How to Draw a Perfect Triangle?October 23, 2007 3:59 PM   Subscribe

My friend myn asks: If you wanted to draw a n equalateral triangle dead center on a rectangle. How would you most efficinently do this using a pen, paper, ruler? THANKS!
posted by fe2dell to Education (23 answers total)

Can we have a compass too?
posted by ikkyu2 at 4:05 PM on October 23, 2007

Measure the length of the bottom of the long side of the rectangle. Find the middle point. Draw a straight line up to the top of the rectangle. Measure this length, it is the height of the rectangle. Let this be "b". "a" will be the length of the sides of your triangles. Now, your equation will be:

(1/2a)2 + b2 = a2

Solve for "a". "a" is the length of each side of the equilateral triangle. Subtract "a" from the length of the long sides of the rectangle. Divide by two and measure in that amount from the corners of the bottom line of the rectangle. Signify where these points are. Draw lines from these two points to the top center point of the rectangle, and you have your equilateral triangle. I think.
posted by billysumday at 4:23 PM on October 23, 2007

posted by chrisamiller at 4:25 PM on October 23, 2007

Need more information. What's the desired relationship between the triangle and the rectangle? If it's supposed to look something like this [/\] then billysunday's method should work, except that instead of solving the quadratic equation yourself, you can just use the general solution from trigonometry: the side length of an equilateral triangle is two divided by the square root of three (that's approximately 1.155) times the height.
posted by flabdablet at 4:30 PM on October 23, 2007

For starters, this is only possible with rectangles of certain dimensions if you want the triangle to remain inside the rectangle. I assume you want the triangle to have it's base be a segment of the long side of your rectangle. So really the problem is how do you draw an equilateral triangle between two parralell lines, such that all it's points lie on these lines.

I'm not sure if this can be done totally compass/straightedge, but here is euclid's method of construction an equilateral triangle generally.
posted by phrontist at 4:32 PM on October 23, 2007

flabdablet, the triangle must be "dead center" in the square. Which means, I think, that the point where the two diagonals of the square meet *is* the center of the triangle, which in turn is the point where the 3 hypotenouses of the triangle meet. This is why billysumday's answer is unique.
posted by carmina at 4:54 PM on October 23, 2007

... but the orientation of the triangle can change, 360 degrees around, no?
posted by carmina at 4:55 PM on October 23, 2007

I think your friend wants you to think outside the box. Mark the midpoints of the long sides. Draw a line bisecting the rectangle and continuing upwards beyond it. Use the ruler to find the point on that line that is one rectangle length from each bottom corner.
posted by weapons-grade pandemonium at 4:58 PM on October 23, 2007

not hypotenouses, dammit, I meant the dichotomouses, stupid greek
posted by carmina at 4:59 PM on October 23, 2007

I just did it without the pen and ruler, by folding the paper.
posted by weapons-grade pandemonium at 5:05 PM on October 23, 2007

He said dead center on a rectangle, right? Not in a rectangle.
posted by weapons-grade pandemonium at 5:16 PM on October 23, 2007

I concede.

Oh, dammit, this is geometry, not linguistics. Pfft.
posted by carmina at 5:22 PM on October 23, 2007

Yes, but even speaking only in geometrical terms it would be virtually impossible to define where such a triangle would be if it were "in" a rectangle. What would be equidistant from what? Are you talking area or points? What points?

So it has to be "on".
posted by weapons-grade pandemonium at 5:42 PM on October 23, 2007

On one interpretation, the centre of the equilateral triangle coincides with the centre of the rectangle, and nothing can be assumed about the size or orientation of the triangle.

Another possibility is that one side of the triangle coincides exactly with one side of the rectangle, and the triangle and rectangle share an axis of symmetry. If that's one of the short sides, the "dead centre" specification looks a bit dodgy unless the rectangle is drawn with the short sides horizontal.

Yet another possibility, which is the one billysunday has addressed, is that one side of the triangle forms part of the long side of the rectangle, the triangle and rectangle share an axis of symmetry, and the height of the triangle is the same as the length of the rectangle's short sides.

