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Vector Theorems - Math Problem Example

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Suppose that ABC is a given triangle as shown in the figure below with vertices given as A= (a1, a2), B= (b1, b2), and C= (c1, c2). Let D, E, and F be the mid-points of sides AB, BC, and AC respectively. By joining these points, a triangle DEF is formed which is called medial triangle.
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XXXXX No. XXXXX MATHEMATICS VECTOR THEOREMS GEOMETRY XYZ 15 November, Theorem: A triangle and its medial triangle have the same centroid.
Suppose that ABC is a given triangle as shown in the figure below with vertices given as
A= (a1, a2), B= (b1, b2), and C= (c1, c2).
Figure 1
Let D, E, and F be the mid-points of sides AB, BC, and AC respectively. By joining these points, a triangle DEF is formed which is called medial triangle. This medial triangle is sometimes also known as auxiliary triangle (Dixon, 1991). The medial triangle DEF consists of the vertices
D= (d1, d2) = ((a1+ b1)/2, (a2+ b2)/2)
E= (e1, e2) = ((b1+ c1)/2, (b2+ c2)/2)
F= (f1, f2) = ((a1+ c1)/2, (a2+ c2)/2)
Now let KLM be the mid points of DE, DF, and EF respectively. By joining these points we get another triangle KLM i.e. the medial triangle of DEF. Therefore,
K= (k1, k2) = ((a1+ b1)/2+ (b1+ c1)/2), (a2+ b2)/2+ (b2+ c2)/2)
= ((a1+ 2b1+ c1)/4, (a2+2b2+ c2)/4)
Similarly, L = (l1, l2) = ((2a1+ b1+ c1)/4, (2a2+b2+ c2)/4), and
M = (m1, m2) = ((a1+ b1+ 2c1)/4, (a2+b2+ 2c2)/4)
Since we have a medial triangle DEF as shown above, for convenience we draw the same triangle separately as shown below. Here D (a, 0, 0), E (0, b, 0), and F (0, 0, c) be the vertices of medial triangle while N is the centroid of the triangle.
By distance formula, N has the coordinates as (a/3, b/3, c/3).
Now DN = √ [(a/3-a)² + (b/3-0)² + (c/3-0)²]
= √ (4a²/9) + (b²/9) + (c²/9)
= √ [(4a² + b² + c²)]/3
3DN = √ [(4a² + b² + c²)] (1)
Let D′ be the mid-point of EF, therefore its coordinates will be (0, b/2, c/2). Now we find the distance between N and D′.
D′N = √ [(a/3-0)² + (b/3-b/2)² + (c/3-c/2)²
= √ (a²/9) + {(b²/9 + b²/4 – 2 (b/3) (b/2)} + {(c²/9 + c²/4 – 2 (c/3) (c/2)}
= √ (a²/9) + (b²/36) + (c²/36)
= √ [(4a² + b² + c²)]/6
6 D′N = 3 DN as by using equation (1)
2 D′N = DN
Similarly 2 E′N = EN, and 2 F′N = FN.
Thus, it means that the centroid is located two thirds of the way from the original vertex to the midpoint of the opposite side of the triangle.
Now, since we have a triangle ABC and E is the middle point of BC, and P is the centroid of ABC. Therefore, by using above theorem we have OP = OA + 2/3 AE. Now we will find the centroid of ABC.
Let P = centroid of ABC = OP = OA + 2/3 AE
Thus P = (a1+ (2/3) [e1- a1], a2+ (2/3) [e2- a2])
Then P = (a1+ (2/3) [(b1+ c1)/2- a1], a2+ (2/3) [(b2+ c2)/2- a2])
P = (a1+ (b1+ c1)/3-(2/3) a1, a2+ (b2+ c2)/3-(2/3) a2)
P= ((a1+ b1+ c1)/3, (a2+ b2+ c2)/3) (2)
Now we will find the centroid of DEF
Let Q = centroid of DEF = OQ = OD + 2/3 DM
Then Q = (d1+ (2/3) [m1- d1], d2+ (2/3) [m2- d2])
Then Q = (d1+ (2/3) ((a1+ b1+ 2c1)/4-(a2+ b2)/2), d2+ (2/3) ((a2+ b2+2c2)/4-(a2+ b2)/2))
Q = (d1+ (2/3) ((-a1- b1+ 2c1)/4), d2+ (2/3) ((-a2- b2+2c1)/4))
Q= ((a1+b1)/2+ (-a1- b1+2c1)/6, (a2+ b2)/2+ (-a1- b1+2c1)/6)
Q= ((2a1+2b1+2c1)/6, (2a2+2b2+2c2)/6)
Q= ((a1+ b1+ c1)/3, (a2+ b2+ c2)/3) (3)
From above equation no.2 and equation no.3 it is evident that P=Q. Thus it has been proved that the centroid of the triangle ABC is equal to the centroid of medial triangle DEF.
Reference
Dixon, R. (1991). Mathographics, New York: Dover, p. 56 Read More
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