Let w be circle passing through PQRS. Let radius of $w_1,w_2,w$ be $r_1,r_2,r$ respectively, center be $O_1,O_2,O$ respectively. Now power of $O_1$ according to circle w is $O_1P$*$O_1Q$=$O_1O^2-r^2$. But $O_1P$*$O_1Q$=$O_1O_2^2-{r_2}^2$, so $O_1O^2=r^2-{r_2}^2+O_1O_2^2$. So the power of O according to $w_1$ is $O_1O^2-{r_1}^2=r^2-{r_2}^2+O_1O_2^2-{r_1}^2$. Similarly, O has same power to $w_2$. So O lies on radical axis XY.

let the centres of the circles be O1,O2,O3. (O3 being the circle around PQRS) let O1 be (x1,y1), O2 be (x2,y2), and O3 (x3,y3) let the radiuses of the three circles be r1,r2,r3. we construct 3 equations, for each circle. by subtracting the equation of O3 from O1, we get the equation of line RS. -(1) by subtracting the equation of O3 from O2, we get the equation of line PQ. -(2) by subtracting the equation of O2 from O1, we get the equation of line XY. -(3) but we have another condition: O1 is on PQ and O2 is on RS. we put in the coordinates of O1 into (1), and the coordinated of O2 into (2). then we add them (or subtract them) we get an equation that says O3 must be on line (3), which is XY. CAN SOMEBODY VERIFY MY SOLUTION?

CatalystOfNostalgia wrote: But the LHS is the power of Z with respect to w_2, while the RHS is the power of Z with respect to w_1. Since these are equal, Z is on the radical axis of w_1 and w_2. nine34 wrote: we construct 3 equations, for each circle. by subtracting the equation of O3 from O1, we get the equation of line RS How would you prove either of these assertions? Thanks for the help.

hmm I thought this was an odd choice for a #1 .. my writeup was done within 15 minutes of the start, but I'm pretty sure I would not have been able to solve this last year.

My general method: Suppose $Z$ was the intersection of $PQ$ and $RS$, and $D$ was the intersection of $XY$ and $O_1O_2$, where $O_1$ and $O_2$ are the centers of $\omega_1$ and $\omega_2$ respectively. Also suppose $C$ is the center of the circle through $PQRS$. Step 1: Prove $C,Z,D$ are collinear. Check. Step 2: Prove $Z$ is on $XY$. Uhhhhhh..... Nope, I had no idea what a radical axis was during the test Too bad, it would've been trivial otherwise. And plus I missed the parallel case like so many other people T-T. So at most a 2 for me on problem 1.

yeah, a easy one, just a radical axis and an orthocentre, solved within 40 minutes.

bzauzmer wrote: nine34 wrote: we construct 3 equations, for each circle. by subtracting the equation of O3 from O1, we get the equation of line RS How would you prove either of these assertions? Thanks for the help. Well, since R and S are both on O3 and O1, by subtracting the equation of O3 from O1, we get a equation with both R and S on them. but this equation is linear, so it is a line, and there is only one line that can go thru R and S. (which is line RS)

Yongyi781 wrote: My general method: And plus I missed the parallel case like so many other people T-T. So at most a 2 for me on problem 1. What was the parallel case?

Let the center of $\omega_1$ be $A$, and the center of $\omega_2$ be $B$. Let the circumcenter of $PQRS$ be $C$. Let $r_1, r_2, r_3$ be the radii of $A, B, C$ respectively. $PQ$ is the radical axis of circles $A$ and $C$, $RS$ is the radical axis of circles $C$ and $B$, and $XY$ is the radical axis of circles $A$ and $B$, because the circles intersect and it is a well known fact that the radical axis of intersecting circles goes through the intersection points. By definition of radical axis, the power of $A$ with respect to circles $B$ and $C$, and similarly for the points $B$ and $C$ with respect to the other 2 circles, we get: $AC^2 - r_3^2 = AB^2 - r_1^2$ $AB^2 - r_2^2 = AC^2 - r_3^2$ Adding the 2 equations gives us: $AC^2 - r_3^2 + AB^2 - r_2^2 = AB^2 - r_1^2 + AC^2 - r_3^2$ $BC^2 - r_2^2 = AC^2 - r_1^2$ Then by defintion,$C$ has equal power with respect to circles $A$ and $B$, so $C$ lies on the radical axis of circles $A$ and $B$, which is $XY$, so we are done. Basically what greentreeroad did, use the powers being equal to find the third circle center having the same power with respect to the other 2 circles. That takes care of the parallel case since the lengths aren't affected if triangle ABC is degenerate, and the powers being equal still holds.

