Homothety

A homothety is a scaling of the whole plane from a fixed center. It explains why tangent circles line up so neatly, why the centroid, circumcenter and orthocenter are collinear, and why chains of circles in an angle form geometric sequences. When a problem has parallel lines and similar figures that share a point, look for the homothety.

Definition and basic properties

Everything follows from the vector formula. If $X \mapsto X'$ and $Y \mapsto Y'$, then $\overrightarrow{X'Y'} = k\,\overrightarrow{XY}$. So a homothety:

• sends every line to a parallel line,
• multiplies every length by $|k|$ and every area by $k^2$,
• preserves angles, and sends a circle with center $C$ and radius $r$ to the circle with center $C'$ and radius $|k|r$.

Conversely, two triangles with corresponding sides parallel (and not congruent by a translation) are related by a homothety: the lines through corresponding vertices all pass through its center.

Homothety centers of two circles

Why. A homothety sending circle $1$ to circle $2$ must send $O_1$ to $O_2$, so its center $X$ is on line $O_1O_2$ with $\overrightarrow{XO_2} = k\,\overrightarrow{XO_1}$ and $|k| = \dfrac{r_2}{r_1}$. There's one solution for each sign of $k$, and then $\dfrac{XO_1}{XO_2} = \dfrac{r_1}{r_2}$. A common tangent line is sent to a parallel line tangent to circle $2$; for a tangent through the center $X$, the image is itself, so every common tangent through $X$ works. Finally, if the circles touch at $T$, then $T$ is on line $O_1O_2$ with $\dfrac{TO_1}{TO_2} = \dfrac{r_1}{r_2}$, so $T$ is a center.

Homothety in the triangle

The medial triangle and the Euler line. The homothety with center at the centroid $G$ and ratio $-\tfrac12$ sends $A$, $B$, $C$ to the midpoints of the opposite sides (because $G$ is two-thirds of the way along each median). It sends triangle $ABC$ to its medial triangle.

The altitudes of the medial triangle are perpendicular to its sides, which are parallel to the sides of $ABC$, and they pass through the midpoints of $ABC$'s sides. So they are the perpendicular bisectors of $ABC$: the orthocenter of the medial triangle is the circumcenter $O$ of $ABC$. Since the homothety sends the orthocenter $H$ of $ABC$ to the orthocenter of the medial triangle,

$\overrightarrow{GO} = -\tfrac12\,\overrightarrow{GH}.$

So $O$, $G$, $H$ are collinear (the Euler line) with $GH = 2\,GO$. In vectors with $O$ as the origin, $H = A + B + C$. The same homothety sends the circumcircle (radius $R$) to the nine-point circle, with radius $\dfrac{R}{2}$.

Tangent lines to the incircle. The line tangent to the incircle and parallel to $BC$ (on the side toward $A$) cuts off a small triangle at $A$, related to $ABC$ by a homothety at $A$. (Notice that the incircle of $ABC$ is an excircle of the small triangle.) The quickest way to get the ratio is with perimeters. The small triangle's perimeter is $2(s - a)$: its third side splits into two tangent segments to the incircle, which match up with the rest of the tangent segments from $A$, so the perimeter equals the two full tangent lengths from $A$. So its ratio to $ABC$ is

$k_A = \frac{s - a}{s}.$

Practice

Practice 1

Circles with radii $4$ and $10$ have centers $21$ units apart. Their two common external tangent lines meet at $P$. Find the distance from $P$ to the center of the larger circle.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 2

Triangle $ABC$ has $AB = 10$, $AC = 17$ and $BC = 21$. A line tangent to the incircle and parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$. Then $DE = \dfrac{m}{n}$ in lowest terms. Find $m + n$.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 3

A sequence of circles $\omega_1, \omega_2, \omega_3, \dots$ lies inside an angle. Each circle is tangent to both sides of the angle, and each is tangent to the next. If $\omega_1$ has radius $4$ and $\omega_2$ has radius $6$, the radius of $\omega_5$ is $\dfrac{m}{n}$ in lowest terms. Find $m + n$.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 4

Triangle $ABC$ has sides $13$, $14$, $15$. Three lines are drawn tangent to its incircle, each parallel to one side of the triangle. They cut off three small triangles at the corners of $ABC$. The sum of the circumradii of the three small triangles is $\dfrac{m}{n}$ in lowest terms. Find $m + n$.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 5

Triangle $ABC$ has $AB = 13$, $BC = 14$ and $CA = 15$, with circumcenter $O$ and orthocenter $H$. Then $OH^2 = \dfrac{m}{n}$ in lowest terms. Find $m + n$.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 6

Triangle $ABC$ has $AB = 11$, $BC = 30$ and $CA = 25$. Circle $\omega$ is tangent to sides $AB$ and $AC$ and externally tangent to the incircle, and lies between the incircle and vertex $A$. The radius of $\omega$ is $\dfrac{m}{n}$ in lowest terms. Find $m + n$.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint (1/2)

Practice 7

Circle $\Gamma$ has radius $13$, and $AB$ is a chord of length $24$. Point $P$ is on $AB$ with $AP = 6$. Circle $\omega$ is tangent to $AB$ at $P$, lies on the same side of $AB$ as the center of $\Gamma$, and is internally tangent to $\Gamma$ at $T$. Then $PT = \dfrac{m}{n}$ in lowest terms. Find $m + n$.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint (1/2)

Real contest practice

These are full past AIMEs. Each has geometry problems with tangent circles or parallel lines where a homothety shortens the work.

• 2006 AIME I
• 2017 AIME II
• 2022 AIME I