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Let $\Delta ABC$ be a triangle. If we place points $D,\ E,\ F$ arbitrarily on the sides $\overline{AB},\ \overline{BC}$ and $\overline{CA}$ respectively, then the circumcircles of the triangles $\Delta ADF,\ \Delta BDE$ and $\Delta CEF$ will all pass through a common center via Miquel's theorem. Let $M$ be this common point, the Miquel point.

If we locate the centers of the circumcircles and name them $P,\ Q,\ R$ respectively, then the resulting triangle $\Delta PQR$ will be similar to triangle $\Delta ABC$. I want to prove that the Miquel point $M$ is the center of the spiral similarity which carries $\Delta PQR$ to $\Delta ABC$         miquel

If somebody could provide a proof or a reference to a proof that would be greatly appreciated. Thank you.

If someone could just provide a proof for the regular similarity between the two triangles, that would also be appreciated. I have looked around online but could not find a proof.

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So does a "spiral similarity with center M" mean that you can perform a combination of rotation about that point and dilation to match the figures? –  rschwieb Oct 4 '12 at 16:15
    
@rschwieb Yes, that's what I mean. A dilation and rotation with center $M$ bringing $\Delta PQR$ to $\Delta ABC$. –  EuYu Oct 4 '12 at 16:44

2 Answers 2

up vote 1 down vote accepted

Let's consider only two different settings in our proof, as shown in figure 1 and 2. You can complete the proof for other settings.

(1) Case ($\angle ERM$ and $\angle MRF$ are adjacent angles).

See figure 1.

MiqueSimi

We know that $\triangle MRE$ is isosceles and $QR$ is bisector of $\angle MRE$. We also know that $\triangle MRF$ is isosceles and $PR$ is bisector of $\angle MRF$.

Let $m(\angle MRE) = 2 \alpha$ and $m(\angle MRF) = 2 \beta$, then we can conclude that $m(\angle ECF)= \alpha + \beta$, since $\angle ECF$ is a inscribed angle and its measure is half the measure of the central angle $\angle ERF$. ($m(\angle ERF) = 2\alpha + 2 \beta$).

So it follows that $$\angle ECF = \angle QRP.$$

By a similar approach we can conclude that

$$\angle DAF = \angle QPR.$$

Therefore $\triangle ABC \sim \triangle QPR$.

(2) Case ($\angle ERM$ and $\angle MRF$ are not adjacent angles).

See figure 2.

In that setting there is a superposition of $\triangle MRE$ and $\triangle MRF$, so $m(\angle ERF) = 2 \beta - 2 \alpha$, and no more $2 \alpha + 2 \beta$.

Note that $m(\angle QRF)= \beta - \alpha$ and that $m(\angle ECF)= \beta - \alpha$, since $m(\angle ECF)$ is still half the measure of $\angle ERF$.

Therefore $$\angle ECF = \angle QFR.$$

As in case 1 (by a similar approach), we can conclude that

$$\angle DAF = \angle QPR.$$

Therefore $\triangle ABC \sim \triangle QPR$, as in case 1.

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Excellent. Thank you. –  EuYu Jan 9 '13 at 20:24
    
@EuYu. You're welcome:) –  RicardoCruz Jan 9 '13 at 20:28

Fact. If $x$, $y$, $z$ are points on the complex plane, then the circumcenter of $\triangle xyz$ is the point

$$ i \frac{x(z\overline{z} - y\overline{y})+ y(x\overline{x}-z\overline{z})+z(y\overline{y}-x\overline{x})}{x(\overline{z} - \overline{y})+ y(\overline{x}-\overline{z})+z(\overline{y}-\overline{x})}$$

where "$\overline{x}$" indicates the complex conjugate of $x$.


Let us consider $\triangle abc$, defining $d$, $e$, $f$ on sides $bc$, $ca$, $ab$, respectively:

$$d := b + \alpha(c-b) \qquad e := c+\beta(a-c) \qquad f := a+\gamma(b-a)$$

for real scalars $\alpha, \beta, \gamma$.

Let $p$, $q$, $r$ be the respective circumcenters of $\triangle aef$, $\triangle dbf$, $\triangle dec$; for simplicity, we take $a$, $b$, $c$ on the unit circle so that $\overline{a}=1/a$, etc. Then,

$$\begin{align} p &= i\;\frac{c(b-a)+ \beta b ( a-c) + \gamma c( a - b )}{b-c} \\[6pt] q &= i\;\frac{a(c-b)+ \gamma c ( b-a) + \alpha a( b - c )}{c-a} \\[6pt] r &= i\;\frac{b(a-c)+ \alpha a ( c-b) + \beta b( c - a )}{a-b} \end{align}$$

We deduce

$$\frac{p-q}{a-b} = \frac{q-r}{b-c} = \frac{r-p}{c-a} =: u$$

which, by taking the modulus, proves $\triangle abc \sim \triangle pqr$. A little algebra reveals that $$u + \overline{u} = 1$$

Now ... A point, $m$, (which may or may not be the Miquel point) is the center of spiral similarity from $\triangle abc$ and $\triangle pqr$, if and only if

$$\frac{a-m}{p-m} = \frac{b-m}{q-m} = \frac{c-m}{r-m}$$

(This characterization is what inspired me to approach the problem in the complex plane.) Solving for $m$ via the first equality gives

$$m = \frac{aq-bp}{(a-b)-(p-q)} = \frac{r - c u}{1-u}$$

whence $$m-r = \frac{u(r-c)}{1-u} = \frac{u}{\overline{u}}(r-c) \qquad\qquad \overline{m}-\overline{r} = \frac{\overline{u}}{u}(\overline{r}-\overline{c})$$ so that $$|m-r|^2 = (m-r)(\overline{m}-\overline{r}) = (r-c)(\overline{r}-\overline{c}) = |c-r|^2$$

indicating that $c$ and $m$ are equidistant from $r$: thus, $m$ is on the circumcircle of $\triangle dec$. By symmetry, it is also on the circumcircles of $\triangle aef$ and $\triangle dbf$; the center of spiral similarity is in fact the Miquel point.

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This result has been a curiosity at the back of my mind for a very long time. Thank you for providing such a nice proof. –  EuYu Jan 10 '13 at 16:51

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