Homothety is the habit of using dilation centers and scale factors to relate circles, triangles, and tangent structures. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
$$h(P) = O + k\,(P - O): \quad \text{a dilation about center } O \text{ with ratio } k \text{ scales all lengths by } |k|$$
A homothety is a scaling from a center point; it maps every figure to a parallel, similar copy with a fixed ratio.
Why It Works
1
A homothety with center $O$ and ratio $k$ sends each point $P$ to $P'$ with $\vec{OP'} = k\,\vec{OP}$.
2
Because $\vec{OP'} \parallel \vec{OP}$, lines map to parallel lines and the image is similar to the original with scale factor $|k|$.
3
Circles map to circles, with radius scaled by $|k|$; in particular the centers of two circles and their external (or internal) center of similitude are collinear.
4
Composing two homotheties gives another homothety (or a translation), and the three centers of similitude of three circles are collinear (Monge's theorem) — the source of many olympiad results.
Worked Examples
Example 1
Two circles are internally tangent at point $T$, with radii $3$ and $7$. A line through $T$ meets the small circle again at $A$ and the large circle again at $B$. Find $TB:TA$.
1
The internal tangency at $T$ means $T$ is the center of the homothety mapping the small circle to the large circle.
2
Its ratio is the ratio of radii, $k = \dfrac{7}{3}$, and it sends $A$ (on the small circle) to $B$ (on the large circle) along the line through $T$.
In $\triangle ABC$, $D$, $E$, $F$ are the midpoints of $BC$, $CA$, $AB$. The medial triangle $\triangle DEF$ is the image of $\triangle ABC$ under a homothety. Find its center and ratio, and the ratio of areas $[\triangle DEF] : [\triangle ABC]$.
1
The medial triangle is the image of $\triangle ABC$ under the homothety centered at the centroid $G$ with ratio $k = -\tfrac12$ (it maps each vertex to the midpoint of the opposite side, on the far side of $G$).
2
Lengths scale by $|k| = \tfrac12$, so each side of $\triangle DEF$ is half the corresponding side of $\triangle ABC$.
3
Areas scale by $k^2 = \left(\tfrac12\right)^2 = \tfrac14$.
Answer:
Center $G$ (centroid), ratio $-\tfrac12$; $[\triangle DEF] : [\triangle ABC] = 1 : 4$.
Contest level
A circle of radius $2$ is internally tangent to a circle of radius $5$. The center of similitude (homothety center) is the tangency point $T$. A chord of the large circle is mapped to a chord of the small circle by the homothety. If a point $P$ on the large circle has $TP = 8$, find the distance $TP'$ to its image on the small circle.
1
Internal tangency makes $T$ the center of the homothety carrying the large circle (radius $5$) to the small circle (radius $2$), with ratio $k = \tfrac{2}{5}$.
2
The image of $P$ lies on ray $TP$ with $TP' = |k|\cdot TP$.
Prove that in any triangle the centroid $G$, the circumcenter $O$, and the orthocenter $H$ are collinear with $HG : GO = 2 : 1$, using the homothety at $G$ with ratio $-\tfrac12$.
1
The homothety $h$ centered at $G$ with ratio $-\tfrac12$ maps each vertex to the midpoint of the opposite side, so it maps $\triangle ABC$ to its medial triangle $\triangle DEF$.
2
An altitude of $\triangle ABC$ from a vertex is perpendicular to the opposite side; the corresponding line in the medial triangle is the perpendicular bisector of that side, because $D, E, F$ are midpoints and the medial sides are parallel to the original sides.
3
Thus $h$ maps the orthocenter $H$ of $\triangle ABC$ (where the altitudes meet) to the point where the perpendicular bisectors meet, which is the circumcenter $O$. So $h(H) = O$.
4
A homothety of ratio $-\tfrac12$ at $G$ satisfies $\vec{GO} = -\tfrac12\,\vec{GH}$, so $G$ lies on segment $HO$ with $HG : GO = 2 : 1$ — the Euler line.
Answer:
$H$, $G$, $O$ are collinear with $HG : GO = 2 : 1$ (the homothety $h(H) = O$ at $G$).
Going Deeper
Generalization: composing two homotheties with ratios $k_1, k_2$ gives a homothety of ratio $k_1 k_2$ (a translation if $k_1 k_2 = 1$), and the three centers of similitude of three circles, taken consistently, are collinear (Monge's theorem). Spiral similarity is the further generalization where the ratio $k$ is allowed to be a complex number (rotation $+$ scaling).
Where it appears: homothety is the cleanest tool for tangent-circle configurations — the tangency point of two tangent circles is automatically a homothety center, instantly giving length ratios (as in the worked examples). It also proves the nine-point circle (image of the circumcircle under $h$ at $H$ with ratio $\tfrac12$) and the Euler line.
Pitfall: a negative ratio $k < 0$ flips orientation — the image lies on the OPPOSITE side of the center. The EXTERNAL center of similitude (positive ratio, $k = +r_2/r_1$) lies outside the segment joining the centers; the INTERNAL center (negative ratio, $k = -r_2/r_1$) lies between them. For two tangent circles the tangency point is one of these centers, so getting the sign of $k$ wrong sends the image to the wrong side and corrupts every length ratio.
Spot the Signal
Look for problems where the key step is using dilation centers and scale factors to relate circles, triangles, and tangent structures.
You can describe the hard part as using dilation centers and scale factors to relate circles, triangles, and tangent structures, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the diagram relation that calls for homothety, then rewrite the givens around it.
Name the relation that makes Homothety legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Homothety just because the surface looks familiar; verify the required condition first.
Applying Homothety because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let homothety reveal the geometric configuration; then compute only what remains.
Try it on:
Practice a contest problem where the key step is using dilation centers and scale factors to relate circles, triangles, and tangent structures.