Resetting Zero On Bushnell Elite 3200

The triangle is a flat surface bounded by three lines that intersect two to two in three points called vértices.El triangle is the polygon with the least number of sides, three sides, three angles and three vértices.El symbol Δ is used to designate a triangle and the plural Δs, are used to name the letters of their vertices or a Roman numeral placed inside the triangle.

TRIÁNGULOSSEGÚN CLASSIFICATION sides

EQUILATERAL
has three equal sides

Isosceles has at least two equal sides. Equal sides form an angle that is contrary to a side called the base. The angle opposite the base is called the vertex angle.

SCALENE
not have equal sides

by Angles

acutangula
has its three angles acute.

RECTANGLE
has a right angle. The sides forming the right angle are called legs and the side opposite this angle is called the hypotenuse.

obtuse
has an obtuse angle.

STRAIGHT AND NOTEWORTHY POINTS OF THE TRIANGLE
In any triangle can observe and draw some lines and points for their properties and characteristics are important, they are:

MEDIUM
segment is drawn from one vertex to the midpoint of the opposite side.

Baricentro
is the point where the medians intersect is the center of gravity of the triangle.

bisector is a perpendicular drawn at the midpoint of each side of the triangle.

circumcenter.

is the point where the bisectors intersect and is also the center of the circle circumscribed to the triangle

bisector is the line that divides an angle into two equal parts

Incentra.

is the point among the three bisectors and is also the center of the circle inscribed in the triangle

HEIGHT.

is the perpendicular drawn from one vertex the opposite side or the extension thereof.

orthocenter

is the point where the three alturas.En Intersept every triangle the centroid, circumcenter and orthocenter are aligned, ie are collinear points as part of the so-called straight Euler.

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Mathematics Olympiad 2007

The northeastern Mexico VI Mathematics Olympiad (NMO) in Monterrey
- by José Muñoz Delgado - XXI

Mexican Mathematics Olympiad 2007

Delegation Tamaulipas Ciudad Victoria, Tamaulipas, UAT
UAMCEH, September 24, 2007


After several months of preparation, selection of WMO Tamaulipas Monterrey came to northeastern Mexico to the math competition.


Tamaulipas adolescents
The 15 highest-performing in math faced competition in a spirit of joyful security, confident that their prior qualifications would help them win.

The good news is that overwhelmingly won Tamaulipas in northeastern Mexico Mathematics Olympiad (NMO). Five of the 9 gold medals went to Tamaulipas, Tamaulipas total score and got 231 points, while Nuevo Leon Coahuila 185 and 180. (And the problem 2 - geometry - the scores were: Tam = 76, NL = 38; Coah = 21.)

ONJ was held on 20, 21 and September 22, 2007 in Monterrey, Nuevo León . Participating states are Nuevo Leon, Coahuila and Tamaulipas. Each of the three adolescents participated with 15 high performance in math (although the selection of Nuevo León included several children - under 12).
Tamaulipas
The delegation welcomes this victory and extend congratulations to all participants and their institutions to which they are assigned. We now famous (for the first time dominated by a stunning Tamaulipas in the ONM) and is a triumph that will be remembered by the community Tamaulipas with a special interest in mathematics.

However, the process does not end the contest northeastern Mexico. We can not stay celebrating the victory and forget that the northeastern Mexico is an intermediate step toward the national competition. So this delegation also had two or three things to say about the expectation of performance in the national contest in November.

First in The northeastern Mexico and the tables were turned not only to Nuevo Leon and Coahuila in previous years. It also turned the tables for preselected Tamaulipas about their record of performance in the 9 screening tests: the favorites were in the queue and the angry won the gold medal. How to explain this strange result?

without it being interpreted as wanting to find a spot to our success, one possible explanation is that the delegation agrees that the review was designed as northeastern Mexico for children were not disadvantaged by a limited tooling background. So what? - Say some - that is good is good with easy problems and difficult problems.

Well, this delegation maintained the hypothesis that, as in this case, when the teenager has just acquired the tools are in a transition period in which still lacks an expert opinion to decide which tool is appropriate for given problem. Consequently, we believe that whoever has more tools (recently acquired) is at a disadvantage - it has to decide which to use.

Instead, he has only two or three, do not have to look far because the problem is solved with whom you have or have not resolved at all. And we are in the event that the same limited resources in their favor in elementary problems.

this human factor - too human - is added the fact that the expectation was that the level of difficulty was high-low as last year -: it would motivate the basic problem is faced with sophisticated tools. They just do not fit in the head to the most gifted of northeastern Mexico that a problem was resolved with an example!

