Given circles ω1 and ω2 intersecting at points X and Y, let ℓ1 be a line through the center of ω1 intersecting ω2 at points P and Q and let ℓ2 be a line through the center of ω2 intersecting ω1 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.
Problem
Source: 2009 USAMO problem 1
Tags: geometry, analytic geometry, circumcircle, symmetry, parameterization, USAMO, 2009 USAMO
30.04.2009 22:14
30.04.2009 22:18
Let w be circle passing through PQRS. Let radius of w1,w2,w be r1,r2,r respectively, center be O1,O2,O respectively. Now power of O1 according to circle w is O1P*O1Q=O1O2−r2. But O1P*O1Q=O1O22−r22, so O1O2=r2−r22+O1O22. So the power of O according to w1 is O1O2−r12=r2−r22+O1O22−r12. Similarly, O has same power to w2. So O lies on radical axis XY.
01.05.2009 00:38
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?
01.05.2009 01:33
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.
01.05.2009 01:34
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.
01.05.2009 01:37
My general method: Suppose Z was the intersection of PQ and RS, and D was the intersection of XY and O1O2, where O1 and O2 are the centers of ω1 and ω2 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.
01.05.2009 01:53
yeah, a easy one, just a radical axis and an orthocentre, solved within 40 minutes.
01.05.2009 01:56
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)
01.05.2009 02:50
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?
01.05.2009 03:00
Let the center of ω1 be A, and the center of ω2 be B. Let the circumcenter of PQRS be C. Let r1,r2,r3 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: AC2−r23=AB2−r21 AB2−r22=AC2−r23 Adding the 2 equations gives us: AC2−r23+AB2−r22=AB2−r21+AC2−r23 BC2−r22=AC2−r21 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.
01.05.2009 03:44
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 ω1 and ω2. 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 ω1 and ω2 be X1,X2, respectively. Proving the power of a point is equal for both circles proves that CX1=CX2, implying that they are the same point, i.e. Y. Then this implies that C,X,Y are collinear. Does this logic work?
01.05.2009 09:03
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]).
01.05.2009 09:51
a problem very easily, I do not think this is the problem USAMO.
01.05.2009 22:25
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.
02.05.2009 08:45
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.
02.05.2009 18:32
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 ω1 have radius r and center (a,b). Let ω2 have radius s and center (c,d). Let ω3 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: ω1:(x−a)2+(y−b)2=r2ω2:(x−c)2+(y−d)2=s2ω3:(x−e)2+(y−f)2=t2 Rearranging terms a bit, we get ω1:x2+y2=2ax+2by+r2−a2−b2ω2:x2+y2=2cx+2dy+s2−c2−d2ω3:x2+y2=2ex+2fy+t2−e2−f2 Now consider the points P and Q. We know that these must necessarily satisfy both ω2 and ω3. Thus, by subtracting equations, we see that P and Q must necessarily satisfy 2cx+2dy+s2−c2−d2=2ex+2fy+t2−e2−f2. 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 ω1 and ω2 and must be collinear with the center of ω2. Thus R and S are coincident with X and Y, and XY forms a diameter of ω2. Thus, in this special case, (e,f) is on XY (since XY forms a diameter of the coincident ω2 and ω3). Aside from the special case, since two distinct point define a line, we have the equation of the line PQ as 2cx+2dy+s2−c2−d2=2ex+2fy+t2−e2−f2. We can employ a similar argument to find the equations of the lines RS (simultaneously satisfying ω1 and ω3) and XY (simultaneously satisfying ω1 and ω2). We then have: PQ:2cx+2dy+s2−c2−d2=2ex+2fy+t2−e2−f2RS:2ax+2by+r2−a2−b2=2ex+2fy+t2−e2−f2XY:2ax+2by+r2−a2−b2=2cx+2dy+s2−c2−d2 Now, finally, we note that (a,b) is on PQ and (c,d) is on RS. So we have the simultaneous equations: 2ac+2bd+s2−c2−d2=2ae+2bf+t2−e2−f22ac+2bd+r2−a2−b2=2ce+2df+t2−e2−f2 We subtract the two equations and simplify to find that (e,f) satisfies the equation for line XY: s2−c2−d2−(r2−a2−b2)=2ae+2bf−(2ce+2df)2ae+2bf+r2−a2−b2=2ce+2df+s2−c2−d2 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.
