Is it possible to CONSTRUCT a triangle given only the lengths of the medians?

Question

All you are given is the lengths of the medians. You don't know what angles they make with each other. Is it possible to construct a triangle with this information.

What about if you are given only the three altitudes? Can you construct the triangle?

Answer 1

This is a very well-known old problem. Let m_a, m_b, m_c are the medians lengths.

Part 1 (Analysis) Take a triangle ABC with medians AA', BB', CC' /A' - mid-point of BC etc., m_a = |AA'| etc./. The centroid M divides each median in the ratio:
|AM| : |MA'| = |BM| : |MB'| = |CM| : |MC'| = 2
Take a mid-point, say D, of CM, we have
|CD| = |DM| = |MC'|
and connect D with B'. The segment B'D is the mid-segment of ACM, so |B'D| = |AM|/2. The triangle B'DM has side lengths:
|B'D| = m_a/3; |B'M| = m_b/3; |DM| = m_c/3,
so we found a very important triangle B'DM, that will help below.

Part 2 (Construction) Divide the given m_a, m_b, m_c into 3 equal parts and construct a triangle (say B'DM) with these side lengths (m_a/3, m_b/3, m_c/3). Expand B'M from B' to M twice, You obtain the vertex B. Expand DM in both sides once, You obtain C and C'. Draw lines BC' and CB', they intersect in A. Ready.

Part 3 (Proof) Almost obviously the triangle, constructed above, is the required one.

Part 4 (Details) Reviewing Part 2, we notice, that the critical moment is the construction of the triangle B'DM, further all goes automatic. That to be possible (B'DM) requires:
(***) m_a + m_b > m_c;
(***) m_b + m_c > m_a;
(***) m_c + m_a > m_b - these conditions are necessary and sufficient. Thus we proved also the following statement: from the medians of arbitrary triangle it's always possible to construct another one (the latter has area 3/4 of the original area; both are nicely presented on the picture Dr Dumbfellow has provided a link to in his answer below)

Now the three altitudes problem - it's also interesting. Having
Area = a*h_a = b*h_b = c*h_c we obtain
a : b : c = (1/h_a) :(1/h_b) :(1/h_c),
h_a : h_b : h_c = (1/a) :(1/b) :(1/c),
that could lead to the following approach, I've encountered in some books: let us construct a triangle with sides h_a, h_b and h_c, then a, b and c are proportional to its altitudes, so the latter 3 altitudes form a triangle, similar to the requested. We can construct it and shrink or expand the sizes homotethically until we reach the altitudes to be equal to the given h_a, h_b and h_c.
Sorry, but this is defective. Did You notice the flaw? Then You can carry out the rest.

P.S.(EDIT)
1. Meanwhile Dr Dumbfellow has deleted his answer.
2. Ksoileau: Your results about the cosines of the required triangle's angles are perfectly correct (I haven't given a low rating of course)
cos γ = (a² + b² - c²)/(2ab) = (1/2)(a/b + b/a - (c/a)(c/b)) =
= (1/2)(h_a/h_b + h_b/h_a - (h_a/h_c)(h_b/h_c)) etc.
and this is a possible approach - to construct a similar triangle with the same angles after You have found them, or, even easier - by h_a, h_b and γ. The point in the common approach I described above is that medians in every triangle ALWAYS satisfy the inequalities (***) above, but the altitudes DO NOT ALWAYS (take a triangle with 2°, 89°, 89° angles), so to construct a triangle with sides h_a, h_b and h_c is not always possible - this is the flaw I mentioned above - surprisingly some authors of schoolbooks have overlooked it!

Answer 2

Given three medians, you can construct a triangle, but it is not unique.
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Ideas to construct the triangle: Start at 2/3 of the three segments of the medians, turn them around until you find a triangle.

Answer 3

I wouldnt think so. This problem has 4 unknowns (x,y) for 2 points on the triangle (set one corner at (0,0) to start). The number of knowns is 3, so it seems like you need one more piece of information to solve uniquely.

