Dark Hair With Highlights 2010

Three babysitting problems (II)

Statement This time, again, the comments I have removed much of the work.

The first question is simple. Indeed, as Matthew has to climb 12 stories, half of road has risen 6, if in the 8 th game is that it has 2, so that Clara must live in the 14 th, that is the answer (c).

The second response, M = 1000 and CD = 400 (since C is 100 and D 400). Since XL = 40 (similar), IX = 9. So we have to MCDXLIX = 1449. Corresponding to the response (a).

Solve the third question is more complicated. In reality, we need to find out what it's worth every letter, and as we have the products of each pair is for me and try to factor it easier to multiply and divide quickly head. Thus, xy = 2 * 3 * 3, xz = 3 and z = 3 * 2. If we multiply the second by the third equality, we have x * y * z * z = 3 * 3 * 2, and if we divide by the value of x * and also have it, is that z * z = 1, ie z = 1, being positive . Quickly obtain that x is 3 and y is 3 * 2 = 6. Ie x + y + z = 10, response (b).

Another approach (faster) could be that since xz = 3 yxyz are positive integers, or x = 3 and z = 1, where y = 6 (and the sum is what we have said), or x = 1 and z = 3, but in this case, and should be worth 2, and could not assert x 18. Therefore, we only have the solution (b).

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The Transmonegrina 2011 beginning on March 27 in the town of Monegrillo (Zaragoza)

Click to go to losmonegros.com / transmonegrina

Transmonegrina 2011 The day begins March 27 in the town of Monegrillo (Zaragoza). The circuit of this edition, which will be the fourth-consists of nine tests.
Transmonegrina
The marches are a set of non-competitive mountain bike cyclists taking place in different locations in the region of Los Monegros. Organize

foundation for the promotion of Youth and Sport of the Comarca de los Monegros , and has the support of the various clubs and associations in the area cyclists, as well as the city council of the host towns .

The organization gives a shirt to anyone who participates for the first time on the circuit this year, and a commemorative jersey to all those runners attending five events this year.

data, registrations and all information losmonegros.com / transmonegrina

2011 March 27 ; Monegrillo

April 17 Lanaja

08 de Mayo                   Sena

29 de Mayo                   Tardienta

06 de Junio                     Farlie

17 de Junio \u200b\u200bCastejón

03 de Julio Frulio

Night Alcubierre

Sariñena September 25

mountain bike ;-) In a wonderful and different way of knowing the Monegros and enjoy unique and wonderful places through which run the tests. Photos
fotopedaleando in monegros here

Subida a San Caprasio and characteristic landscape of the Monegros

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There

Statement several ways to solve these problems without resorting to formal algebra. The idea is to play with watermelons and melons, giving values \u200b\u200bto their weights, putting more together, and so on.

A nice way to solve is this: if three and four melons, watermelons weigh 13 and four watermelons and three cantaloupes weigh 15, to put it all together we have seven and seven melons watermelons weigh 28. Now, if we make seven parts, each couple have a watermelon and a cantaloupe weigh 4.

Having this idea in mind, we return to a track in the beginning. As three and four melons watermelons weigh 13, and is very close to being matched quantities, try to reach a similar figure.

now know that three watermelons and three cantaloupes are three couples as we have done before, which in total weighed 4 + 4 + 4 = 12. That means that the melon is most in the initial track weighs 13 to 12 = 1kg, so that each watermelon weighs 3kg.

We found that both tracks are true, ie three and four melons, watermelons weigh 9 + 4 = 13 and four watermelons and three cantaloupes weigh 12 + 3 = 15.

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did not know her too. Enough to know that nothing in life was easy, on certain afternoons walked to the lighthouse with his childhood friend and sometimes he was accompanied by a cold coffee on the terrace of the port. No one remembers if fortune ever walked at his side. What if one day the chance was wrong, and we were graced by the lottery all, she had lost her ticket. That "s" the dictionary was probably the only place you ever found luck.

did not know her too, but I suspected that nothing good could mean that red scarf hiding his bald head, his face was a clear image of the terrible bill that was paying for being alive and that his silence was no longer so shy as hopeless resignation.

I did not know much, but I was happy the day I saw that meager clump of hair on his head, later on I discovered that at some unknown place he found the strength to overcome cancer, when I held that their commendable efforts were rewarded for once.

did not know her too. I lost all chances to do so. This morning was hit by a tractor. Poured out his life on the cobbles. I am ashamed to be alive, if lucky, lucky to be breathing that she never enjoyed. Buy at whatever price a manual for designing offshore. I urgently need to build a heaven for her.


