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Next Sunday March 27 starts at Monegrillo-Zaragoza-La Ruta BTT Transmonegrina

Click to go to the 2011 Transmonegrina

The Transmonegrina 2011 begins next Sunday, March 27 in the town of Monegrillo -Zaragoza .- This first gear have two routes: a 48 km departing from the football field, and another 30 km from the municipal swimming pools. Departure at 09:00 hours.

The registration ends on Wednesday before the test, you can register the day of the march, with an increase in the fee of five dollars, and does not give the gift of choice jersey or food

The Transmonegrina are a set of gears mountain bike cyclists uncompetitive taking place for various localities in the region of Los Monegros. The circuit of this edition, which will be the fourth-consists of nine tests

Gift of a T-shirt to anyone who participates for the first time on the circuit this year, and a commemorative jersey to all those riders who attend five evidence on this issue.

data, registrations and all information losmonegros.com / transmonegrina 2011

Organize foundation for the promotion of Youth and Sport of the Monegros , and collaboration features of the various clubs and associations in the area cyclists, as well as the city council of the host towns.

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Statement

To attack this problem well there are several approaches, but I have been very useful by the fact that the centroid is exactly twice as far from the vertex to the midpoint of the opposite side.

This fact is also used to show that the three medians intersect at a single point.

The idea is to draw the segments perpendicular to the line, not only from the vertices, if not from the midpoints of the sides.

line and perpendicular Baricentro

In our picture, the vertex that is separated from the other by the line I call A, and two B and C. Obviously, A 1, which is the midpoint of segment AB is on the same side as B and C and B and C _{1 1} are on the same side as A, because if the line could not pass the cutoff.

Now, by similarity, we have the segment from C is twice that of C ₁ , that of B is twice that of B ₁ and A is twice that of ₁ .

As the midpoints of the segments divided into two equal parts on the sides, the segments defined with respect to our line also play a special relationship. Indeed, it is easy to see that the segment unless the segment of B ₁ is exactly like the B segment _{1 More} segment C. Likewise, the segment of B minus 1 A segment worth the same as the segment of A ₁ less than C, and the segment of A less than C ₁ worth the same as the of B plus C ₁ .

Let us now try to show the equality that we demand. We start from the sum of the segments B and C. As we have seen that the segment of B plus C ₁ is equal to A less than B ₁ , we need the segment of C is equal to the segment unless the segment twice ₁ B . Therefore, the sum of the segments B and C is equal to the sum of B plus A least twice the segment B 1 . But we also know that the segment of B is exactly twice that of B ₁ therefore conclude that the sum of the segments B and C matches that of A.