Algebracus

An easy, but perhaps a little unsophisticated approach is to check perfect squares modulo 100. Since (50n + k)^2 and (50n - k)^2 are both congruent to k^2 for every natural numbers n and k, it is enough to check the numbers 0, 1, 2, 3, ..., 25. We can of course check all these 25 numbers, but since we are only interested in the endings 01, 04, 09, we check only the numbers that gives perfect squares ending in 1, 4 or 9, that is, 1, 2, 3, 7, 8, 9, 11, 12, 13, 17, 18, 19, 21, 22 and 23. The endings of these are 1, 4, 9, 49, 81, 21, 44, 69, 89, 44, 24, 61, 41, 84 and 29, none of them being among 01, 04 or 09, giving the desired result. The result is stronger than what you asked for, but that is not exactly a loss.

I have already proved that the upper limit is 1, so your belief is wrong.

As uncool already has pointed out, the triangle is a scaled 3-4-5-triangle, and surely it must be a right triangle. You can easily prove it by using Pytagoras: A triangle with sides a, b and c, c >= a, b, is a right triangle if and only if c^2 = a^2 + b^2. (If and only if, as the Law of Cosine would show, or alternatively the simple observation that when you know three sides of a triangle, the triangle is uniquely given.)

You possibly used the addition- and subtraction-formulas for tan(x + y) and tan(x - y), and you have possibly learned those formulas at some instance. This would not be all that bad, but when it comes to those formulas, it is more than enough to learn the formulas for sinus and cosinus. To find the formulas for the tangent, just substitute tan x = sin x/cos x, and multiply with cos x cos y/cos x cos y, that is, 1. So, what I did was just to use the same method as I do when I prove those addition- and subtraction- formulas. Together with the philosophy of first trying to find sweet solution, this is dynamite. And for the sake of it, competing in several contests and math olympiads, I have been lucky to sharpen my skills of problem solving. The ability to find short solutions is the strongest weapon in such environments.

We are trying to find [MATH]\frac{\tan(\frac{\pi}{4} + a) - \tan(\frac{\pi}{4} - a)}{\tan(\frac{\pi}{4} + a) + \tan(\frac{\pi}{4} - a)}[/MATH] which equals [MATH]\frac{\frac{\sin {x}}{\cos {x}} - \frac{\sin {y}}{\cos {y}}}{\frac{\sin {x}}{\cos {x}} + \frac{\sin {y}}{\cos {y}}}[/MATH], where [MATH]x = \frac{\pi}{4} + a[/MATH] and [MATH]y = \frac{\pi}{4} - a[/MATH]. Here is what follows: [MATH]\frac{\frac{\sin {x}}{\cos {x}} - \frac{\sin {y}}{\cos {y}}}{\frac{\sin {x}}{\cos {x}} + \frac{\sin {y}}{\cos {y}}}[/MATH][MATH]= \frac{\sin {x}\cos{y} - \sin {y}\cos {x}}{\sin{x}\cos{y} + \sin {y}\cos {x}}[/MATH][MATH]=\frac{\sin (x - y)}{\sin (x + y)} = \sin {2a}[/MATH].

Let us say that this happens x seconds after 1 o'clock. Then the minute hand has moved (x seconds)/(3600 seconds in an hour)*(360 degrees per hour) = x/10 degrees, and the hour hand has moved (x seconds)/(3600 seconds in an hour)*(30 degrees per hour) = x/120 degrees. We have x/10 = x/120 + 120 (90 + 30, since we started at 1 o'clock) (+360k, k an integer,but in this case k = 0, since we are interested in the first case) and therefore, 11x/120 = 120, or x = 120*120/11 = 1309 + 1/11. Now, 1309 = 60*21 + 49, so it has passed 21 minutes and 49,090909... seconds.

It is not enough that a species reproduce. It is also necessary that the offsprings are up and healthy. The possibilty of this happening dwelves on the general capability of adjusting oneself to the nature and the changes in the nature. This ability is on the other side bases on the ability to eliminate flaws and carry on advantages, that is, it is based on a fast reproduction and therefore a rather short life span. In other words, the "ideal" life span is balanced, not too short, neither too long-lasting. If your first argument were correct (that a great life span is indeed a great advantage), then it would not necessarily imply the possibility of immoratal species. More naturally, we would have two possibilities: (1) During the ages, the maximum life span converges to a certain limit, based on certain limitations that does not directly base itself on the biological evolution (for instance, the fact that the earth is limited). (2) During the ages, the maximum life span diverges. Every individual dies sooner or later, but it takes longer and longer time before they die. This is surely not the same as immortality. Also, if immortality in fact was a possibility, at some stage we would have "the first (theoretically) immortal individual". But were came the immortality from? It cannot been inherited, so several mutations then must have occured, but how? I cannot imagine how an individual could be immortal, when encountering for instance the energy laws (especially the law of an increasing unorder), but it is one the other side much easier to imagine an individual being very, very old (like 3000 years old ore something). (But then, I really don't know what life and death is all about, and we all know the hypothesis that death is only an illness that can be avoided, with the right diet and medical assistance. In this case, the question of how the species became theoretically immortal is more or less irrelevant.)

