Euclidean Geometry - Homothety

A homothety, also known as a dilation, is an affine transformation of the plane, determined by a point $P$ and a ratio $k\neq 0$ that sends any point $A$ to a point $A'$ $($called the image of $A)$ such that $k\vec{AP}=\vec{A'P}.$

• If $|k| >1,$ this transformation is known as an expansion.
• If $k=1,$ the homothety is the identity transformation.
• If $|k| <1,$ this transformation is known as a contraction.

Homothety is a simple concept with surprisingly sophisticated consequences. It is useful for proving collinearity, concurrency, determining ratios, and constructing points.

There are several ways to express a homothety. Switching between these different representations confer different benefits and can greatly simplify the problem, as will be observed later.

1) Geometric interpretation
A homothety is just an expansion or contraction. The facts which apply to those situations also apply here. Here are examples of

• $H (A, 2)$ applied to a triangle $ABC:$
  
• $H \big( T, \frac{1}{2} \big)$ applied to a circle $\Gamma$, where $T$ is a point on the circumference:
• $H(P, -1)$ applied to the line $XY$, where $P$ does not lie on $XY:$
  

2) Euclidean geometry (simplified)

• In the case that the center of homothety is the origin, an expansion by $k$ would bring the point $(x,y)$ to $(kx, ky)$. This is the simplest case and often makes calculations very direct.
• In the case that the center of homothety is the point $(a, b),$ an expansion by $k$ would bring the point $(x,y)$ to $\big(k(x-a) + a, k(y-b) + b \big)$. This expression is slightly more complicated, which is why setting $(a,b) = (0,0)$ is often more helpful.

3) Complex numbers
In the argand plane, if the center of homothety is represented by the complex number $p$, then an expasion by $k$ would bring the complex number $z$ to $z' = p + k ( z - p )$.

4) Vectors
As stated above, if the center of homothety is represented by the vector $\vec{P}$, then an expansion by $k$ would bring the vector $\vec{A}$ to $\vec{A'} = \vec{P} + k \vec{ PA }$.

5) Abstract identification
Given familiarity with the applications of homothety, we can identify a homothetic transformation and apply the results.

Proof of the results can be viewed by clicking on the "Proof" button.

1. The points $O, X,$ and $X'$ form a straight line.
   

   Proof
   
   

   This follows directly from the geometric interpretation. $_\square$

2. The slope of a line is preserved under homothety. Hence, parallel lines are preserved.
   

   Proof
   
   

   We will use the simplified Euclidean geometry approach, where we let the center be the origin.
   Suppose we have the points $A = ( x_a, y_a)$ and $B = ( x_b , y_b)$.
   Then, $A' = ( k x_a, ky_a)$ and $B ' = (kx_b, ky_b)$.
   We can verify that the slopes are equal:
   $(\text{Slope of } AB) = \frac{ y_b - y_a } { x_b - x_ a} = \frac{ k ( y_b - y_a ) } { k ( x_b - x_a) } = \frac{ k y_b - k y_a } { k x_b - k x_ a } = (\text{Slope of } A'B').\ _\square$ $$
   Note: The general case can easily be proven using vectors or complex numbers to keep the mathematics simple.
   Corollary: Since the slope of a line is preserved, if two lines are parallel, then their images are parallel to the original lines, and thus the images are parallel.

3. Angles are preserved, meaning that $\angle ABC = \angle A'B'C'$.
   

   Proof
   
   

   Since the slopes of lines are preserved (2), the angle between $AB$ and $BC$ is equal to the angle between $A'B'$ and $B'C'$. $_\square$

4. Every homothety has an inverse: $H(O, k) ^ { -1 } = H \big( O , \frac{1}{k} \big)$.
   

   Proof
   
   

   This follows directly from the geometric interpretation. $_\square$

5. The ratio of lengths satisfy $\big|A'B'\big| = k |AB|$.
   

   Proof
   
   

   We will use the simplified Euclidean geometry approach, where we let the center be the origin.
   Suppose we have the points $A = ( x_a, y_a)$ and $B = ( x_b , y_b)$.
   Then, $A' = ( k x_a, ky_a)$ and $B ' = (kx_b, ky_b)$.
   Hence, the distances obey $\big| A'B'\big| = \sqrt{ ( k x_a - k x_b) ^2 + ( k y_a - k y_b)^2 } = |k| \sqrt{ ( x_a - x_b) ^2 + ( y_a - y_b)^2 } = |k| \times | AB |.\ _\square$ $$
   Note: Using the interpretation of directed lengths, we have $\big|A'B'\big| = k |AB |$.