02.05.2009 20:49
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 )
02.05.2009 23:10
How does the orthocenter solution work in the projective plane?
02.05.2009 23:24
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.
24.06.2021 10:58
Let T=PQ∩RS, (PRQS)=ω3, the center of ω3 be O3, and K=XY∩O1O2. We first assume O1, O2, and O3 are non-collinear. Notice Powω1(T)=TR⋅TS=Powω3(T)=TP⋅TQ=Powω2(T)⟹T∈XY(the Radical Axis of ω1 and ω2). Since RS is the Radical Axis of ω1 and ω3, and PQ is the Radical Axis of ω2 and ω3, we know the intersection of the perpendicular from O1 to RS and the perpendicular from O2 to PQ is O3. Looking at △O1O2O3, it's obvious that T is its orthocenter. Now, let the foot of the altitude from O3 to O1O2 be K′. Then, ∠TK′O1=90∘. Using Radical Axis properties, however, we know ∠TKO1=90∘. Thus, K=K′ because they both lie on O1O2. Hence, O3,T,K are collinear or O3∈XY as desired. If O1, O2, and O3 are collinear, then we know PQ∥RS∥XY. Claim: If O1, O2, and O3 are collinear, then O3=K=XY∩O1O2. 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. ◻ ◼ Remark: In order for the problem to hold when the centers are collinear, O3∈O1O2 and O3∈XY has to hold, implying O3=XY∩O1O2=K must be true, motivating the last claim.
24.06.2021 18:55
11.12.2021 05:04
Let the center of the circle ω passing through P,Q,R,S be O, and let the length of its radius be r. Let the center of ω1 be O1 and let the center of ω2 be O2, and let the lengths of their radii be r1,r2 respectively. It suffices to show that O1O2−r21=O2O2−r22. But O1O2−r2=O1P⋅O1Q=O1O22−r22O2O2−r2=O2R⋅O2S=O1O22−r21so we are done.
19.12.2021 19:09
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.
31.01.2022 03:07
Let ω3 and O3 be (PQRS) and its center, respectively. Notice O1O22−r22=powω2(O1)=powω3(O1)=O1O23−r23and O2O23−r23=powω3(O2)=powω1(O2)=O2O21−r21.Adding yields powω2(O3)=O2O23−r22=O1O23−r21=powω1(O3)so O3 lies on ¯XY. ◻
11.02.2022 00:27
11.02.2022 00:33
imagine doing amo problems "for fun" orz
22.02.2022 15:02
Let O1,r1 and O2,r2 be the centers and radii of w1 and w2, respectively. Also, let w3 be the circle containing P,Q,R,S with a center at O3 and a radius of r3. Since O1 is on the radical axis of w2 and w3, we have: Poww2(O1)=Poww3(O1)O1O22−r22=O1O23−r23Similarly, since O2 lies on the radical axis of w1 and w3: Poww1(O2)=Poww3(O2)O1O22−r21=O2O23−r23Subtracting these two equations gives the following equation, which we continue to simplify. (O1O22−r22)−(O1O22−r21)=(O1O23−r23)−(O2O23−r23)r21−r22=O1O23−O2O23O2O23−r22=O1O23−r21Poww2(O3)=Poww1(O3)Therefore, O3 lies on the radical axis (line XY) of w1 and w2. ◻
24.04.2022 04:41
Let O1,O2 be the centers of the circles w1, and w2 with radius r1 and r2 respectively. Furthermore, let O3 be the center of the circle w3 passing through P,Q,R, and S with radius r3. Then, the problem is equivalent to proving that P(O3,w1)=P(O3,w2)⇔O3O21−r21=O3O22−r22. Note that P(O1,w2)=P(O1,w3)⟹O1O23−r23=O1O22−r22 and P(O2,w1)=P(O2,w3)⟹O2O23−r23=O2O22−r21. Subtracting the two equations gives O1O23−O2O23=r21−r22⟹O3O21−r21=O3O22−r22 which is what we wanted to prove.
