Tough geometry problem about triangles, proof

[physics user1]

Homework Statement

let be ABC a generic triangle, build on each side of the triangle an equilater triangle, proof that the triangle having as vertices the centers of the equilaters triangles is equilater

Homework Equations

sum of internal angles in a triangle is 180, rules about congruency in triangles

The Attempt at a Solution

i know two ways to demontrate that a triangle is equilater, congruency of the sides or same angles, but i have no clue on how to do it, i didn t found any relation between the triangles, only some opposite angles

 

[QuantumQuest]

If you haven't already done so, create a good diagram. Then see what does the center of the equilateral triangles give you regarding properties and see if it's better to work with angles or sides for the requested triangle.

 

[physics user1]

i still can't find a way

 

[QuantumQuest]

By now, I assume you have a good diagram. Now, have you learned about cevians?

 

[physics user1]

I didn't get in deep with the argument, I just know the definition of it, a segment that connects a vertex with the opposite side

 

[berkeman]

@Cozma Alex -- please post your sketch of the problem or your thread will be deleted.

 

[jedishrfu]

As a suggestion, if you have a smartphone with a camera, you can hand draw a clean well labeled diagram, take a picture and upload the image to this thread.

 

[physics user1]

I got the image of the problem

 

[physics user1]

Since the triangle ABC in the given image of the problem doesn't look generic, and following yours advice I also drawn one by myself

(Sorry for the double posting of the image, it was because of the lag)

I think a way is to use similar triangles and then get to "same angles way"

 

[QuantumQuest]

I think a way is to use similar triangles and then get to "same angles way"

Do you know which theorem is this thing you're trying to prove?

 

[physics user1]

Similar triangles leads to equals angles between those triangles

 

[QuantumQuest]

It has a specific name. Anyway, have you learned / studied the properties of equilateral triangles? If not take a look here https://en.wikipedia.org/wiki/Equilateral_triangle. This is absolutely necessary. Second, do you know the fundamentals of trigonometry ( including law of sines and cosines)? I am asking this, because my hints so far, go towards a geometric direction of the proof involving a little fundamental trigonometry.

 

[physics user1]

Yeah I know trig

 

[QuantumQuest]

OK, so I am assuming that you know well both things I pointed out in my previous post. Now, going to your sketch, can you see any way to relate a side of the inner triangle (the one you want to prove equilateral) with some known angles from the outer equilateral triangles? As an extra hint, look at cevians which are among other things, angle bisectors.

 

[physics user1]

I see that every side is opposite to an angle: 60+ angle of the ABC

Side opposite to angle reminds me sine theorem but that works only in the same triangle and those are different

 

[QuantumQuest]

Side opposite to angle reminds me sine theorem but that works only in the same triangle and those are different

So, you want to find a triangle formed by one side of the inner triangle and two cevians of the outer triangles, in order to have one triangle to work on. Also, what about cosines law as well?

 

[physics user1]

I can find the length of AQ and AS in function of AC and AB and with cosine law find the length of a side of the inner triangle using the angle between AQ and AS is that the way?

 

[QuantumQuest]

...and with cosine law find the length of a side of the inner triangle using the angle between AQ and AS is that the way?

Yes. Form the law of cosines and see how you can find the length for each side included in the expression, in order to substitute.

 

[physics user1]

Ok thanks for your help and patience :)

 

[QuantumQuest]

Ok thanks for your help and patience :)

You 're welcome. Now, can you see how to proceed further? There are some more steps to do in order to be done.

 

[physics user1]

There is still a problem, how can I prove that these relations connecting each side found are the same?

Here are the relations:

 

[berkeman]

There is still a problem, how can I prove that these relations connecting each side found are the same?

Here are the relations:

Sorry Alex, your image is unreadable. Can you try again, or type your work into the forum window? Thanks.

 

[physics user1]

Here

 

[QuantumQuest]

There is still a problem, how can I prove that these relations connecting each side found are the same?

Isn't it easier to work with one of them e.g. the first? Try substituting what you previously didn't. Then look at your sketch again. There is one more triangle of interest down the line...

 

[physics user1]

Sorry, I can't find the other triangle of interest :( also what can I do with the first relation?

 

[QuantumQuest]

Sorry, I can't find the other triangle of interest :( also what can I do with the first relation?

For the second question, is it better to have $\cos(60 + \alpha)$ or expand it? If you do, you'll have $\cos$ or $\sin$ separately. Think about that.
For the first question, you have used the three external equilateral triangles and the inner that you want to prove equilateral. There is exactly one triangle, that you have not used yet.

EDIT: Don't forget your goal. You must somehow end up with a relation that concludes about sides of the inner triangle.

 

[physics user1]

Are you referring to ABC?

 

[QuantumQuest]

Are you referring to ABC?

That's right. This has an interesting angle regarding your previous calculations and some interesting sides. But what can you do on this triangle? You have again to connect angles with sides, so you know what to do by now. Your goal is substituting things into the first relation you initially wrote. But there is something more, regarding ABC, that you can use.

 

[physics user1]

Ok, thanks for the help, I am going to use the sine and the cosine theorem on ABC to find something
I can go alone by now, thanks again for the patience
:)