6. The image is similar to the original object.
   

   Proof
   
   

   Since angles are preserved (3) and the ratio of lengths is a constant $k$ (5), the image is similar. $_\square$

7. The ratio of areas satisfy $\big[A'B'C'\big] = k^2 [ABC]$.
   

   Proof
   
   

   Since the image is similar and the ratio of lengths is $k$, the ratio of areas is $k^2$. $_\square$
   $$
   Note: Using the interpretation of directed area, we have $\big[A'B'C'\big] = k^2 [ABC]$. In particular, orientation is preserved.
   

8. $H(O,1)$ is the identity transformation.
   

   Proof
   
   

   This follows directly from the geometric interpretation. $_\square$

9. $H(O,-1)$ is a reflection through the center, and is also a $180^ \circ$ rotation about the center.
   

   Proof
   
   

   This follows directly from the geometric interpretation. $_\square$

10. $H(O, k \neq 1)$ has 1 fixed point.
    

    Proof
    
    

    We will use the complex number interpretation. Suppose that $z$ is a fixed point, meaning that $z = z' = k( z - a) + a$. Then we have $(k-1) ( z-a) = 0$. Since $k \neq 1$, $z = a$ is the only fixed point. $_\square$

11. The set of lines that remain invariant under $H(P, k \neq 1 )$ are lines through the center $P$.

    Proof
    
    

    We will use the vector interpretation with homothety $H ( A, k )$. Suppose that the line $l = \vec{T} + t \vec{S}$ is preserved. Since slopes are preserved, this means that the point $\vec{T}$ must be mapped onto this line, so there exists some $t$ such that $\vec{A} + k \vec{ AT} = \vec{T} + t \vec{S} \Rightarrow (k-1) \vec{AT} = t \vec{S}$. Hence, $\vec{AT}$ is parallel to $l$, meaning that $A$ lies on the line.
    $$
    Conversely, given any line through the center of homothety, if follows from the geometric interpretation that the line is preserved. $_\square$

12. A homothety is defined uniquely by where any 2 points are mapped to.
    

    Proof
    
    

    Geometric interpretation: Suppose that the points $A, B$ are mapped to $A', B'$. Then, since the center $O$ lies on both $AA'$ and $BB'$, this can be determined by the intersection of the lines. After that, the scaling factor $k$ can be determined by $\frac{ \big|OA'\big| } { |OA| }$.
    $$
    Complex number interpretation: Suppose that the complex numbers $z_1, z_2$ are mapped to $z_1 ', z_2 '$. Then, we need to find a complex number $p$ and a real number $k$ such that
    $\begin{cases} z_1' = p + k ( z_1 - p ) \\ z_2 ' = p + k (z_2 - p ). \end{cases}$ If this system has no solutions, then we do not have a homothety.
    If this system has a unique solution, then that uniquely defines the homothety.
    If this system has infinitely many solutios, then this must be the identity, which is unique. $_\square$
    $$
    Note: We can solve this system to obtain $p = \frac{ z_1 ' z_2 - z_1 z_2 ' } { z_1' + z_2 - z_1 - z_2' }$ and $k = \frac{ z_2' - z_1' } { z_2 - z_ 1 }$.

13. Any 2 (non-congruent) circles have 2 centers of homothety. The positive scaling factor corresponds to the direct homothety, and the negative scaling factor corresponds to the indirect homothety.
    

