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Medians of Triangles Proof


Date: 05/29/2000 at 10:34:52
From: Jenny
Subject: Medians of triangles

Prove that in any triangle, the sum of medians is more than 3/4 of its 
perimeter, but less than the whole perimeter.


Date: 05/29/2000 at 17:49:02
From: Doctor Floor
Subject: Re: Medians of triangles

Hi, Jenny,

Thanks for writing.

The proof that the sum of the medians is more than 3/4 of its 
perimeter is not difficult using the triangle inequality. 

Let us consider triangle ABC, with centroid G. Let ma, mb and mc be 
the medians from A, B and C respectively. Then, for instance, AG = 
2/3.ma, since G divides the medians in ratio 2:1.

We can apply the triangle inequality to triangles ABG, AGB and GBC, to 
find:

     2/3.ma + 2/3.mb > c (= AB)
     2/3.mb + 2/3.mc > a (= BC)
     2/3.mc + 2/3.ma > b (= AC)

Adding these three inequalities we find:

     4/3.(ma + mb + mc) > a + b + c
     ma + mb + mc > 3/4.(a + b + c)

For the proof that the sum of the medians is less than the perimeter 
we use Ptolemy's theorem. This theorem states that if ABCD is a 
noncyclic quadrilateral, then AB.CD + AD.BC > AC.BD. See, for 
instance, "Ptolemy's Theorem" from the Dr. Math archives:

   http://mathforum.org/dr.math/problems/breitling9.7.97.html   

Let Ma, Mb and Mc be the midpoints of sides a, b, and c respectively. 
We will apply Ptolemy's theorem to the quadrilateral AMcBC. This 
gives:

     AMc.BC + AC.BMc > CMc.AB
     c/2.a + b.c/2 > mc.c       [divide by c]
     a/2 + b/2 > mc

In the same way we find that b/2 + c/2 > ma and a/2 + c/2 > mb. Adding 
these three inequalities we get the desired result:

     a + b + c > ma + mb + mc

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/29/2000 at 21:25:17
From: Jenny
Subject: Re: Medians of triangles

Prove that in any triangle, sum of medians is more than 3/4 perimeter 
but less than whole perimeter.

I have already received the following hints:
 
To show it's less than the whole perimeter, try making the triangle 
into a parallelogram (there are three different ways to do this; do 
them all) and then extend the medians to be diagonals of the 
parallelograms. Then you can use the triangle inequality to show that 
the extended medians, twice the original medians, add up to less than 
twice the perimeter of the original triangle.

For the more than 3/4 part, in the original triangle ABC with medians 
AL, BM, and CN meeting at point G in the middle, you can use the 
triangle inequality to find that, for example, AG + GB > AB. Do that 
three times, then use a relationship between AG and GL ...

I still do not understand the "less than the whole perimeter" part. I 
followed your suggestions, but I am still lost. Can you please help 
me? 

Thank you


Date: 05/30/2000 at 18:00:25
From: Doctor Schwa
Subject: Re: geometry-medians and perimeter of triangles

Refer to the following figure:



just to make sure we are thinking of the same thing.

In this picture, we want to show that the medians are SHORTER than the 
whole perimeter; so the medians need to go on the short end of the 
triangle inequality.

We need to get the medians on the short end of the inequality, so in 
my picture we have for example C'C < C'B + BC, and using the 
parallelogram you can relate C'B distance to one of the sides of the 
original triangle, and you can also prove that C'C is twice as long as 
CN.  Putting all that together, the net result is that the sum of the 
medians is less than the perimeter.

Give that hint a try and please write us back if you'd like more help.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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