Letting the circle containing $P,Q,R,S$ have center $C$, I proved that $C$ had a power of a point equal with respect to $\omega_1$ and $\omega_2$. But instead of citing radical axes (I only have a vague idea about them, mostly that it is the line $XY$, so I was careful), I said to consider the extension of line $CX$ and its two intersections with $\omega_1$ and $\omega_2$ be $X_1, X_2$, respectively. Proving the power of a point is equal for both circles proves that $CX_1=CX_2$, implying that they are the same point, i.e. $Y$. Then this implies that $C,X,Y$ are collinear. Does this logic work?

Yeah that looks like it works (and it's pretty much how you prove the existence of radical axes [in the case of intersecting circles]).

a problem very easily, I do not think this is the problem USAMO.

math10 wrote: a problem very easily, I do not think this is the problem USAMO. What do you mean? This was problem #1 on the 2009 USAMO.

I feel there is a problem here example : +) problem 1,4 is too easy compared to the year other +)Problem 2 Use the items on the mathematics journals AMM This problem never happened.

nine34's approach is nice as it doesn't require knowledge of the "radical axis" concept and isn't plagued with an insidious parallel case. I realize that the "creative geometers" would balk at said use of analytic geometry, but this one seems pretty straightforward: Let $\omega_1$ have radius $r$ and center $(a,b)$. Let $\omega_2$ have radius $s$ and center $(c,d)$. Let $\omega_3$ with radius $t$ and center $(e,f)$ be the circle containing $P$, $Q$, $R$, and $S$. We can express the equations of the three circles as follows: $$\begin{align*} \omega_1: (x - a)^2 + (y - b)^2 & = r^2 \\ \omega_2: (x - c)^2 + (y - d)^2 & = s^2 \\ \omega_3: (x - e)^2 + (y - f)^2 & = t^2 \end{align*}$$ Rearranging terms a bit, we get $$\begin{align*} \omega_1: x^2 + y^2 & = 2ax + 2by + r^2 - a^2 - b^2 \\ \omega_2: x^2 + y^2 & = 2cx + 2dy + s^2 - c^2 - d^2 \\ \omega_3: x^2 + y^2 & = 2ex + 2fy + t^2 - e^2 - f^2 \end{align*}$$ Now consider the points $P$ and $Q$. We know that these must necessarily satisfy both $\omega_2$ and $\omega_3$. Thus, by subtracting equations, we see that $P$ and $Q$ must necessarily satisfy $2cx + 2dy + s^2 - c^2 - d^2 = 2ex + 2fy + t^2 - e^2 - f^2$. Note that this is the equation of a line unless $(c,d) = (e,f)$ and $r = s$. In the special case where $(c,d)=(e,f)$ and $r=s$, we know that $R$ and $S$ must simultaneously be on $\omega_1$ and $\omega_2$ and must be collinear with the center of $\omega_2$. Thus $R$ and $S$ are coincident with $X$ and $Y$, and $XY$ forms a diameter of $\omega_2$. Thus, in this special case, $(e,f)$ is on $XY$ (since $XY$ forms a diameter of the coincident $\omega_2$ and $\omega_3$). Aside from the special case, since two distinct point define a line, we have the equation of the line $PQ$ as $2cx + 2dy + s^2 - c^2 - d^2 = 2ex + 2fy + t^2 - e^2 - f^2$. We can employ a similar argument to find the equations of the lines $RS$ (simultaneously satisfying $\omega_1$ and $\omega_3$) and $XY$ (simultaneously satisfying $\omega_1$ and $\omega_2$). We then have: $$\begin{align*} PQ: 2cx + 2dy + s^2 - c^2 - d^2 & = 2ex + 2fy + t^2 - e^2 - f^2 \\ RS: 2ax + 2by + r^2 - a^2 - b^2 & = 2ex + 2fy + t^2 - e^2 - f^2 \\ XY: 2ax + 2by + r^2 - a^2 - b^2 & = 2cx + 2dy + s^2 - c^2 - d^2 \end{align*}$$ Now, finally, we note that $(a,b)$ is on $PQ$ and $(c,d)$ is on $RS$. So we have the simultaneous equations: $$\begin{align*} 2ac + 2bd + s^2 - c^2 - d^2 & = 2ae + 2bf + t^2 - e^2 - f^2 \\ 2ac + 2bd + r^2 - a^2 - b^2 & = 2ce + 2df + t^2 - e^2 - f^2 \end{align*}$$ We subtract the two equations and simplify to find that (e,f) satisfies the equation for line XY: $$\begin{align*} s^2 - c^2 - d^2 - (r^2 - a^2 - b^2) & = 2ae + 2bf - (2ce + 2df) \\ 2ae + 2bf + r^2 - a^2 - b^2 & = 2ce + 2df + s^2 - c^2 - d^2 \end{align*}$$ Nice work nine34. P.S. IMHO, missing the parallel case is a much bigger deal than a 1 point deduction that many people are assuming. Its solution is not a trivial extension of the power-of-a-point argument.