    Proof
    
    

    We know that the centers of the circle must be mapped onto each other, and thus the scaling factor must satisfy $|k| = \frac{ r_2} { r_1 }$. The center of homothety lies on the line containing these circles. For the negative ratio, the center of homothety will lie on the line segment between the 2 centers. For the positive ratio, the center of homothety will lie outside of the line segment between the 2 centers. $_\square$

14. If $k_1 k_2 \neq 1$, then $H (O_1, k_1) \circ H (O_2, k_2)$ is a homothety.
    

    Proof
    
    

    We will use the complex number interpretation: $H (z_1, k_1) \circ H (z_2, k_2) ( z ) = H (z_1, k_1) \big( z_2 + k_2 (z - z_2 ) \big) = z_1 + k_1 ( \big( z_2 + k_2 (z - z_2 ) \big) - z_1 .$ In order to make this a homothety, we must find $H ( z_3, k_3 ) (z) = z_1 + k_1 \big( z_2 + k_2 (z - z_2 ) \big) - z_1$.
    Equating coefficients of $z$, we obtain $k_3 = k_1 k_2$.
    Equating the constant term, we obtain $( 1 - k_1 k_2 ) z_3 = z_1 + k_1 z_2 - k_1k_2 z_2 - k_1 z_1$. Since $k_1 k_2 \neq 1$, we can set $z_3 = \frac{ z_1 + k_1 z_2 - k_1k_2 z_2 - k_1 z_1 } { 1 - k_1 k_2 }$. $_\square$
    $$
    Note: If we used the geometric interpretation, then this (and the following) result would not be immediately obvious.

15. If $k_1 k_2 = 1$, then $H (O_1, k_1) \circ H (O_2, k_2)$ is a translation.
    

    Proof
    
    

    We will use the complex number interpretation. Since $k_1k_2 = 1$, the previous expression looks like $\text{(constant)} + z$. Thus, this is a translation by the complex number $z_1 + k_1 z_2 - k_1k_2 z_2 - k_1 z_1$. $_\square$

16. In general, it need not be true that $H (O_1, k_1) \circ H (O_2, k_2) = H (O_2, k_2) \circ H (O_1, k_1)$. A specific case where this is true is $O_ 1 = O_2$.

    Proof
    
    

    We will use the complex number interpretation: $\begin{aligned} H (O_1, k_1) \circ H (O_2, k_2) (z) &= z_1 + k_1 z_2 - k_1k_2 z_2 - k_1 z_1 + k_1k_2 z \\ H (O_2, k_2) \circ H (O_1, k_1) &= z_2 + k_2 z_1 - k_2k_1 z_1 - k_2 z_2 + k_1k_2 z. \end{aligned}$ The coefficient of $z$ is equal to $k_1 k_2$. Comparing the constant terms, we have equality if and only if (after some simplification) $(z_1-z_2) ( k_1 - 1) ( k_2 - 1) = 0$. This correspeonds to 1) the 2 centers of homothety are identical, 2) the first homothety is the identity, 3) the second homothety is the identity, $_\square$

17. If 3 non-congruent sets of points are pairwise homothetic, then the (direct) centers of homothety are collinear.

    Proof
    
    

    Suppose the homotheties are $H ( O_i, k_i )$ for $i = 1, 2, 3$.
    By (12), it follows that $H ( O_1, k_1 ) \circ H ( O_2, k_2 ) = H ( O_3, k_3 )$.
    The line $O_1 O _ 2$ is invariant under $H ( O_2, k_2 )$, and again under $H ( O_1, k_1 )$, and hence it is invariant under $H ( O_3, k_3 )$.
    By (11), it follows that $O_3$ lies on the line $O _ 1 O_2$.
    Thus these 3 points are collinear. $_\square$
    $$
    Hint: Find an invariant line.

1. Consider 2 non-intersecting, non-congruent circles.
   $\hspace{5mm}$ Show that the external tangents intersect at the external center of homothety.
   $\hspace{5mm}$ Show that the internal tangents intersect at the internal center of homothety.
   

   Proof
   
   

   Let the external tangents intersect at $O$.
   Let an external tangent touch circles $\Gamma_A , \Gamma_B$ at points $A, B$ respectively.
   Consider the external homothety $H \Big( O, \frac{ r_b}{r_a}\Big)$ applied to point $A$.
   It has to be mapped to some point on $\Gamma_B$.
   Consider the external tangent at $A.$ By (11), since the line passes through $O$, it is invariant under the map. Hence, the only possible point that it gets mapped to is $B$. Hence, by (1), points $O, A, B$ lie on a straight line. $_\square$
   $$
   Note: The internal version of the problem follows using a similar argument.