Juan Luis Blanco
22/2/2011

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Ruta BTT Zaragoza, Alfajarín, Castle Ruins, Toro, Montes de Villamayor Villamayor GPS Track

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Area route - View route and track in Wikiloc - details and location maps

The route has departed from the Ebro dam and continued under the bridge of Ronda de la Hispanidad on the River. Thus, for the road and enjoying Alfranca proximity Cantalobos Soto up the walkway in front of the neighborhood is the monastery, where we crossed the Ebro Alfajarín direction.

Once in that area, visit the ruins of Castle ninth century Muslim, of which only left the tower. From this vantage you can see clearly the stark contrast of the landscape north of Los Monegros and its famous Mont Blanc and the valley south of the Ebro is striking aside extreme aridity and fertility vega other.

addition, as one would expect, visit the monumental metal bull (photo) , where there are magnificent views of the river, as well as Zaragoza ( photo ) with Moncayo in the background, which we again going through the neighborhood of Santa Isabel.

More photos of this bike in Picasa here

Click on the photos to enlarge

Alfajarín Castle Ruins

Admiring the metallic bull

Zaragoza View from Mount Alfajarín, with the snowy background Moncayo

Resting under the monumental Toro

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Summer Fruit Four whole or irrational

Statement

The idea is that there can be equal to a proper fraction, ie it is not a regular number.

To demonstrate this statement, we will support the demonstration of the irrationality of √ 2, which normally must all have seen in class.

In case you have not seen I do a quick check. We proceed by contradiction, assume that the value of this number is a fraction a / b, a and b coprime, because if they were not, simplify the fraction. After taking equality and denominator roots are removed and the expression is reached ^{2b 2 = a} 2 , so that 2 is even. Because of that, a must be even, as if it were odd, 2 should also be odd. Then a = 2k for some k, and substituting in the previous equality, we come to ^{2b 2 = 4k} 2, so 2 b ² = 2k, where we also b is even, which is absurd, since we had assumed that the fraction a / b could not be further simplified. Hence, √ 2 can not be rational.

applied to the problem at hand, we can start proving that for all natural n, √ n must be either integer or irrational. As

proceed by reductio ad absurdum. If not an integer or irrational, √ \u200b\u200bn is of the form a / b, and we can simplify the fraction to a and b are relatively prime. Also, b must be greater than 1, because if not, √ n would be over.

Now suppose that p is a prime that divides n (n = pk). If we remove the root squaring and denominators we arrive at the following equation with integers, nb = PKB ² 2 = a 2 , then p divides 2 aa, and therefore must divide a. That means that a = pt, and the equality becomes PKB 2 = p t ^{2 2} . Simplifying kb 2 = pt 2 . As p does not divide ab, he was a cousin, you must divide k, so k = ps, so ² sb = t ² , and we can repeat reasoning. That means that all prime factors of p are repeated twice, which means that its root is full. Which was discarded from the outset.

can also be reasonable to assume that we have found the smallest n for which this expression is given, and the method we have built, you come to another value s less than n is of this form.

Once seen, we can build another similar reasoning to cube roots, as the method to remove a cube root is simply to raise the bucket, and would get another expression of the form ^{nb 3} = PKB ³ = to ³ and reasoner in the same way. In fact, it can be shown for roots of generic index.

Now imagine that we have natural nym. If the expression √ n + √ m 3 = a / b, a and b mutually prime and b> 1, then √ n can not be whole because we could clear ³ √ m in the form of an irreducible fraction, we know that is impossible. Nor ³ √ m can be whole for the same reason, ie, we could clear √ n would be a fraction. However, equality ^{√ n + √ 3} m = a / b we can remove and clear denominators, leaving b * 3 √ m = a - b √ n. Raise the bucket to remove the cube root, and get 3 mb ^{= a 3} - 3rd ^{2 √ n + b} 3anb 2 - 3 nb * √ n (this can be checked if we develop cube subtraction, multiplying a polynomial or three times a - b √ n and applying properties of the square root). Grouping

roots, we have the expression ³ mb = a + 3anb ³ 2 - (3a ² b + nb 3) √ n, so we can solve for √ n, becoming equal to ³ + 3anb 2 - 3 mb = (3a ² b + nb 3) √ √ n and thus n = (a + 3anb ³ 2 - 3 mb) / (the 3rd ² b + nb 3). Thus, √ n should be rational or integer, which can not by reasoning that we have done before. Then the sum

√ n + √ m 3 is either an integer, or is an irrational number, we wanted to prove.

