I love geometry

1. May 16, 2005

juef

I love geometry :)

Hey all!

This is one problem I haven't been able to solve. I've had a class on theorems relative to circles and triangles but I still can't figure this one out.

3 identical, equilateral triangles are placed as seen on the attachment. Their sides measure 28cm, and the contact between them is "perfect". A circle could be drawn (see attachment) around the 3 triangles. What's the shortest possible diameter for the circle?

I absolutely have no idea. The center of the circle doesn't seem to be an important point here... and I can't think of an applicable theorem.

Thank you very much for looking! :)

Juef

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2. May 16, 2005

jdavel

juef,

I don't understand the question. Don't you have to know the length of the contact between the top and left triangle? Or is that what you're supposed to find to minimize the radius of the circle?

Also, why don't you think finding the center of the circle will be important. It seems crucial to me, no matter what is being asked.

3. May 16, 2005

Moo Of Doom

Almost sounds like a calculus problem... :P

What happens to the tangent points as you slide the top triangle along the other one? These tangents form a triangle, right? How do you find the circumscribed circle of a triangle? How do you find its radius?

Hope this helps.

4. May 17, 2005

juef

The circumscribed circle is formed by the intersection of the right bisectors of the sides of the triangles. I had no idea about its radius, but after a quick search on google I found that it's something like abc/4A, where a,b,c are the lenghts of the sides of the triangle, and A its area.

I can only find the lenght of one side of the triangle. I think I see what you mean by telling me to move the top triangle and everything, but... I still don't get it I guess. Sorry :/

Thanks a lot by the way :)

5. May 17, 2005

juef

I don't really know if we have to find it, but we sure don't have it...
You're probably right about the center after all, it seems we can't get the radius otherwise.

6. May 17, 2005

PeteSF

The center of the circle will always be on the line AB (see attached image).

If triangle 1 were aligned with triangle 2 (F coincident with A), then the center of the circle would be at point A.

If triangle 1 were slid right up until G were coincident with A, then the center of the circle would again be at point A.

When triangle is midway between these points, the center of the circle is somewhere along A-B... but that's all I've got so far.

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7. May 17, 2005

juef

Yup, you're certainly right, because AB is on the right bisector of the circumscribed triangle... Now I was thinking of finding some sort of relation between the distance AG and the radius of the circle, but since AG isn't really defined, I don't see what we could get out of it.

I also thought of using Ceva's theorem (because the three right bisectors intersect at the circle's center), but since it involves right bisectors, it doesn't give us anything new.

I also tried to use Héron's formula to simplify the formula I previously stated (abc/4A), but even after the substitution of the lenght of one of the sides, is it still impossible to find out the answer.

8. May 18, 2005

PeteSF

Consider the attached diagram (which I made too small. Sorry!)
H is the midpoint of CE
I is the center of the circle (where the perpendicular bisector of CE meets AB).
J is where a perpendicular dropped from H meets CF.

Now:
AC = 28cm = HJ.tan(b) + HJ/tan(b) + x.tan(b) + x.tan(60°)
where:
HJ = half of height of triangle = 28cm.sqrt(3)/2/2 = 7√3 cm = 12.1cm
b = angle ACE
x = perpendicular distance of I from AC

Now you can manipulate this equation to give x in terms of tan(b).
Once that's done, you need to manipulate tan(b) to maximize x (this will minimize the radius IC), and from there calculate what the radius is!

There's probably a more elegant way, but that should work.

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9. May 18, 2005

PeteSF

After a touch more work, it looks like you need basic calculus to find the minimum in that equation... Have you done derivatives using the quotient rule?

10. May 18, 2005

juef

Yeah, I've done all that stuff. Anyway Maple is here to help me :D

Are you sure b = angle ACE? It's angle CHJ on your diagram. Anyhow, I'm not sure I get that:

AC = HJ.tan(b) + HJ/tan(b) + x.tan(b) + x.tan(60°).

You're adding all the parts of the AC segment, right? In other words, CJ + JG + GZ + ZA? Then, I can understand CJ = HJ.tan(b) but not the rest of the equation. Maybe I'm just bad at trigonometry though, I haven't done that in a while.

Anyway, I did what you suggested about giving x in terms of tab(b) and plotting it with b from 0 to Pi/2 (thank you Maple! :D). But with these tan's in the denominator of the expression, I get asymptots which don't make much sense in this situation. Maybe I'm doing something wrong?

Hey, I appreciate your help a lot!! Thank you.

11. May 18, 2005

PeteSF

It's more likely that I've got mistakes in there (you picked one up already - angle b is indeed CHJ, as on the diagram).

Yes, the idea was to add all the bits of the AC segment. See if you can go through and add them all yourself, without worrying about how I tried to do it.
Start by drawing a larger diagram, label all the points, and mark all the angles you can.