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having produced your equilateral triangle HERE

you now have to put inside three equal size circles touching each other and two sides of the triangle - you have only a compass - none of the circles overlap each other r triangle... Nice snug fit!

Also easy peasy,

Edited by Lost in space
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Since all we have is a compass, I would begin with the corners of the triangle, as they are the only points that are defined. From there, I would create an arc from that point that has a radius tangent to the opposite side. Repeat from the other two points, creating 3 points where the arcs intersect. I suspect these points will be the center points of your circles, but I haven't found the language to describe why as of yet.

Given the available tools, and since you said it was easy, this seems the most logical method. All that's left is the prooving. :(

Edited by Grayven
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I haven't been able to get to a compass to verify this but I believe:

If you bisect all three angles you will have the incentre of the equilateral triangles, and each bisector will create two right angled triangles. Mark the midpoint of the equilateral and then proceed to find the incentre of 3 of the right angled triangles. Use these new centres as the centres of the circles and set the compass radius to the centre of the original equilateral triangle.

//EDIT: scratch that. . . it's wrong :lol:

//EDIT: I think I'm nearly there, just missing one thing <_<

Using the method in the first spoiler box, get the right angled triangles (bisect the angles) and then draw the incircles for each triangle. Tah Dah! Only problem is getting the radius of said circles without a little trial and error.

Edited by Exelcior
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We're not told that we have a straight-edge or ruler at our disposal, but it will be necessary to drop perpendicular lines from the corners to the midpoints of the three sides of the triangle in order to construct the circles (or at least, that's how I'd do it). Fold the triangle in half so that the opposite corners of an edge touch and mark the midpoint of the edge. Then, place your compass on the midpoint and draw an arc with a radius equal to half the length of a triangle side. The point where the arc touches your fold line is the center of the circle you're looking for. Place the point of your compass there and draw a circle tangent to the two closest edges of the triangle. If you've already made the other two folds, you'll find that the circle is also tangent to those fold lines. Turn the triangle and repeat twice to get the other circles.

post-6822-1230741940.gif

Edited by HoustonHokie
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We're not told that we have a straight-edge or ruler at our disposal, but it will be necessary to drop perpendicular lines from the corners to the midpoints of the three sides of the triangle in order to construct the circles (or at least, that's how I'd do it). Fold the triangle in half so that the opposite corners of an edge touch and mark the midpoint of the edge. Then, place your compass on the midpoint and draw an arc with a radius equal to half the length of a triangle side. The point where the arc touches your fold line is the center of the circle you're looking for. Place the point of your compass there and draw a circle tangent to the two closest edges of the triangle. If you've already made the other two folds, you'll find that the circle is also tangent to those fold lines. Turn the triangle and repeat twice to get the other circles.

post-6822-1230741940.gif

having produced your equilateral triangle HERE

you now have to put inside three equal size circles touching each other and two sides of the triangle - you have only a compass - none of the circles overlap each other r triangle... Nice snug fit!

Also easy peasy,

You can probably fold, but no other tools are available. I think your picture illustrates how my solution fails, however. I think I need another step or two in the middle somewhere.

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We're not told that we have a straight-edge or ruler at our disposal, but it will be necessary to drop perpendicular lines from the corners to the midpoints of the three sides of the triangle in order to construct the circles (or at least, that's how I'd do it). Fold the triangle in half so that the opposite corners of an edge touch and mark the midpoint of the edge. Then, place your compass on the midpoint and draw an arc with a radius equal to half the length of a triangle side. The point where the arc touches your fold line is the center of the circle you're looking for. Place the point of your compass there and draw a circle tangent to the two closest edges of the triangle. If you've already made the other two folds, you'll find that the circle is also tangent to those fold lines. Turn the triangle and repeat twice to get the other circles.

post-6822-1230741940.gif

That's nearly the same as my method, just with folding instead of bisections :P

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1. Starting at any point extend compass to touch opposite line and draw an arc to touch other 2 lines.

2. Measure the points the arc touches the lines and divide the distance in half.

3. Place the pencil at the point and needle towards the opposite line draw a circle, repeat 2 more times

4. The points at which the circles intersect are the center of your circles.

5. The radius of the circles is from the dot to the line

triangle.bmp

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1. Starting at any point extend compass to touch opposite line and draw an arc to touch other 2 lines.

