Easy Geometric Problem

[stup_id]

Well I got 5 at first so I'm right and you're all wrong! I was doing it all in my head and made a pretty silly assumption...

I got the equilateral triangle, and broke it down eventually to a right angled triangle with 30 and 60 degree angles in the other two corners. Nice round numbers. One side of the triangle was length 3. Nice and round again.

At this point I wrongly guessed that the triangle was the old 3,4 & 5 sided triangle that regularly crops up in simple maths problems. Therefore I skipped straight to r=5...

If you need any more wrong answers you know where I am.

PS. After thinking about it, I got 3.4something as above.

Maybe there's where the mistake of the organisers come from...

 

[The Don]

It should be obvious that the answer is not 5.

Try drawing the figure as it would have to be, for OM and ON to each be 6, but for all three points to be on a circle with radius of 5.

There's no way angle MON can be 60°.
Obviously, someone screwed up.

As a suggestion, try working back from the answer of 5, to figure out what angle MON would need to be on the original circle for this to work.

That may give some indication of where they went wrong?

The way I figure it..... 106.26^o

 

[wollery]

Shrinker's answer is correct for the problem as it is stated in the OP.

My solution

----------------------------------------
If angle MON is 60 degrees then it can be considered an equilateral triangle.
You then have two right angle triangles, OPN, where P is the midpoint of MN, and PCN, where C is the centre of the new circle. Both 30/60/90 triangles. From there it's simple trigonometry.
The Hypoteneuse of OPN (ON) is 60 which gives OP a length of 5.196.
NP is 3 which gives PC a length of 1.732.
OP - PC gives OC, the radius of the new circle, which is 3.464.
----------------------------------------

edited for clarity
re-edited for a major cock-up in my notation!!

 

[sphenisc]

At this point I wrongly guessed that the triangle was the old 3,4 & 5 sided triangle that regularly crops up in simple maths problems. Therefore I skipped straight to r=5...

That's brilliant Shrinker, I think it's quite likely that was the mistake that was made.

 

[Almo]

I got 3.464 as well. That's a cool problem.

 

[I'll_buy_that]

You are all thinking to hard on this one.

Draw a new circle including points MNO on its circumfrence.

Bisect the triangle that makes the slice of the new circle to point A.
Draw a line from M to A.
The new triangle is a simple 30 60 90 triangle which as pointed out can be solved through the 3,4,5 rule.

the measure of MO = 6, therefore the measure of OA=10.
OA is the diameter of the new circle, 1/2 OA = 5

 

[chipmunk stew]

Is the question

"What is the radius of the circle which has points M, N and O on its circumference ?"

edited to add....

If that's the question, then I get an answer of 3.46

This is the answer I got.

 

[chipmunk stew]

The new triangle is a simple 30 60 90 triangle which as pointed out can be solved through the 3,4,5 rule.

Actually, it can't.

A 30 60 90 has sides x, x-sqrt-3, and 2x.

Which negates the rest of your solution.


eta: btw, the angles of a 3,4,5 are about 37, 53, and 90 degrees.

 

[wollery]

I'll_buy_that, try a rough scale drawing before you pronounce, may save embarassment in future.



ps, if it were a 3,4,5 triangle MO would be the 4 side, not the 3 side, which would be MA. But it isn't a 3,4,5 triangle so it's a moot point!