There are probably other constructions that could plausibly fit the criteria as well. So, what exactly is it you want us to tell you how to draw?
posted by flabdablet at 6:13 PM on October 23, 2007 [1 favorite]

w-pg. I think that you can construct any number of equilateral triangles which have the same center as the rectangular and a height a*sqrt(3)/2. (This is what flabdablet is saying using other words). "a" is a random number here. If "a" is large enough, then you have a triangle that is bigger than the rectangle. If "a" is small enough, then the triangle is smaller than the rectangle.
posted by carmina at 6:21 PM on October 23, 2007

I mean, from the poster's wording, the *only* requirement is that the triangle and the rectangle share centers. Right? It doesn't say they have to share sides too!
posted by carmina at 6:30 PM on October 23, 2007

flabdablet is correct. The OP needs to specify what 'on' means.
posted by number9dream at 6:32 PM on October 23, 2007

Well, since finding the centre of a rectangle is trivial (find where the diagonals cross) then the problem reduces to drawing, using only pen, paper and a ruler, an equilateral triangle whose centre is given. This is, as ikkyu2 hints, very easy given a compass as well.

Unless, of course, the rectangle doesn't already exist, and we're being asked to construct, with pen, paper and ruler, both it and the triangle.

Or maybe the paper is the rectangle.

Also, it's odd to find the paper itself specified as one of the allowable tools, which leads me to suspect that this is actually a puzzle question rather than a geometric one, and that the known solution involves folding paper.

posted by flabdablet at 6:45 PM on October 23, 2007

I mean, from the poster's wording, the *only* requirement is that the triangle and the rectangle share centers. Right?

Wrong. Although an infinite number of such triangles exist, the poster asks for the one that can be drawn most efficiently. Elegance is a big constraint.
posted by weapons-grade pandemonium at 7:14 PM on October 23, 2007

Hah! Wrong. Efficiency is number of outcomes per try!
posted by carmina at 7:22 PM on October 23, 2007 [1 favorite]

If the paper size is A4 or in fact any of the other ISO A-series sizes, you can take advantage of its dimensions.

The ratio of the short side to the long side of any A-series sheet is 1:sqrt(2). This is so that when you cut an A-series sheet in half, you get the next A size with no wastage: an A4 sheet is exactly half an A3 sheet, and has the same proportions; an A5 sheet is half an A4 and so on.

If we call the short side of an A-series sheet 1 unit, the long side will be sqrt(2) units. By Pythagoras's theorem, that makes the length of the diagonal sqrt(3) units.

So, if you crease an A4 sheet along both diagonals to find its centre, the length from any corner to the centre will be sqrt(3)/2 units long. I'm going to call that Distance H.

If you then get another sheet and crease it in half lengthwise, and use your first sheet to measure Distance H along that crease, you can mark a point on the centreline of the sheet that's sqrt(3)/2 units away from a short edge. Running creases or ruling lines from that point to the corners of that edge makes an equilateral triangle.

Now that you have one of those, you have a ready supply of 60 degree angles, which you can use in conjunction with your pen and ruler to plot more equilateral triangles anywhere that pleases you. You may also be able to make use of your 1-unit and sqrt(3)/2 -unit edges directly in some cunning way.

If you don't have A-series paper to hand, you can still make an equilateral triangle for measurement purposes from other sizes: crease one sheet in half lengthwise, then lay the short edge of a second sheet on the first so that one corner is over the first sheet's corner and the second corner is on the midline. This is essentially Euclid's construction, using the fixed-length sheet edge as a cheat's compass.
posted by flabdablet at 8:33 PM on October 23, 2007

OK, I've been messing about with folding paper and I think I've found a pretty slick procedure.

Take a rectangular sheet of paper. Doesn't have to be an A-series sheet. Label the corners A, B, C, D, where AB and CD are the short sides and BC and DA are the long sides.

Fold A to B and C to D, creating a midline crease parallel to the long edges.

Fold A to touch the midline, in such a way that one end of the resulting crease is at B. The point where the other end of that crease hits side DA we'll now call E. ABE is now exactly half of an equilateral triangle.

Put in another fold, or rule a line, that continues the folded edge EA across to side BC, call the point of intersection F, and you now have an equilateral triangle BEF with a handy midline BA.

Fold B to E, crease and unfold; where the resulting crease joins BA is the centre of the triangle BEF.

You can generate any number of smaller equilateral triangles by folding B in to meet any point on BA, and find their centres similarly.

It's also very easy to lay the centre of such a triangle over the centre of any given rectangle on a second sheet (fold the triangle in half, line up the centres, open it out) and then use a pen to mark its corners, and a pen and straight-edge to rule lines between them.
posted by flabdablet at 8:15 AM on October 25, 2007

I was confused as to how a compass would help you accurately draw an equilateral triangle until I realised we were talking about pairs of compasses.
posted by nthdegx at 2:58 AM on November 3, 2007

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