Angle of 60° - 2

The best angle from which to approach any problem is the try-angle. — Author Unknown

## Level Information

### Instruction:

Construct a straight line through the given point that makes an angle of 60° with the given line.

## 3L solution

Let A be the given point.

1. Construct a circle with center A and an arbitrary radius, intersecting the line at B and C
2. Construct the circle with center B and radius AB, intersecting circle A at D and E (to match the orientation in explore mode, both B and D should be on the left)
3. Construct the perpendicular bisector of CD

### Reflection approach

Let B be an arbitrary point on the line

1. Construct circle B with radius AB, intersecting the line at point C
2. Construct circle C with radius AC, intersecting circle B at point D
3. Construct circle D with radius CD, intersecting circle C at points E and F (E on the left to match the orientation in explore mode)
4. Construct line AE

### Isosceles approaches

Here, there are multiple related ways to reach 4E from a common start; to make explanation easier and match explore mode, each construction of a pair of points places the first letter on the left, and a letter is reused between constructions only if it is the same point.

1. Construct a circle with center A and an arbitrary radius, intersecting the line at B and C
2. Construct the circle with center B and radius AB, intersecting circle A at D and E, and intersecting the line at F and G

* Choice of:
	1. Construct the circle with center F and radius DF, intersecting circle B at H and circle A at J
	2. Construct line AJH
* or:
	1. Construct the circle with center D and radius DF, intersecting circle A at J
	2. Construct line AJ
* or:
	1. Construct the circle with center G and radius DG, intersecting circle B at H and circle A at L
	2. Construct line HAL
* or:
	1. Construct line EC, intersecting circle B at H
	2. Construct line AH
* or:
	1. Construct line EF, intersecting circle A at J
	2. Construct line AJ
* or:
	1. Construct line EG, intersecting circle A at L
	2. Construct line AL

## 2V solution

From the 3L solution (total 4L):

1. Construct the perpendicular bisector of CE

From the Reflection 4E solution (total 5E):

1. Construct line AF

From the Isosceles 4E solution (total 6E):

1. Repeat any of the choices for the final two steps, but swapping the use of D and E, and of F and G. Just as the first variant created one or two of the collinear points HJAL, the swapped points will create one or more of the collinear points IKAM.

## Explanation

The 4E reflection solution starts by using circle B and C to reflect point A across the line BC to point D, with AD perpendicular to the line. Then circle D creates the equilateral triangle CDE by construction, so the length of arc ED in circle C is 60°. By the inscribed angle theorem,

∠

{\displaystyle \angle}

EAD is thus 30°. If point G is the intersection of AE with the original line, and H the midpoint of AD, we have just constructed a 30-60-90 triangle, with

∠

{\displaystyle \angle}

AGH the desired 60°.

The 3L solution uses the same analysis as the isosceles 4E solution; either way, the first two steps create the isosceles triangle ABC and equilateral triangles ABD and ABE. The choice of remaining steps produce several points of interest: point H on circle B is the reflection of D across the line (via circle F or circle G),

∠

{\displaystyle \angle}

DHE is 60° since it inscribes arc DAE of circle B with AH as the angle bisector. If both circles D and F are constructed, it is easy to show that triangle DFJ is equilateral, and that

∠

{\displaystyle \angle}

DHJ is 30° since it inscribes arc DJ of circle F. This is enough to prove that D, J, and A are collinear, even if only one of D or F is constructed. Harder to prove is that triangle CDH is equilateral; but if that can be shown, it explains the 3L solution (the perpendicular bisector of base CD is the angle bisector of

∠

{\displaystyle \angle}

DHE, constructing line AH); as well as showing that the constructed line is indeed the desired 60° angle from line BC (also an altitude of the equilateral triangle since it is perpendicular to DH).