Lesson · Homothety
Transformations
A hypothesis that ties together far-apart parts of a figure rarely yields to a chase from vertex to vertex. The stronger move is to pick a transformation - a mirror, a half-turn, a rotation, a translation, a scaling - and move a whole piece of the configuration at once. Every length and angle it carries comes along for free, and lands next to the thing it must be compared with.
Moving the whole configuration
Principle
Reflections, half-turns, rotations and translations preserve every length and every angle; a homothety multiplies all lengths by one fixed ratio and keeps every direction. So instead of proving a dozen small equalities one by one, apply a single map that is guaranteed to preserve them all, and read the conclusion off the image. The skill is choosing the map - and the hypothesis usually names it.
The dictionary is short enough to memorise:
| The data contain | Reach for |
|---|---|
| a sum of distances \(PA + PB\), a path to straighten | reflection in a line |
| a midpoint, a median, a factor of \(2\) | half-turn about the midpoint |
| two equal segments out of one point, a fixed angle between them | rotation about that point |
| two equal parallel segments in different corners | translation |
| midpoints or ratios linking two parallel copies of a figure | homothety |
The isometries: mirror, half-turn, rotation, translation
Reflection in a line preserves distances and is the only isometry that swaps the two sides of its mirror. That makes it the tool for sums of distances: a bent path with one bend on the mirror unfolds into a straight one.
Example
A point \(P\) runs along a line \(\ell\); the points \(A\) and \(B\) lie on the same side of \(\ell\). Reflect \(B\) in \(\ell\) to get \(B'\). Then \(PB = PB'\) for every position of \(P\), so \(AP + PB = AP + PB' \ge AB'\), with equality exactly when \(P\) is the point where the segment \(AB'\) crosses \(\ell\). The same unfolding, repeated, straightens billiard paths and broken lines - a chain is never shorter than the segment joining its endpoints, and reflections let you choose which endpoints.
Central symmetry - the half-turn about a point \(M\) - sends \(X\) to the point \(X'\) for which \(M\) is the midpoint of \(XX'\). It carries every segment to a parallel segment of equal length, so a midpoint in the data is a standing invitation to apply it.
Example
Let \(AM\) be a median of the triangle \(ABC\). The half-turn about \(M\) swaps \(B\) and \(C\) and sends \(A\) to the point \(D\) with \(AD = 2\,AM\). The diagonals of the quadrilateral \(ABDC\) bisect each other at \(M\), so \(ABDC\) is a parallelogram - in particular \(BD = AC\). The side \(AC\) has been transported next to \(AB\), and every hypothesis about the median can now be read inside the triangle \(ABD\), whose side \(AD\) is the doubled median.
Rotation about a centre \(O\) through an angle \(\varphi\) is the map to use when two equal segments emanate from one point at a known angle: the rotation carrying one onto the other transports everything attached to them. In equilateral and square configurations the centre and the angle are on the table already.
Example
A point \(P\) inside an equilateral triangle \(ABC\) satisfies \(PA = 3\), \(PB = 4\), \(PC = 5\). Rotate the plane by \(60^\circ\) about \(B\), in the sense carrying \(C\) to \(A\), and let \(P'\) be the image of \(P\). Then \(BP' = BP = 4\) and \(\angle PBP' = 60^\circ\), so the triangle \(PBP'\) is equilateral and \(PP' = 4\); and \(P'A = PC = 5\), because the rotation carries the segment \(PC\) onto \(P'A\). The triangle \(APP'\) has sides \(3, 4, 5\), so \(\angle APP' = 90^\circ\) and \(\angle APB = 90^\circ + 60^\circ = 150^\circ\). The rotation transplanted the length \(PC\) across the figure, into one triangle with the other two.
Translation moves every point by the same vector and is the quiet member of the family: use it when two equal parallel segments live in different corners. The standard instance is the trapezoid \(ABCD\) with \(AB \parallel CD\): translating the diagonal \(DB\) by the vector \(\overrightarrow{DC}\) places it next to the diagonal \(AC\), and the two diagonals become two sides of a single triangle whose third side is \(AB + CD\) - ready for the triangle inequality.
Warning
Name the map completely - mirror line for a reflection, centre for a half-turn, centre and angle for a rotation, centre and ratio for a homothety - and then prove where the images land. "The half-turn about \(M\) sends \(B\) to \(D\)" defines \(D\); that \(D\) also lies on the line you want is a claim with a proof, not something to read off the picture. Transformation solutions lose points on unproved image claims, never on the idea.
Homothety: scaling from a centre
Definition
The homothety with centre \(O\) and ratio \(k \neq 0\) sends each point \(X\) to the point \(X'\) on the line \(OX\) with \(\overrightarrow{OX'} = k\,\overrightarrow{OX}\). It multiplies every length by \(|k|\), maps every line to a parallel line and every circle to a circle, and for negative \(k\) it sends the figure through the centre to the other side - the half-turn is the case \(k = -1\).
Two signals call for it. First, midlines: the midpoints of the sides of a triangle form the medial triangle, which is the image of the original under the homothety with centre the centroid \(G\) and ratio \(-\tfrac{1}{2}\) - every midline figure is a scaled parallel copy of a figure you already have, so any statement proved for one transfers to the other. Second, whole circles: a homothety is the only elementary map that carries a circle to a circle of a prescribed different radius, so when midpoints of segments through one point \(D\) appear near a circle, the homothety centred at \(D\) moves the entire circle, centre and all, in one stroke.
Example
Two parallel lines crossed by several lines through one point \(F\): the homothety with centre \(F\) carrying one parallel onto the other sends each intersection point on the first line to the corresponding one on the second. All ratios along the first line therefore reappear, unchanged, along the second - a midpoint on one parallel forces a midpoint on the other. Whenever a picture contains two parallels and a bundle of concurrent lines, this homothety is already drawn; it only remains to say so.
Warm-up
Example
In a triangle \(ABC\), let \(M\) be the midpoint of \(BC\). Prove that \(AM < \tfrac{1}{2}\,(AB + AC)\).
The target hides a factor of two at a midpoint
Read the claim as \(2\,AM < AB + AC\). A factor of two next to the midpoint \(M\) asks for the half-turn about \(M\): it sends \(A\) to the point \(D\) with \(AD = 2\,AM\), and swaps \(B\) with \(C\).
Name the parallelogram
The diagonals \(AD\) and \(BC\) of the quadrilateral \(ABDC\) bisect each other at \(M\), so \(ABDC\) is a parallelogram and \(BD = AC\). All three lengths of the target now sit on the single triangle \(ABD\).
One triangle inequality
In the triangle \(ABD\), \(AD < AB + BD = AB + AC\), that is, \(2\,AM < AB + AC\). The inequality is strict: equality would force \(B\) onto the segment \(AD\), and \(A\), \(B\), \(D\) collinear would make \(A\), \(B\), \(M\) collinear, degenerating the triangle.