A ray extends from the midpoint of one side of a triangle...

A ray extends from the midpoint of one side of a triangle at a specified angle. The other two sides of the triangle meet on the ray.

What is the formula that relates the lengths of the two sides that meet on the ray?

This is a problem I came up with myself; I'm just trying to figure some things out. I asked it in the Math Community on Google+ and got no responses. I'm pretty sure I've stated it with all the necessary information. Looking forward to any responses! (Hi)

Re: A ray extends from the midpoint of one side of a triangle...

Quote:

Originally Posted by

**mackerm** A ray extends from the midpoint of one side of a triangle at a specified angle. The other two sides of the triangle meet on the ray. *It seems to me that you are talking about the median.*

What is the formula that relates *(?)* the lengths of the two sides that meet on the ray?

This is a problem I came up with myself; I'm just trying to figure some things out. I asked it in the Math Community on Google+ and got no responses. I'm pretty sure I've stated it with all the necessary information. Looking forward to any responses! (Hi)

Hello,

have a look here: Median (geometry) - Wikipedia, the free encyclopedia

Re: A ray extends from the midpoint of one side of a triangle...

Quote:

Originally Posted by

**earboth** It seems to me that you are talking about the median.

Well, yes, the ray is a median, but I mean it to determine two sides of the triangle, not the other way around.

Usually when you say "median" it means that you are given a triangle, and you draw a segment from an angle to the opposite side. I'm saying that one side of the triangle is given, and the ray (which functions as a median) is given via a specified angle.

What is the relationship between the two sides of the triangle which are NOT given?

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Re: A ray extends from the midpoint of one side of a triangle...

Quote:

Originally Posted by

**mackerm** Well, yes, the ray is a median, but I mean it to determine two sides of the triangle, not the other way around.

Usually when you say "median" it means that you are given a triangle, and you draw a segment from an angle to the opposite side. I'm saying that one side of the triangle is given, and **the ray** (which functions as a median)** is given via a specified angle.**

What is the relationship between the two sides of the triangle which are NOT given?

Let us assume that the length of the side c is given **and the length of $\displaystyle m_c$**

1. From the article of Wikipedia you can get:

$\displaystyle m_c=\sqrt{\frac{2a^2+2b^2-c}4}~\implies~2m^2+\frac12 c = a^2+b^2$

At the LHS of the equation are the lengthes of the known parts and at the RHS of the equation are those unknown sides of the triangle.

2. If $\displaystyle m_c$ has to be determined only by an angle (which angle do you refer to?) then this problem can't be done because you can't determine a complete triangle by the length of a side and the value of an angle. You need to know 3 (independent) parts of the triangle.

Re: A ray extends from the midpoint of one side of a triangle...

Thank you for adding that drawing. It should help.

I don't want the length of http://latex.codecogs.com/png.latex?m_c. I want a function that relates the lengths of sides *a* and *b*. Something in the form of:

*b=a+1 *

(This formula is obviously wrong. I'm just showing the form of the solution I want.)

Referring to your drawing, the only given values are angle *M* and side *c*, and the fact that point *M* is the midpoint of side *c*.

Re: A ray extends from the midpoint of one side of a triangle...

Quote:

Originally Posted by

**mackerm** ...

Referring to your drawing, the only given values are angle *M* and side *c*, and the fact that point *M* is the midpoint of side *c*.

Here is what I did so far, but I don't know if this is what you are looking for:

1. Let $\displaystyle \mu$ denote the angle $\displaystyle \angle(CMA)$.

2. Using Cosine rule with the triangle $\displaystyle \Delta(AMC)$ you'll get:

$\displaystyle b^2 = \left(\frac12 c\right)^2+m_c^2-2\cdot \frac12 c \cdot m_c \cdot \cos(\mu) \\ \implies~m_c=\frac c2 \cos(\mu) \pm \frac12\cdot \sqrt{4b^2-c^2 \cdot (\sin(\mu))^2}$

3. Using Cosine rule with the triangle $\displaystyle \Delta(MBC)$ and the property $\displaystyle \cos(\pi-\mu) = -\cos(\mu)$ you'll get:

$\displaystyle a^2 = \left(\frac12 c\right)^2+m_c^2+2\cdot \frac12 c \cdot m_c \cdot \cos(\mu) \\ \implies~m_c=-\frac c2 \cos(\mu) \pm \frac12\cdot \sqrt{4b^2-c^2 \cdot (\sin(\mu))^2}$

4. Since $\displaystyle m_c$ and $\displaystyle \mu$ are equal in both triangles I (tried to) solve the equation for b:

$\displaystyle \frac c2 \cos(\mu) \pm \frac12\cdot \sqrt{4b^2-c^2 \cdot (\sin(\mu))^2} = -\frac c2 \cos(\mu) \pm \frac12\cdot \sqrt{4b^2-c^2 \cdot (\sin(\mu))^2}$

Finally I got:

$\displaystyle b^2 = a^2+c^2\cdot (\cos(\mu))^2 + c\cdot \cos(\mu) \cdot \sqrt{4·a^2 - c^2 \cdot (\sin(\mu))^2}$

5. Now b² depends on a² and the value of angle $\displaystyle \mu$ - but again, I'm not sure that you were looking for such an equation. (Thinking)

Re: A ray extends from the midpoint of one side of a triangle...

Just an update. I've been fooling around with WolframAlpha's ability to solve simultaneous equations. I've clearly done something wrong, but here's what I've come up with so far. I'm using the same designations as before, with a couple additions:

a b c are sides of a triangle with c as the base.

m is the median of side c.

d is the altitude to side c.

e is a segment from point A to altitude d.

WolframAlpha Link to solution so far

I made the first equation m=1.1d just as a quick and dirty way to set an angle.

As I said, I've done something wrong. I intended it for it to have *a* be a function of *b*, and all the other variables would cancel out.