Your analytical solution is equivalent to greentreeroad's pythagorean-based solution (see post #3). (I found this solution during the test but decided to write up with CatalystOfNostalgia's orthocenter-based solution...I can only hope the grader thinks in a projective plane now )

How does the orthocenter solution work in the projective plane?

Hey guys, for karma's sake I don't want to take credit for the solution I posted, it was shown to me by someone at my school (I didn't actually solve the problem during the test ). On another note, I still don't see how it's possible for the three centers can be collinear, can someone explain this? I'm probably missing something obvious. [edit] never mind.

Let $T = PQ \cap RS$, $(PRQS) = \omega_3$, the center of $\omega_3$ be $O_3$, and $K = XY \cap O_1O_2$. We first assume $O_1$, $O_2$, and $O_3$ are non-collinear. Notice $$Pow_{\omega_1}(T) = TR \cdot TS = Pow_{\omega_3}(T) = TP \cdot TQ = Pow_{\omega_2}(T) \implies T \in XY$$ (the Radical Axis of $\omega_1$ and $\omega_2$). Since $RS$ is the Radical Axis of $\omega_1$ and $\omega_3$, and $PQ$ is the Radical Axis of $\omega_2$ and $\omega_3$, we know the intersection of the perpendicular from $O_1$ to $RS$ and the perpendicular from $O_2$ to $PQ$ is $O_3$. Looking at $\triangle O_1O_2O_3$, it's obvious that $T$ is its orthocenter. Now, let the foot of the altitude from $O_3$ to $O_1O_2$ be $K'$. Then, $\angle TK'O_1 = 90^{\circ}$. Using Radical Axis properties, however, we know $\angle TKO_1 = 90^{\circ}$. Thus, $K = K'$ because they both lie on $O_1O_2$. Hence, $O_3, T, K$ are collinear or $O_3 \in XY$ as desired. If $O_1$, $O_2$, and $O_3$ are collinear, then we know $PQ \parallel RS \parallel XY$. Claim: If $O_1$, $O_2$, and $O_3$ are collinear, then $O_3 = K = XY \cap O_1O_2$. Proof. The claim is equivalent to showing $KP = KQ = KR = KS$ (in this configuration). This is trivial via the Pythagorean Theorem, however, so we're done. $\square$ $\blacksquare$ Remark: In order for the problem to hold when the centers are collinear, $O_3 \in O_1O_2$ and $O_3 \in XY$ has to hold, implying $O_3 = XY \cap O_1O_2 = K$ must be true, motivating the last claim.

Let the center of the circle $\omega$ passing through $P,Q,R,S$ be $O$, and let the length of its radius be $r$. Let the center of $\omega_1$ be $O_1$ and let the center of $\omega_2$ be $O_2$, and let the lengths of their radii be $r_1,r_2$ respectively. It suffices to show that $O_1O^2-r_1^2=O_2O^2-r_2^2$. But $$O_1O^2-r^2=O_1P\cdot O_1Q=O_1O_2^2-r_2^2$$ $$O_2O^2-r^2=O_2R\cdot O_2S=O_1O_2^2-r_1^2$$ so we are done.

By Radical Center theorem we know O1Q and O2S and XY meet at single point Z. Let O3 be center of SPRQ. we have O1O3 ⊥ SR and O2O3 ⊥ PQ and O1O2 ⊥ XY so Z is orthocenter of O1O2O3 and as it lies on XY we have O3 lies on XY as well. we're Done.

