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The Incenter/Excenter Lemma - Evan Chen

The Incenter/Excenter LemmaEvan Chen August 6, 2016In this short note, we ll be considering the following very useful a triangle with incenterI,A-excenterIA, and denote byLthemidpoint of arcBC. Show thatLis the center of a circle throughI,IA,B, is just angle chasing. LetA= BAC,B= CBA,C= ACB, and notethatA,I,Lare collinear (asLis on the angle bisector). We are going to show thatLB=LI, the other cases being , notice that LBI= LBC+ CBI= LAC+ CBI= IAC+ CBI=12A+ , BIL= BAI+ ABI=12A+ ,4 BILis isosceles. SoLB=LI. The rest of the proof proceeds along , let s see where this Lemma has come up before.

midpoint of arc BC. Show that L is the center of a circle through I, I A, B, C. A BC I L I A Proof. This is just angle chasing. Let A = \BAC, B = \CBA, C = \ACB, and note that A, I, L are collinear (as L is on the angle bisector). We are going to show that LB = LI, the other cases being similar. First, notice that \LBI = \LBC + \CBI = \LAC ...

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Transcription of The Incenter/Excenter Lemma - Evan Chen

1 The Incenter/Excenter LemmaEvan Chen August 6, 2016In this short note, we ll be considering the following very useful a triangle with incenterI,A-excenterIA, and denote byLthemidpoint of arcBC. Show thatLis the center of a circle throughI,IA,B, is just angle chasing. LetA= BAC,B= CBA,C= ACB, and notethatA,I,Lare collinear (asLis on the angle bisector). We are going to show thatLB=LI, the other cases being , notice that LBI= LBC+ CBI= LAC+ CBI= IAC+ CBI=12A+ , BIL= BAI+ ABI=12A+ ,4 BILis isosceles. SoLB=LI. The rest of the proof proceeds along , let s see where this Lemma has come up before.

2 ChenThe Incenter/Excenter Lemma1 Mild EmbarrassmentsProblem 1(USAMO 1988).TriangleABChas incenterI. Consider the triangle whosevertices are the circumcenters of4 IAB,4 IBC,4 ICA. Show that its circumcentercoincides with the circumcenter 2(CGMO 2012).The incircle of a triangleABCis tangent to sidesABandACatDandErespectively, andOis the circumcenter of triangleBCI. Prove that ODB= 3(CHMMC Spring 2012).In triangleABC, the angle bisector of Ameetsthe perpendicular bisector ofBCat pointD. The angle bisector of Bmeets theperpendicular bisector ofACat pointE. LetFbe the intersection of the perpendicularbisectors ofBCandAC.

3 FindDF, given that ADF= 5 , BEF= 10 andAC= 4(Nine-Point Circle).LetABCbe an acute triangle with orthocenterH. LetD,E,Fbe the feet of the altitudes fromA,B,Cto the opposite sides. Show that themidpoint ofAHlies on the circumcircle Some Short-Answer ProblemsProblem 5(HMMT 2011).LetABCDbe a cyclic quadrilateral, and suppose thatBC=CD= 2. LetIbe the incenter of triangleABD. IfAI= 2 as well, find theminimum value of the length of 6(HMMT 2013).Let triangleABCsatisfy 2BC=AB+ACand have incenterIand circumcircle . LetDbe the intersection ofAIand (withA,Ddistinct). ProvethatIis the midpoint 7(Online Math Open 2014/F19).

4 In triangleABC,AB= 3,AC= 5, andBC= 7. LetEbe the reflection ofAoverBC, and let lineBEmeet the circumcircle ofABCagain atD. LetIbe the incenter of4 ABD. Compute cos 8(NIMO 2012).LetABXCbe a cyclic quadrilateral such that XAB= XAC. LetIbe the incenter of triangleABCand byDthe foot ofIonBC. GivenAI= 25,ID= 7, andBC= 14, Intermediate ExamplesProblem an acute triangle such that A= 60 . Prove thatIH=IO,whereI,H,Oare the incenter , orthocenter, and 10(IMO 2006).LetABCbe a triangle with incenterI. A pointPin theinterior of the triangle satisfies PBA+ PCA= PBC+ thatAP AI, and that equality holds if and only ifP= 11(APMO 2007).

