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How can you fit a equilateral triangle on three arbitrary parallel lines with an edge and compass?

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  • $\begingroup$ IIRC this was a Math Olympiad problem (IMO or USAMO) from the 1970s or 1980s. I can remember working on it as a practice problem when I was a senior in high school in spring 1991. $\endgroup$ – Jason S Nov 7 '17 at 21:13
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Well, I'm late to this, but I've been obsessing over this far too much to back out now. So: Label your parallel lines $a$, $b$, and $c$ from bottom to top. Construct a line $d$ with a "positive slope" that crosses $b$ such that the top right angle of their intersection is $60^\circ$. Extend the line so that is crosses $c$. Call the intersection of $d$ and $c$ point $A$. Construct another line $e$ such that it lies to the left of $d$, is parallel to $d$, and is the same distance from $d$ as $a$ is from $b$. Call the intersection of $e$ and $b$ point $B$. Draw an arc centered at $B$ and having radius $AB$, and crossing $a$. Call the intersection of the arc with $a$ point $C$. $\triangle ABC$ is your triangle.

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Why does this work? Here's a hint: Start with an equilateral triangle $\triangle ABC$. Draw any line through $A$, and then draw a parallel line through $B$. Now identify the center $O$ of $\triangle ABC$. You have now have a triangle and two lines. Rotate the entire configuration $120^\circ$ around $O$. What do you get? Rotate another $120^\circ$. What do you get now?

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  • $\begingroup$ Beautiful. Both answers are good, but yours is better explained (with scaffolding attached), so I have accepted this one. Thank you ! $\endgroup$ – qed May 6 '13 at 14:46
  • $\begingroup$ Thanks! BTW, a variation would be to construct a third line $f$, parallel to $d$ and the same distance from $d$ that $b$ is from $c$. The intersection with $a$ would give point $C$. This might be conceptually simpler, but requires much more work than just drawing $\angle ABC$. $\endgroup$ – bob.sacamento May 6 '13 at 15:26
  • $\begingroup$ "Construct a line dd with a "positive slope" that crosses bb such that the top right angle of their intersection is 60∘" -- this isn't a valid compass and straightedge construction, at least not directly. (though you can if you break it down into a few more steps) Neither is constructing a line "e" the same distance from "d" as "a" is from "b". $\endgroup$ – Jason S Nov 7 '17 at 21:18
  • $\begingroup$ @JasonS I skipped details to keep my post from getting longwinded. Both constructions can be accomplished with compass and straightedge. $\endgroup$ – bob.sacamento Nov 7 '17 at 22:38
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    $\begingroup$ Anyway it's a nice solution. $\endgroup$ – Jason S Nov 8 '17 at 4:51
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For brevity, I will not go into details about how to construct perpendiculars and 30-degree angles using straightedge and compass. Same for doubling lengths. Translating those operations to compass-and-straightedge primitives is left to the reader. Given the three parallel lines and a vertex C arbitrarily fixed on one of those lines, I will find the two locations of another vertex, named E and E', that are suited for the fit, and I will not lose words about the rest because that should be obvious.

In C, construct a line perpendicular to the three parallel lines, and name the new intersection points A and B. Now have a look at the attached figure, which is symmetric with respect to the perpendicular line AC, and assume that all filled triangles are equilateral. In this demonstration, |AC| >= |AB| >= |BC| for tidyness, but it turns out that algebraically there is no such restriction for the scheme to work.

Figure: Fitting an equilateral triangle on three parallel lines

Rotate the kite CEFE' 60 degress around E to find that it is congruent to DEE'G. Therefore |DG|=|DE|=|DC|, so |CG|=2|CB|. Using this you can locate G and then branch off CG in G at angles of +/-30 degrees, intersecting with the parallel line through A to find E resp. E'.

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  • $\begingroup$ Could you please explain a bit how you came up with the solution (i.e the scaffolding) ? Thanks! $\endgroup$ – qed May 3 '13 at 18:15
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    $\begingroup$ @CravingSpirit: How I came up with the solution? Brace yourself. I actually calculated a suitable rotation angle (ECA in the attached figure) using trigonometry. Then I tried to interpret the result geometrically, which led me to define the point G by |CG| = 2|CB| and assert that the angle EGA is 30 degrees, in other words, that the triangle GEE' must be equilateral. This suggested that a simple geometric proof was within reach, and I found it via congruence resp. rotation by 60 degrees. You see, the first way found is rarely presentable. The figure has been made with Geogebra. $\endgroup$ – ccorn May 3 '13 at 19:25
  • $\begingroup$ What do you mean a suitable rotation angle, and how did you calculate it? Thanks~ $\endgroup$ – qed May 4 '13 at 14:05
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I've found a different invariant that solves this problem: two circles bisecting a perpendicular between any two parallels cut each other at points that are exact middles of the side opposite to the vertex on the third parallel.

