I am in the middle of solving a problem and stuck at the following link:-

Below is a standard diagram showing three circles [namely, $\omega=(ABCD) \; (CDPG) \;and\; (BCGQ)$] intersecting each other so that DCQ and BCP are straight lines. Let M be the midpoint of PQ. MC produced cuts $\omega$ at R.

enter image description here

If PR cuts $\omega$ at L, prove that L, D, G are collinear. Or equivalently, if we assume that PR and GD produced meet at L, prove that L is a con-cyclic point of $\omega$.


Lemma 1: Given a pair of circles $c_1$ and $c_2$ there is a unique special line $l_{12}$, called the radical axis of $c_1$ and $c_2$, with the property that the center of any circle that is simultaneously perpendicular to both circles $c_1$ and $c_2$ lies on $l_{12}$. Conversely, any circle whose center is on the radical axis $l_{12}$ and is perpendicular to one of the circles $c_1$ or $c_2$ is necessarily perpendicular to the other circle too. The radical axis $l_{12}$ is always perpendicular to the line connecting the centers of $c_1$ and $c_2$. Moreover, if $c_1$ and $c_2$ intersect, the radical axis $l_{12}$ passes through the intersection points of $c_1$ and $c_2$.

Lemma 2: Let $c_0$ be a circle with center $O_0$ and $c$ be another circle with center $O$ so that the center $O_0$ of $c_0$ lies on the circle $c$. Then the inversion of circle $c$ with respect to $c_0$ is a line $l$ perpendicular to the line $OO_0$ and $l$ is in fact the radical axis of the pair of circles $c_0$ and $c$.

Radical Axis Theorem: Given three circles, every pair of circles has a unique radical axis. Since there are three pairs of circles, there are three radical axes. These radical axes intersect in a common point.

Solution: Draw the circles $c_P$ and $c_Q$ such that they are centered at the points $P$ and $Q$ respectively and are perpendicular to the circle $\omega = c(ABCD)$.

enter image description here

Denote by $T_1$ and $T_2$ the intersection points of circle $c_P$ with $\omega$. Consider the three circles $\omega, \, c_P$ and $c(CDPG)$. They form three pairs and there is one radical axis for each pair. By Lemma 1 the radical axes are $$l\big(\omega, \, c(CDPG)\big) = CD, \,\,\,\, l(\omega, \, c_P) = T_1T_2, \,\,\,\, l\big(c_P, \, c(CDPG)\big) = AB \,\,\, \text{ by Lemma 2}$$ Now, by the Radical Axis Theorem, the three radical axes $CD, \, T_1T_2$ and $AB$ intersect in a common point, which is point $Q$. This, in particular implies that $Q$ lies on the radical axis $T_1 T_2$. Now, by Lemma 1, since the center $Q$ of $c_Q$ lies on the radical axis $T_1T_2$ and by construction $c_Q$ is perpendicular to $\omega$, the circle $c_Q$ is perpendicular to the circle $c_P$. This means that if $E_1$ and $E_2$ are the intersection points of $c_P$ and $c_Q$ then $\angle \, PE_1Q = PE_2Q = 90^{\circ}$. Therefore, if $c_M$ is the circle with center $M$ and diameter $PQ$, it passes also through the points $E_1$ and $E_2$. However, by Lemma 1, $E_1E_2$ is the radical axis of $c_P$ and $c_Q$ and again by Lemma 1, $E_1E_2$ is in fact the radical axis of, for example, $c_P$ and $c_M$. Thus, by Lemma 2, since $\omega$ is perpendicular to $c_P$ it must be also perpendicular to $c_M$.