Let $\omega_3$ and $O_3$ be $(PQRS)$ and its center, respectively. Notice $$O_1O_2^2-r_2^2=\textit{pow}_{\omega_2}(O_1)=\textit{pow}_{\omega_3}(O_1)=O_1O_3^2-r_3^2$$ and $$O_2O_3^2-r_3^2=\textit{pow}_{\omega_3}(O_2)=\textit{pow}_{\omega_1}(O_2)=O_2O_1^2-r_1^2.$$ Adding yields $$\textit{pow}_{\omega_2}(O_3)=O_2O_3^2-r_2^2=O_1O_3^2-r_1^2=\textit{pow}_{\omega_1}(O_3)$$ so $O_3$ lies on $\overline{XY}.$ $\square$

imagine doing amo problems "for fun" orz

Let $O_1, r_1$ and $O_2, r_2$ be the centers and radii of $w_1$ and $w_2$, respectively. Also, let $w_3$ be the circle containing $P, Q, R, S$ with a center at $O_3$ and a radius of $r_3$. Since $O_1$ is on the radical axis of $w_2$ and $w_3$, we have: $$Pow_{w_2}(O_1)=Pow_{w_3}(O_1)$$ $$O_1O_2^2-r_2^2=O_1O_3^2-r_3^2$$ Similarly, since $O_2$ lies on the radical axis of $w_1$ and $w_3$: $$Pow_{w_1}(O_2)=Pow_{w_3}(O_2)$$ $$O_1O_2^2-r_1^2=O_2O_3^2-r_3^2$$ Subtracting these two equations gives the following equation, which we continue to simplify. $$(O_1O_2^2-r_2^2)-(O_1O_2^2-r_1^2)=(O_1O_3^2-r_3^2)-(O_2O_3^2-r_3^2)$$ $$r_1^2-r_2^2=O_1O_3^2-O_2O_3^2$$ $$O_2O_3^2-r_2^2=O_1O_3^2-r_1^2$$ $$Pow_{w_2}(O_3)=Pow_{w_1}(O_3)$$ Therefore, $O_3$ lies on the radical axis (line $XY$) of $w_1$ and $w_2$. $\square$

Let $O_1, O_2$ be the centers of the circles $w_1,$ and $w_2$ with radius $r_1$ and $r_2$ respectively. Furthermore, let $O_3$ be the center of the circle $w_3$ passing through $P, Q, R,$ and $S$ with radius $r_3$. Then, the problem is equivalent to proving that $P(O_3, w_1) = P(O_3, w_2) \Leftrightarrow O_3O_1^2 - r_1^2 = O_3O_2^2 - r_2^2$. Note that $P(O_1, w_2) = P(O_1, w_3) \implies O_1O_3^2 -r_3^2 = O_1O_2^2 - r_2^2$ and $P(O_2, w_1) = P(O_2, w_3) \implies O_2O_3^2 - r_3^2 = O_2O_2^2 - r_1^2$. Subtracting the two equations gives $O_1O_3^2 - O_2O_3^2 = r_1^2 - r_2^2 \implies O_3O_1^2 - r_1^2 = O_3O_2^2 - r_2^2$ which is what we wanted to prove.

Let $(PQRS)=\omega_3$ and let $r_i$ and $O_i$ denote the radius and center of circle $\omega_i$ for $i\in[1,2,3]$. Since $O_1$ is on the radical axis of $O_2$ and $O_3$ we have $$\text{Pow}_{\omega_2}(O_1)=\text{Pow}_{\omega_3}(O_1) \iff O_1O_2^2-r_2^2=O_1O_3^2-r_3^2.$$ Since $O_2$ is on the radical axis of $O_1$ and $O_3$ we have $$\text{Pow}_{\omega_1}(O_2)=\text{Pow}_{\omega_3}(O_2) \iff O_2O_1^2-r_1^2=O_2O_3^2-r_3^2.$$ Subtracting yields $$r_1^2-r_2^2=O_1O_3^2-O_2O_3^2 \iff \text{Pow}_{\omega_1}(O_3)=\text{Pow}_{\omega_2}(O_3)$$ and we are done.