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Zaragoza Zaragoza Ruta BTT, Alfajarín, Toro, Montes de Villamayor Villamayor, Zaragoza

View route and track in Wikiloc - details and location maps

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Statement

This problem I found fascinating. I had not found a problem with a very interesting challenge.

The idea I continued to explore all the possibilities is to draw circles. Imagine that over the point from which more equal lines go. At least, you have two equal, since they start three of them. The idea that occurred to me to explore all distributions of points from there.

distribution

First

The first possibility is that the three points are at equal distance from it (in a circle). If we consider one of the points, think you also have the other two points the same distance (on another circle). This situation determined by the intersection of two circles of same radius) the position of the points. The form is that of a diamond, and only a distance will be different (and longer).

Second and third distribution

The second possibility is that the three points are on the circumference, but only two of them are within radio distance (and thus make him an equilateral triangle) . The third point, which can not be the same distance from the center of one of the other points, it would repeat the same arrangement as before, will be equidistant from the two, so they form an isosceles triangle is say, will be on the bisector. There are two possibilities, obviously. The first has greater distances than the original radio, and the other children. The picture one after the other.

Fourth

distribution

Once we have exhausted this possibility, we can only assume that none of the three points is the same distance from another central point. but that means that the three are at the same distance, forming an equilateral triangle. This is the fourth distribution points.

Therefore, we now assume that only two points are the same distance from one of them (because we have already explored all such distributions.) We got to the section more difficult. Imagine

central point of the circle. On the circle can only have two points. And the distance between them and can not be equal to the radius, as if it were, form an equilateral triangle, and the third point would be the same distance from the three (which we have seen) or again be at a distance of radius of one of the points, so that point would be the same distance from all three.

Now, the distance between those two points will be different. If the fourth item is out of the circle, its distance from the center matches the distance you are the other two points. Again, can not be that the missing point is at the same distance of the two new girth, then again form an equilateral triangle can not be. Thus, we have two possibilities: either the last point is a radius distance of the circumference of the two is either a radio distance and other distance of the other.

From here in the first case is easy to understand that form a square, as all four sides "outside" are equal, and both interiors are equally between them. This is the figure that came as an example in the statement.

sixth and final distribution

The latest figure is more difficult to visualize. Three sides are equal, and the fourth is like the distance to the cross (diagonal). Form in this case no fewer than four isosceles triangles between points, the same pairs. If we study the relations that exist between the angles, we find that this ring is neither more nor less than a trapeze, and not just any, if not an isosceles trapezoid whose largest angle is 108 degrees.

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Three points kangaroo problems

Statement

In a competition against the clock, it's almost over speed matter that rigor in reasoning, but it is important to find strategies to check and discard solutions which are not.

If we listen to the points given for each, the ideal would be to spend a little less than 2 minutes in the easy, 2 ½ minutes in the middle and just over 3 minutes difficult.

In the first, it is clear that 6 is the sum of two symbols Δ, so each must be 3. If we have to waste time changing the value proposed by check, you can also do, but be quick. The correct answer is D.

The second quickest way is to look at the two more like, 20 * 10 + 20 * 10 and 20 * 10 + 10 * 20. It is clear that they are equal, so we estimate its value (400) and try to find one that do not provide the same result. The single is 20/10 * 20 + 10, which gives 50. Therefore the correct answer is E.

In the third, I see it easier to divide the weight of moving between two on each branch, so that half will be worth 112 / 2 = 56. In the area where the star, divide again by 2, and comes 56 / 2 = 28 (square). Then divide again between 2 and is 14 (the triangle) and then we get the weight of the star again divide by 2, so it is worth 7 (response A).

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Clavelitos, Clavelitos,
and a glass hammer, nailing
Clavelitos,
Clavelitos.


Juan Luis Blanco
13/2/2011

Will Running Change My Body

Night 12


If you want the guitar sing, ringing I do not know what. If you want the dream to stay, is not it better to be a spectator? If life goes, if space is delayed if the weather is spilled, what matters is going to hold? If the unexpected ending to happen, what matters to me not getting it right. If the horizon is not the end, I'm dwarfing. If they are 300,000 kilometers, okay. But it may be too fast. Why all fled? Why all again? Who the hell we caught the illusion of permanence? If my mind is clearer with my reading glasses, will the future less fuzzy with my glasses off? No. There are all blind. Thank goodness that never comes. In five minutes it will be midnight. The music does not stop in the street. Dust cabinet remains oblivious to the weather forecasts. Not blind, but more myopic. And yet happy. I was tired of so clearly illusory. Limits upstarts. Underground car parks for dreams. The minutes are still filled with the music. And do not let the darkness to kill contours. They say that tomorrow is another day. If not I do not see it.


