Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, June 1, 2019

### Geometry Problem 1438: Triangle, Incenter, Centroid, Circumcenter, Parallel, 90 Degree, Art, Poster

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90,
angle,
centroid,
circumcenter,
degree,
geometry problem,
incenter,
parallel,
perpendicular,
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Extend AI to meet the circle in M, the midpoint of smaller arc BC. Further, Join M to O and produce it to meet the circumcircle in N. Let the in-circle touch AB in L. From M draw a perpendicular to BC meeting the latter in P. Now we can see that Tr. AIL///Tr. NBM; AI*MB=IL*NM=2r*R. We further observe that MP=r since GI//BC and thus r/MB=sin(A/2) as also r/AI=sin(A/2) which implies that AI=MB & therefore AI²=2r*R. We now recall by Euler's Formula : OI²=R² -2r*R or OI²=R²-AI² or AI²+OI²=R² or /_OIA =90° by the converse of Pythagoras Theorem.

ReplyDeleteBorrowing from Ajit’s proof

ReplyDeleteWhen we show that MP = r (not clear how this is so) AI = MI = BM

So I is the mid point of chord AM and so OI is perpendicular to AM

Ajit - I would like u to clarify how MP = r

Sumith,

DeleteI'm afraid my proof is incomplete. Must figure out a way to prove my assumption. I hope Antonio will help.

In this triangle, 2a = b + c

ReplyDeleteAdding to Ajit's proof,

ReplyDeleteLet touch point of incircle at at AC is T.

Then AT=s-a=(b+c-a)/2, since b+c=2a, AT=a/2

AT=BP=a/2,Angle IAT=Angle MBP=A/2, and Angle ATI =Angle BPM=90

Hence Triangle ATI is congruent to Triangle BPM. And AI=BM=CM=MI,

Hence I is midpoint of chord AM and Angle AIO=90.

Pradyumna indeed as I have mentioned above b+c = 2a but it must first be proved

DeleteYes, thanks Pradyumna. AT =a/2; that is the crux of the matter. But where and how have we utilised the fact that IG//BC

DeleteThanks, Pradyumna.

DeleteLet AI meet BC at U and the circle at M.Let MO cut BC at P, the midpoint of BC

ReplyDeleteLet BM = BI = p (since < MBI = < MIB = (A+B)/2)

Let AI = 2q so that IU = q since GP = AG/2 and IG//UP

Now AB/BU = AI/IU = 2q/q = 2 so that BU = c/2

Triangles MBU & ABM are similar so AM/BM = AB/BU hence (p+2q)/p = c/(c/2) = 2

Therefore p + 2q = 2p and so p = 2q

Hence AI = MI and I is the midpoint of AM and OI is perpendicular to AI

Sumith Peiris

Moratuwa

Sri Lanka

dear Pradyumna Agashe.

ReplyDeletereferring to line 3 of your comment, I wonder how do you get b+c= 2a.

please explain

Peter Tran

DeleteHiPeter,

Here is proof for b+c=2a

Let altitude AH to BC is equal to h, If we draw perpendicular to BC from G, then it is equal to (1/3)h, since IG is parallel to BC, perpendicular to BC from I is r=h/3,

r=S/s

h/3=(1/2)ah/s, we get 2s=3a or a+b+c=3a

Or b+c=2a.