WEEK 20

 

6.4 Constructions

6..4.1 Use the properties of a circle in constructions (1)
6..4.2 Use the properties of a rhombus in constructions (1)

This final set of tasks involves geometric constructions. Most require fairly standard techniques with ruler and compass, but some are mathematically quite challenging.

Monday: This is a fairly straightforward construction task, which can be solved in a variety of ways. The most obvious method, perhaps, but not the most efficient, is to construct a line through C perpendicular to AB and then construct another line through C perpendicular to that line.

A more efficient method, and perhaps mathematically more interesting, is to choose two points on AB (for example A itself and a point P) and to construct a point Q such that APQC is a parallelogram. A more efficient version of this is to choose P such that AP = AC, so that APQC is a rhombus. These two methods are shown below, left and below, right respectively.

Two versions of another method are shown below. These are mathematically interesting though not very efficient. The numbers show the order in which the construction was carried out. I leave it to the reader to discern the nature of each step!

Tuesday: This is similar to Monday's task, though in one way it is simpler: rather than some students coming up with the idea of constructing a parallelogram to produce the desired parallel line, we are told to construct a parallelogram, although this time there are three possible positions for the missing vertex.

Wednesday: It is possible to construct a copy of ABCD efficiently using a ruler, protractor and compasses, or a ruler and protractor only, or a ruler and compasses only.

Here is an efficient way of producing a copy, A'B'C'D', of ABCD using a ruler, protractor and compasses:
draw the 10 cm line segment A'B';
measure the internal angle at B and draw a line at that angle from B';
measure BC and mark a point C' on the line that distance from B';
draw an arc with radius AD, centre A';
draw an arc with radius CD, centre C';
mark the point D' where the two arcs meet and join D' to A' and C'.

Here is an efficient way of producing a copy, A'B'C'D', of ABCD using a ruler and protractor only:
draw the 10 cm line segment A'B';
measure the internal angle at B and draw a line at that angle from B';
measure BC and mark a point C' on the line that distance from B';
measure the internal angle at A and draw a line at that angle from AB';
measure AD and mark a point D' on the line that distance from A';
join D' to C'.

Here is an efficient way of producing a copy, A'B'C'D', of ABCD using a ruler and compasses only:
draw the 10 cm line segment A'B';
draw an arc with radius BC, centre B';
draw an arc with radius AC, centre A';
mark the point C' where the two arcs meet and join C' to B'.
draw an arc with radius AD, centre A';
draw an arc with radius CD, centre C';
mark the point D' where the two arcs meet and join D' to A' and C'.

 

Students might well come up with other variants of these methods, of which there are several. 

Thursday: The mirror line that is asked for here lies on the perpendicular bisector of AB. Students will probably know the standard way to construct this, but if not it provides quite a nice challenge.
The desired centre also lies on this line, but where exactly? (There are two possibilities here.)

Friday: This is a challenging task....

One way to solve this uses the idea of geometric enlargement:
Start by constructing (by any known method....) a line parallel to AB. Then, using a fixed compass, mark off 4 equally spaced points C, D, E and F, on that line. Draw a line through A and C and through B and F. Mark the point (G) where these two lines meet. Draw lines from G through D and from G through E, and mark the points (H and K) where these two lines meet AB. H and K cut AB into three equal parts. This construction is shown in the diagram below, left.
The diagram below, right is in some ways a simplified version, but involves the construction (by any known method....) of two parallel lines, in order to produce similar triangles.

Bonus: This bonus task has an elegant solution, but may require a flash of insight!