The rules were obviously not written clearly enough: I suppose that when you go from a word to another, the rule was supposed to be that only one letter can be changed, counting a rearrangement of the letters as a change of at least two letters. If a rearrangement is not counted as a change, then Newtonian surely is better off. Doing it in only five steps, not allowing rearrangements, is impossible, given the rule that every accepted word has to be encountered at http://dictionary.reference.com (I find such a rule much better than a rule talking about "real words". What does that mean?): Black - bhack, wlack, blick, blatk or blace Of these, only blick is allowed. Blick - wlick, bhick, blitk or blice Of these, not a single one is allowed.

If you had been from Norway, then everything had been easier. First you go through a couple of highly elementary stuff, and then the final awaits. If you then are as lucky as I am, you get a high score, win or rank among the top three in as many subjects you like, and suddenly the problem is not to qualify for an olympiad, but to find out which olympiad attracts you the most.

The third problem: We have alternating rows of 100 and 99 circles, and the vertical distance between the centres of circles in neighbouring rows is [MATH]\frac{\sqrt3}{2}[/MATH]. Let n be the number of rows. Then [MATH]1 + \frac{\sqrt3}{2}(n - 1) < 100 < 1 + \frac{\sqrt3}{2}n[/MATH], or [MATH]n - 1 < \frac{198}{\sqrt3} < n[/MATH], giving [MATH]n = 114[/MATH]. By this, we have 57 rows with 100 circles and 57 rows with 99 circles, totally 11343 circles, 1343 more circles than in the first packing.

I was wondering about the more general case in which we substitute 2002 with n and 3k + 1 with [MATH]mk + a[/MATH]. First, we have the trivial observations that at the end the a - 1 first numbers are not erased, meaning we have too many numbers left for a > 2, and for a = 2, the last erased number is surely 1 (since we for each round having more than one number left surely decrease the number of numbers, erasing the second number, and therefore at some moment the second number also is the last number). The only case of interest is therefore a = 1. The simple algorithm to solve the problem is the same as before, and based on the same simple result: Suppose the last erased number for a particular n, letting m be unchanged, is [MATH]f(n)[/MATH]. Then, we first write [MATH]n = mn' + r, r = 1, 2, ..., m.[/MATH]. After the first round of erasing, we have [MATH]n - n' - 1 = n - int[n/m] - 1 = int[\frac{m - 1}{m}n][/MATH] numbers left, and we find that [MATH]f(n) = int(\frac{m}{m - 1}f(int[\frac{m - 1}{m}n]).[/MATH]

A function is not necessarily a formula, it is just a correspondance between elements of one set and elements of another set, with the only restriction that every element in the first set can only be attached to one element in the other set. We are surely taking about a function here (even if the answer is something like "the n th random natural number in he interval from 1 to 20 that The Thing works out using a specific calculator, beginning at some specific moment).

I stopped at 13 elimination rounds because I found it easier to just start working with nine numbers than going on until I only had one number left. The deduction for nine numbers is immeadiate: 1 2 3 4 5 6 7 8 9 2 3 5 6 8 9 3 5 8 9 5 8 8 OK, I will do the rest of it, then (not because it is necessary, but why not?) 9 = 3*2 + 3, so after 14 eliminations we have 9 - 3 = 6 numbers left. 6 = 3*1 + 3, so after 15 eliminations we have 6 - 2 = 4 numbers left. 4 = 3*1 + 1, so after 16 eliminations we have 4 - 2 = 2 numbers left 2 = 3*0 + 1, so after 17 eliminations we have 2 - 1 = 1 number left. Among 1 number the last one singled out is, naturally, 1. Among 2 numbers the 1st not in the form 3k + 1 is 2. Among 4 numbers the 2nd not in the form 3k + 1 is 3. Among 6 numbers, the 3rd not in the form 3k + 1 is 5. Among 9 numbers, the 5th not in the form 3k + 1 is 8.

