Prove at least one angle PAB, PBC, or PCA is at most 30 degrees

[bogdan]

Let ABC be a triangle and P a point inside it...
Prove that at least one of the angles PAB, PBC or PCA measures less or equal to 30 degrees...

 

[enslam]

Aº+Bº+Cº = 180
Aº=Bº=Cº=60º is the largest value for the smallest angle.
(60º)/2=30º

If any of the angles (Aº,Bº or Cº) are greater than 60º then the smallest angle is less than 60º. Thus PAB, PBC or PCA measures less or equal to 30 degrees.

 

[bogdan]

Why 60/2 ?

 

[enslam]

Because 60 is the greatest possible value that the smallest angle can be.

60+60+60=180 smallest 60
59+61+60=180 smallest 59
30+60+90=180 smallest 30

Its impossible for the smallest value to be greater than 60

 

[bogdan]

Take a look...
http://www.angelfire.com/pro/fbi/tri.bmp

 

[enslam]

PCA is less that 30 in that pic.
No matter where the point P is there will always be an angle equal to (in the case of an equalateral) or less than (in all other cases) 30.

 

[bogdan]

Yeah...but I need a demonstration...

 

[enslam]

What do u mean by demonstration.

 

[bogdan]

proof...logical...mathematical...

 

[enslam]

Aº+Bº+Cº = 180
Aº=180-Bº-Cº where Aº is the smallest angle.
<PAC+<PAB = Aº
either <PAC or <PAB is less than or equal to 30
Is that mathematical enough

 

[KLscilevothma]

another IMO question

This question appears in IMO 1991

 

[bogdan]

Of course...but it "looks" simple enough to be solved by anyone...

 

[marcus]

Originally posted by KL Kam
This question appears in IMO 1991

would this be online and if so could you post the link?

is there a website with past IMO?

 

[marcus]

Originally posted by marcus
would this be online and if so could you post the link?

is there a website with past IMO?

no need to post it, thanks anyway
I got it from google