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January 14th, 2012, 12:23 PM  #1 
Newbie Joined: Jan 2012 Posts: 9 Thanks: 0  Help with some pretty hard geometry issues
Hi, our teacher gave us these two issues as a homework, I'm quite weak in geometry, can someone help me? Nr.1 We need to find the lengths of a, b and c sides of the triangle. All we know is: 1 The length of the sides of a triangle are in meters, they're also integers. 2 the perimeter of the triangle is 72m 3 The intersection of the longest side of the triangle and its incircle (inscribed) divides the longest side 3:4 Nr.2 All we know is: 1 The triangle is an isosceles triangle (two sides are equal in length) We need to find the radius R of the circumscribed circle of the triangle and the radius r of the inscribed circle of the triangle with the use of variables Sorry for my bad English, it's not my native language if someone will really help me, I'll be really THANKFUL!!! 
January 14th, 2012, 07:01 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,613 Thanks: 2071 
Nr 1. The triangle's sides are either 16m, 21m and 35m or 20m, 24m and 28m. Are you allowed to use trigonometry to find these?

January 15th, 2012, 12:53 AM  #3 
Newbie Joined: Jan 2012 Posts: 9 Thanks: 0  Re: Help with some pretty hard geometry issues
Wow, thanks! Yes, we are. Can you at least write some hints on how should I begin with the solution of this, please? 
January 15th, 2012, 02:07 AM  #4 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  Re: Help with some pretty hard geometry issues
Trigonometry is not needed. Post what YOU did.

January 15th, 2012, 03:51 AM  #5 
Newbie Joined: Jan 2012 Posts: 9 Thanks: 0  Re: Help with some pretty hard geometry issues
I didn't get to the conclusion, all I could do was make a sketch. Then I realised, that the intersection of the angle bisectors is the center of the inscribed circle. Also, since the bisectors intersect with the center of the circle, means that we can apply the Thales' theorem here, meaning we can make up some rightangled triangles inside the inscribed circle (by connecting the diameter points with the intersections of other bisectors and the inscribed circle). In these little rightangled triangles, we can use trigonometry, but then again, we lack the information about the angles... That's a dead end for me...:/ Am I on the right way? Or are the steps for this issue absolutely different?

January 15th, 2012, 04:38 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: Help with some pretty hard geometry issues
> 3 The intersection of the longest side of the triangle and its incircle (inscribed) divides the longest side 3:4 That should be "the tangent point of the longest ........." Let c = AB = longest side of triangle ABC; let b = AC and a = BC (a < b). Let M = incircle center. Place point D on AB such that MD is perpendicular to A. (so D is tangent point!). Then AD = 4c/7 and BD = 3c/7. Start as above...you'll find you don't need angles. 
January 16th, 2012, 10:35 AM  #7 
Newbie Joined: Jan 2012 Posts: 9 Thanks: 0  Re: Help with some pretty hard geometry issues
Thank you for the reply, Denis! I don't get one thing, though. The last step you stated = place the D point on AB such that MD is perpendicular to AD. I have done that, but how will this exactly help me? If I do it so, then the D point won't be (or doesn't have to be) an intersection of AB and the incircle. Either way, then I would use tgMAD= radius/x, in which x>4c/7, what good would that do to me? Sorry, my stupidity in this problem might be annoying, but I appreciate every hint given by you all 
January 16th, 2012, 01:06 PM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: Help with some pretty hard geometry issues Quote:
In this case, circle is TANGENT to AB at point D. Let tangent point on BC be E, on AC be F: then draw 2 more perpendiculars: ME and MF BE = 3c/7 ; so CE = a  3c/7 AF = 4c/7 ; so CF = b  4c/7 Since CE = CF then a  3c/7 = b  4c/7 : simplifies to 7(b  a) = c That's all I'll give you; you need to find a triangle that reflects above: there are more than 1, so there is not a unique solution. Anyway, the fact you didn't know what an incircle is makes me think you're not ready for this.  
January 16th, 2012, 02:21 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,613 Thanks: 2071 
Tangency is a form of intersection.

January 17th, 2012, 11:09 AM  #10 
Newbie Joined: Jan 2012 Posts: 9 Thanks: 0  Re: Help with some pretty hard geometry issues
Oh I get it now! I got totally confused by the english math terminology and translated TANGENT wrong! Hahaha sorry, now I finally get the picture, I'll try to solve it now! Thank you very much for your help! 

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