Given circles $ω_1$ and $ω_2$ intersecting at points $X$ and $Y$, let $ℓ1$ be a line through the center of $ω_1$ intersecting $ω_2$ at points $P$ and $Q$ and let $ℓ_2$ be a line through the center of $ω_2$ intersecting $ω_1$ at points $R$ and $S$. Prove that if $P, Q, R,$ and $S$ lie on a circle then the center of this circle lies on line $XY$.

$O_1,O_2,O_3$ are the correspending centers of circles $\omega_1,\omega_2,\omega_3$.

We want to show that $Pow_{ω_2}(O_3) = Pow_{ω_1}(O_3)$. Let $r_1, r_2, r_3$ denote the raddi of $ω_1$, $ω_2$, and $ω_3$, respectively, where $\omega_3$ is the circle that the points $P, Q, R,$ and $S$ lie on.

Claim: If $O_1 O_3$ bisects $\angle SO_3R$ and $O_2 O_3$ bisects $\angle PO_3Q$ then we're done.

Proof: We have $$Pow_{ω_1}(O_3) = O_1 O_3^2 − r_1^2=O_1S^2$$ and $$Pow_{ω_2}(O_3) = O_2 O_3^2 − r_2^2=O_2Q^2$$ but since $S$ and $Q$ both lie on $\omega_3$ we're done

But how can I show that actually $O_1 O_3$ bisects $\angle SOR$ and $O_2 O_3$ bisects $\angle POQ$

  • 1
    $\begingroup$ There are some unclear notations. What are $O_1,O_2,O_3$? And what is $O$? What is Pow${}_{\omega_1}(O_3)$? Is $S$ on $\omega_1$ with center $O_1$? In this case i would expect $O_1S=r_1$. Please insert a picture. $\endgroup$
    – dan_fulea
    Mar 7, 2022 at 23:46
  • $\begingroup$ I edited the question. For the picture, the construction is tough to draw. @dan_fulea $\endgroup$
    – PNT
    Mar 8, 2022 at 11:57
  • $\begingroup$ There is still some problem for me with the relation $O_1O_3^2-r_1^2=O_1S^2$, since as said above i would expect $O_1S=r_1$, so $O_1O_3^2-r_1^2=r_1^2$ cannot have been the intention. I will drop a solution... $\endgroup$
    – dan_fulea
    Mar 10, 2022 at 18:29

1 Answer 1


Drawing the picture, one gets more insight about the position of the lines $PQ$, $RS$, $XY$, this insight solves the problem.

Here is the pic:

mathematics stackexchange problem 4396011

The statement of the problem is so that points are not introduced in the "right order". But let us resume, to see what we have.

First statement: The following data is given. There are three circles $\omega_1,\omega_2,\omega_3$ with centers in $O_1,O_2,O_3$ respectively.

  • The circles $\omega_1, \omega_2$ intersect in $X,Y$.
  • The circles $\omega_2, \omega_3$ intersect in $P,Q$.
  • The circles $\omega_1, \omega_3$ intersect in $R,S$.

As a coincidence, the point $O_1$ is given to be on the line $PQ=:l_1$, and $O_2$ is given to be on the line $RS=:l_2$. We have to show that $O_3$ is on the third line $XY=:l_3$.

Proof: See the restatement below.


We restate equivalently, so that the given coincidence is a part of a construction, which is a consequence of the given data. So here is an equivalent way to restate. We observe that $O_1O_3$ is the side bisector of the segment $RS$, so in particular it is perpendicular on $RS=l_2$. Similarly, $O_2O_3$ is perpendicular on $PQ=l_1$. We use this information first.

Second (equivalent) statement: The following data is given. There is a triangle $\Delta O_1O_2O_3$, and let $H$ be its orthocenter. We draw the heights (as lines) $l_1,l_2,l_3$ through $H$. Let $\omega_3$ be a circle centered in $O_3$, which (has a sufficiently big radius so that it) intersects $l_1$ in $P,Q$, and $l_2$ in $R,S$.

  • Then $O_1$ is on the side bisector of $RS$, so there is a circle $\omega_1$ centered in $O_1$ passing through $R,S$.
  • And similarly $O_2$ is on the side bisector of $PQ$, so there is a circle $\omega_2$ centered in $O_2$ passing through $P,Q$.

We furthermore assume that $\omega_1,\omega_2$ intersect, let $X,Y$ be the points of intersection.

Then the line $XY$ coincides with the line (height) $l_3$.

Proof: (The same figure above applies.)

See the restatement below.


We remove objects as much as possible from the picture. We need only the triangle $\Delta O_1O_2O_3$ and its heights. Having the height through $O_1$, we do not need both points $P,Q$, one of them is enough. So we keep only one point in each pair. We no longer draw the two circles through $P$, it is enough to know the property relating $P$ and $R$, namely $O_3P=O_3R$. So we restate:

Third (equivalent) statement: In $\Delta O_1O_2O_3$, we project the vertices on opposite sides into $K_1,K_2,K_3$, and let $H=O_1K_1\cap O_2K_2\cap O_3K_3$ be orthocenter.

We pick points $P$ on $O_1K_1$, $R$ on $O_2K_2$, and $T$ on $O_3K_3$, so that two of the three relations are holding: $$ \begin{aligned} O_3R &= O_3P\ ,\\ O_2P &= O_2T\ ,\\ O_1T &= O_1R\ . \end{aligned} $$ Then the third relations is also true.

math stackexchange problem 4366011

Proof of the equivalence of the second and third statements: In the situation from the second statement, keep $\Delta O_1O_2O_3$, and the points $P,R$ on the heights from $O_1,O_2$, with same distance to $O_3$, so the first relation above is satisfied. Construct $T$ on the third height satisfying a second relation among the above three. Then the third is also satisfied. So $T$ is in the intersection $\omega_1\cap\omega_2$, being thus either the point $X$ or the point $Y$ from the second statement. So $X$ or $Y$ is on the $O_3$-height, so from $XY\perp O_1O_2$ the other point is on it, too.


Proof of the third statement: We have $O_3P^2-O_3H^2=(O_3K_1^2+PK_1^2)-(O_3K_1^2+HK_1^2)=PK_1^2-HK_1^2$. Writing similar relations also for $R,T$, we obtain: $$ \begin{aligned} O_3R^2 - O_3P^2 &= (RK_2^2-HK_2^2) - (PK_1^2-HK_1^2)\ ,\\ O_2P^2 - O_2T^2 &= (PK_1^2-HK_1^2) - (TK_3^2-HK_3^2)\ ,\\ O_1T^2 - O_1P^2 &= (TK_3^2-HK_3^2) - (RK_2^2-HK_2^2)\ . \end{aligned} $$ Adding the relations, we obtain zero on the R.H.S. - so if two of the three relations from the statement are true, the third one is also true.

  • $\begingroup$ The solution was given in way the search for it was leading the words. In the given initial situation there was less symmetry, the third version of the problem, as displayed in the last picture, comes with its own symmetry, best one should also use $P_1$, $P_2$, $P_3$ instead of $P,R,T$. There is still a small issue about the existence of $T$, but i will let this untouched here, i hope the idea above is transparent. I would have to rewrite everything, hiding the idea, to cover also this point. $\endgroup$
    – dan_fulea
    Mar 10, 2022 at 18:35

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