The lesson should be learned here is that high-achieving students should be able to assess the level of difficulty of the problem before solving it. But that is precisely what they need to competitors novices - though you had a good training in problem solving - because there is never enough time for such instruction.

In short, at least two problems (1 and 3) benefiting the most vulnerable, in the sense that postulated above. Problem 2 - geometry - although it was elementary (two or three applications of the theorem of the midline and one of Thales and already was) was where they felt more prepared for the Delegation Tamaulipas. And that's because it required the proper application of two theorems, and they were basic.

In any case, any way you see, the selection Tamaulipas dominated northeastern Mexico and possibly contest this fact alone is becoming a legend - In our favor. So the results can be interpreted and reinterpreted - which would make the selection most legendary Tamaulipas 2007 - and yet, as there are hard data that proves our success. (Remember, even for once, Tamaulipas that the university has no power to do have Mathematics as Nuevo Leon and Coahuila.)

The important point here is that this delegation effort invested in training and - -even more important - hard work and tenacity of pre-selected, paid a fruit. And if - as expected - the mathematical community members in training, give some credit the good opinion of the delegate, it is asking a vote of confidence to include in the shortlist to two elements (added to the 10 that were announced on 22 in Monterrey), who - despite maintaining a high performance during training - collapsed in northeastern Mexico: This is Alexis and Brandon.

This delegation makes a bet on them to stay in the fight for one of the six sites Tamaulipas Math Team that will compete in November 2007 in Saltillo Coahuila in WMO XXI. The bet Brandon is learning speed, and that fact alone, be a promise of Alexis is for his career in the screening, which but only twice was way up on the scores remained always above average. (Say, by the way, who participated in the 2006 National.) At risk of sounding sentimental, this delegation does believe that Alexis deserves a second chance.

Alexis and Brandon with the preselection is left with 12 members. The question now is why did not the other three? Well, the answer I have is pragmatic - and gives priority to the expectations of high performance in the national competition. The anger that are already among the top 10 in northeastern Mexico will remain a natural right, but those who are cholera in northeastern Mexico and either oscillating or consistent downward selective, this delegation can not bet on them ... (In any case, are freed from hard work to stay within the pre-selection for the month of October.)

Another, even more pragmatic reason - is that the leaders in northeastern Mexico were cholera in the selective pressure may not hold your fighting in the next four selective, since they already have other interests or do not have enough time to devote to WMO - one of the reasons why, let us recall once again the pre Tamaulipas began with more than 20 .

In any case, the pragmatic approach of this delegation is that the contest northeastern Mexico should not deviate from the goal - drawn from beginning - a high performance in the national competition. We have won in northeastern Mexico and that is indisputable, and is a source of pride for all of Tamaulipas, but in the national competition is incomparably greater competition and the difficulty level much higher. That's why the selection Tamaulipas 2007 should not depend on the results in a contest of moderate difficulty. That said, Tamaulipas screening becomes: Screening

Tamaulipas (Post-NOM)

Guzmán Navarrete L Brandon
Alexis Briones Garza Gutiérrez Aguilar N

Gabriela Ramirez Maldonado J A López Romero

Honorio Del Ángel José Medellín Luis
Iván Roberto López López Martínez J Armando Garza

Magaña Sergio Vargas A
Oscar López García Saldaña A Alvizo
Thomas A
Cortez Adriana Tinoco

screening tests Calendar postnorestense

September 29,
October 6, October 1920
and
October 1927.

In each of these selective shortlisted will be deleted 2: the two lowest scores. All selective, except possibly one, will be by email. (Is pending Pierre Fermat contest date that would have to be selective.)
selective in the email, the review would be sent pre-selected on Friday afternoon and return by the same route on Saturday before 12 pm. The electronic form of selective efficacy was tested in the selective 6 and 9, so there should be no problem this time. Each

selective review between Sunday and Monday, and sent to the shortlisted the result by Tuesday afternoon - resulting in at least two members short listed. The process is repeated until the fourth selective, after which the list of the six selected will be sent to the national committee of the WMO by the end of October. Concluding Remarks


This delegation considers that the 12 members of the Tamaulipas pre 2007 and have acquired sufficient maturity to study mathematics as their own. So, from now on, you may just have a workout. The goal - I repeat - is to choose the best 6 of the 12 remaining. The selective will consist of three problems of national competitions (not necessarily from Mexico). The list of national problems (WMO) will be sent later. Either way, we are on line if you have any questions ...

Congratulations again and keep their position, because now follow the forced-light at the end of the tunnel is in sight now ... in November ...

