -fold rotations $+$ mirrors identify this as the wallpaper group $p4m$. $✓$□

Solution. the presentation is $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, with $r$ a $90^{\circ }$ rotation and $s$ a reflection. Step 2 — the rotations are $r^0=1,r,r^2,r^3$ (i.e. $0^{\circ },90^{\circ },180^{\circ },270^{\circ }$); that is $4$ elements. Step 3 — the relation $srs=r^{-1}$ lets us push every $s$ to the right, so any word reduces to the form $r^k$ or $r^{k}s$ with $k\in \{0,1,2,3\}$. The four $r^{k}s$ are the four reflections. Step 4 — total $4+4=8$ elements, matching $|D_4|=2n=2\cdot 4=8$. $✓$ Sanity check on the twist relation: $srs=r^{-1}=r^3$. Geometrically, reflecting, rotating a quarter-turn, then reflecting back looks like a quarter-turn the other way — consistent. $✓$□

Solution. both have length $2$: $AB=\sqrt{2^2+0^2}=2$ and $CD=\sqrt{0^2+2^2}=2$. Equal, so a congruence should exist. Step 2 — translate $A$ to $C$ by $v=C-A=⟨1,1⟩$. Then $A\mapsto (1,1)=C$ and $B\mapsto (2+1,0+1)=(3,1)$. Step 3 — rotate about $C=(1,1)$ by $+90^{\circ }$ to swing the (now horizontal) image onto the vertical segment $CD$. The point $(3,1)$ is at offset $⟨2,0⟩$ from $C$; rotating that offset $90^{\circ }$ counterclockwise gives $⟨0,2⟩$, so it lands at $(1,3)=D$. Conclusion: the composition (translate by $⟨1,1⟩$, then rotate $90^{\circ }$ about $C$) is a direct congruence taking $AB$ exactly onto $CD$. Equal length was the only thing we needed. $✓$□

Solution. align a side. Because $AB=DE$, choose the isometry that sends $A\mapsto D$ and $B\mapsto E$ (translate $A$ to $D$, then rotate about $D$ until the image of $B$ meets $E$; the equal lengths guarantee it reaches $E$). Step 2 — track the third vertex. Let $C^{\prime }$ be the image of $C$. Isometries preserve distance, so $D{C}^{\prime }=CA=FD$ and $E{C}^{\prime }=BC=EF$. Step 3 — pin it down. The point $F$ satisfies $DF=FD$ and $EF=EF$; the set of points at distance $FD$ from $D$ and distance $EF$ from $E$ is the intersection of two circles — at most two points, mirror-images across line $DE$. So $C^{\prime }$ is either $F$ or its reflection $F^{\ast }$. Step 4 — if $C^{\prime }=F$, done (a direct congruence). If $C^{\prime }=F^{\ast }$, follow up with the reflection over line $DE$: it fixes $D$ and $E$ and swaps $F^{\ast }$ with $F$, so the composite carries $C$ to $F$. Conclusion: in either case an isometry maps $△ABC$ onto $△DEF$, so the triangles are congruent. This is the modern, physics-free version of Euclid's superposition. $✓$□

Solution. check orientation of $ABC$. Going $A\to B\to C$ turns counterclockwise (from pointing along the $+x$ direction to the $+y$ direction): positive orientation. Step 2 — check orientation of $A^{\prime }{B}^{\prime }{C}^{\prime }$. Going $A^{\prime }\to B^{\prime }\to C^{\prime }$ goes from the $+y$ direction to the $+x$ direction: that is clockwise, the opposite sense. Step 3 — since the two labelings have opposite orientation, no direct (orientation-preserving) congruence can match them; you need an opposite one. Indeed the reflection over the line $y=x$ sends $(0,0)\mapsto (0,0)$, $(1,0)\mapsto (0,1)$, $(0,1)\mapsto (1,0)$ — exactly $A\mapsto A^{\prime }$, $B\mapsto B^{\prime }$, $C\mapsto C^{\prime }$. Conclusion: they are congruent, but only via an opposite congruence (a reflection). Mirror-image labelings force a flip. $✓$□