For each construction challenge solved, you "get" a star. If you find a construction that uses the minimum number of steps, you get another star, and yet another one if you find an optimal straightedge and slack compass construction. Searching for these optimal solutions is usually more difficult than merely solving the puzzle, but it can be instructive. For example, puzzle 2.7:

Given a lineMy first solution was based on standard techniques to bisect an angle, or to perpendicularly bisect a segment. However, knowing that there is a three-step construction using straightedge and slack compass got me thinking. Here is what I came up with:land a pointPonl, construct a perpendicular tolthroughP.

Step 1: Draw a circle through through P, with center

*O*not on

*l*. It meets

*l*again at a point

*Q*.

Step 2: Draw a line through

*O*and

*Q.*It meets the circle at a new point

*R*.

Step 3: The line through

*R*and

*P*is the desired perpendicular to

*l*.

This could lead to a great lesson: why does this work? will it always work? To answer this question students need to know that the sum of the angles in a triangle is 180°, plus the isosceles triangle theorem and some basic algebra. If they have trouble, you may offer the hint: "when working with circles, listen to the radii!" Indeed, analyzing the figure below should lead to Thales' theorem about an angle intercepting a half-circle:

(All figures created in GeoGebra.) |

*Geometry Labs (*free download). The reason is that the labs give students practice solving problems of this type with actual numbers before tackling the general case. (These labs require circle geoboards, or circle geoboard paper.)

Anyway, back to Euclidea. I was not always able to find optimal solutions. My first failure in this regard is on 1.7, inscribing a square in a circle, given one vertex on the circle, in seven straightedge and slack compass steps. My best attempts required eight steps.

In any case, I was able to solve all the puzzles one way or another, until 10.6:

Construct a circle throughPthat is tangent to both sides of the angle

*l*and a point

*P*not on

*l*, construct a circle through

*P,*tangent to

*l*. Both puzzles are part of the construction unit I assigned year after year to my geometry class, and the second is the underlying strategy for the construction of a parabola with focus

*P*and directrix

*l*.

And yet, I was not able to crack Euclidea 10.6. I did it in GeoGebra:

As you see, there are two solutions. I used the fact that the center of the desired circle is equidistant from

*P*and each side of the angle. Therefore it lies on the parabolas with focus*P*and the sides as directrixes. So it must be at the intersections of these parabolas:
This was straightforward, as GeoGebra has a

* Available tools for 10.6: straightedge, compass (slack or rigid), perpendicular bisector, perpendicular line, angle bisector, and parallel line.

**Parabola**tool, but I still have no idea how to do this in Euclidea. The creators of the app claim this can be done in six steps (using any Euclidea tools*) or 11 steps (using only straightedge and slack compass.) If you figure it out, I would love a gentle hint, as Euclidea is not allowing me to proceed any further until I solve this.* Available tools for 10.6: straightedge, compass (slack or rigid), perpendicular bisector, perpendicular line, angle bisector, and parallel line.

[

**Breaking news**: The Euclidea developer gave me a big hint via Twitter. The approach involves dilation. I am embarrassed I didn't think of it, as I had used a very similar strategy to construct a tangent to two circles.]

--Henri

PS: for my thoughts on the mathematics and pedagogy of geometric construction, see this recent post, (and for more, follow the links therein.)

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