5 In triangleABC, we haveAB > ACand A= 60 . LetIandHdenote the incenter and orthocenter of the triangle. Show that 2 AHI= 3 ChenThe Incenter/Excenter LemmaProblem 12(ELMO 2013, Evan Chen).TriangleABCis inscribed in circle . A circlewith chordBCintersects segmentsABandACagain atSandR, respectively. SegmentsBRandCSmeet atL, and raysLRandLSintersect atDandE, respectively. Theinternal angle bisector of BDEmeets lineERatK. Prove that ifBE=BR, then ELK=12 13(Online Math Open 2012/F27).LetABCbe a triangle with circumcircle . Let the bisector of ABCmeet segmentACatDand circle atM6=B.

6 Thecircumcircle of4 BDCmeets lineABatE6=B, andCEmeets atP6=C. Thebisector of PMCmeets segmentACatQ6=C. Given thatPQ=MC, determine thedegree measure of Harder TasksProblem 14(Iran 2001).LetABCbe a triangle with the midpoint of arcBCnot containingA, and letNdenote the midpoint ofarcMBA. LinesNIandNIAintersect the circumcircle ofABCatSandT. Prove thatthe linesST,BCandAIare 15(Online Math Open 2014/F26).LetABCbe a triangle withAB= 26,AC= 28,BC= 30. LetX,Y,Zbe the midpoints of arcsBC,CA,AB(not containingthe opposite vertices) respectively on the circumcircle ofABC. LetPbe the midpointof arcBCcontaining pointA.

7 Suppose linesBPandXZmeet atM, while linesCPandXYmeet atN. Find the square of the distance 16(Euler).LetABCbe a triangle with incenterIand circumcenterO. ShowthatIO2=R(R 2r), whereRandrare the circumradius and inradius of4 ABC, 17(IMO 2010).LetIbe the incenter of a triangleABCand let be itscircumcircle. Let the lineAIintersect again atD. LetEbe a point on the arcBDCandFa point on the sideBCsuch that BAF= CAE <12 , letGbe the midpoint ofIF. Prove thatDGandEIintersect on .5 Bonus ProblemsProblem 18(Russia 2014).LetABCbe a triangle withAB > BCand circumcircle .PointsM,Nlie on the sidesAB,BCrespectively, such thatAM=CN.

8 LinesMNandACmeet atK. LetPbe the incenter of the triangleAMK, and letQbe theK-excenterof the triangleCNK. IfRis midpoint of arcABCof then prove thatRP= a triangle with circumcircle , and letDbe any point onBC. We draw acurvilinear incircletangent toADatL, toBCatKand internallytangent to . Show that the incenter of triangleABClies ChenThe Incenter/Excenter Lemma6 Hints to the isO? the circumcenter of4 ABC. Who areDandE? is the incenter of4 DEF? What is theD- excenter ? thatAC= Ptolemy s isC? Ptolemy s BHC= BIC= BOC= 120 , pointsHandOnow lie on the magiccircle too.

9 SoIH=IOis just an equality of certain the angle condition to show thatPalso lies on the magic pointHlies on the magic circle. So IHC= 180 need to do quite a bit of angle chasing. Show thatRis the incenter isB? arc midpoints. (Why?) show thatS,T,I,IAare concyclic, say byNI NS=NM2=NIA the incenterI. LineMNis a in pointL, the midpoint of arcBC. By Power of a Point, it s equivalent toproveAI IL= 2Rr, which can be done with similar a homothety with ratio 2 atI. This sendsGtoFandDto arc midpoints on the circumcircles of both4 AMJand4 CNK. Usespiral similarity the tangency point to beT, letMbe the midpoint of arcBC, and let linesKLandAMmeet atI.

10 Show thatM,K,Tare collinear. Show thatALIT iscyclic. Prove thatMI2=MK MT=MC2=


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