**, Zona Centro, Congruence of Triangles**

Problem 1 - the Regional Centre Zone Tamaulipas-

Problem 1 - the Regional Centre Zone Tamaulipas-

jmd 03/21/2006

Pre-selective examination on Saturday, 18 March at the CBTIS 236 has some lessons for the 16 shortlisted candidates to integrate the selection state of the Mexican Mathematics Olympiad. Above all, training suggestions. Then I will comment on the suggestions of training that we left the problem of geometry.

The problem in the legs AC and BC of triangle ABC have been built (outside the triangle) and BMEC ADKC squares. Points D and M are lowered perpendicular DH and MP on (sic) the continuation of the hypotenuse AB. Show that AB = DH + MP.

Elemental

but not both. How to link MP and DH with AB? The problem certainly requires scouting.

The first association is Pythagoras, but after a little pondering is that PB and HA could not be removed from the resulting chain of equalities. In any case, not only Pythagoras.

similarity is another possibility. To the trained eye it is easy to see (or at least suspect) that the three triangles are similar.

*Solution:*

Denoting a, b, c to the sides opposite the vertices A, B, C, respectively, we can use the previous plan to use Pythagoras (A ^ 2 + b ^ 2 = c ^ 2) and replace b by similar triangles. Clearly the triangles

BMP and DHA are both similar to ABC (the angles CAB and MBP are the same as being relevant, and the same is true of CBA and HAD). Hence

DH / b = b / c MP / a = a / c. That is, a ^ 2 = b ^ 2 = CMP HRC, and substituting in Pythagoras gives the result: c (MP + DH) = c ^ 2 and you're done.

*Comments to the solution (from the perspective of the learner)*

This problem no contestant solved it. But why is it so difficult? Try seeing it for the parties to respond.

1. Before you even have a chance to resolve the learner should be able to figure (not included in the review). Most of the contestants drawn, so the difficulty is not there. It should, however, included in the training exercises nontrivial geometric trace. (For example, draw a triangle with a ruler and compass given an angle, the opposite side and adjacent one side.)

2. The evocation of Pythagoras is not problem-in fact the first thing that occurred to the majority.

3. The difficulty seems to lie in the recognition of similarity of triangles and the development of the settlement plan that combines Pythagoras and likeness. This points to an intensive training in similar triangles with many exercises, but also training in strategies for solving geometric problems in a "reading between the lines" of data and the figure, an interpretation that allowed the development of a settlement plan.

4. Note on training perspective of the writer:

1) Training is training, and therefore to adopt a position of letting the learner alone with their own creativity is to adopt a populist perspective (but also a contradiction. .. that is the point of training?)

2) Unfortunately, this perspective is present even among mere leaders of the WMO-an example: Illanes said in his book Problems of Olympiad included in the 2 nd edition a new chapter of induction "almost against my will as I have always thought that what should be assessed ... is the ability of students to solve problems and ingenuity put into solving them "-

3) The theoretical results are always useful as equip the learner with a tool box ready to use when drawing up the plan solution, and do not take away creativity and ingenuity but rather the stronger. Workaround

This solution is subtle, it requires training to see the movement of figures through the eyes of the mind.

A dynamic geometry training (with CABRI, for example), would provide the apprentice with a wider menu of creative ideas when preparing a solution plan. Might guess, for example, that the triangles at the ends may be rotated and ...

BMO Turning the triangle on the center B, 90 degrees is obtained BCP triangle. " Turning -90 DHA on A gives the ACH. "

And the settlement plan is almost ... (You're seeing the result, still need to prove it.)

If we call C 'at the foot of the perpendicular to AB lowered from C, the plan would be to demonstrate congruence between pairs of triangles BMP and CBC', ADH and CAC. "

cognitive power acquired by the apprentice training in geometric transformations is enormous. And this is particularly true in the development of the settlement plan. (Note, incidentally, to begin an official solution "is C 'the foot of the perpendicular ..." In other words, lies precisely what the learner needs the most: the methods of reasoning that lead to the development of the settlement plan.)

JMD in VL

greets