Miquel circles and a beautiful similarity

pohoatza

Points $A_{1}$ , $B_{1}$ , $C_{1}$ are chosen on the sides $BC$ , $CA$ , $AB$ of a triangle $ABC$ respectively. The circumcircles of triangles $AB_{1}C_{1}$ , $BC_{1}A_{1}$ , $CA_{1}B_{1}$ intersect the circumcircle of triangle $ABC$ again at points $A_{2}$ , $B_{2}$ , $C_{2}$ respectively ( $A_{2}\neq A, B_{2}\neq B, C_{2}\neq C$ ). Points $A_{3}$ , $B_{3}$ , $C_{3}$ are symmetric to $A_{1}$ , $B_{1}$ , $C_{1}$ with respect to the midpoints of the sides $BC$ , $CA$ , $AB$ respectively. Prove that the triangles $A_{2}B_{2}C_{2}$ and $A_{3}B_{3}C_{3}$ are similar.

Comment

**Posted:** Thu Jun 28, 2007 10:57 am

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Cosmin Pohoata, Bucharest, Romania

Little Gauss

This solution was proposed by Yulhee Nam in Korean Team Intensive Training.

Define a point $X$ s.t. a quadrilateral $CA_{1}XB_{3}$ be a parallelogram.
Then $BXB_{3}A_{3}$ and $AXA_{1}B_{1}$ be a parallelogram, too.
$\Rightarrow \angle CC_{2}A_{1}=\angle CB_{1}A_{1}=\angle B_{3}AX$
$\Rightarrow$ Lines $AX$ and $C_{2}A_{2}$ meet on the circumcircle of triangle $ABC$ at $Q$ .
$\Rightarrow \angle CBQ=\angle CC_{2}Q=\angle CAQ=\angle A_{1}XQ$
$\Rightarrow$ $BXA_{1}Q$ is cyclic.
$\Rightarrow \angle B_{3}A_{3}C=\angle XBC=\angle XQA_{1}=\angle C_{2}CA$ .
Therefore, we get following amazing result :
$\angle B_{3}A_{3}C=\angle C_{2}CA$ .
We can easily prove that $A_{2}B_{2}C_{2}$ and $A_{3}B_{3}C_{3}$ are similar by this result.

We can also use complex numbers.
Because $C_{2}AB_{1}\sim C_{2}BA_{1}$ , $\frac{c_{2}-b_{1}}{c_{2}-a}=\frac{c_{2}-a_{1}}{c_{2}-b}$ .
So we can easily express $a_{2}, b_{2}, c_{2}$ by $a,b,c,a_{1},b_{1},c_{1}$ .
Now, with some calculation, we get the result.

**Posted:** Sun Jul 01, 2007 6:24 am

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Soo-Hong, LEE

benjamin

This one was proposed by Bodo Lass, a mathematician from Lyon in France :
$A_{2}$ is the center of the spiral similarity which maps $BC_{1}$ on $CB_{1}$ . So we have
$\frac{A_{2}B}{A_{2}C}= \frac{BC_{1}}{CB_{1}}= \frac{AC_{3}}{AB_{3}}$ , that is the triangles $A_{2}BC$ and $AC_{3}B_{3}$ are similar. Working with angles mod 180° we get $(A_{2}B_{2},C_{2}B_{2}) = (A_{2}B_{2},BB_{2})+(BB_{2},C_{2}B_{2}) = (A_{2}C,BC)+(BA,C_{2}A) = (AC,C_{3}B_{3})+(A_{3}B_{3},CA)...$ , Quite Easily Done Razz

Benjamin

**Posted:** Sun Jul 01, 2007 3:22 pm

pohoatza

**Posted:** Sun Jul 01, 2007 3:33 pm

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Cosmin Pohoata, Bucharest, Romania

nobody1 · P versus NP Offline Joined: 20 Dec 2006 Posts: 41

**Posted:** Sun Jul 01, 2007 6:20 pm

silouan

**Posted:** Mon Jul 02, 2007 2:02 am

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Προετοιμάζομαι για το καλύτερο και είμαι έτοιμος για το χειρότερο

iura

Lemma: Consider a triangle $ABC$ and points $A',B',C'$ on $BC, CA, AB$ . The perpendiculars from $A'$ to $BC$ , $B'$ to $AC$ and $C'$ to $AB$ intersect in points $A_{1}, B_{1}, C_{1}$ (correspondingly). The lines $AA_{1},BB_{1},CC_{1}$ meet the circumcircle of $ABC$ in $A_{2},B_{2},C_{2}$ . Then $A_{2}B_{2}C_{2}$ is similar to $A'B'C'$ .