Perform inversion in circle $c_M$. As $\omega$ is perpendicular to $c_M$, it is inverted into itself and moreover point $C$ is mapped to point $R$. By the properties of inversion $$MC \cdot MR = MP^2$$ which translates into $$\frac{MC}{MP} = \frac{MP}{MR}$$ which combined with $\angle \, CMP = \angle \, PMR$ shows that triangles $\Delta\,CMP$ and $\Delta \, PMR$ are similar and hence $$\angle \, CPM = \angle \, PRM = \alpha$$ Furthermore, by assumption, quad $ABCR$ is inscribed in circle $\omega$, hence $$\angle \, ARM = \angle \, ARC = 180^{\circ} - \angle \, ABC = \angle \, QBC = \angle \, QBP = \beta$$ In triangle $\Delta \, BQP$ $$\angle \, BQP = 180^{\circ} - \angle \, BPQ - \angle \, QBP = 180^{\circ} - \alpha - \beta$$ Thus in quad $AQPR$ $$\angle \, PRA + \angle \, AQP = (\angle \, PRM + \angle \, ARM) + \angle \, BQP = (\alpha + \beta) + (180^{\circ} - \alpha - \beta) = 180^{\circ}$$ The quad $AQPR$ is inscribed in a circle.

Because of the perpendicularity between $c_P$ and $\omega$, when inverted in circle $c_P$, point $A$ is mapped to point $D$, point $C$ is mapped to point $B$ and point $R$ is mapped to point $L$. By Lemma 2, circle $c(CDPG)$ is mapped to the line $AB$ and since $G$ is from $c(CDPG) \, \cap \, PQ$ and $Q$ is on $AB$, point $G$ is inverted into point $Q$. By inversion in $c_P$ and Lemma 2, since the circumcircle $c(AQPR)$ of quad $AQPR$ passes through the center $P$ of $c_P$, it is inverted into a straight line that contains the inverse images of points $R, \, A, \, Q$, which happen to be $L, \, D, \, G$. Therefore the points $L, \, D, \, G$ are collinear.

Remark (outside the scope of the problem): The segment $OG$ is perpendicular to $PQ$ because: first of all, the point $O$ lies on the radical axis $E_1E_2$ of the three circles $c_P, c_Q$ and $c_M$ due to the fact that $\omega$ is perpendicular to all three of them. Then, after inversion in $c_P$, the line $PQ$ is mapped to itself and the circle $c_M$ is mapped to the straight line $E_1E_2$ (Lemma 1) and $E_1E_2$ is perpendicular to $PQ$ (because of Lemma 1 or also because $c_M \, \perp \, PQ$ and so should be its image $E_1E_2$). However, we have already established that $Q$ is inverted into $G$, i.e. $G$ also lies on $E_1E_2$ so $E_1E_2$ intersects $PQ$ in $G$ and therefore $OG \equiv E_1E_2$ is perpendicular to $PQ$.

  • $\begingroup$ Your explanation is detailed but is a bit hard to visualize. Can you supply a picture about it? $\endgroup$ – Mick Mar 27 at 16:39
  • $\begingroup$ @Mick I posted a picture. $\endgroup$ – Futurologist Mar 29 at 0:37
  • $\begingroup$ Thanks. wWll study it. $\endgroup$ – Mick Mar 29 at 19:04

Below, there is a longer story, it became long as i had to change notation, and explain the new, alternative notation. It would be impossible to use barycentric coordintes w.r.t. $\Delta APQ$ from the OP, without changing notations so that it becomes $\Delta ABC$. Also, i had to change the "mess" of rather randomly chosen letters for the many involved points, so that similar triangles get similar notations. Else typing would be an adventure. This is conform to one of the first principles i learned from my old school teachers, often, the notation is part of finding a solution and having comfort while doing it.

Since i plan to use this example as a link, and since barycentric computations may be an effective, almost algorithmic weapon in Olimpiad contents, details are provided also for the "linked audience". A good reference is:

Barycentric Coordinates for the Impatient, Evan Chen and Max Schindler

First of all, it would be nice to have a construction point after point of the given configuration. Instead of having points as in a figure, given in the same time, so that some conditions are satisfied. So let us analyze the given situation, then restate. We have

  • $\widehat{PGD}=\widehat{PCD}=180^\circ-\widehat{BCD}=\widehat{BAD}=\hat A$,
  • $\widehat{QGB}=\widehat{QCB}=180^\circ-\widehat{BCD}=\widehat{BAD}=\hat A$,

making the three triangles similar: $$ \Delta APQ\sim \Delta GPD\sim \Delta GBQ\ . $$ math stackexchange 4076259

So it is natural to construct the given situation by starting with $\Delta APQ$, then considering an arbitrary point $G$ on $PQ$, then building in $\Delta APQ$ the "anti-parallels" $GB$, $GD$ through $G$ for the two sides in $A$, considering the existing intersection $C$ of the circles $\omega:=(ABD)$, $(PGD)$, $(QGB)$, and lines $PB$, $QD$, then showing $L=\omega\cap RP\cap GD$. There is also a cousin point, $K=\omega\cap RQ\cap GB$. As a bonus properties, $R$ is on the circle $(APQ)$.

