In triangle ABC, the incircle, with center I, touches the sides BC at point D. Line DI meets AC at X. The tangent line from X to the incircle (different from AC) intersects AB at Y. If YI and BC intersect at point Z, prove that AB=BZ. Proposed by Hooman Fattahimoghaddam
Problem
Source: 4th Iranian Geometry Olympiad (Advanced) P1
Tags: geometry, Iran, IGO
15.09.2017 13:33
WLOG let AC≥AB. Note that ∠DXC=90−∠C⟹∠YXD=90−∠C. So from quadrilateral BYXD we get ∠BYZ=∠ZYX=90+∠C−∠B2.Thus we have ∠YZB=∠A/2. Hence △AIF≅△ZID implying that ZD=AF⟹AB=BZ.
15.09.2017 21:29
I is the excenter of △ABC, thus ∠BYI=ˆC+ˆA2, hence ∠BZI=ˆA2=∠BAI, done. Best regards, sunken rock
06.01.2018 16:46
@bgn ,can you post a link to IGO website?
06.01.2018 20:30
liekkas wrote: @bgn ,can you post a link to IGO website? http://igo-official.ir/?lang=en
06.01.2018 20:54
Also, the problems of 2017 IGO: https://artofproblemsolving.com/community/c520386_2017_iranian_geometry_olympiad
07.01.2018 10:26
Thanks above
20.08.2018 11:26
03.09.2018 14:02
Can someone explain this: Solution. In triangle AXY , the point I is excenter so <XIY=90-<A/2 It s a part of the solution IGO published.
03.09.2018 15:56
http://mathworld.wolfram.com/notebooks/PlaneGeometry/NeubergCubic.nb
30.05.2020 11:58
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/26.%203.%20Segments.pdf p. 9-10. Sincerely Jean-Louis
25.11.2020 01:32
Our end-goal is to prove that DF∥AZ Let T be the second tangent from X onto the incircle. Throw the configuration onto the complex plane, and let every point in the Euclidean plane be represented by its' lowercase letter and the incircle is the unit circle. We claim W.L.O.G. that d=1. Then we have that: a=2efe+fb=2ff+1c=2ee+1 We have that x belongs to the real number line, thus we have that x=¯x Since we have that e−a¯e−a=e−x¯e−x, this implies that x=2e1+e2 By definition and from x, we have that t=1e, this implies that y=2fef+1. Since we have that z must satisfy these conditions, y¯y=y−z¯y−z and b−c¯b−c=b−z¯b−z, we have that z=2ff+e By easy computation we see that: z−a¯z−a=d−f¯d−fthis implies that DF∥AZ, from which we have that AB=BZ
03.07.2021 10:50
Let T be the point on AC such that ZT is tangent to the incircle. By Brianchon's, T is on BI, so A and Z are symmetric about BI.
03.11.2021 16:38
Drop a tangent from Z to the incircle and let it meet AC at E. By biranchon on the degenerate hexagon BYXEZD we have E,I,B colinear and since ∠AEB=∠ZEB we have A,Z symetric w.r.t. BI thus AB=BZ as desired.
Edit: 600th post yay!!!
03.11.2021 17:11
This is very easy. The quadrilateral BYXC has an incircle and the angle bisectors concur at I.Some trivial angle chasing (or angle halfing and doubling) gives that ∠IZB=A2⟹AIZC is cyclic .So ∠AZC=90+B2, as desired.
27.12.2021 16:07
∠XIY = 90 - ∠A/2 ---> ∠IZB = ∠A/2 now triangles IZB and IAB are congruent so AB = BZ.
12.05.2022 03:24
Notice ∠IZB=90−∠DIZ=∠IZX+∠YXI−90=12(∠BYX+∠YXC−180)=12∠Aso △ABI≅△ZBI. ◻
06.02.2024 16:24
Let E be the incenter of ΔAXY and F be the tangency point of AB and the incircle of ΔABC. The incenter of ΔABC is the A-excircle of △AXY. So, by Incenter-Excenter Lemma, EXIY is cyclic. ∠AIY=∠EIY=∠EXY=12∠AXY=12(180∘−2∠CXD)=12(180∘−(180∘−2∠C))=12⋅2∠C=∠ACZ⟹AIZC is cyclicFinally, ∠AZI=∠ACI=∠ZCI=∠ZAI⟹IA=IZ⟹△AIF≅△IDZ⟹AF=DZ⟹AF+BF=BD+DZ⟹AB=BZas desired.
02.07.2024 13:01
Angle chase and some isométrique triangle