Ok, you can do it. I think you can start by letting the first corner be at (0,0) and the second bottom corner at (x_2,0), i.e. take any triangle, shift it so one corner is at (0,0) and then rotate it so that one edge lies along the x-axis. Then you only have to solve for the x_2 coordinate of this point and the coordinates (x_1,y_1) for the last corner. This has 3 unknowns and you have 3 equations for the lenghts of the medians, so you can solve it. The same should be true of the altitudes.

Answer 4

I´lltry to explain a geometric construction
Take a point M and draw a circunference radio m1(C1)
Take any radius a on it and take the point distant 1/3 m1 from M which would be the center of the triangle Z.The other end of the radius is vertex C
with center at Z and radius 2/3 m2 and 2/3 m3 draw two circunf.
C2 and C3
Take the symmetrical of C2 with respect to M.
Where this symmetrical cuts C3 you have one vertex A
Join this vertex with M and cut C2 and you get B

Answer 5

i think like if you turn them so that they cross 2/3 down the middle of each median and then build a triangle around those 3 medians but it's not definite b/c you can use any angles you want

Answer 6

I would think it is possible, but may only apply to right triangles. Constricting the length of two medians really limits the possiblities of what trinagles can be constructed.

Answer 7

Given
3in
and
4in
and
5in
Now u can construct the triangle

Answer 8

The length of the median of side a of a triangle is

Ma = sqrt(2b^2 + 2c^2 - a^2)/2

So let's assume we are given Ma, Mb and Mc for the length's of the median's of sides a b and c respectively.

So
Ma = sqrt(2b^2 + 2c^2 - a^2)/2
Mb = sqrt(2a^2 + 2c^2 - b^2)/2
Mc = sqrt(2a^2 + 2b^2 - c^2)/2

Ma^2 = (2b^2 + 2c^2 - a^2)/4 or
4Ma^2 = 2b^2 + 2c^2 - a^2

Likewise
4Mb^2 = 2a^2 + 2c^2 - b^2
4Mc^2 = 2a^2 + 2b^2 - c^2

4Mb^2 - 4Mc^2 = (2a^2 + 2c^2 - b^2) - (2a^2 + 2b^2 - c^2)
4Mb^2 - 4Mc^2 = 3c^2 - 3b^2

4Mb^2 + 2(4Ma^2) = (2a^2 + 2c^2 - b^2) + 2(2b^2 + 2c^2 - a^2)
4Mb^2 + 8Ma^2 = 6c^2 + 3b^2

(4Mb^2 + 8Ma^2) + (4Mb^2 - 4Mc^2) = (6c^2 + 3b^2) + (3c^2 - 3b^2)

8Mb^2 + 8Ma^2 - 4Mc^2 = 9c^2
4(2Mb^2 + 2Ma^2 - Mc^2) = 9c^2

c = 2sqrt(2Mb^2 + 2Ma^2 - Mc^2)/3

So likewise,
a = 2sqrt(2Mc^2 + 2Mb^2 - Ma^2)/3
b = 2sqrt(2Ma^2 + 2Mc^2 - Mb^2)/3

Now, knowing the length of the sides, we can by the law of cosines (and optionally the law of sines) find the angles.

Law of cosine:
cos(A) = sqrt((b^2 + c^2 - a^2)/(2bc))

so once we know the length of the 3 sides, we can find any of the three angles. Alternatively, just find one angle, and then use the law of Sines.

sin(A)/a = sin(B)/b = sin(C)/c

Answer 9

Dr D. I have just today taught this construction to my students. Its really very interesting. We can use the following link to explain the whole construction.

http://jwilson.coe.uga.edu/EMT669/Student.Folders/Cronic.Drew/untitled%20folder/Drew's%20Constructions

Answer 10

Given the altitudes a1, a2, and a3, by the Law of Cosines and the area formula for a triangle it follows that

cos(t1)=1/2*a2*a3
*(1/a2^2+1/a3^2-1/a1^2)
cos(t2)=1/2*a1*a3
*(1/a1^2+1/a3^2-1/a2^2)
cos(t3)=1/2*a1*a2
*(1/a1^2+1/a2^2-1/a3^2)

where
t1, t2, and t3 are the angles which include altitudes a1, a2, and a3 respectively.

EDIT Would someone who gave me a thumbs down point out my error? LOL