19.07.2022 05:28
Let (PQRS)=ω3 and let ri and Oi denote the radius and center of circle ωi for i∈[1,2,3]. Since O1 is on the radical axis of O2 and O3 we have Powω2(O1)=Powω3(O1)⟺O1O22−r22=O1O23−r23.Since O2 is on the radical axis of O1 and O3 we have Powω1(O2)=Powω3(O2)⟺O2O21−r21=O2O23−r23.Subtracting yields r21−r22=O1O23−O2O23⟺Powω1(O3)=Powω2(O3)and we are done.
28.01.2023 19:58
Let O1,O2,O3 be the centers of ω1,ω2, 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 O3 lies on the perpendicular bisector of both PQ and RS. Thus, we have that O1O3 is perpendicular to RS, and O2O3 is perpendicular to PQ. Therefore, C is the orthocenter of △O1O2O3. This means that O3C is perpendicular to O1O2. Since C lies on XY, we are done.
10.07.2023 21:48
:unamused: Denote ω3:=(PQRS). Because O1∈PQ, Powω2(O1)=Powω3(O1)⟹O1O22−r22=O1O23−r23;and since O2∈RS, Powω1(O2)=Powω3(O2)⟹O1O22−r21=O2O23−r23.Subtracting, r21−r22=O1O23−O2O23⟹O1O23−r21=O2O23−r22,i.e., Powω1(O3)=Powω2(O3) or O3∈XY, as desired. ◻
19.12.2023 03:49
Let M=¯SR∩¯PQ. Let O1,O2,O3 be the centers of ω1,ω3 and (SPRQ). Now, it is well known that the radical axis of two circles is perpendicular to the line joining their centers. Then, SR⊥O1,O2 and PQ⊥O2O3. This means that M must be the orthocenter of △O1O2O3. This means, O3M⊥O1O2 as well. Clearly M lies on XY (MP⋅MQ=MR⋅MS so it lies on the radical axis of ω1 and ω2). But then, XY is the line passing through M and perpendicular to O1O2. This means that indeed O3 lies on the line XY and we are done.
21.02.2024 05:15
Let O1 be the center of ω1 and let O2 be the center of ω3. Denote the center of (PRQS) to be O3 and let the intersection of PQ and SR be C. Now note that SR, PQ, and XY are the three pairwise radical axes of ω1, ω2, and (PRQS). Denote the length of O1O2 to be a and let the radii of ω1, ω2, and (PRQS) be r1, r2, and r3, respectively. Since O2 is on the line PQ, by the Power of a Point definition of the radical axis, we have that a2−r21=O2O21−r21=Powω1(O2)=Powω3(O1)=O2O23−r23,which gives that O2O3=√a2−r21+r23. Similarly, since O1 is on RS, using Power of a Point on O1 with respect to O2 and O3, we get that O1O3=√a2−r22+r23. Finally, using these lengths and Power of a Point one last time, Powω1(O3)=O1O23−r21=a2+r23−(r21+r22)=O1O22−r22=Powω2(O3),which implies that O3 is on the radical axis of ω1 and ω2, which is XY, finishing the problem.
15.12.2024 00:08
By radical axis, the problem basically asks us to prove that for any three circles ω1,ω2,ω3 then Pow(O1,ω2)=Pow(O1,ω3),Pow(O2,ω1)=Pow(O2,ω3)⟹Pow(O3,ω1)=Pow(O3,ω2)But the first two equalities give us (O1O2)2−r22=(O1O3)2−r23(O1O2)2−r21=(O2O3)2−r23and subtracting equations gives the desired result.