2. Circles $\Gamma_1$ and $\Gamma_2$ are tangential at $T$. Draw a line through $T$ cutting the circles at $A$ and $B,$ respectively.
   Show that the tangent at $A$ is parallel to the tangent at $B$.

   Proof
   
   

   Let the center of $\Gamma_1$ be $P$ and the center of $\Gamma_2$ be $Q$. Because $T$ is the center of homothety between the two circles, $AT$ : $BT$=$r_1$ : $r_2$=$AP$ : $BQ$=$PT$ : $TQ$. Therefore, $\angle PAT$=$\angle TBQ$. Since the angles the radii make with the tangents are both 90 degrees, the angles that line $AB$ makes with the tangents to both $\Gamma_1$ and $\Gamma_2$ are equal by simple angle subtraction. By transversal properties, this shows that the tangents at $A$ and $B$ are parallel. $_\square$

3. Using a straightedge and compass, in triangle $ABC$ inscribe a square whose edge lies on $AB$.

   Proof
   
   

   Construct square $ABPQ$ such that $PQ$ is constructed on the side of $AB$ opposite to point $C$. Let $D$ be the point where $CP$ intersects $AB$ and let $E$ be the point where $CQ$ intersects $AB$. Then, construct rectangle $DEFG$ inscribed in triangle $ABC$. Since triangles $CFG$ and $CAB$ are similar with the same orientation, points $A$ and $B$ can be obtained by applying $H \Big( C, \frac{FG}{AB}\Big)$ to points $F$ and $G$ respectively. Similarly, since triangles $CDE$ and $CPQ$ are similar with the same orientation, points $P$ and $Q$ can be obtained by applying the same homothety, $H \Big( C, \frac{FG}{AB}\Big)$, to points $D$ and $E$ respectively, noting that $FG$=$DE$ and $AB$=$PQ$. Since rectangle $FGDE$ is homothetic with square $ABPQ$, $FGDE$ is a square, and we are done. $_\square$

4. Any 2 non-congruent triangles whose pairs of sides are parallel to each other and are oriented similarly can be mapped by a homothety.
   Hint: First show that these triangles are similar. Then guess the center of homothety.
   Corollary: [Centroid] Draw the medians $AD, BE, CF$ of triangle $ABC$, where points $D, E, F$ lie on the opposite sides.
   Show that the medians of a triangle are concurrent at $G$.
   Show that $AG:GD = 2 : 1$.

   Proof
   
   

5. Consider 3 non-congruent circles.
   Show that the (possibly indirect) centers of homothety are collinear if and only if the scaling factors satisfy $k_{ab} \times k_{bc} \times k_{ca} = 1$.
   Corollary: [Monge's theorem] Given 3 non-congruent, non-overlapping circles, the 3 intersection points of the external tangents taken pairwise are collinear.

   Proof
   
   

6. In circle $\Gamma$, draw chord $AB$. Circle $\Gamma_1$ is internally tangent to $\Gamma$ at $T$, and tangential to $AB$ at $C$. Let $D$ be the midpoint of arc $AB$ that does not contain $T$.
   Show that $TCD$ is a straight line.

   Proof
   
   

   Call the center of $\Gamma$ $P$, the radius of $\Gamma$ $r_2$, the center of $\Gamma_1$ $Q$, and the radius of $\Gamma_1$ $r_1$. Since $D$ is the midpoint of arc $AB$, the tangent at $D$ is parallel to $AB$. Also, it is clear that the homothety $H \Big( T, \frac{r_1}{r_2}\Big)$ maps circle $\Gamma$ to circle $\Gamma_1$. Therefore, $T$, $Q$, and $P$ are collinear. Since $QC$ and $PD$ are perpendicular to parallel lines, $QC$ and $PD$ are parallel. Using line $TP$ as a transversal, $\angle TQC$=$\angle TPD$. Also, $TQ$ : $TP$ = $QC$ : $PD$ = $r_1$ : $r_2$. By SAS similarity, triangle $TQC$ is similar to triangle $TPD$. Since $T$, $P$, and $Q$ are collinear and $\angle CTQ$ = $\angle DTP$, $T$, $C$, and $D$ must be collinear as well. $_\square$
   

7. In a large circle $\Gamma$, draw 2 smaller circles $\Gamma_A$ and $\Gamma_B$ that are internally tangent to $\Gamma$ at points $A$ and $B,$ respectively. Draw an external tangent of $\Gamma_A$ and $\Gamma_B$ intersecting these circles at $C$ and $D,$ respectively.
   Show that $AC$ and $BD$ intersect on the circumference of $\Gamma$, which we will denote by $P$.
   Show that $ABCD$ is a cyclic quadrilateral and that $PC \times PA = PD \times PB$. In fact, $P$ is the radical center of which 3 circles?
   