Proof: $m(\angle BAA_{2})=m(\angle C'AA_{1})=m(\angle C'B'A_{1})$ as $A_{1}B'C'A$ is cyclic. Analogously $m(\angle BCC_{2})=m(\angle A'CA_{1})=m(\angle A'B'A_{1})$ as $A_{1}B'C'C$ is cyclic. Therefore $m(\angle C_{2}B_{2}A_{2})=m(\angle BCC_{2})+m(\angle BAA_{2})=$ $m(\angle C'B'A_{1})+m(\angle A'B'A_{1})=m(\angle A'B'C')$ . So $m(\angle A_{2}B_{2}C_{2})=m(\angle A'B'C')$ . Together with the analogously deduced equalities this means that $A_{2}B_{2}C_{2}$ and $ABC$ are similar.

Let's return to the problem. Let $O$ be the circumcircle of $ABC$ . Let $A', B', C'$ be diametrally opposite to $A,B,C$ . Also pick up $A_{4}$ be the intersection of the perpendicular from $B_{2}$ to $AC$ and $C_{2}$ to $AB$ , and analogously define $B_{4}, C_{4}$ . $A_{2}$ is symmetric to $A$ wrt $OS$ where $S$ is the midpoint of $AA_{4}$ (the circumcenter of $AB_{2}C_{2}$ (, therefore $A_{2}$ is the intersection of $A'A_{4}$ with the circumcircle of $ABC$ .

If we let the perpendicular from $B_{3}$ to $AC$ intersect the perpendicular from $C_{3}$ to $AB$ in $A_{4}'$ then clearly $A_{4}$ and $A_{4}'$ are symmetric with respect to $O$ .

Finally if we let $A_{5},B_{5}, C_{5}$ be symmetric to $A_{3}B_{3}C_{3}$ with respect to $O$ then the perpendicular from $B_{5}$ to $A'C'$ meets the perpendicular from $C_{5}$ to $A'B'$ in a point which is symmetric to $A_{4}'$ with respect to $O$ (by symmetry), therefore in $A_{4}$ .

It remains to apply the lemma for triangle $A'B'C'$ and points $A_{5},B_{5},C_{5}$ on its sides, as $A_{5}B_{5}C_{5}$ is congruent to $A_{3}B_{3}C_{3}$ by symmetry.

**Posted:** Tue Jul 03, 2007 12:58 am

April

We have
$\left\{\begin{array}{c}\widehat{A_{2}BC_{1}}=\widehat{A_{2}CB_{1}}\\ \widehat{A_{2}C_{1}B}=180^\circ-\widehat{A_{2}A_{1}A}=18...$
$\Rightarrow\triangle A_{2}BC_{1}\sim\triangle A_{2}CB_{1}\Rightarrow\frac{C_{1}B}{B_{1}C}=\frac{A_{2}B}{A_{2}C}\Rightarrow\fr...$
$\Rightarrow\triangle AC_{3}B_{3}\sim\triangle A_{2}BC\Rightarrow\widehat{AC_{3}B_{3}}=\widehat{A_{2}BC}$

Similarly, we also have $\widehat{BC_{3}A_{3}}=\widehat{B_{2}AC}$
Thus, $\widehat{A_{3}C_{3}B_{3}}=180^\circ-\widehat{AC_{3}B_{3}}-\widehat{BC_{3}A_{3}}=180^\circ-\widehat{A_{2}BC}-\widehat{B_{2}AC}$

On the other hand,
$\begin{eqnarray*}\widehat{A_{2}C_{2}B_{2}}&=&\widehat{AC_{2}C}-\widehat{AC_{2}A_{2}}-\widehat{B_{2}C_{2}C}\\ &=&a...$
Therefore $\widehat{A_{3}C_{3}B_{3}}=\widehat{A_{2}C_{2}B_{2}}$
And similarly for the other angles, we get $\triangle A_{3}B_{3}C_{3}\sim\triangle A_{2}B_{2}C_{2}$

Source: http://imocompendium.com/phpBB1/viewtopic.php?t=113

**Posted:** Mon Jul 23, 2007 3:44 pm

Leonhard Euler

lemma 1: Let $P$ be a Miquel point of four line $l_1,l_2.l_3,l_4$ . Simson line of $P$ wrt triangle that formed by $l_1,l_2,l_3,l_4$ (of course, there are four triangle and four Simson line are coincide) is perpendicular to the Gauss line of $l_1.l_2,l_3,l_4$ .
proof) Since Aubert line is perpendicular to the Gauss line(see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=842), it is suffice to show Simson line is parallel to Aubert line but it is true since Simson line pass midpoint of $PH_i(i = 1,2,3,4)$ where $H_i$ is orthocenter of four triangle that formed by four line $l_1,l_2,l_3,l_4$ .

lemma 2: Let $P,Q$ are points on circumcircle of triangle $ABC$ and $l_1,l_2$ be Simson line of $P,Q$ wrt $\triangle ABC$ . Then $\measuredangle (l_1;l_2) = - \measuredangle PAQ$ (directed angle mod 180).
proof) Let $X,Y$ are on circumcircle of triangle $ABC$ such $PX\perp BC,QY\perp BC$ . It is known that $AX\parallel l_1,AY\parallel l_2$ . Hence $\measuredangle (l_1;l_2) = \measuredangle XAY = \measuredangle XPY = - \measuredangle PYQ = - \measuredangle PAQ$ .