Let $O_1,O_2,O_3$ be the centers of $\omega_1,\omega_2$, and $(PQRS)$. Let $C$ be the intersection of $PQ$ and $RS$. Note that $C$ also lies on $XY$ since the pairwise radical axes of the three circles are concurrent. Now, note that $O_3$ lies on the perpendicular bisector of both $PQ$ and $RS$. Thus, we have that $O_1O_3$ is perpendicular to $RS$, and $O_2O_3$ is perpendicular to $PQ$. Therefore, $C$ is the orthocenter of $\triangle O_1O_2O_3$. This means that $O_3C$ is perpendicular to $O_1O_2$. Since $C$ lies on $XY$, we are done.

:unamused: Denote $\omega_3:=(PQRS)$. Because $O_1\in PQ$, $$\text{Pow}_{\omega_2}(O_1)=\text{Pow}_{\omega_3}(O_1)\implies O_1O_2^2-r_2^2=O_1O_3^2-r_3^2;$$ and since $O_2\in RS$, $$\text{Pow}_{\omega_1}(O_2)=\text{Pow}_{\omega_3}(O_2)\implies O_1O_2^2-r_1^2=O_2O_3^2-r_3^2.$$ Subtracting, $$r_1^2-r_2^2=O_1O_3^2-O_2O_3^2\implies O_1O_3^2-r_1^2=O_2O_3^2-r_2^2,$$ i.e., $\text{Pow}_{\omega_1}(O_3)=\text{Pow}_{\omega_2}(O_3)$ or $O_3\in XY$, as desired. $\square$

Let $M=\overline{SR} \cap \overline{PQ}$. Let $O_1,O_2,O_3$ be the centers of $\omega_1,\omega_3$ and $(SPRQ)$. Now, it is well known that the radical axis of two circles is perpendicular to the line joining their centers. Then, $SR \perp O_1,O_2$ and $PQ \perp O_2O_3$. This means that $M$ must be the orthocenter of $\triangle O_1O_2O_3$. This means, $O_3M\perp O_1O_2$ as well. Clearly $M$ lies on $XY$ ($MP\cdot MQ = MR\cdot MS$ so it lies on the radical axis of $\omega_1$ and $\omega_2$). But then, $XY$ is the line passing through $M$ and perpendicular to $O_1O_2$. This means that indeed $O_3$ lies on the line $XY$ and we are done.

Let $O_1$ be the center of $\omega_1$ and let $O_2$ be the center of $\omega_3$. Denote the center of $(PRQS)$ to be $O_3$ and let the intersection of $PQ$ and $SR$ be $C$. Now note that $SR$, $PQ$, and $XY$ are the three pairwise radical axes of $\omega_1$, $\omega_2$, and $(PRQS)$. Denote the length of $O_1O_2$ to be $a$ and let the radii of $\omega_1$, $\omega_2$, and $(PRQS)$ be $r_1$, $r_2$, and $r_3$, respectively. Since $O_2$ is on the line $PQ$, by the Power of a Point definition of the radical axis, we have that $$a^2-r_1^2=O_2O_1^2-r_1^2=Pow_{\omega_1}(O_2)=Pow_{\omega_3}(O_1)=O_2O_3^2-r_3^2,$$ which gives that $O_2O_3=\sqrt{a^2-r_1^2+r_3^2}$. Similarly, since $O_1$ is on $RS$, using Power of a Point on $O_1$ with respect to $O_2$ and $O_3$, we get that $O_1O_3=\sqrt{a^2-r_2^2+r_3^2}$. Finally, using these lengths and Power of a Point one last time, $$Pow_{\omega_1}(O_3)=O_1O_3^2-r_1^2=a^2+r_3^2-(r_1^2+r_2^2)=O_1O_2^2-r_2^2=Pow_{\omega_2}(O_3),$$ which implies that $O_3$ is on the radical axis of $\omega_1$ and $\omega_2$, which is $XY$, finishing the problem.

By radical axis, the problem basically asks us to prove that for any three circles $\omega_1, \omega_2, \omega_3$ then $$\text{Pow}(O_1, \omega_2) = \text{Pow}(O_1, \omega_3), \text{Pow}(O_2, \omega_1) = \text{Pow}(O_2, \omega_3) \implies \text{Pow}(O_3, \omega_1) = \text{Pow}(O_3, \omega_2)$$ But the first two equalities give us $$(O_1O_2)^2 - r_2^2 = (O_1O_3)^2 - r_3^2$$ $$(O_1O_2)^2 - r_1^2 = (O_2O_3)^2 - r_3^2$$ and subtracting equations gives the desired result.