Square by Opposite Midpoints

For the first condition to be fulfilled in Mathematics is to be exact; the second is to be as clear and simple as possible. — Lazare Carnot

## Level Information

### Instruction:

Construct a square, given two midpoints of opposite sides.

## 6L solution

Let A and B be the given midpoints.

1. Construct line AB
2. Construct the perpendicular to AB from A
3. Construct the circle with center A and radius AB, intersecting the perpendicular at C and D
4. Construct the perpendicular bisector of AC
5. Construct the perpendicular bisector of AD
6. Construct the perpendicular to AB from B

## 10E solution

Let A and B be the given midpoints.

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at C and D
3. Construct line AC, intersecting circle A at E
4. Construct line BE, intersecting circle B at F and G
5. Construct the circle with center D and radius DF, intersecting circle B at H and circle A at I and J, with I the reflection of H through CD and J the reflection of F through CD
6. Construct line BH
7. Construct line AI
8. Construct line FJ
9. Construct the circle with center D and radius DG, intersecting circle A at K, the reflection of G through CD
10. Construct line GK

Alternate 10E solution:
Let A and B be the given midpoints.

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at C and D
3. Construct line AC, intersecting circle A at E
4. Construct line BE, intersecting circle B at F and G
5. Construct the circle with center F and radius BF, intersecting circle A at H and circle B at I and J
6. Construct line FH
7. Construct line GI
8. Construct line BJ
9. Construct the circle with center F and radius AF, intersecting circle A at K
10. Construct line AK

### Explanation

The 6L construction is fairly straightforward; the two sides through A and B are perpendicular to segment AB. A bit harder is the remaining two lines; but since circle A has radius 1, and A is on the midpoint of the side, the distance from A to its two nearest vertices is 1/2, which the perpendicular bisector is able to produce.

The 10E constructions are a bit more involved. In both variations, the first two steps create the points C and D that would form the perpendicular bisector of AB, where triangles ABC and ABD are equilateral. Constructing line ACE then produces point E as a reflection of D across the side of the target square through A, such that DE is parallel to AB, with triangle ADE an equilateral triangle as well. Next, line BE is the perpendicular bisector of segment AD, and therefore the angle bisector of AED; so point F divides arc AD of circle B in half; with that, we know ∠XBF (where X is the vertex of the square below B) is 60°. Segments AC and BF have the same length (both equal to AB), and the distance from B to the desired vertex is half of that, sufficient to prove that B, F, and the vertex form a 30-60-90 triangle with F on the side of the square perpendicular to the side through A; this also puts G on the line through the other side of the square.

From there, the first variant constructs two circles through D to reflect points across CD, the perpendicular bisector of AD. The first circle D with radius DF gives the points needed for three sides: I is on the angle bisector of ∠DAE (side AI) for the side through A; BH is its reflection across CD for the side through B, and J as the reflection of F gives the perpendicular side. Another circle through D, plus a reflection of G to K, completes the square. The second variant instead constructs the first circle through F to form two equilateral triangles BFI and BFJ with FH as the perpendicular bisector of BJ, and second circle through F to reflect A across FH into K.

### 3rd Alternative

There is a third 10 E alternative that might be simpler to understand than the previous ones. They are based on perendiculars line and perpendicular bisectors.

We start as if we were to draw a square with side [AB].

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at C and D.
3. Construct the circle with center C and radius CB
4. We draw the line (AC) that crosses the circle of center C in D.
5. We draw DB, which is perpendicular to AB

Then we draw a perpendicular bisector to BE, using the regular methods by drawing the circle of center E and radius EB, and the line between the two circle intersection.

We know that F is the corner of the square because BE is size of the square, so BF is half of the size, which corresponds to the size from the center to the corner.

And we do the same on the other side, for BG, by drawing the circle of center G and radius GB. H is the second corner of the square.

    Finally we see that AK is parallel to BG (not proven here). So we draw the last side of the square with the line AK.