In above problem, Let $X,Y$ be midpoint of $BB_1,CC_1$ and $l_1$ be Simson line of $A_2$ wrt $\triangle ABC$ . By lemma 1, $l_1\perp XY$ .Let $M$ be midpoint of $B_1C_1$ .

Then $MX = \frac {1}{2}C_1B = \frac {1}{2}AC_3$ and $MY = \frac {1}{2}AB_3$ . Hence triangle $AC_3B_3$ and $MXY$ are similar and we obtain $B_3C_3\parallel XY$

But $l_1\perp XY$ . So $B_3C_3\perp l_1$ . Let $l_2$ be Simson line of $B_2$ wrt $\triangle ABC$ . Then we also have, $l_2\perp C_3A_3$ . By lemma 2, $\measuredangle (l_1,l_2) = - \measuredangle A_2C_2B_2$ . But since $l_1\perp B_3C_3,l_2\perp C_3A_3, \measuredangle (l_1,l_2) = \measuredangle B_3C_3A_3$ . Hence

$\measuredangle A_3B_3C_3 = \measuredangle A_2B_2C_2$ and we are done.

**Posted:** Sun Feb 03, 2008 6:11 am

ArbelosYS · New Member Offline Joined: 09 Feb 2009 Posts: 11

For convenience, let's call A instead of " Yulhee Nam's solution "

Can A be finished solution? I think that she's solution is just a cool Theorem,

but It can't cover all of the solution, isn't it? Huh?

**Posted:** Tue Feb 10, 2009 10:45 pm

livetolove212

From here, we have two impacts:
(1): Given triangle $ABC$ with circumcenter $O$ . If $A_1, B_1, C_1$ lie on $BC, CA, AB$ and $A_2$ is the reflection of $A_1$ through midpoint of $BC$ , similar for $B_2, C_2. M_1, M_2$ is Miquel points of triangle $ABC$ wrt $(A_1,B_1,C_1)$ and $(A_2,B_2,C_2), \angle M_1A_1C = \alpha$ then $OM_1 = OM_2$ and $\angle M_1OM_2 = 180^o - \alpha$ .
(2): Given triangle $ABC$ . A circle $(O) \cap BC = \{A_1, A_2\}, \cap AC = \{B_1,B_2\}, \cap AB = \{C_1,C_2\}$ . Let $M_1, M_2$ be Miquel points of triangle $ABC$ wrt $(A_1,B_1,C_1)$ and $(A_2,B_2,C_2), \angle M_1A_1C = \alpha$ then $OM_1 = OM_2$ and $\angle M_1OM_2 = 180^o - \alpha$ .
Back to our problem:
$\angle B_2A_2C_2 = \angle B_2AC_2 = \angle B_2AB + \angle BCC_2$
$= 180^o - \angle B_2BA - \angle BB_2A + \angle A_2M_1C_2 = C + \angle A_2M_1C_2 - \angle C_1M_1B_2$
$= \angle C + \angle B_2M_1C_2 - \angle C_1M_1A_1 = \angle C + \angle B - 180^o + \angle B_2M_1C_2 = \angle B_2M_1C_2 - \angle...$
On the other side, $\angle C_2A_3B_3 = \angle BM_2C - \angle A$ . So we need to show that $\angle B_2M_1C_2 = \angle BM_2C (*)$
Let $AA_2, BB_2, CC_2$ intersect each other and make triangle $A'B'C', M_1, M_2$ be Miquel points of triangle $ABC$ wrt $(A_1,B_1,C_1)$ and $(A_3,B_3,C_3)$
Put $\angle M_1A_1C = \alpha$
We have $\angle M_1B_2B = \angle BC_1M_1 = \angle M_1A_1C = \angle M_1A_2C$ then $B_2M_1C_2A'$ is cyclic quadrilateral. Similarly we get $M_1$ is Miquel point of triangle $A'B'C'$ wrt $(A_2,B_2,C_2)$
From $(1): OM_1 = OM_2$ and $\angle M_1OM_2 = 180^o - \alpha (3)$
Let $M'_2$ be Miquel point of triangle $A'B'C'$ wrt $(A,B,C)$ . From $(2)$ we obtain $OM_1 = OM'_2$ and $\angle M_1OM'_2 = 180^o - \angle M_1C_2C = 180^o - \alpha (4)$
From $(3)$ and $(4), M'_2\equiv M_2$ so $M_2$ is Miquel point of triangle $A'B'C'$ wrt $(A,B,C)$
$M_2BA'C$ is cyclic thus $\angle BM_2C = 180^o - \angle C'A'B' = \angle B_2M_1C_2$ .
Therefore $(*)$ is true. We are done!

**Posted:** Sat Sep 05, 2009 8:37 pm

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