Juan Luis Blanco
12/2/2011

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Sowing


Tunisian. Sounds like a diminutive, a tiny thing. As a seed.


Juan Luis Blanco
7/2/2011

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The region of La Hoya de Huesca updates hikers and mountain bike routes. MTB Route

Go to senderos.hoyadehuesca.es

The region of Hoya de Huesca has improved, and adequate signposted mountain bike routes, paths and trails offered by this magnificent territory.

signaling pathways for MTB have followed the international Signage mountain bikes, established by the International Mountain Bike. (IMBA).

routes are available on www.senderos.hoyadehuesca.es where you can download them for follow through your GPS device. It also has a search by modality and an interactive viewer, as well as information of walks: map, profile and sheet.

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Wording

Shaded area have already explained many of these issues. The trick is to calculate this figure as addition or subtraction of figures known.

Basically, if we look at the figure, it may be the sum of three squares least a triangle, or half of a rectangle least two small rectangles.

If we interpret as the sum of squares less than a triangle, we could calculate the area of \u200b\u200bsquares (16 + 9 + 1 = 26) and subtract the area of \u200b\u200bthe triangle (base of 4 + 3 + 1 = 8, height 4), which would match up base 2, ie, 26 - 8 * 4 / 2 = 26 - 16 = 10.

If we interpret it as a medium rectangle least two small rectangles, notes that the two small rectangles measure both 3 by 1, ie, the two have area 6, while the large rectangle measuring 8 by 4, ie, has area 32 . Its half is area 16, and if we take 6, is exactly 10, as before.

Another resource could be grid-paper and count squares, matching those that are cut with others who also are in the same way, but would be harder to see with precision.

If you know Pick's theorem, a graph can count the points that lie within and are on the edge of the figure and apply the formula (interior - edge / 2 - 1). Works where the vertices are on the grid.

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Sierra de Alcubierre: Leciñena, Alto de Puig Thief, Mount of Pajero, Leciñena

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Area route map and track View - details and location maps

The route through the rugged Sierra de Alcubierre has started and completed in Leciñena at the door pools in this town.

Go to cross the Sierra de Alcubierre mountain bike never disappoints. Its imposing watchtowers to spot on the side of the Ebro valley and the other the plains monegrinas are excellent. On a clear day you can spot from the Pyrenees Moncayo. In addition, its impressive pine forests, ravines and shady areas are always worth seeing.

today's itinerary has been somewhat slower than expected, for presenting the tour some parts with mud, ice and water. Along the way, cross the A-129, road between Leciñena with Alcubierre.

Click on pictures to enlarge

Views from Mount Puig Thief in the Sierra de Alcubierre

icy road

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Gps Track route through the Sierra de Alcubierre: Leciñena, Puig Thief, Mount of the Pajero, Leciñena

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View route and track - details and location maps

Track

GPS route through the Sierra de Alcubierre: Leciñena, Puig Thief, Mount of the Pajero, Leciñena

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shaded area in a decagon

Statement

I'll start by posting a solution that has provided me Peiró Ricard, which, playing with cabri and Ptolemy's Theorem , has found a solution.

regular decagon

Obviously, we begin by drawing a shape (like a) a regular decagon. If we draw with ruler and compass, first drawing a regular pentagon and divide into two parts from the center of their faces. Drawing a circumcircle, easily find the cutoffs of the decagon.

In any case, there are many possible segments that can play the role of A 1 4 A and A ₁ A 2. In the drawing Ricard, has found a ring formed with sides and diagonals of a decagon, using the vertices A 1, A 2, A and A ₄ 6.

is easy to see that A ₁ A ₆ is a diameter of the circle, that A ₁ A 2 is a side of decagon, that A ₂ 4 A and A ₄ A ₆ are corresponding sides of the pentagon, A ₁ A ₄ is one of segments appear in the problem, and A 2 6 A is a diagonal of the pentagon.

If you have studied some geometry of a regular pentagon, you know that the ratio between a diagonal and the side is called the golden ratio, represented by the Greek letter φ = (1 + √ 5) / 2. In our case, In a decagon there is also another very special relationship as the central angle of each side is 360/10 = 36 degrees, drawing two radios will form an isosceles triangle with unequal angle 36 degrees, which is similar to that formed between one side Pentagon and two diagonals. This means that a radius divided by the side of the pentagon is also equal to φ. Among the points we have chosen, this means that the product of the diagonals equals the sum of the product from its opposite sides. In fact, a registered arc derived result and is very useful when working with circles and segments. In our ring, applying this theorem means that ie, Romans and Carthaginians a couple of the other side, then a trip to take two Greeks who complete the group. However, at a given time to do go back a group of people for the further implementation of the rules. Obviously, the boat can not go no one to row, and you should always be busy. One way, as suggested by a comment from Claudio Meller our facebook group would be as follows, where, for short, is named for the initials soldiers (and a number representing the amount, if more than one) there on one side in the boat or the other side. RCG Cruzan in the boat, leaving 3R3C3G on the shore. Dejan on the other side and returns RC G in the boat, gathering 3R3C4G. Cruzan

