Square by Adjacent Midpoints

Thinking is the hardest work there is, which is probably the reason so few engage in it. — Henry Ford

## Level Information

### Instruction:

Construct a square, given two midpoints of adjacent sides.

## 7L solution

Let A and B be the given midpoints.

1. Construct line AB
2. Construct the perpendicular to AB from A
3. Construct the angle bisector of ∠BAC with C an arbitrary point on the perpendicular (to match the orientation of explore mode, choose C above A)
4. Construct the perpendicular to that angle bisector through B, intersecting AC at D
5. Construct the circle with center B and radius BD, intersecting BD at E and AC at F
6. Construct the perpendicular bisector of BE
7. Construct the perpendicular to the angle bisector of ∠BAC through F

## 10E solution

Let A and B be the given midpoints.

1. Construct line AB
2. Construct the perpendicular bisector of AB, intersecting AB at C
3. Construct the circle with center C and radius AC, intersecting the perpendicular bisector of AB at D and E (D above AB to match orientation of explore mode)
4. Construct line AD
5. Construct line BD
6. Construct the circle with center E and radius DE, intersecting AD at F, DE at G and BD at H
7. Construct line FG
8. Construct line GH

## 2V solutions

From the 7L solution:
Let G be the intersection of line AB and the line of the square side through F

1. Construct the line DG
2. Construct the line BF, intersecting circle B at point H
3. Construct the perpendicular bisector of BH
4. Construct the perpendicular bisector of BF (total 11L 26E)

Alternative construction to minimize E moves:

1. Construct circle A with radius AB
2. Construct circle B with radius AB, intersecting circle A at C and D
3. Construct line CD
4. Construct line AB, intersecting CD at E, circle A at F, and circle B at G
5. Construct circle E with radius AE, intersecting CD at H and I
6. Construct line AH, intersecting circle B at J
7. Construct line BH, intersecting circle A at K
8. Construct line AI, intersecting circle B at L
9. Construct line BI, intersecting circle A at M
10. Construct line FM
11. Construct line GL (first square in 11E)
12. Construct line FK
13. Construct line GJ (total 13L 13E)

## Explanation

The key to starting the construction is realizing that the sides of the squares through A and B are at a 45° to the line AB. The different constructions then minimize the number of steps to get that angle (the 7L construction by using the angle bisector of a perpendicular at point A, the 10E construction by using circle C at the midpoint of AB to construct a 45-45-90 triangle ACD). From there, the constructions use a different choice of circle to determine where remaining vertices of the square will fall. The 7L solution constructs circle B with a radius that matches the length of a side of the square, such that F is the reflection of D across AB and found on the midpoint opposite from B, and E is the reflection of D across BF and twice as far as the final vertex. The 10E solution instead constructs circle E at twice the size of circle C, where E is the midpoint of the final square and intersects the remaining three vertices.

The 2V solutions reflect the original solution over AB. The only challenge here is minimizing the number of L and E steps by reusing already constructed points. The 7L solution can be reflected with just four more lines and 8E steps. Starting from the 10E solution also requires four more lines; however, the alternative 2V solution demonstrates that by manually constructing the perpendicular bisector of AB, rather than using the tool, it is possible to get a first square in 11E moves by using 2 lines of the second square, such that the overall minimum 2V solution takes 13L 13E.