   Proof
   
   

   Let $O$ be the center of $\Gamma$, $R$ be the center of $\Gamma_A$, and $Q$ be the center of $\Gamma_B$. Since $\Gamma_A$ is internally tangent to $\Gamma$, $O$, $R$, and $A$ are collinear. Construct a tangent line to $\Gamma$ that is parallel to $CD$. Let the point of tangency be $T$. Since the constructed tangent line and $CD$ are parallel, $\angle ARC$ = $\angle AOT$. Also, it is clear that $AR$:$AO$=$RC$:$OT$, since $AR$=$RC$ and $AO$=$OT$. Therefore, by SAS similarity, $\triangle ARC$ is similar to $\triangle AOT$. Therefore, $\angle ACR$=$\angle ATO$, and it follows that $A$, $C$, and $T$ are collinear by transversal properties. We can similarly show that $B$, $D$, and $T$ are collinear, and therefore $AC$ and $BD$ intersect on the circumference of $\Gamma$, and $T$ is $P$. Afterthought: Another way to show that $\triangle ARC$ is similar to $\triangle AOT$ would be through homothety. $_\square$

8. Given 2 non-overlapping circles, let $AB$ be the internal tangent with $A$ on $\Gamma_A$ and $B$ on $\Gamma_B$. Let $O$ be the indirect center of homothety. Draw diameters $AC$ and $BD$.
   Show that $COD$ is a straight line.
   

   Proof
   
   

   Let $P$ be the center of $\Gamma_A$ and $Q$ be the center of $\Gamma_B$. If $r_1$ is the radius of $\Gamma_A$ and $r_2$ is the radius of $\Gamma_B$, the ratio of all homothetic transformations of points between the two circles is $r_1$:$r_2$. Since $O$ is the center of homothecy, $PO$:$QO$ = $CO$:$DO$ = $r_1$:$r_2$ = $CP$:$DQ$. Therefore, $COP$ is similar to $DOQ$. Therefore, $\angle COP$ = $\angle DOQ$. Therefore, $COD$ is a straight line. $_\square$
   

9. Consider a semicircle with diameter $AB$. A circle is internally tangent to the circumference at $T$ and touches the diameter at $C$.
   Calculate $\angle BTC$.
   

   Proof
   
   

   Call the midpoint of $AB$, also the center of the semicircle, $O$. Call the center of the internally tangent circle $P$. Let us call the measure of $\angle TBO$ $x$. We know that $\angle TOA = 2x$, and that $\angle BTO = \angle TBO = x$. Since $\angle BCO$ is right, $\angle CBO = 90-2x$. Since the smaller circle is internally tangent to the larger semicircle at point $T$, $T$, $P$, and $O$ are collinear. Therefore, $\angle TPC = 180-90+2x = 90+2x$. Since triangle $TPC$ is an isosceles triangle, $\angle PTC = 45-x$. Finally, $\angle BTC = \angle BTO + \angle PTC = 45$ degrees. $_\square$

10. In a large circle $\Gamma$, draw 2 smaller circles $\Gamma_A$ and $\Gamma_B$ that are internally tangent to $\Gamma$ at points $A$ and $B$. Let the direct center of homothety of $\Gamma_A$ and $\Gamma_B$ be $C$.
    Show that $ABC$ is a straight line.
    Note: Does the indirect center of homothety lie on this line?