3C in the boat, leaving on one side 3R4G and the other (when they) r4c. Back R in the boat, leaving the other side 4C and the starting 4R4G. Now cross 3R 1R4G leaving and joining the other side, where there are now 3R4C. Now comes the difficult step, because they must return with the boat full. Any other move us back to the same place we were, or breach of any rule. R2C are back in the boat, leaving the other side 2R2C. When coupled to the edge of departure, are 2R2C4G. Now cross

2G, leaving 2R2C2G and joining an identical amount on the other side. Again

must return the boat filled with RCG, leaving the other side of RCG and the initial side 3R3C3G. back across the boat with 3G, reaching the final edge RC4G and leaving the starting 3R3C. The boat returns with a C. Now there is in the starting 3R4C and the arrival R4G. Cruzan R2C, leaving the starting 2R2C and the arrival 2R2C4G. back the boat to RC, leaving the RC4G arrival and the departure 3R3C. Cruzan 3C now, leaving the starting 3R and the arrival R4C4G. Back _R now, leaving the finishing 4C4G, and the starting 4R. _Cruzan 3R, leaving the R and conducted an in 3R4C4G arrival. Back R in the boat, leaving the 2R4C4G arrival and the departure 2R. Cruzan the last two in the boat.

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Surprise Crossing the River

_Statement

_{Approximating π} Like many problems in geometry, we can draw it with a certain scale (for example, radio is worth 1) and then change the scale. The key measures we have to build is to observe that these measures appear in triangles we see in the picture that accompanies these lines. If we call D to point to A and B form an isosceles right triangle, the distance AD \u200b\u200bis equal to the distance AH. The other important point is E, which also forms a triangle with A and D, and the distance AE is equal to AI. Applying the Pythagorean Theorem, the distance AD \u200b\u200b

2 = 1 + 1 = 2, so AD = √ 2. And the distance AE is also a hypotenuse, so ₂ AE = 1 + 2 AD = 1 + 2, so AE = √ 3. So the distance IH = √ 2 + √ 3, and This is about half the length of the circumference, which is half of 2π, ie π. Try to see the match: √ 2 + √ 3 is worth approximately 3.1462 ... and π is worth 3.1415 ... It is clear that the difference is 0.0046 ... which is about a percentage, a 0.15% higher. course, if the radio was another number, the length multiplied by this factor would, so the difference would be r * 0.0046 ..., that is, that the approach would still be approximately 0.15% higher than the real value length half-circle.

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GPS Track MTB route Zaragoza , Cadrete, La Plana, Barranco del Montanes, Zaragoza Liga Aragonesa

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Route Area - View route and track - Details and location maps

The place of departure of the classic route today was the Great Park. The final, on the bike path Hispanic Round, at the roundabout which gives access to the mall Puerto Venecia (Ikea)

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route

Zaragoza Area, Cadrete, La Plana, Barranco del Montanes, Zaragoza

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Guidance Calendar 2011 Mountain Bike-O-BM

Click to go to FARO

La Liga Aragonesa 2011 Mountain Bike orientation O-BM begins March 27 in Alcaniz, have a total of 8 tests.
Organizers: FARO, Clu Ibón And Peña Guara
All information en orientaragon.com y en clubibon.es/ calendario

27 de Marzo               Alcañíz

17 de Abril                 Villalengua

 

15 de Mayo              Laspuña

4 and June 5

Huesca

June 19 Urrea de Gaén

September 18

Leciñena
October 1 ; Rogaine - Fuendetodos

Guideline Mountain Biking: Cycling and adventure in nature. FEDO - English Orienteering Federation.

MTB Calendar 2011 Guidance Aragon

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mountain bike: Zaragoza, Juslibol, La Alfocea, Zaragoza

Click on the photos to extend that advantage

sunlight lasts more and sunset later, on my bike from the Expo Zaragoza to Alfocea the mountains of Juslibol. Upload

low you: trialeras, trails, cold, galachos, puddles, rocks, gullies, strong gale, beats the edge and balconies with beautiful views. Ultimately
almost an hour and half mountain bike, which is something.

Juslibol Galachos of the river Ebro and the bottom