Find the size of a triangle given only 2 medians?

Question

In a triangle with vertices ABC and center point O, there exists 3 medians. AMa, BMb, and CMc. If AMa = 156 and CMc = 204, can the triangle be solved and how? I say it cannot be solved but my teacher says it can

If it cannot be solved, then what are at least 2 triangles that use those medians so I can show it has at least more than one answer.

I can solve the medians into parts:
AO= 104
OMa = 52
CO = 136
OMc = 68
But that is about it

I have looked at working with a parallelogram involving OMa and a/2 seeking to use a^2 + b^2 + c^2 + d^2 = x^2 + y^2 where a, b, c, d are the parallelogram sides and x and y are the parallelogram diagonals. I have tried to use the equal areas of the six small triangles also with no luck. I solved AOMc and COMa for cos theta and solved the equations against each other but that left me with 2 unknowns

I thought JD was on to something at first, but medians are not bisectors so applying the Law of Sines to A/2 or B/2 cannot be right.

I guess I will have to draw out santmann2002's answer as I don't quite follow it. However, he seems close to providing proof of nonexistance.

I hope to undertand this soon, because it is driving me batty. And to answer the question, "Is there something missing?" No, I have this on a printout and that is all the information I was given.

Answer 1

I've looked at this very carefully, checking out the case for special triangles, such as right triangles and isoceles. There isn't a general symmetric solution in terms of 2 medians, meaning that the ORDER of the medians matters. For example, in the case of right triangles, the solution is symmetric only if the medians bisects the orthogonal sides, not the hypotenuse. That prohibits a general solution for any triangle, so I think something got left out of this problem. I'll look at it some more, but I don't see anything.

Are you sure there isn't another fact left out, like maybe the medians bisects opposite sides of an isoceles triangle?

As an simple example of why given only 2 medians, one cannot determine an unique area, imagine that the 2 medians are equal, and bisected the sides of an isoceles triangle. A moment's thought will show that more than one isoceles is possible with such given medians, having different areas.

In vector geometry, the cross product of the 2 medians gives you twice the area of the triangle, but in order to compute this value, you still need information which comes out to knowing the angle they subtend. Give it up, this problem is short one card of a deck of 3.

Answer 2

Alas you teacher is correct. I'm not sure this is the easiest way but here is how I did it.

Place the tip of you triangle at (0,0) of a co-ordinate system, and the base along the x axis. Then attach the two medians to this base of unknown length X. I am calling the right side of the triangle Y, and the left Side Z, the right angle will be A the left B.

We know the two triangle side meet at the top thus we get two equations:

y*sin(A) = z*sin(b)
y*cos(a) = x-z*cos(b)

we know because it is a median that M1 goeas from (0,00 to intersect Z at exactly half its length so:

M1*sin(A/2) = Z/2 *sin(b)
M1*cos(A/2) = X-z/2*cos(B)

and from the other median

y/2*sin(a) =M2*sin(b/2)
y/2*cos(a)=x-m2*sin(b/2)

Also you know that the altidues at which the medians intesect thier respective sides are equal so:
y/2*sin(a)=z/s*sin(b)
m1*sin(a/2) = m2*siin(b/2)

So you now have 8 equations, and 5 unknowns, the three sides, and two bottom angles. Pick any 5 and solve.

Alas five equations and five unknowns is then a lot of arithmetic, so there might be a simpler way but I don't know it.

Answer 3

Given 3 medians, you are able to build a triangle, even nevertheless it is not unique. -------- ideas to construct the triangle: initiate at 2/3 of the three segments of the medians, turn them around till you hit upon a triangle.

Answer 4

Lets Draw the median AMa and take the point of this median distant 1/3 of its mesure from Ma. This is the point of intersection of the three median.Call it D.With center at D draw a circunference with radium 2/3 of CMc.This Circunference contains vertix C and its symetrical with respect to point Ma contains vertix B. (cf(1)
The point Mc is on a circunference center D and radium 1/3 of CMc.
Take its homothetic with center of homothetia A and ratio 2 cf(2)
The intersection of cf(1) and cf(2) gives you the vertix B.
and its symetrical respect Ma is C

Both cfs can intecept at two one or none point so there can be two one or no solution.

This is a solution without calculus only with ruler and compass.
I hope you can understand it