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This paper aims at a difficult though exciting target: the mathematics teachers' interest for computers as teaching tools. In spite of the increasing number of computer endowed schools and a lot of students owning home computers, the classroom remains, to a large extent, a computer free area. Amongst various factors responsible for this sad phenomenon, the lack of professional software cannot be neglected. There is a lot of applications devoted to primary school students, particularly in what concerns their counting skill and ability to compose and solve elementary equations. Also high school algebra benefits from many good quality programs. However, all this stuff is seldom available in languages other than English, partially German and French. As for geometry software, it is scarce and, except for a few well known examples, unprofessional.
    Geometry provides two important educational tools: shape and motion. Both of them can be modeled on the computer screen. Traditionally, the secondary school students are being taught geometry in two steps: plane geometry, then solid geometry. This order seems natural though has its drawbacks. Dienes [1]  remarks: "All really existing things have a single legitimate representative: three dimensional solids. So it looks evident that the study of geometry should begin with real three dimensional objects. The plane does not exist, therefore one cannot really experience the two dimensional geometry" (quoted from the Hungarian translation, p. 242-243). The last sentence looks disputable. Of course, all we see we see in space, though all we measure we measure in plane. Indeed, all common geometric instruments such as ruler, compass, setsquare, protractor are designed to be pressed against a flat surface while being used. Even a tailor putting his tape measure around some tender waist actually acts in an imaginary planar cross-section.
    We are going to discuss here a geometry software named POLYHEDRON, which was made at the University of Timisoara. It is basically intended to remove some of the artificial obstacles that persist on the geometry's way toward the child's brain and heart and so often determine the child to reject geometry. It particularly concerns the mentioned above perceptive difference between the ways plane and solid geometry are being taught in school, unnecessary conceptual distinction between calculation and construction problems, and the domination of the deductive element over the intuitive one.
    Now that the software has been briefly introduced, the reader is invited to answer a short test just to let him or her decide whether it is worth going on with this paper or change it for a more useful reading.

This test, which, at a first sight, looks like a simple joke, had a double purpose: first, to attenuate the reciprocal distrust between the supporters and opponents of the computer as teaching tool and second, to inoculate the reader with a good dose of skepticism, which is indispensable to any serious and discerning reading.

How can one choose a problem to solve?

POLYHEDRON was primarily intended to run in the classroom. That is why the first choice is the teacher's. There are two basic modes in which a POLYHEDRON session runs: independent and supervised. In the former case, the student undertakes the teacher's role and thus is free to choose a subject by himself and interrupt the resolution whenever he or she wants. In the latter case, the subject is selected by the teacher as follows: problem type (calculation or construction), shape (one of the six above), topic (distance, angle, perpendicularity, parallelism, area, volume, other), and complexity level (from 1 to 6). As the student finishes solving a problem and gets displayed the computer's verdict, it is up to the teacher either to select another subject or run the resolution's film asking for details when necessary.

The student provides an answer thereby telling the computer the resolution has been finished. The computer evaluates the answer, compares it to the correct one, and displays the answer's quality: correct, approximate or wrong.

The problems have been composed in such a way that it is practically impossible to guess the answer. On the other hand, even the resolution of the simplest problems assumes a series of measuring's, constructions, and calculations. Usually, there are several equally efficient algorithms for the same problem. The probability that an erroneous algorithm conducts to the same result as a correct one is negligible. However, there is but one way to achieve this certainty: practicing with POLYHEDRON.

No, of course. That is why we better say subjects but problems when talking about POLYHEDRON. Beside of the problem itself (name it better hypothesis), a subject includes a set of available tools and allowed basic constructions (some may be forbidden, for didactic reasons) and a polyhedron model on the screen. The student  is free to find any length or angle by measuring, there is no need to give them in the hypothesis. In difficult problems all tools are usually available.

POLYHEDRON was conceived as a computer based realization of the so called "carpenter workshop" philosophy. Imagine a carpenter workshop. You are given a chunk of wood shaped as in Figure 1 and asked to find its volume. Obviously, nothing can prevent you from measuring the edges and computing the volume in your notepad. But you must agree that it is going to be a tough computation if the tetrahedron lacks any "regularity". A saw would relieve you. You draw the perpendiculars DE on AB and EF on AB then cut the chunk along DE and EF. Once the freshly cut surface shows up, you draw the perpendicular DH on EF, measure it and finish the resolution with an elementary calculation.

You can equally do it with its image on the screen. Moreover, you can switch from the initial, opaque, model to a transparent one (like the drawing in Figure 1) and conversely. You may order the vertices to be labeled, for the sake of measuring and drawing accuracy, or you may hide the labels, to do some drawing by hand. You can remove a faulty or unnecessary drawing by the eraser. The intermediate results may be written down on screen. A calculator is available for simple calculations. Lastly, you can ask at any moment for context sensitive help.

Problem 1. A straight line passes through vertex C of this regular pyramid (Figure 2a) and makes equal angles with the edges issuing from the same vertex. Find the second point in which the line intersects the pyramid's surface.

