A Nice Problem Trivialized By a Single Lemma!

(CGMO 2012) The incircle of a triangle ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of triangle BCI where II denotes the incenter of ABC\bigtriangleup ABC. Prove that ODB=OEC\angle ODB = \angle OEC


Proof\textbf{Proof}

We will first prove this lemma.


Lemma\textbf{Lemma} Let ABCABC be a triangle with incenter II , A-excenter IaI_a, and denote the midpoint of arc BCBC to be LL, then points B,I,C,IaB, I, C, I_a are concyclic with LL as the circle center.

(We will denote angle AA , BB, CC, as α,β,\alpha, \beta, and γ\gamma, respectively.)

Also note that α+β+γ=180\alpha + \beta + \gamma = 180

Proof: First notice that LL is the intersection of the angle bisector of AA and perpendicular bisector of BCBC.

First we prove that LB=LI=LCLB = LI = LC.

Since the perpendicular of LL to BCBC is the perpendicular bisector of BCBC, we know that LB=LCLB = LC.

Because BI,AIBI, AI are angle bisectors, we know that IAB=α2\angle IAB = \frac{\alpha}{2} and ABI=IBC=β2\angle ABI= \angle IBC = \frac{\beta}{2} . By Exterior Angle Theorem, we know that BIL=α+β2\angle BIL = \frac{\alpha + \beta}{2} . If we prove that IBL=α+β2\angle IBL = \frac{\alpha + \beta}{2} we have proved the segments are equal.

Notice that CBL=LAC=α2\angle CBL = \angle LAC = \frac{\alpha}{2} because they substend the same arc. Therefore this implies that IBL=IBC+CBL=α+β2=BIL\angle IBL = \angle IBC + \angle CBL = \frac{\alpha+\beta}{2} = \angle BIL . Thus, by isosceles triangles we get that LB=LI=LCLB = LI = LC . This tells us that B,I,CB, I, C are concyclic with LL as a center.

Now, we must prove that IaI_a also lies on this circle.

Recall that the IaI_a is the intersection of the exterior-angle bisectors of BB and CC. Therefore if you extend ABAB and ACAC and construct the angle bisector of the exterior angle, it intersects at IaI_a.

Denote MM as a point on ray ABAB past B.

By Exterior Angle Theorem, we have that MBC=α+γ\angle MBC = \alpha + \gamma .

Since BIaBI_a is the angle bisector, we have that IaBC=α+γ2\angle I_aBC = \frac{\alpha + \gamma }{2} . Since LBC=α2\angle LBC = \frac{\alpha}{2} , this implies that IaBL=γ2\angle I_aBL = \frac{\gamma}{2} .

We proved that LBI=LIB=α+β2    BLI=γ\angle LBI = \angle LIB = \frac{\alpha + \beta}{2} \implies \angle BLI = \gamma .

Again by exterior angle theorem, since LBIa=γ2\angle LBI_a = \frac{\gamma}{2}, we have that BIaL=γ2\angle BI_aL = \frac{\gamma}{2} which implies that BLIa\bigtriangleup BLI_a is isoscles which implies that BL=LIa    BL=LIa=IL=LCBL = LI_a \implies BL = LI_a = IL = LC . Therefore IaI_a must be concyclic with B,I,CB, I, C because it's radius to the center, LL, is the same. Therefore, we have proved the lemma.


Okay good. Now let's have fun.


Restatement: The incircle of a triangle ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of triangle BCF where FF denotes the incenter of ABC\bigtriangleup ABC. Prove that ODB=OEC\angle ODB = \angle OEC


By the lemma, we know that the circumcenter, OO, of BFC\bigtriangleup BFC is actually the midpoint of arc BC of the circumcircle of ABC\bigtriangleup ABC.

To prove ODB=OEC\angle ODB = \angle OEC , it suffices to prove that ADO=AEO\angle ADO = \angle AEO .

Since OO is on the angle bisector, we know that DAO=OAE\angle DAO = \angle OAE. We also know that AD=AEAD = AE because they are both tangents to the incircle. Additionally, DAO\bigtriangleup DAO and AOE\bigtriangleup AOE share side AOAO. Therefore, by SASSAS, we have that ADOAEO\bigtriangleup ADO \cong \bigtriangleup AEO .

This implies that ADO=AEO    ODB=OEC\angle ADO = \angle AEO \implies \angle ODB = \angle OEC and we are done.

Note by Alan Yan
4 years, 1 month ago

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First, let's clarify that point I and point F are the one and the same, which is the incenter of triangle ABC. It seems to me that it's immediately obvious that the quadrilateral AEOD is a kite, so that angles \angleODA = \angleOEA, and therefore angles \angleODB = \angleOEC? Am I missing something here?

Michael Mendrin - 4 years, 1 month ago

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It is true that AEOD is a kite, but can you prove it?

Alan Yan - 4 years, 1 month ago

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If, given two points on a circle, the tangents through are drawn so that they intersect at a 3rd point, the distances from the 3rd point to the first two points are the same. Any point on the line that passes through this 3rd point and the center of the circle forms the 4th point of the kite, by reason of symmetry.

Michael Mendrin - 4 years, 1 month ago

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@Michael Mendrin Yes...that works. But note both arguments rely on the fact that O lies on the angle bisector of A, and that was the main concept I was trying to convey.

Alan Yan - 4 years, 1 month ago

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@Alan Yan I think by using the Lemma you've discussed, a harder geometry problem probably can be devised. Let me think on it for a while on how that can be done.

Michael Mendrin - 4 years, 1 month ago

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@Michael Mendrin I agree. Please do so :)

Alan Yan - 4 years, 1 month ago

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