Stretching The Labia Before And After

northeastern Mexico Conjectures given geometric configuration Tamaulipas

The northeastern Mexico VI Mathematics Olympiad (NMO) in Victoria
- by José Muñoz Delgado -

Event and Context

In an unprecedented symbolic act, the opening (Thursday, 21 September) ONM VI was the backdrop to announce the interest of the Ministry of Education of Tamaulipas (SET) and the Universidad Autonoma de Tamaulipas (UAT) in institutional strengthening school mathematics and its teaching in the state through the inter-agency project called "Learning Mathematics for Life."

The ONM is an event held every year as a contest prior to (and preparedness) the national competition for the Mexican Mathematics Olympiad (OMM). ONJ involved in Coahuila, Nuevo Leon and Tamaulipas, each with 15 teenagers shortlisted, and training for possible participation in the national contest of WMO.

The three delegations stayed at the Hotel Holiday Inn Express, the opening ceremony was held at the Centre of Excellence for the UAT, and the competition itself was on the premises of the UAM of Science, Education and Humanities (UAMCEH) for the TSU.

The Mathematics and Its Context Cares

For some reason (possibly sentimental) human beings are attracted to certain poles of attraction and communities formed around them (and, arguably, tribes). And following that human law (human too) a significant number
school mathematics teachers in Tamaulipas have gathered around the Math Olympiad devoting their time and effort to preparing teenagers for the competitions, no more pay than the mere satisfaction
see them rise in their level of knowledge and skills involved in math.

very particular in my case, I have dedicated much of my free time to the mathematics of competition since it started in Mexico, the organization of the Olympics Math (this year is 20, so take out the calculator ...) with some periods
forays into other fields (such as two and half years - 2003 to 2005 - I spent to get a master's degree in communication University of Guadalajara more expensive, thanks to PROMEP). But this year 2006 - and since late 2005 - I sold my watchtower math "with the slogan" The UAT should serve the state mathematically gifted adolescents. " Y. .. well ... against all odds ... a server and fellow teachers who joined my effort we convinced
UAT UAT and convinced the SET and the result is the interagency project launched at the opening of ONJ.
I think we're on the plastic moment in which the community of mathematicians Tamaulipas, related and supporters should join the project to flourish.


problems
contest the contest day I was walking in an effort to apply (to the Dra.Radmila Bulajich, president of the WMO Organizing Committee, who kindly accompanied us during the days of the event) transfer DEGETI Delegation to the UAT and therefore had not planned to be on the jury. But to get that day's contest UAMCEH facilities where the examination took place, the current company convinced me to be present at the jury, at least as an observer. Troubleshoot

decide the evaluation criteria for each test review, discuss the scores of cases unclear and / or controversial, casting hub scores, allocation of places (the medals) to competitors according to their performance, and report preparation. These are the tasks I was involved with the other jurors during the Friday, September 22 - from dawn to dusk and a little more.

For that the reader can form an idea of \u200b\u200bthe level of difficulty of the problems and the type of problems that the contestants have to solve, then the transcribed (the wording is not mine) and comment. Also transcribe the official solution, and drafted an alternative solution in the style of reasoning can of adolescents to address these difficult problems.

Problem 1. Show that there are infinite "m" such that n ^ 4 + m is a composite number for all n positive integer.

Problem 2. C1 and c2 are two circles that intersect at two points A and B. The tangent at A intersects c2 for c1 in C and the tangent to c1 and c2 A intersects D. A ray passing through A, inside the angle CAD, in M \u200b\u200bintersects c1 to c2 in N and the triangle ACD circuncírculo P. Prove that AM = NP.

Problem 3. It has a circumference perimeter 210. We have marked in the same 20 points, P1, P2, ..., P20, following the direction of clockwise so that the arc joining the point P1 to P2 has length 1, the arc connecting the point P2 P3 has the length 2, ... and so on, the arc joining the point P20 to P1 has length 20. Find all pairs of points scored such that the segment that joins them is the diameter of the circle.

General Comments on the Problems

The easy problem is 3. And this in the sense of being more visual and approachable by brute force - enough to know the meaning of "diameter." However, I should say that the problem 1 is easier that 3, but in another sense: the contestant must be convinced that intimidating statements on issues of competition tend to be easily "broken" translating into a language less esoteric.

In Problem 1, this translation could be "express m in terms of an integer r, such that n ^ 4 + f (r) is factorizable for any positive integer r." This translation has been to seek to complete the perfect square trinomial, difference of squares require and obtain the m requested. - The method could be called "backward inference" and illustrated in the alternative solution of this problem shown below.