The required point O is obviously located within face ABD. Since triangles COA and COB are congruent, O is equidistant from A and B and thus belongs to the perpendicular DN to AB. Since the sought line CO makes equal angles with CA and CD, it lies in a plane p that passes through the bisector of angle ACD and is perpendicular to face ACD. Therefore O is the intersection of p and DN.
    POLYHEDRON tool bag supplies suitable features to realize the plan outlined above. First, one bisects angle ACD. Let CE be the bisector. Second, one draws (with the setsquare) the altitude DF in triangle ACD and cuts the pyramid through B, F, D into two congruent parts (Figure 2b). On either of them (e.g. [DBFC]) one draws the perpendicular BG to FD. p is parallel to BG and passes through a point that corresponds to J in which CE intersects DF in Figure 2a. By means of a special POLYHEDRON feature, one marks the intersection point J, measures (with the graduated ruler) DJ, lays off (with the compass) a same length segment [DK] on DF, and draws the perpendicular KL to DF.
    Note that to fulfil the last construction one has to switch to the opaque model. That is because the command "Draw a perpendicular in K to DF" is ambiguous: such a perpendicular may be drawn in either FDB or FDC. POLYHEDRON will detect the ambiguity and ask one to choose one of the two faces, which may be done with a mouse click on the corresponding face in the opaque model. [KL] once drawn, one may switch back to the transparent model.
    Obviously, KL is the intersection of p and DFB. To locate L's alias on the initial tetrahedron, one measures DL and lays it off on DB thus getting DM = DL. One can see that p = CEM. Lastly, one draws EM, the perpendicular DN to AB and marks the intersection O of EM and DN. CO is the sought straight line. One types O in the answer dialogue box and gets back "Correct solution". The resolution is finished.

Problem 2. This wooden cube (Figure 3a) must be cut off to a cylindrical top whose diameter makes a quarter of the cube's edge. What is the maximal length of the top?

One may remark that, by virtue of symmetry, the sought cylinder should have one of the cube's diagonals as its axis. Therefore, one can reduce the problem to a planar one by considering one of the cube's diagonal sections in which the cylinder is represented by a rectangle that is symmetrical about the cube's diagonal and touches two of its opposite faces (Figure 3b). So one cuts the cube through B, D, H, draws the diagonal BH and proceeds locating the centers I, K of the cylinder bases on [BH]. Since the cylinder's diameter is AB/4, its radius, i.e. IJ, is equal to AB/8.
    Then HI = (AB/8)cotBHF. So one measures length AB, then angle BHF (with the protractor) and compute HI on the built-in calculator. Or one may notice cotBHF = HF/BF = ABsqrt(2)/AB = sqrt(2) thus easing a bit one's task. Now one goes straight to the end: since HI = BK, one has IK = HB-2HI. The length IK is the solution.

Problem 3. Draw on the upper base of this truncated pyramid (Figure 4a) a line parallel to AB so that the distance between it and AB is equal to the length of edge [AD].

Since DE is parallel to AB, the sought line is parallel to DE. Therefore it forms together with [AB] the bases of a trapezoid, which lies inside the truncated pyramid and whose altitude equals AD. To enable the construction of this altitude, one cuts the pyramid with a plane perpendicular to AB. For example, one may draw the perpendiculars FG to DE, GH to AB and cut through F, G, H. The pyramid is not regular, so that of the two resulted pieces (Figure 4b) one is a truncated pyramid, too, and the other is not. One selects the latter and seeks for a point J on FG such that HJ = AD.
    GJ can be found in triangle HJG by solving the quadratic equation

JH² = GH² + GJ²- 2GH×GJcosHGJ

where GJ is the only unknown, but it would be against POLYHEDRON's motto: Less computation, more geometry. So one measures the altitude h of trapezoid IHGF (dropped, e.g., from F), then the length of [AD], and compute HK = sqrt(JH² - h²) = sqrt(AD² - h²). Now K may be actually located by laying off HK from H. Finally, one raises the perpendicular KJ from IH thus locating J and draws the perpendicular JL to FG in the upper base. JL is the solution.

Problem 4. Cut this prism (Figure 5a) with a plane such that the section is a rectangle.

This problem is perhaps one of the most representative for POLYHEDRON due to its suppleness and elegance. Same is the resolution below. (It does not involve a single calculation!)
    First one tries to guess where such a section might be constructed. Any plane parallel to the prism's bases gives a triangular section. By cutting in the lateral edge's direction one generally gets a parallelogram, since the given prism is not a right one. There are infinitely many parallelogram-shaped sections. Let's have a closer look at them. Let us choose one of the three lateral edges, for instance FC, and figure out a plane that is "anchored" at that edge but is able to rotate about the corresponding axis (i.e. FC). The section provided by such a plane varies from FEBC to FDAC. Is there a position in which the section turns out rectangular? One measures angles EFC and DFC and finds both are obtuse. Therefore all intermediate sections have obviously obtuse angles at vertex F.
    Hard luck. Never mind, there are other two lateral edges to experiment with. The lateral faces that join at edge EB look, in Figure 5a, as if their angles at E are both acute. Is that true? One turns the prism vertically as depicted in Figure 5b. This time angle BED looks obtuse. (Notice that angle BEF is certainly acute, since angle EFC proved obtuse.) One measures it with the protractor and gets, indeed,
BED > p/2. Thus, there exists, for continuity reasons, a plane that passes through FC and gives a rectangular section.
    Notice that angle BED might happen to be acute, too, but then EDA < p/2 and FDA > p/2 hold, so one has to take edge [AD] instead of [BE] to draw a rectangular section through.
    Thus one only has to find a point H on DF such that EH is perpendicular to EB. H lays in a plane p that passes through E and is perpendicular to EB. p is also perpendicular to DA, hence it contains the perpendicular EG to DA and the perpendicular raised from G in face ADFC. One draws the perpendiculars EG to DA, then GH to AD. H is the sought point. One just have to cut through B, E, H to get a rectangular section.

   The POLYHEDRON package (English and Romanian versions) is available on the World Wide Web for free download here.
    A POLYHEDRON manual [2] is available at the Timisoara University Library. [3] contains an outline of the "carpenter workshop" approach and describes how POLYHEDRON appeared and developed
    The readers may wish to resolve Problems 1-4 in a real POLYHEDRON session. They should select, in the "Problem type" page, the following options: