In $\triangle ABC$, $\angle B=90^\circ$, segment $AB>BC$. Now we have a $\triangle A_iBC(i=1,2,...,n)$ which is similiar to $\triangle ABC$ (the vertexs of them might not correspond). Find the maximum value of $n+2018$.
Problem
Source: 2019 Taiwan APMO Preliminary
Tags: geometry
hsiangshen
23.08.2020 14:10
Original Chinese Version
Attachments:
2019Taiwan APMO primary.pdf (117kb)
franzliszt
10.11.2020 23:40
Why can't $n$ get any larger than $2029$ or $2030$? I don't see why there is a upper bound.
hsiangshen
22.11.2020 06:54
Oh thanks! Fixed the answer The answer is n+2018=2030 or 2029
v_Enhance
22.11.2020 07:16
franzliszt wrote: Why can't $n$ get any larger than $2029$ or $2030$? I don't see why there is a upper bound. I think the problem author means for the points $A_i$ to be distinct (and lie in the same plane). (This doesn't appear to be a translation error as I do not see a "distinctness" present in the original Chinese text either.)