Easy Geometric Problem

stup_id

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Since I specialize in using this forum to all kind of things that aren't suppose to relate with it, i'm going to give you some math problem.

You'll see, I'm training kids for the math olimpyad (currently just junior high school level) and the official answer of this problem given by the University seems wrong to me...

so.. we have this figure:

MON is a sector from a circle, with angle MON = 60° or 1/3pi rad suit yourselves... ON = OM = 6,

what is the measure of the radius of the circumference that includes points M, O and N?

The answer is 3.33...., but the official answer says 5


ETA Correct answer
 

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Man, am I rusty! I've been at this one for a few minutes without much progress. Lots of scribbles, though. I'm determined to get it, though, without looking at the hint.
 
I can't understand your question. What do you mean by the 'radius of the circumference that includes M O and N?'

At first I just thought you meant the length of the circle segment MN but that would be 2*pi, and the answers you give in the spoilers are simple rationals.

This is me: :confused:
 
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
 
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this means, that you should asumme you draw an smaller circumference containing MON, and then calculate the radius of this given circumference i'm attaching another draw...
 

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The Don Is the question

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

Yes, that's the right sintaxis of the question.. sorry about the ambiguity...
 
You mean, If points O,M and N all lie on a circle, what is its radius?
2 sin 30 = 1
NM= 2Sin30 *r = 6
(Which is the only way the 3 points could lie on a circle.)
So the answer would be 3.

I think you meant "What is the radius of the Circle on which O.M an N lie"

ETA Dammit. Read the question. No it isn't.
 
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Did you see the figure i attached?


The Don got the right sintaxis of the question,

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

In the figure, we want to determine the measure of the red line, which goes from a point equidistant from N,M,O (thus a radius).

Remember that any 3 points determine a circumference so I don't get what you mean by
(Which is the only way the 3 points could lie on a circle.)
Click to expand...

There was a mistake in my original solution here it is corrected.
The answer is [latex]$$\frac{6\sqrt3}{3} = 3.46....$$[/latex]

I'm posting the complete solution as a spoiler also:


LOOK THE ATTACHED FIGURE
Since MON = 60°, and MO = ON, then MON is an equilateral triangle
(by Side-Angle-Side criteria of congruency of triangles),
and thus MN = MO = ON = 6

The circumcenter of a given equilateral triangle, lies in the intersection of its heights/bisecters/mediatrices
so, let us take the middle point of ON, A and joint it to M, the line MA is perpendicular to NO, thus we got
the right triangle, MAO with hypotenuse MO = 6, and cathete AO = 3.

MO^2 - AO^2 = MA^2
36 - 9 = MA^2
27 = MA^2
from pythagoras it follows that MA = 3*(sqrt[3])

Now the heights of the equilateral triangle, intersect at circumcenter C in a 2:1 ratio, so MC/CA = 2/1,
MC=2CA, but MC + CA = MA = 3*(sqrt[3]) thus
CA = {6*(sqrt[3]) } / 10 - MC

we subsitute that and get:

MC = 2*([3*(sqrt[3])] -MC)
MC = {6*(sqrt[3]) } -2MC
3MC = 6*(sqrt[3])
MC = 6*(sqrt[3])/3 = 3.4641...



ETA : corrections to the answer
 

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now.. the thing is precisely to find mistakes in my solution.. cause the "official one" seems different to me.. so i'm open to suggestion to find the real value...
 
that's true... thank you.. haha I always make that sort of mistakes... anyway the answer is then even more farfetched... :S not in a milion years 5...

Or what do you think?
 
No I don't see any figures and I can't get the spoiler to work either.
I'm on a pc at work which blocks all sorts of things .
I just figured NM as 6, then stopped thinking. An old vice.
 
The Don said:
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
Click to expand...


You have the same answer than me...
 
Looks to me like it's the square root of 12 (or, rounded, 3.46, like the others are saying).

That subtraction mistake seems to be your only mistake; once it's fixed, I think you're getting the square root of 12 as well.
 
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?
 
Here's my answer, same as the Don:


OC^2 = OA^2 + AC^2
with:
OA = OM/2
AC = OC/2

thus:
OC^2 = (OM/2)^2 + (OC/2)^2
OC^2 = (OM^2)/4 + (OC^2)/4
(3/4) OC^2 = (OM^2)/4
OC^2 = (OM^2 )/3
OC = sqrt(12) ~= 3.46
 
so is settled.. i'll have to write to the national organizers.. of this.. it is not an small mistake.. because this was easily one of the hardest problems of the test (the test is for youngers than 15 years old) and making it impossible to answer can affect the whole perfomance for the kids...

by the way.. if you want to have some fun and understand some of spanish (it is not very important since math are universal...) here are the links to the PDF's of the exams and the answers:

Official page: http://www.amc.unam.mx/programas/c_pprimavera.htm
direct link to test level 1 (younger than 13 years) = http://www.amc.unam.mx/programas/exa/primer_nivel06.pdf
direct link to test level 2 (younger than 15 years) =
http://www.amc.unam.mx/programas/exa/segundo_nivel06.pdf
And finally the answers to both.. (where if you are enough curious you can check the mistake...) =
http://www.amc.unam.mx/programas/exa/Respuestas_Primera_Etapa_2006.pdf

have fun! And thanks a lot!!!! :D
 
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.
 
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