Proposed Problem

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Complete Problem 321

Level: High School, SAT Prep, College geometry

## Sunday, July 12, 2009

### Problem 321: Triangle, Incenter, Excircle, Altitude, Collinearity

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Can you give me Hint????

ReplyDeleteSolution:

ReplyDeleteNote that A, I and E are collinear, since they all lie on the bisector of angle BAC.

Let AE and BC intersect at N. Since AH || EF, triangles NEF and NAH are similar. Thus, AH/FE = AN/NE, so AM/FE = AN/2NE.

Let the incircle have radius r, and the excircle have radius R. Then r/R = AI/AE = IN/NE, so we have

AI * NE = AE * IN

AI * NE = (AI + IN + NE) * IN

2 AI * NE =

= (AI + IN + NE) * IN + AI * NE

= IN^2 + AI * IN + NE * IN + AI * NE

= (IN + AI) (IN + NE)

= AN * IE

Therefore, AM/FE = AN/2NE = AI/IE. Since the angles IAM and IEF are equal, the triangles IAM and IEF are similar. Thus, the angles AIM and EIF are equal, so F, I, M are collinear.

https://photos.app.goo.gl/FqYWXp2JHPKLi7Zh2

ReplyDeletesee sketch for location of points K, L and N

denote 2p= perimeter of triangle ABC.

Note that AK= p-a

KL= BK+BL= BN+BK= BC= a

And AM//EF => ∠ (MAI)= ∠ (IEF)

Calculate area of triangle ABC in 2 ways

S(ABC)= ½ AH.BC= AM.BC

S(ABC)= (p-a).EF=AK.EF= AM.BC

So EF/AM= BC/AK= KL/AK=IE/AI

So triangle AMI similar to EFI ( case SAS) => ∠ (AIM)= ∠ (FIE)=> M, I, F are collinear