Let ABC be a triangle. Suppose that D, E, and F are points on segments ¯BC, ¯CA, and ¯AB respectively such that triangles AEF, BFD, and CDE have equal inradii. Prove that the sum of the inradii of △AEF and △DEF is equal to the inradius of △ABC. Aprameya Tripathy
Problem
Source: ELMO Shortlist 2024 G2
Tags: geometry, ELMO Shortlist
23.06.2024 02:32
Let A0, C1, and B2 be the intouch points of AEF opposite A, E, and F, respectively, and define the intouch points of BFD and CDE similarly. Let IA, IB, and IC be the incenters of AEF, BFD, and CDE, respectively. Let r, r′, and r0 denote the inradii of ABC, DEF, and AEF, respectively, and let p, p′, pA, pB, and pC denote the perimeters of ABC, DEF, AEF, BFD, CDE, respectively. By equal tangents, p′=DC0+C0E+EA0+A0F+FB0+B0D=DA2+B1E+EB2+C1F+FC2+A1D=A1A2+B1B2+C1C2=IBIC+ICIA+IAIB,so p′ is the perimeter of IAIBIC. We have rp=2[ABC]=2([DEF]+[AEF]+[BFD]+[CDE])=r′p′+r0(pA+pB+pC)=r′p′+r0(p+p′),so it suffices to show rp=(r−r0)p′+r0(p+p′)⟺p′p=r−r0r. But notice that the homothety at I with scale factor r−r0r maps ABC to IAIBIC, as desired. ◻
09.07.2024 10:39
Let X, Y and Z denote the incenters of △AEF, △BFD and △CDE, respectively. Let r denote the common inradius of these three triangles. By chasing tangents, the perimeters of △DEF and △XYZ are equal. Let p be this common perimeter, and let P be the perimeter of △ABC. We have [ABC]−[XYZ]=r(p+P)2=[ABC]−[DEF],so the areas of △XYZ and △DEF are equal as well. Thus, their inradii are equal. Since the incenter of △XYZ coincides with the incenter of △ABC, it is clear that the inradius of △ABC is the sum of the inradius of △XYZ and r, which is in turn equal to sum of the inradius of △DEF and r.