The simplest solution i saw first is using barycentric coordinates. This comes with the need of an adapted notation, so that standard notations are applied in a natural way. I was anyway very unhappy with the given notation which uses random letters and denies the knowledge of the symmetries in the picture, below we restate, keeping notations as much as possible, and then give proofs.

Alternative notations: Please use the parallel between the pictures to have the translation dictionary.

math stackexchange 4076259 restated

Restating with the above notations, and introducing objects one by one:

Claim: Let $\Delta ABC$ be a triangle. Consider on the line $BC$ the mid point $M$, and a further point $A'$. Construct the "antiparallels" $A'B'$, $A'C'$ to the sides $AB$, $AC$ as in the picture, i.e. $B'\in AC$, $C'\in AB$, and the angles are equal (in measure): $$ \widehat{BAC}=\widehat{BA'C'}=\widehat{B'A'C} \ . $$

Then $(1)$ the circles $(AB'C')$, $(A'BC')$, $(A'B'C)$ and the lines $BB'$, $CC'$ intersect in a point. Let this point be denoted $X$.

We construct $R$ as the second intersection of $MX$ with the circle $\omega:=(AB'XC')$.

Then $(2)$ we have:

  • $(2.1)$ The lines $RB$, $A'C'$, and the circle $\omega$ intersect in a point, denoted by $L$.
  • $(2.2)$ The lines $RC$, $A'B'$, and the circle $\omega$ intersect in a point, denoted by $K$.
  • $(2.3)$ Bonus: $R$ is a point on the circumcenter $(ABC)$ of the triangle we were starting with.
  • $(2.4)$ Bonus: $R$ is a point on the circle $(AA'M)$.


$(1)$ Let $X$ be the intersection of the circles $(BA'C')$ and $(CA'B')$. We compute the angle $$ \begin{aligned} \widehat{BXB'} &= \widehat{BXA'}+ \widehat{A'XC}+ \widehat{CXB'} \\ &= \widehat{BC'A'}+ \widehat{A'B'C}+ \widehat{CA'B'} \\ &= \hat C + \hat B+\hat A=180^\circ\ , \end{aligned} $$ so $B,X,B'$ collinear. By symmetry, the same holds for $C,X,C'$. Moreover, from $$ \widehat{BAC}= \widehat{BA'C'}= \widehat{BXC'}= 180^\circ-\widehat{C'XB'} $$ we see that $AB'XC'$ is inscriptible. This solves $(1)$. Now we finally in position to attack the OP.


