Euclidea 4.9

Euclidea 4.9 Square by midpoints on opposite edges


Given the mid-points on opposite edges, $${\rm A} $$, $${\rm B} $$, make a square.

Goal: 6L, 10E 

Solution 6L
It is very easy:

Connect \overline{\rm AB};

Draw perpendicular from \odot B to \odot A ;

Center {\rm B}, radius \overline{\rm AB} draw circle;

Draw the bisector of \overline{\rm AC}, where \odot C is the other end of the diameter of the circle of center B ; intersecting at points {\rm D} and {\rm E} ;

Draw the bisectors of \overline{\rm BD} and \overline{\rm BE} to finish.

Solution 10E (a)
There are two different 10E solutions, both very interesting. Let's start with one.




 * 1) Center $$ {\rm A} $$, radius $$ \overline{\rm AB} $$ draw circle;
 * 2) Center $$ {\rm B} $$, radius $$ \overline{\rm BA} $$, draw circle to intersect $$ \odot {\rm A} $$ at points  $$ {\rm C, D} $$;
 * 3) Connect $$ \overline{\rm CB} $$, intersect $$ \odot B $$ at $$ {\rm E} $$;
 * 4) Connect $$ \overline{\rm EA} $$, intersect $$ \odot A $$ at $$ {\rm F, G} $$;
 * 5) Center $$ {\rm D} $$, radius $$ \overline{\rm DG} $$, draw circle to intersect $$ \odot {\rm B} $$ at points  $$ {\rm H, J} $$, and $$ \odot {\rm A} $$ at $$ {\rm K} $$;
 * 6) Connect $$ \overline{\rm KA} $$, $$ \overline{\rm JB} $$, and $$ \overline{\rm HG} $$, intersecting at points $$ {\rm L} $$ and $$ {\rm M} $$  (3E);
 * 7) Center $$ {\rm L} $$, radius $$ \overline{\rm LM} $$, draw circle to intersect  $$ \overline{\rm KA} $$ at $$ {\rm N} $$;
 * 8) Connect $$ \overline{\rm NF} $$ to finish.

Why it works?
This is pretty amazing, isn't it? Every intersection point is used in an unexpected way.

To see how it works, the key is to understand this circle $$ \odot {\rm D} $$, with radius $$ \overline{\rm DG} $$. The following picture is helpful.



From the construction, it is not hard to see that $$ \overline{\rm CD} $$ is the perpendicular bisector of $$\overline{\rm AB} $$, and that

$$ \angle {\rm CEG} = \frac{1}{2} \angle {\rm CBA} = 30^{\circ}$$

Also, $$ \triangle {\rm ABE} $$ is Isosceles, so $$ \angle {\rm BAG} = 150^\circ $$.

This means if we could draw a line $$ \overline{\rm KA} $$ perpendicular to $$ \overline{\rm AB} $$, we would have $$ \angle {\rm KAG} = 60^\circ $$. Moreover, if we could further draw $$ \overline{\rm GL} \perp \overline{\rm KA}  $$, then we would have a triangle $$ \triangle {\rm GLA} $$ as a nice $$ 30^\circ - 60^\circ $$ right triangle, which implies that

$$ \overline{\rm LA} = \frac {\overline{\rm GA}}{2} = \frac{\overline{\rm AB}}{2} $$,

as desired.

This is where the circle $$ \odot {\rm D} $$ work for multiple purposes. By symmetry along $$\overline{\rm CD} $$, we see that points $$ {\rm H, J} $$ are mirror images of $$ {\rm G, K} $$. This gives

$$ \overline{\rm HG} \parallel \overline{\rm AB}$$

Also $$ {\rm K, D} $$ are mirror images along $$\overline{\rm AD} $$, so

$$ \angle {\rm GAD} = \angle {\rm KAD} = 360^\circ - \angle {\rm GAB} - \angle {\rm BAD} = 360^\circ - 150^\circ - 60^\circ = 150^\circ $$

That gives the beautiful fact that $$ \angle {\rm KAG} = 60^\circ$$ and thus $$ \overline{\rm KA} \perp \overline{\rm AB} $$. This as mentioned before, ensures that $$ {\rm L} $$ as the right point for a corner of the square. Similarly, by symmetry, $${\rm M} $$ is another desired corder.

At this point, the circle $$\odot {\rm L} $$ with radius $$\overline{\rm LM} $$ makes perfect sense. Its radius is the edge length of the desired square. Immediately, it finds another corner at $${\rm N} $$.

This is where another small trick happens. In order to avoid drawing another circle at $$ {\rm M} $$, we directly connect $$ \overline{\rm NF} $$ to get the last edge. To see why this works, note we have $$\overline{\rm AF} = \overline{\rm AG}, \overline{\rm AL} = \overline{\rm AN} $$, and by SAS, the two triangles $$ \triangle {\rm ALG}, \triangle {\rm ANF} $$ are congruent. Thus $$ \overline{\rm FN} \perp \overline{\rm AN} $$.