Constructions and Tilings · Lesson 7 of 12
Related Constructions
“Combine bisection, hexagons, circles, and perpendiculars to solve a gallery of geometric design tasks.”
• Construct 30° and 15° by successive bisection. • Explain the equilateral points of a six-pointed star. • Plan designs by identifying repeating supports and arc centres. • Recreate the source’s construction gallery and optical illusion. • Construct a perpendicular from a point outside a line.
Construction of 30° and 15° Angles
One exact construction can become the starting point for another. The 60° angle came from an equilateral triangle. Bisecting that angle gives 30°, and bisecting 30° gives 15°. Each step uses the same equal-side reasoning you already know.
Problem
Construct a 15° angle from a starting ray.
- 1.Construct an equilateral triangle on the ray to obtain 60°.
- 2.Use a vertex-centred arc and equal intersecting arcs to bisect that angle. The resulting ray makes 30° with the starting ray.
- 3.Bisect the 30° angle in the same way. The new ray makes 15° with the starting ray.
- 4.Keep each construction stage labelled so the second bisection is applied to 30°, not to the original 60° again.
6-Pointed Star
A regular hexagon supplies six equally spaced vertices. Connecting alternate vertices gives an equilateral triangle, and connecting the other three gives another. Their overlap forms a six-pointed star with six matching points.
Why are the six small point triangles equilateral? Each outer tip is an angle of one of the large equilateral triangles, so it is 60°. The triangle edges occur in three directions with 60° acute angles between them. At each small triangle’s base, the appropriate edge directions again make 60°. Thus each point triangle has three equal angles and is equilateral.
The equal arrangement repeats after a turn of 60°, so the star has rotational symmetry. Rotational symmetry means the outline matches itself after a turn smaller than a full turn. It is different from needing the same orientation before and after any arbitrary rotation.
Problem
Construct a six-pointed star using a regular hexagon ABCDEF.
- 1.Construct the regular hexagon first, keeping its equal side lengths and 60° central sectors.
- 2.Join A to C, C to E, and E to A. Join B to D, D to F, and F to B.
- 3.The alternate-vertex triangles are equilateral and congruent because the hexagon divides the circle equally.
- 4.Trace the outer star boundary or retain both triangles to show the construction. Their edge directions justify the 60° angles at the six points.
An Inflexed Arc and Circular Designs
The following designs combine methods rather than introducing unrelated formulas. Before drawing, identify a scaffold: equal segments, a regular hexagon, a circle, or a repeated angle. Draw the support first and use it to locate matching centres.
For the inflexed arch, begin with the symmetric pointed framework from the previous lesson, then continue the side directions as straight lower supports. Copy the centre choices on both sides. For the six-petal circle design, arrange six equal support directions around a centre and draw matching circular curves. Its circle-only challenge is an investigation: equal-circle intersections can locate the centres without a drawn straightedge scaffold.
The hexagon-in-a-circle design reuses the equal-radius stepping method directly. For six surrounding circles, place the six centres at equally spaced points on a suitable circle and choose equal radii to obtain the intended contacts. The triangular-mesh design begins with a regular hexagon; divide its sides equally in the shown pattern and draw the three families of parallel lines. Each small triangular unit must agree in size with its neighbours.
To add the three-spoke detail shown inside each small triangular unit, construct the midpoint of two of its sides and join each midpoint to the opposite vertex. Use their intersection as the interior junction, then join that junction to the three vertices. Repeat the same procedure in the congruent units. For a simple inflexed arch, draw two equal quarter-circle arcs from a top point to symmetric shoulders: their centres lie to the left and right of the top on a horizontal support line. Continue downward from both shoulders with equal vertical segments. This specifies the rounded shoulders and pointed top of the displayed design.
Optical Illusion
A geometric arrangement can suggest a boundary that has not actually been drawn. Look at the next figure before reading the explanation. Do you seem to see a pale triangle between the dark curved shapes and the incomplete edge segments?
The shape you seem to see comes from the alignment of separate pieces. Your visual system tends to complete an outline when fragments strongly suggest one. To recreate the effect, keep the three corners and incomplete edge directions consistent with an equilateral arrangement. Do not add the missing triangle boundary, because the absence of that boundary is part of the illusion.
A Perpendicular from a Point Outside a Line
Previously the specified point was on the line. Now let P lie outside line l. We can still use a perpendicular bisector if we first find two points of l that are equally far from P. A circle centred at P provides those points.
- Choose a compass radius greater than the shortest distance from P to l, so the circle crosses l twice.
- Draw the arc or circle centred at P, meeting l at X and Y. Then PX = PY.
- Construct the perpendicular bisector of XY using equal-radius arcs.
- P lies on that bisector because PX = PY. The bisector is therefore the perpendicular to l through P.
Problem
An arc centred at P crosses l at X and Y. Explain how to find the perpendicular foot M.
- 1.Construct the perpendicular bisector of XY and call its crossing with l point M.
- 2.The crossing divides XY equally and makes right angles with l.
- 3.Since PX = PY, P lies on this bisector, so P, M, and the bisector direction are on one straight line.
- 4.PM is the perpendicular from P, and M is its foot on the line.
A circle centred at P that merely touches l gives only one crossing. Choose a larger radius to obtain X and Y. For a design gallery, also avoid guessing unseen centres from the outline: establish the supporting geometry and test that the chosen arcs pass through the required points.
Check Your Understanding
Use these questions to check the conditions behind each method, as well as the result. For a diagram-based question, follow the labels and given information rather than judging by appearance.
Quiz
How is 15° obtained from a constructed 60° angle?
What does joining alternate vertices of a regular hexagon produce?
Which turn matches the six-pointed star with itself?
Why does the external point P lie on the bisector of XY?
What is essential to the optical illusion task?
Practice Problems
- Construct 30° and 15° and show the two bisection stages.
- Construct the six-pointed star and explain why its point triangles are equilateral.
- Recreate each of the five gallery designs: inflexed arch, six-petal circle, hexagon in a circle, surrounding circles, and triangular mesh.
- Recreate the optical illusion without drawing the apparent pale triangle boundary.
- Construct the star within a hexagon shown in the gallery and identify its repeated directions.
- Draw a line l and an external point P. Construct the perpendicular through P and explain why it passes through P.
60° ÷ 2 = 30°; 30° ÷ 2 = 15°. Each bisection uses equal vertex distances, equal intersection radii, and SSS.
Key Takeaways
• Successive bisection gives 30° and 15° from 60°. • A regular hexagon organises a six-pointed star. • Exact design work begins with supports, centres, and repeated equal distances. • The source’s optical illusion suggests a boundary that is not fully drawn. • A circle centred at an external point finds two equidistant points on a line. • Their perpendicular bisector is the required perpendicular through that point.