If you want more explanations and sketches please go to:

http://romanyandronov.elementfx.com/pse/ryapserac03.html

Sorry for providing the link - it's a small part of a bigger thing. In my articles I'm interested not in the answers themselves but rather in the ways one can find them. You may find other things of interest there.

This is an eleven-step construction, but I think that number can be reduced.

Outline: random point on any parallel, perpendicular through it, two circles bisecting any segment between any two lines, line through their intersection point and a point on the remaining parallel, perpendicular to that last line locates two remaining vertexes:

Constructing equilateral triangle on three parallel lines

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  • $\begingroup$ Your website is absolutely cool! $\endgroup$ – qed Oct 20 '13 at 12:16
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My previous construction took 11 steps to accomplish. Here's an 8-step one. My reasoning is here:

http://romanyandronov.elementfx.com/pse/ryapserac04.html

Invariant this time - an equilateral triangle built on the outer parallels locates two vertexes one of which is on the inner parallel:

[img]http://romanyandronov.elementfx.com/pse/imgs04/pserac0403.png[/img]

In the sketch above triangle AGH locates A and C. That observation lead me to the following 8-step construction (the construction lines are numbered):

[img]http://romanyandronov.elementfx.com/pse/imgs04/pserac0405.png[/img]

Algebraically (C() is circle, L() is line):

$$O_1, O_2 \in L_1$$ $$C(O_1, O_1O_2)$$ $$C(O_2, O_2O_1) \cap C(O_1, O_1O_2) = A, A_1$$ $$L(O_1, A) \cap L_2 = B, L_3 = C$$ $$C(O_1, O_1C) \cap L_1 = D$$ $$C(B, BD) \cap L_3 = E$$ $$L(B, D)$$ $$L(D, E)$$ $$L(E, B)$$

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I'm not going to go into the details of a ruler-and-compass construction, but I'll indicate the main idea, which is very simple.

Referring to the figure in the question, $CDE$ being equilateral is equivalent to saying that $D$ is obtained by rotating $C$ by an angle of $60^{\circ}$ about $E$.

Therefore, start by picking an arbitrary point $E$ on the third line. A suitable point $D$ can (and must) be obtained by rotating the first line by $60^{\circ}$ about $E$ and letting $D$ be the intersection of the second line with the rotated line. You get two triangles for a given $E$ this way, one by rotating clockwise, the other counterclockwise.

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  • $\begingroup$ This is beautiful! A true aha moment, and much simpler than the other solutions. $\endgroup$ – Roar Stovner Apr 24 '18 at 10:06
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Here is a simple 10-step construction. Looking at the diagram given in the OP, denote the uppermost line by $l$, the middle line by $m$, and the lowermost line by $n$.

On $l$ choose a point $A$ and construct a line $t_1$ through $A$ which forms a $60$ degree angle with $l$ and has negative slope. Let $t_1$ intersect $m$ at $B$. Through $B$ construct a line $t_2$ which forms a $60$ degree angle with $m$ and has positive slope. Let $t_2$ intersect $n$ at $C$. Using a compass mark point $D$ on $m$ such that $AD=AC$ and $D$ is on the opposite side of $AC$ as $B$. I claim triangle $ACD$ is equilateral.

Note that $\angle ABC=120$, so it suffices to prove that quadrilateral $ABCD$ is cyclic. We will do this by showing that $\angle ADB=\angle ACB$. By Law of Sines on triangle $ABD$, we have $AB/AD=\sin(\angle ADB)/\sin(60)$. By Law of Sines on triangle $ABC$, we have $\sin(\angle ACB)=(AB/AC)\sin(120)$. But $AC=AD$ and $\sin (120)=\sin(60)$, so combining the equations give $\sin(\angle ADB)=\sin(\angle ACB)$. $\angle ADB$ is less than $120$ and $\angle ACB$ is less than 60, so their sum is less than $180$. Thus $\angle ADB=\angle ACB$, as desired.

Sorry for the lack of a diagram.

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Here is a very simple construction: Suppose the parallel lines are called a,b,c. A transversal crosses a at A, b at B and c at C. Let XYZ be an arbitrary equilateral triangle. Find point E on XY so that the ratio EX over EY is the same as BA over BC. From X and Y draw lines parallel to ZE. This is almost the answer. Now, we only need to proportionally blow it up or down.

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  • $\begingroup$ This is an question which already has a well-accepted answer. You are not contributing anything new. $\endgroup$ – Shailesh Jan 6 '16 at 7:46
  • $\begingroup$ This is a lot simpler than the constructions I see here and also it can be generalized to any triangle. $\endgroup$ – Ali N Jan 6 '16 at 9:07

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