2. Measure the points the arc touches the lines and divide the distance in half.

3. Place the pencil at the point and needle towards the opposite line draw a circle, repeat 2 more times

4. The points at which the circles intersect are the center of your circles.

5. The radius of the circles is from the dot to the line

triangle.bmp

Nice solution. Except step #2 uses a tool not provided in the OP. We have ONLY a compass.

Welcome to the Den :D

P.S. I think your picture helps me fill in the gap. We'll see.

Edited by Grayven
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1. Starting at any point extend compass to touch opposite line and draw an arc to touch other 2 lines.

2. Measure the points the arc touches the lines and divide the distance in half.

3. Place the pencil at the point and needle towards the opposite line draw a circle, repeat 2 more times

4. The points at which the circles intersect are the center of your circles.

5. The radius of the circles is from the dot to the line

triangle.bmp

We can't measure anything, we don't have a measuring device, only a compass.

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That's nearly the same as my method, just with folding instead of bisections :P
Yes, but don't you need a straight edge to complete the bisection of an angle? As I recall, the process goes like this:

1. With compass point on angle vertex, swing an arc from side to side of angle.

2. With compass point on intersection of arc and side of angle, swing an arc with radius of at least half the distance between the two intersections interior to the arc. Repeat on the other side.

3. The intersections of the two new arcs will be on the bisecting line, which you can connect to the original angle vertex with a straight edge.

And the OP doesn't say that we have a straight edge... So I folded instead. Of course, the OP doesn't say that our triangle is on paper, either - but it does say that we got the triangle from our work on bonanova's topic, which did specify paper (and allowed use of a straight edge, but I think that tool has been taken away).

If I had a way to guarantee that my bisection point was at the center of the desired circles, then I'd say that bisection would work. Otherwise, I'm going to fold. B))

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Yes, but don't you need a straight edge to complete the bisection of an angle? As I recall, the process goes like this:

1. With compass point on angle vertex, swing an arc from side to side of angle.

2. With compass point on intersection of arc and side of angle, swing an arc with radius of at least half the distance between the two intersections interior to the arc. Repeat on the other side.

3. The intersections of the two new arcs will be on the bisecting line, which you can connect to the original angle vertex with a straight edge.

And the OP doesn't say that we have a straight edge... So I folded instead. Of course, the OP doesn't say that our triangle is on paper, either - but it does say that we got the triangle from our work on bonanova's topic, which did specify paper (and allowed use of a straight edge, but I think that tool has been taken away).

If I had a way to guarantee that my bisection point was at the center of the desired circles, then I'd say that bisection would work. Otherwise, I'm going to fold. B))

Ahh very true <_< I didn't think about that :P your method wins so far then!

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1. Starting at any point extend compass to touch opposite line and draw an arc to touch other 2 lines.

2. Measure the points the arc touches the lines and divide the distance in half.

3. Place the pencil at the point and needle towards the opposite line draw a circle, repeat 2 more times

4. The points at which the circles intersect are the center of your circles.

5. The radius of the circles is from the dot to the line

triangle.bmp

If I concede that your inexpensive compass has markings on it which let you divide the radius in half (I'm just jealous), I still have an issue with your approach: in step 3, how do you make sure that the needle is on the perpendicular between the angle point and the opposite side without being able to use a straight edge or folding in some way - neither of which are specified in your solution? In other words, how do you find the center of those circles and guarantee they won't be oriented like this? The diagram is exaggerated, but you get my point.

post-6822-1230747184.gif

Edited by HoustonHokie
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I have to agree that there can be errors made in my solution. If you reverse the compass and draw a small arc, the lowest point of the arc is the center. Again this leaves room for error, the larger the triangle the easier it would be to induse an error.

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1. Starting at any point extend compass to touch opposite line and draw an arc to touch other 2 lines.

2. Measure the points the arc touches the lines and divide the distance in half.

3. Place the pencil at the point and needle towards the opposite line draw a circle, repeat 2 more times

4. The points at which the circles intersect are the center of your circles.

5. The radius of the circles is from the dot to the line

triangle.bmp

Your solution is very nice, can you tell me how did you come to this solution, I mean the logic in it? For example why are you dividing the distance. I assume your solution is correct, but can you prove it with math? So that I'll understand the solution thoroughly. I have no idea where the centre of circles should be geometrically.

A small objection: At step 3 after placing the pencil, you place the needle towards the opp. line, but practically how can you be sure that the needle is at exactly the closest point to opp. line?