Problem 1, however, Easy IMO the 1969 and would show that the judges considered it IMO international level. (See www.kalva.demon.co.uk/imo/ )

The really difficult problem is 2, a circle geometry problem. It is difficult in two respects. First you have to do (or generate - see alternative below) the binding idea and use it to build the solution, which requires familiarity with instances of use of the geometry of the circle - in particular, angles enrolled and semi-attached - to establish the similarities between pairs of triangles in the configuration. And this is assuming that the competitor and dominates geometric terminology (tangent, circuncírculo, etc.).

A lesson could be learned from the problems 1 (of the IMO 1969) and 2 - and usable for teens interested in math competition - is "if you can solve at home is better." At home, ie during pre-contest preparation.

Official Solutions

As in the textbooks that the solution of a problem is too compressed to see the official solution to the problems of a mathematics olympiad examination is often frustrating for the adolescent and their coaches, because it very often, enigmatic -To say the least. Is no exception for solutions to problems brought NOM jurors prepared.

then present them and remind readers that, in both math textbooks and in the issues of competition, you always need an intellectual
additional work to fully understand the solutions.

Solution to Problem 1.

n ^ 4 + 4r ^ 4 = (n ^ 2 + 2RN + 2r ^ 2) (n ^ 2 - 2RN + 2r ^ 2). Clearly the first factor is greater than 1, the second factor is also greater than 1 for r greater than 1. So we can take m = 4r ^ 2.

Solution to Problem 2.

Consider without loss of generality that the ray is inside the angle BAC. Then angMCA + angCAM angCMP = = = angCAD angMAD + angCAM. Moreover, angCPM = angCAD by subtend the same arc. Then, as the triangles ACD and MCP are similar, we have MC / AC = MP / AD. In addition, angACM = angNAD and angCAM = angADN. Therefore, the ACM and DAN triangles are similar. So AN / AD = MC / AC, and hence MP = AN and AM = NP as intended.




Solution to Problem 3.

couples looking all points (Pj, Pk), with 0 less than k jmenor below 21, that meet the segment that joins them is diameter. O equivalently, the length of the arc that joins them is 105. But as the length of the arc that goes from Pj to Pk is

j + (j +1) +...+ (k-1) = (k-1) k / 2 - (j-1) j / 2 ,

we should look for partners who meet k-1) k / 2 - (j-1) j / 2 = 105.

As (k-1) k / 2 - 105 should be positive integer or zero, and k = 15 is zero, then (P1, P15) is one of the couples wanted. The rest are giving ak value greater than 15 and verify if the result can be achieved with some j (j-1) j / 2. The partners sought are (P1, P15), (P6, P16) and (P12, P19).

Alternative Solutions and / or Constructive Feedback

constructive solution to problem 1

Note first that you want to factor n ^ 4 + m. That is, we n ^ 4 + m = (?)(?). The easiest thing is to complete the perfect square trinomial, and in the process, trying to guess how m. First, notice that we should make m = a ^ 2. So we have to do is complete the square in n ^ 4 + a ^ 2. Let's see:

n ^ 4 + a ^ 2 = n ^ 4 + 2 (n ^ 2) a + a ^ 2 - 2 (n ^ 2) a = (n ^ 2 + a) ^ 2 - 2 (n ^ 2) a.

However, for this expression is factorable, it is easiest to see as the difference of squares. Thus, a simple 2 (n ^ 2) a = r ^ 2.

But ... No. No. Here I have one 2 ... So I can improve my building by 2 (n ^ 2) a = (2r) ^ 2. But ... No. No. Not that either. The n ^ 2's I can leave out because it is a square. So it is better to 2a = (2r) ^ 2. Therefore, 2a = 4r ^ 2 or ^ 2 = 2r. So a ^ 2 = 4r ^ 4.

(Note that taking an initial idea as a perfect square m in general, it was a perfect square shape generated more specific the obligations of the solution is sought.)

After phase search and exploration of earlier (the discovery phase) and can be passed to the justification phase in the solution shown official. It would, leaving few traces of the search phase, something like


n ^ 4 + 4r ^ 4 = (n ^ 2) ^ 2 + (2r ^ 2) ^ 2 + 4n ^ 2 (r ^ 2 ) - 4n ^ 2 (r ^ 2) = (n ^ 2 + 2r ^ 2) ^ 2 - (2NR) ^ 2.

reader is left to the pleasure of completing the argument.

constructive solution to problem 2

The search for a solution to problem 2 below shows that is instructive of how to proceed in troubleshooting contest proves to be very effective in breaking the enigma that is the question. The statement is: Sean


c1 and c2 two circles intersecting at two points A and B. The tangent to by A intersects c2 c1 in C and the tangent to c1 and c2 A intersects D. A ray passing through A, inside the angle CAD, in M \u200b\u200bintersects c1 to c2 in N and the triangle ACD circuncírculo P. Prove that AM = NP.


