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/
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