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I know that but a compass has numbers and lines on them. At least the $1 ones that I buy my kids do.

That's funny - the high-powered one I got for my college drafting courses doesn't have numbers! Maybe I should have gotten the cheap kind! :lol:

Almost the compasses I've seen have numbers on.... except the really cheap ones which just have N,S,E and W.

If we were talking about a 'pair of compasses' on the other hand...

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There is more than one way to achieve straight line - folding is easiest, what of this was on heavy gauge steel ???
Assuming the pencil lead in my compass could draw on heavy gauge steel, this is another possibility.
I've already assumed that I can draw an arc tangent to a straight line with a compass and not miss, so I guess it's not too much of a stretch to think I can draw an arc tangent to another arc. With that in mind, here's a solution that doesn't need to draw any straight lines - just arcs. This time, I've put an X at the radius point of my arc so that you can see where the compass needle goes. The first arc is drawn tangent to the opposite edge of the triangle so that I can find the midpoint of that edge for the next arc. The final radius point is the tangent intersection of two arcs. No measuring, no folding, no dividing. Just a compass and a really good eye.

post-6822-1230933349.gif

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Positive theory and 'technically it works, though in practice I would not use it - try it and see.

It was one of the answers I submitted for an exam (41 years ago). It was an acceptable answer, but it bothered me the same way that it does for HH. It is reliant on 'perfect' tangents.

Is that the final answer? What's need is a way to define the exact points and not rely on tangents alone.

Is it possible to find another way outside the box/triangle?

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Positive theory and 'technically it works, though in practice I would not use it - try it and see.
You're right. I did try it before I submitted the post and it failed. My circles were too small and didn't touch each other. One problem is that it is almost impossible to keep the compass perpendicular to the paper. The other, of course, is being able to pick out the tangent point of a circle by eyeballing it. This solution relies on both perfect circles and perfect tangents. If it were one or the other, it would probably have worked. Almost pulled the post, but I figured it was still closer to the solution you were looking for. I love a good geometry problem, so I'll keep looking. B)) Edited by HoustonHokie
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can be accomplished using only a compass and a series of iterations. Starting with the bottom side of the triangle, put the point of the compass on one end of the line and set the compass to a size just smaller than the length of that line. Strike an arc inside the triangle, and move the point to the other end of the line. Without changing the compass setting, strike another arc intersecting the first arc. Adjust the compass to a slightly smaller setting and repeat. Continue until you have several intersecting arcs inside the triangle. From there you should be able to connect the intersections with a straight line (you'll have to freehand it, since no straightedge was allowed, but if you have enough pairs of arcs, it should be fairly easy). Then repeat the whole process for the other two sides to get your three bisecting perpendiculars, from which you can follow HH's method of creating the 3 circles. I hope my explanation was clear, for I have no good way to illustrate what I'm talking about with this computer. Perhaps tomorrow I'll sketch something at work.

^_^
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Assuming the pencil lead in my compass could draw on heavy gauge steel, this is another possibility.
I've already assumed that I can draw an arc tangent to a straight line with a compass and not miss, so I guess it's not too much of a stretch to think I can draw an arc tangent to another arc. With that in mind, here's a solution that doesn't need to draw any straight lines - just arcs. This time, I've put an X at the radius point of my arc so that you can see where the compass needle goes. The first arc is drawn tangent to the opposite edge of the triangle so that I can find the midpoint of that edge for the next arc. The final radius point is the tangent intersection of two arcs. No measuring, no folding, no dividing. Just a compass and a really good eye.

Your solution is very simple to understand. But to reach that solution you need to know that:

post-10474-1231149026.jpg

You assume that the red line (height of a circle centre) is at half length of the base blue line.

Sure it's correct but how could you calculate this?

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Your solution is very simple to understand. But to reach that solution you need to know that:

post-10474-1231149026.jpg

You assume that the red line (height of a circle centre) is at half length of the base blue line.

Sure it's correct but how could you calculate this?

The arc drawn in the third diagram which has the red line as the radius has a diameter equal to the blue line. The radius is half the diameter.

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The arc drawn in the third diagram which has the red line as the radius has a diameter equal to the blue line. The radius is half the diameter.

I didn't mean that. Of course the red line is 1/2 of blue line.

But my question is that:

How do you know that centre of the circle need to be at that point? Which calculation gave you the information that the centre is "1/2 of blue line" apart from blue line?. Sorry for wording.

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