Before we start looking at the problems, let me give a quick note. At early years of Math Olympiads, the geometry were mostly constructions, and 3D geometry locus problems. I will not be going through those type of problems. In the recent years, the style of olympiads has changed significantly. These new type of problems started around 1988. The first such 'new type' problem appeared on the 1988 USAMO, as problem 4. This problem was the first USAMO problem I ever solved in my life.
Now, at that time, Internet was relatively new, and the usual exposition for Olympiad Geometry was "Geometry Revisited", which unfortunately did not contain the Incenter-Excenter Lemma I talked about in the last math stuff of the week post before the CCMSL Contest. The reason I did not prove it on the post, is because today, we are going to do the proof in the problem, because the 1988/4 is trivial by the I-E Lemma.
1988/4 USAMO
Let D be the circumcenter of IAB
In fact, all we need to prove is that DB=DI=DC in the I-E lemma, because that would mean that D is the circumcenter of BIC, and D is on the circumcircle of ABC, and the same goes for E and F by symmetry.
Proof
By the Definition of D, we note that DB=DC. So, we just have to prove DB=DI.
So, angle(DBI)=angle(DIB), meaning DB=DI=DC, and D is indeed the circumcenter of IAB, so D=A', and A' is on the circumcircle of ABC. The results for B' and C' can be proven similarly.
Before we start looking at the problems, let me give a quick note. At early years of Math Olympiads, the geometry were mostly constructions, and 3D geometry locus problems. I will not be going through those type of problems. In the recent years, the style of olympiads has changed significantly. These new type of problems started around 1988. The first such 'new type' problem appeared on the 1988 USAMO, as problem 4. This problem was the first USAMO problem I ever solved in my life.
Now, at that time, Internet was relatively new, and the usual exposition for Olympiad Geometry was "Geometry Revisited", which unfortunately did not contain the Incenter-Excenter Lemma I talked about in the last math stuff of the week post before the CCMSL Contest. The reason I did not prove it on the post, is because today, we are going to do the proof in the problem, because the 1988/4 is trivial by the I-E Lemma.
1988/4 USAMO
Let D be the circumcenter of IAB
In fact, all we need to prove is that DB=DI=DC in the I-E lemma, because that would mean that D is the circumcenter of BIC, and D is on the circumcircle of ABC, and the same goes for E and F by symmetry.
Proof
By the Definition of D, we note that DB=DC. So, we just have to prove DB=DI.
angle(DBI)=angle(DBC)+angle(IBC)=angle(DAC)+angle(ABC)/2=angle(BAC)/2+angle(ABC)/2.
angle(DIB)=angle(ABI)+angle(BAI)=angle(ABC)/2+angle(BAC)/2=angle(DBI).
So, angle(DBI)=angle(DIB), meaning DB=DI, and so we are done.
1988/4 USAMO Formal Solution Write-up (My Original Write-Up when I first solved it)
Let the three circumcenters be C', A', and B' respectively. We claim that A' is the midpoint of arc BC on the circumcircle of ABC.
Let D be the midpoint of arc BC on the circumcircle of ABC. Then, we note that DB=DC.
angle(DBI)=angle(DBC)+angle(IBC)=angle(DAC)+angle(ABC)/2=angle(BAC)/2+angle(ABC)/2.
angle(DIB)=angle(ABI)+angle(BAI)=angle(ABC)/2+angle(BAC)/2=angle(DBI).
So, angle(DBI)=angle(DIB), meaning DB=DI=DC, and D is indeed the circumcenter of IAB, so D=A', and A' is on the circumcircle of ABC. The results for B' and C' can be proven similarly.
Q.E.D.