10.4: Method of inversion
- Page ID
- 23643
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Here is an application of inversion, which we include as an illustration; we will not use it further in the book.
Let \(ABCD\) be an inscribed quadrangle. Assume that the points \(A,B,C\), and \(D\) appear on the circline in the same order. Then
\(AB \cdot CD + BC \cdot DA = AC \cdot BD.\)
- Proof
-
Assume the points \(A,B,C,D\) lie on one line in this order.
Set \(x = AB\), \(y = BC\), \(z = CD\). Note that
\(x \cdot z + y \cdot (x + y + z) = (x + y) \cdot (y + z).\)
Since \(AC = x + y\), \(BD = y + z\), and \(DA = x + y + z\), it proves the identity.
It remains to consider the case when the quadrangle \(ABCD\) is inscribed in a circle, say \(\Gamma\).
The identity can be rewritten as
\(\dfrac{AB \cdot DC}{BD \cdot CA} + \dfrac{BC \cdot AD}{CA \cdot DB} = 1.\)
On the left hand side we have two cross-ratios. According to Theorem 10.3.1a, the left hand side does not change if we apply an inversion to each point.
Consider an inversion in a circle centered at a point \(O\) that lies on \(\Gamma\) between \(A\) and \(D\). By Theorem Theorem 10.3.2, this inversion maps \(\Gamma\) to a line. This reduces the problem to the case when \(A, B, C\), and \(D\) lie on one line, which was already considered.
In the proof above, we rewrite Ptolemy’s identity in a form that is invariant with respect to inversion and then apply an inversion which makes the statement evident. The solution of the following exercise is based on the same idea; one has to make a right choice of inversion.
Assume that four circles are mutually tangent to each other. Show that four (among six) of their points of tangency lie on one circline.
- Hint
-
Label the points of tangency by \(X, Y , A, B, P\), and \(Q\) as on the diagram. Apply an inversion with the center at \(P\). Observe that the two circles that tangent at \(P\) become parallel lines and the remaining two circles are tangent to each other and these two parallel lines.
Note that the points of tangency \(A'\), \(B'\), \(X'\), and \(Y'\) with the parallel lines are vertexes of a square; in particular they lie on one circle. These points are images of \(A, B, X\), and \(Y\) under the inversion. By Theorem Theorem 10.3.1, the points \(A, B, X\), and \(Y\) also lie on one circline.
Assume that three circles tangent to each other and to two parallel lines as shown on the picture.
Show that the line passing thru \(A\) and \(B\) is also tangent to two circles at \(A\).
- Hint
-
Apply the inversion in a circle with center \(A\). The point \(A\) will go to infinity, the two circles tangent at \(A\) will become parallel lines and the two parallel lines will become circles tangent at \(A\); see the diagram.
It remains to show that the dashed line (\(AB'\)) is parallel to the other two lines.