part of the request. Ask you to show AM = NP. After a mental search for possible general strategies, one can conclude that similar triangles should be used. The strategy of similarity is almost infallible in such cases where there is a geometric configuration with overlapping circles. So like ... But what triangles ... Many! What we will do next is to find the right triangles, ie, pairs of similar triangles that would lead to the solution.

first thing you have to do is a good figure. The visual aid is valuable. Secondly we explore the possibilities of triangles to each side of the equation:

AM AMC's side of the triangle (note that triangle is the only configuration that AM is right). Now we focus on AMC triangle sides looking to link with others in the configuration: MC side of the triangle is MCP, and AC is the side of the triangle ACD (and ACP).

Now see the other side of the equation AM = NP. Note that here, as before, there is only one possibility - which makes us task. Now, as before, we focus on NPD to find the triangle linking with other triangles sides of the setup: DN DNA side of the triangle, and PD is next to the PDC.

To facilitate finding the appropriate pairs of triangles we form a crosstab: Side Triangle



AM AMC


AC MC MCP
ACD

NP NPD
DN DNA


PD PDC

This table and the figure should reach (after a while of searching) a pair of triangles ACD, NPD and AMC, PDC. For the purpose of developing a guess that we should look similar enough in Fig. Because there are other similarities that might link the two sides of the equality we want to reach, it is worthwhile to explore whether pairs and found lead us to something solid.

If NPD were similar to MCP, we would have NP / MC = PD / CP (note that we seek to NP and MC are present in the ratio). And if AMC would like DPC, would have AM / DP = MC / PC (note that we now seek to flirt as much as possible with the above.)

So, if you give these similarities, we would have NP / MC = PD / CP = AM / MC and arrive at NP = AM, as intended. So now we have to do is demonstrate that these similarities actually exist and we will finish. Left the reader the task. Only give the key suggestion to observe that the tangents from A to their respective circles are the key to see the similarities of angles required to establish the similarities.

Alternative method for guessing the similarities appropriate for problem 2

new parties of the request to show AM = NP, but now we will be forming ratios and proportions that do not affect equality, but at the same time be ligand triangles in the configuration:

AM = NP

AM / MC = NP / MC (the first reason and is, necessary to resolve the second)

AM / MC = NP / MC = (NP / PA) / (MC / CP) (the denominator and is, missing numerous)

(NP / CP) = (NP / PS) / (CP / PD) (and you are also numerous)

In summary, we have AM = NP is equivalent, in an algebraic sense, the AM / MC = [(NP / PS) / (CP / PD)] / (MC / CP) . And now look, visually aided by the figure, which corresponds to right reason on the left, again in a visual relationship of similarity. It is easy to see that AMC is like DPC and what remains is (NP / PS) / (MC / CP), which must be unity, ie, should be met similarity between triangles NPD and MCP. EUREKA!

Note: Again we reach the same conjecture, and, again, need to prove that the similarities do exist. There must also say that both procedures are heuristic, in the sense that it ensures that reaches a solid guess. And
also be emphasized that similarities in this search could continue to dead ends, before this thing to do is retrace the steps already taken and find another route. Conclusion



We wanted here to account for the Northeast VI Olympiad held in Cd Victoria Tamaulipas facilities UAMCEH-UAT on Friday September 22, 2006. The event was the importance added it was a symbolic act that inaugurated an era of government support for school mathematics in Tamaulipas. Additionally, it has been a pleasure to comment on the problems and their solutions, as well as a method of guessing similarities in geometry problems and useful for competition issues, allowing parties to focus gerométrica settings appropriate for the solution of the problem and avoid the teen "throw in the towel prematurely before the chaos of lines and points he has before his eyes could not decide on a promising path to solution.

Jolteon Stuffed Animal

Selection for OMM 2006

In the bottom photo shows members of the selection to represent Tamaulipas in the national competition for the Mexican Mathematics Olympiad. The last review was selective northeastern Mexico Math Olympiad in which participated the presets of Coahuila, Nuevo Leon and Tamaulipas. The photos are from the closure on Saturday 23, the day after the test.




Thomas appears in the second picture as the seventh member, even if they compete at the national level. But he won the trip for winning silver in the ONM. The gift is what makes the state government as a reward for their effort. Good detail of the authorities, because Thomas is a promise, you still have two chances to go national - is first semester of high school. Radmila Bulajich accompanied us during the days of the event, thanks be given him.


These days I post the details of the contest northeastern Mexico, especially the problems and their solutions. Sincerely


jmd in VL