We use barycentric coordinates. The given constellation is determined by $\Delta ABC$ with sides $a,b,c$ as usual, and the position of $A'$ on $BC$. We use weights $s,t$ with $s+t=1$ so that $A'=(0,s,t)$. Then we compute, and algorithmically solve: $$ \begin{aligned} A &=(1,0,0)\ ,\\ B &=(0,1,0)\ ,\\ C &=(0,0,1)\ ,\\ M &=\left(0,\frac 12,\frac12\right)=[0:1:1]\ ,\\ A'&=(0,s,t)\ ,\qquad\text{ and }A'B=ta\ ,\ A'C=sa\ ,\\ B'&=[sa^2:0:b^2-sa^2]\ ,\qquad\text{ from }\frac{B'C}{sa}=\frac{B'C}{A'C}=\frac{BC}{AC}=\frac ab\ , \\ C'&=[ta^2:c^2-ta^2:0]\ ,\\ \color{blue}{ X:=BB'\cap CC' } & \color{blue}{ =[sta^2\; :\; sc^2-sta^2\; : \; tb^2-sta^2] } \\[3mm] (AB'C')\ :\ 0&=-a^2yz-b^2zx-c^2xy+a^2(x+y+z)(ty+sz)\ , \\ \color{gray}{(BA'C')\ :\ 0} & \color{gray}{ =-a^2yz-b^2zx-c^2xy+(x+y+z)(u_Bx+w_Bz)\ , } \\ &\color{gray}{\qquad\text{ with } u_B=c^2 -ta^2\ ,\ w_B= sa^2\ ,} \\ \color{gray}{(CA'B')\ :\ 0} & \color{gray}{=-a^2yz-b^2zx-c^2xy+(x+y+z)(u_Cx+v_Cy)\ , } \\ & \color{gray}{\qquad\text{ with } u_C=b^2 -sa^2\ ,\ v_C= ta^2\ .} \end{aligned} $$ (The equations of the last two circles is not needed, but shown, so that we can ponder the amount of a possible barycentric solution to $(1)$ obtained by checking them for the above coordinates of $X$.)

Now let us compute the intersection $R$ of the circle $(AB'C')$, the circumcircle $(ABC)$ with the line $MX$. We are solving for $$ \left\{ \begin{aligned} 0&= \begin{vmatrix} 0&1&1\\ sta^2 & sc^2-sta^2 & tb^2-sta^2\\ x&y&z \end{vmatrix} &:&\ (MX)\ , \\ 0&=-a^2yz-b^2zx-c^2xy &:&\ (ABC)\ , \\ 0&=-a^2yz-b^2zx-c^2xy + (ty+sz)(x+y+z) &:&\ (AB'C')\ , \\ 1&=x+y+z &:&\text{ weights norming.} \end{aligned} \right. $$ We solve, then forget about norming (and the involved denominators), obtaining: $$ \color{blue}{ R=\Big[\ sta^2\ :\ s^2c^2-stb^2\ :\ t^2b^2-stc^2\ \Big]\ . } $$ It is easy to check for the above point, now that we have it, that the displayed homogeneous equations are satisfied. In particular, we also record that $R$ is on the "simple line" $ty+sz=0$. For instance, up to some norming factor in $\Bbb Q^\times$: $$ \begin{aligned} &\begin{vmatrix} 0&1&1\\ sta^2 & sc^2-sta^2 & tb^2-sta^2\\ x_R&y_R&z_R \end{vmatrix} \overset{\Bbb Q^\times} \sim \begin{vmatrix} 0&1&\boxed1\\ sta^2 & sc^2-sta^2 & tb^2-sta^2\\ sta^2 & s^2c^2-stb^2 & t^2b^2-stc^2 \end{vmatrix} \\ &\qquad\overset{\Bbb Q^\times}\sim \begin{vmatrix} 0&0&1\\ 1 & sc^2-tb^2 & *\\ 1 & (s^2+st)c^2-(t^2+st)b^2 & * \end{vmatrix} = \begin{vmatrix} 1 & sc^2-tb^2 \\ 1 & sc^2-tb^2 & \end{vmatrix} \qquad\text{ by using }s+t=1 \\ &\qquad=0\ , \\[2mm] &ty_R+sz_R \\ &\qquad\sim t(s^2c^2-stb^2 ) + s(t^2b^2-stc^2 )=0\ , \\[2mm] &a^2y_Rz_R+b^2z_Rx_R+c^2x_Ry_R \\ &\qquad= a^2(s^2c^2-stb^2)(t^2b^2-stc^2) \\ &\qquad\qquad\qquad +sta^2\Big(\ b^2(t^2b^2-stc^2)\ +\ c^2(s^2c^2-stb^2)\ \Big) \\ &\qquad\overset{st\;a^2}{\ \sim\ } (sc^2-tb^2)(tb^2-sc^2) +\Big(\ t^2b^4-2stb^2c^2+ s^2c^4\ \Big) \\ &\qquad=0\ . \end{aligned} $$ Now we compute the coordinates $(x_L,y_L,z_L)$ of the point $L$ by solving the system in $(x,y,z)$ given by the norming condition $x+y+z=1$, and by the line equations, expressed as a vanishing of two determinants: $$ 0 = \begin{vmatrix} 0&1&0\\ sta^2 & s^2c^2-stb^2 & t^2b^2-stc^2 \\ x&y&z \end{vmatrix} = \begin{vmatrix} 0&s&t\\ ta^2 & c^2-ta^2 & 0\\ x&y&z \end{vmatrix} \ . $$ We solve, and also transpose the result for $K$: $$ \color{blue}{ \begin{aligned} L=\Big[\ sta^2\ :\ st(b^2+c^2-a^2)\ :\ t^2b^2-stc^2\ \Big]\ , \\ K=\Big[\ sta^2\ :\ s^2c^2-stb^2\ :\ st(b^2+c^2-a^2)\ \Big]\ . \end{aligned} } $$ Plugging in the above components of $L$ in the two determinants leads easily to two zero results. For instance: $$ \begin{vmatrix} 0&s&t\\ sta^2 & sc^2 - sta^2 & 0\\ sta^2 & st(b^2+c^2-a^2) & t^2b^2-stc^2 \end{vmatrix} \sim \begin{vmatrix} 0&s&t\\ 1 & sc^2 - sta^2 & 0\\ 1 & st(b^2+c^2-a^2) & t^2b^2-stc^2 \end{vmatrix} \sim \begin{vmatrix} 0 & s & t\\ 1 & * & *\\ 0 & stb^2+(st-s)c^2 & t^2b^2-stc^2 \end{vmatrix} \sim \begin{vmatrix} s & t\\ stb^2-s^2c^2 & t^2b^2-stc^2 \end{vmatrix} =0\ . $$

