MHB What is the solution to this vector geometry problem?

[Saitama]

Problem:

The median AD of the $\Delta$ ABC is bisected at E. BE meets AC in F. Find AF:AC.

Attempt:

Let point E divide BF in the ratio $\mu : 1$ and let F divide the line AC in the ratio $\lambda : 1$.

I take A as the origin. Then,

$$\vec{AD}=\frac{1}{2}(\vec{AB}+\vec{AC})$$
$$\vec{AF}=\frac{\lambda \vec{AC}}{\lambda+1}$$
$$\vec{AE}=\frac{\mu\vec{AF}+\vec{AB}}{\mu+1}$$
Also,
$$\vec{AE}=\frac{1}{2}\vec{AD}=\frac{1}{4}(\vec{AB}+\vec{AC})$$

I can substitute the above in the 3rd equation but I don't see what to do with $\vec{AB}$ as I have to find the ratio AF:AC.

Any help is appreciated. Thanks!

 

[Opalg]

Problem:

The median AD of the $\Delta$ ABC is bisected at E. BE meets AC in F. Find AF:AC.

Attempt:

Let point E divide BF in the ratio $\mu : 1$ and let F divide the line AC in the ratio $\lambda : 1$.

I take A as the origin. Then,

$\vec{AD}=\dfrac{1}{2}(\vec{AB}+\vec{AC})$

$\vec{AF}=\dfrac{\lambda \vec{AC}}{C+1}$ (1)

$\vec{AE}=\dfrac{\mu\vec{AF}+\vec{AB}}{\mu+1}$ (2)

Also,
$\vec{AE}=\dfrac{1}{2}\vec{AD}=\frac{1}{4}(\vec{AB}+\vec{AC})$ (3)

I can substitute the above in the 3rd equation but I don't see what to do with $\vec{AB}$ as I have to find the ratio AF:AC.

Any help is appreciated. Thanks!

You are on the right lines. Substitute the expression for $\vec{AF}$ from (1) into (2). Together with (3), that will give you two separate expressions for $\vec{AE}$ as a combination of $\vec{AB}$ and $\vec{AC}$. By comparing the coefficients of $\vec{AB}$ and $\vec{AC}$ in those two expressions, you will get two equations for $\lambda$ and $\mu$. Then once you know $\lambda$ you can find the ratio AF:AC.

 

[Saitama]

Hi Opalg!

You are on the right lines. Substitute the expression for $\vec{AF}$ from (1) into (2). Together with (3), that will give you two separate expressions for $\vec{AE}$ as a combination of $\vec{AB}$ and $\vec{AC}$. By comparing the coefficients of $\vec{AB}$ and $\vec{AC}$ in those two expressions, you will get two equations for $\lambda$ and $\mu$. Then once you know $\lambda$ you can find the ratio AF:AC.

Thanks a lot, I always forget about comparing the coefficients. Thank you very much. :)

I was wondering if there is a way to solve this problem without the use of vectors? If there is a way, can you please give me a few hints?

Many thanks!

 

[johng1]

Hi,
The attached solution just uses analytic geometry; it is in many respects similar to your vector solution, but it is different. I tried unsuccessfully to think of a synthetic solution.

View attachment 1745

 

[Saitama]

Hi,
The attached solution just uses analytic geometry; it is in many respects similar to your vector solution, but it is different. I tried unsuccessfully to think of a synthetic solution.
<image>

Thanks johng for the alternative method. :)

 

[johng1]

HI again,
I guess I liked your problem. Attached is a "purely" synthetic proof of the result.

View attachment 1750

 

[Saitama]

HI again,
I guess I liked your problem. Attached is a "purely" synthetic proof of the result.
<image>

Awesome! I could never really think of it. Thanks a lot johng. :)

Can you please share some motivation behind the construction? I am really interested to know how you approach such problems. Thanks!

 

[johng1]

Hi Pranav,

I like geometry a lot, but I'm not very good at it. So my thoughts probably aren't worth much.

I knew the answer for the unknown point. So it occurred to me that maybe it's the centroid of some triangle. I started casting about for such a triangle. Pretty soon the construction became apparent. Not much help, I bet.

Thanks for your kind words.

 

[caffeinemachine]

Hi Opalg!
Thanks a lot, I always forget about comparing the coefficients. Thank you very much. :)

I was wondering if there is a way to solve this problem without the use of vectors? If there is a way, can you please give me a few hints?

Many thanks!

Hey.

Construct a line $l$ passing through point $D$ and parallel to $BF$, and say this line intersects $AC$ at $M$.

Argue that $M$ is the mid-point of $FC$. This gives $AF/FM=2\lambda$

Also, by similarity of $\Delta AEF$ and $\Delta ADM$, we have, $1=AE/ED=AF/FM=2 \lambda$.

 

[johng1]

CaffeineMachine, very nice simple proof. I modified your proof slightly to give the attached generalization. I had reached this generalization previously with my more complicated solution, but it got kind of messy.

View attachment 1756

 

[caffeinemachine]

CaffeineMachine, very nice simple proof. I modified your proof slightly to give the attached generalization. I had reached this generalization previously with my more complicated solution, but it got kind of messy.

https://www.physicsforums.com/attachments/1756

Thanks. :)
What software do you use to make these diagrams?

 

[Saitama]

I am really sorry, I saw the posts but forgot to reply. :(

Great and short method caffeinemachine, I really like it, thanks! :)

And nice generalisation johng, thanks a lot for the help. :)