    Proof
    
    

    Let the center of $\Gamma$ be $O$, the center of $\Gamma_A$ be $P$, and the center of $\Gamma_B$ be $Q$. Without loss of generality, let $\Gamma_B$ be closer to point $C$ than $\Gamma_A$. Draw ray $CB$ until it intersects $\Gamma_A$ for the first time. Let us call this intersection point $J$. Since $C$ is the direct center of homothety, $\angle CBQ = \angle CJP$. Since circles $\Gamma_A$ and $\Gamma_B$ are internally tangent to $\Gamma$, points $O$, $P$, and $A$, and points $O$, $Q$, and $B$ are collinear. By transversal properties, $OB$ $($which contains $Q)$ is parallel to $PJ$ . Therefore, $\angle AOB = \angle APJ$. Therefore, by SAS similarity, triangle $APJ$ is similar to triangle $AOB$. Therefore, $\angle AJP = \angle ABO = 180 - \angle CBO$. Since $\angle CJP = \angle CBO$, $\angle AJP + \angle CJP = 180$, and therefore ray $CB$ contains $A$. $_\square$

11. $ABCD$ is a quadrilateral. The line through $A$ parallel to $CD$ meets $BD$ at $E$, and the line through $D$ parallel to $AB$ meets $AC$ at $F$. Prove that $EF$ is parallel to $BC$.
    Hint: Let the homothety that maps $BA$ to $DF$ be $H_1$, and the homothety that maps $CD$ to $AE$ be $H_2$. Show that $H_1 \circ H_2 ( BC ) = EF$.
    

    Proof
    
    

    Let the intersection of the diagonals of the quadrilateral be $G$. Then, triangles $GDF$ and $GBA$ are similar, and straight lines connect corresponding points, so there exists a homothety $H_1$ that maps $F$ to $A$ and $B$ to $D$. In the same way, through the similarity of triangles $AEG$ and $GDC$, we can find a homothety $H_2$ that maps $A$ to $C$ and $D$ to $E$. It follows that the homothety $H_1 \circ H_2$ maps $C$ to $F$ and $B$ to $E$. Since homothetic transformations of lines end up parallel to the original lines, $EF$ is parallel to $BC$. $_\square$

12. Two circles are externally tangent at $A$. Their common external tangents meet at $X$, and the circles are homothetic via $H (X, k)$. Take a point $B$ on the circumference of the circle, not on the line $AX$.
    Calculate $\angle B A B '$.

    Proof
    
    

    Let us call the center of the circle closer to $X$ $P$, and the center of the other circle $Q$. Without loss of generality, let point $B$ be on the circle closer to $X$, and $B '$ be on the other circle. First, we know that $X$, $P$, $A$ and $Q$ are collinear. Properties of homothetic transformations tell us that $B'Q$ is parallel to $BP$. Set $\angle B'BA = x$ and $\angle ABP = y$. Since triangle $PBA$ is isosceles, $\angle BPA = 180-2y$. Therefore, by transversal properties, $\angle B'QA = 2y$. Since triangle $B'QA$ is isosceles, $\angle QB'A = 90-y$. Since $\angle B'BP = x+y$, by transversal properties, $\angle BB'Q = 180-x-y$, and therefore $\angle AB'B = 90-x$ by simple angle subtraction. We now know two angles in triangle $BB'A$, and we can easily find that $\angle BAB'$ measures 90 degrees. $_\square$

13. In a large circle $\Gamma$, there are 3 circles $\Gamma_A, \Gamma_B, \Gamma_C$ that are internally tangent at points $A, B, C$ respectively, where $A, B, C$ lie on the circumference in that order. Suppose that $\Gamma_A, \Gamma_B$ are externally tangent with tangent $\ell_{AB}$ and $\Gamma_2, \Gamma_3$ are externally tangent with tangent $\ell_{BC}$.
    Show that $\ell_{AB}$ and $\ell_{BC}$ intersect on $\Gamma$ if and only if $\Gamma_A, \Gamma_B, \Gamma_C$ share a common exterior tangent.
    

    Proof
    
    

14. In triangle $ABC$, let the incircle be tangential to $AC$ at $D$. Let $DE$ be the diameter of the incircle. Let $BE$ intersect $AC$ at $F$.
    Show that $AF = CD$.
    

    Proof
    
    

15. (IMO 1978) In isosceles triangle $ABC$, $AB = AC$. A circle is tangent internally to the circumcircle of $ABC$, and also to sides $AB$ and $AC$ at $P$ and $Q,$ respectively.
    Show that the midpoint of segment $PQ$ is the incenter of $ABC$.

    Proof