It remains for the OP to check that $L$ is on $\omega=(AB'C')$. We use components $x_L,y_L,z_L$ of $L$ obtained from the above after elimination of the factor $t$, and have: $$ \begin{aligned} & a^2y_Lz_L + b^2z_Lx_L + c^2x_Ly_L \\ &\qquad= a^2\; s(b^2+c^2-a^2)\;(tb^2-sc^2) + b^2\;(tb^2-sc^2)\; sa^2 + c^2\;sa^2\;s(b^2+c^2-a^2) \\ &\qquad =sa^2 \Big(\ (b^2+c^2-a^2)(tb^2-\color{green}{sc^2}) + b^2(tb^2-sc^2) + \color{green}{sc^2}(b^2+c^2-a^2)\ \Big) \\ &\qquad= sa^2 \Big(\ (b^2+c^2-a^2)tb^2 + b^2(tb^2-sc^2) \ \Big) \\ &\qquad= sa^2b^2 \Big(\ t(2b^2+c^2-a^2) -sc^2 \ \Big) \ , \\ &x_L+y_L+z_L \\ &\qquad= sa^2\ +\ s(b^2+c^2-a^2)\ +\ tb^2-sc^2\ \\ &\qquad= s(b^2+c^2)\ +\ tb^2-sc^2\ \\ &\qquad= sb^2\ +\ tb^2 \\ &\qquad= b^2\ , \\ &ty_L+sz_L \\ &\qquad= st(b^2+c^2-a^2)\ +\ s(tb^2-sc^2) = st(2b^2+c^2-a^2)\ +\ -s^2c^2\ . \end{aligned} $$ And the homogeneous equation defining the circle $(AB'C')$ is satisfied. This finishes the OP, i.e. $(2.1)$ and $(2.2)$, and also the bonus $(2.3)$.


We still have to deal with the bonus point $(2.4)$. The equation of the circle $(AA'M)$ is $$ (AA'M)\ :\qquad 0=-a^2yz-b^2zx-c^2xy +\frac 12 a^2(x+y+z)(ty+sz)\ , $$ since the points $A=(1,0,0)$, $A'=(0,s,t)$, $M=[0:1:1]$ satisfy it. It remains to check that $R$ is also satisfying. Since $R$ is on $(ABC)$, the part $-a^2yz-b^2zx-c^2xy$ vanishes when plugging in its coordinates. We have also seen that $R$ is on the "simple line" $ty+sz=0$.



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