2002 = 3*667 + 1, so after one elimination we have 2002 - 668 = 1334 numbers left. Going on, 1334 = 3*444 + 2, so after 2 eliminations we have 1334 - 445 = 889 numbers left. 889 = 3*296 + 1, so after 3 eliminations we have 889 - 297 = 592 numbers left. 592 = 3*197 + 1, so after 4 eliminations we have 592 - 198 = 394 numbers left. 394 = 3*131 + 1, so after 5 eliminations we have 394 - 132 = 262 numbers left. 262 = 3*87 + 1, so after 6 eliminations we have 262 - 88 = 174 numbers left. 174 = 3*57 + 3, so after 7 eliminations we have 174 - 58 = 116 numbers left. 116 = 3*38 + 2, so after 8 eliminations we have 116 - 39 = 77 numbers left. 77 = 3*25 + 2, so after 9 eliminations we have 77 - 26 = 51 numbers left. 51 = 3*16 + 3, so after 10 eliminations we have 51 - 17 = 34 numbers left. 34 = 3*11 + 1, so after 11 eliminations we have 34 - 12 = 22 numbers left. 22 = 3*7 + 1, so after 12 eliminations we have 22 - 8 = 14 numbers left. 14 = 3*4 + 2, so after 13 eliminations we have 14 - 5 = 9 numbers left. Among 9 numbers, the 8th is the last one erased. Among 14 numbers, the 8th not in the form 3k + 1 is 12. Among 22 numbers, the 12th not in the form 3k + 1 is 18. Among 34 numbers, the 18th not in the form 3k + 1 is 27. Among 51 numbers, the 27th not in the form 3k + 1 is 41. Among 77 numbers, the 41st not in the form 3k + 1 is 62. Among 116 numbers, the 62nd not in the form 3k + 1 is 93. Among 174 numbers, the 93rd not in the form 3k + 1 is 140. Among 262 numbers, the 140th not in the form 3k + 1 is 210. Among 394 numbers, the 210th not in the form 3k + 1 is 305. Among 592 numbers, the 305th not in the form 3k + 1 is 458. Among 889 numbers, the 458th not in the form 3k + 1 is 687. Among 1334 numbers, the 687th not in the form 3k + 1 is 1031. Among 2002 numbers, the 1031th not in the form 3k + 1 is 1547. So my answer is 1547. This might be wrong, but the underlying algorithm is correct and should not be too hard to see. More to say: Let f(n) be the answer in the general case (n instead of 2002). Then f(n) = int(1.5*f[int(2n/3)]).

Since nobody have replied yet, I will give hints to two fundamentally different solutions to the problem: (1) Let x = b + c, y = bc, find limitations for x and y, and then find a lower limit for a + x. (This is my own solution to the problem.) (2) Just try to figure something out relying on facts such that (a - b)^2 + (b - c)^2 + (a - c)^2 >= 0 and (a + b + c)^2 = ... (The official solution.) It is possible that a solution based on for instance Vietas formulas would work too, but I have not found any. In the contest the whole lot of three people managed to find a proof, so it should not be all that difficult.

Here is a problem which was given in a Norwegian mathematical contest: Suppose [MATH]a[/MATH], [MATH]b[/MATH] and [MATH]c[/MATH] are real numbers such that [MATH]ab + bc + ca > a + b + c > 0[/MATH]. Show that [MATH]a + b + c > 3[/MATH].

Instead of the old problem, I want to first solve the general problem of finding a short-form of [MATH]S(n) = \sum^{n}_{k=0} (-1)^{k} \binom{2n+1}{2k}[/MATH], where [MATH]n[/MATH] is a natural number of any kind. The result of interest is [MATH]S(49) - 1[/MATH]. We start looking at the expression [MATH](1 - i)^m + (1 + i)^m[/MATH]. By the binomial formula, this can be written as [MATH]\sum^{m}_{k = 0} ((-i)^k + i^k) \binom{m}{k}[/MATH]. For all odd [MATH]k[/MATH] we have [MATH](-i)^k + i^k = 0[/MATH]. For all [MATH]k = 4l[/MATH] we have [MATH](-i)^k + i^k = 2[/MATH]. For all [MATH]k = 4l + 2[/MATH] we have [MATH](-i)^k + i^k = -2[/MATH]. Therefore, the sum we look upon can be rewritten as [MATH]2\binom{m}{0} - 2\binom{m}{2} + 2\binom{m}4 + ...[/MATH] Therefore, [MATH]2S(n) = (1 - i)^{2n + 1} + (1 + i)^{2n + 1}[/MATH]. Furthermore, we know that [MATH]1 - i = \sqrt{2} e^{\frac{7\pi}{4} i}[/MATH] and [MATH]1 + i = \sqrt{2} e^{\frac{\pi}{4} i}[/MATH], so [MATH]2S(n) = \sqrt{2}^{2n + 1}(e^{\frac{7(2n + 1)\pi}{4} i} + e^{\frac{(2n + 1)\pi}{4} i})[/MATH] [MATH]=\sqrt{2}^{2n + 1}(e^{\frac{-(2n + 1)\pi}{4} i} + e^{\frac{(2n + 1)\pi}{4} i})[/MATH]. We divide the cases into two; [MATH]n[/MATH] odd and [MATH]n[/MATH] even. In the first case, [MATH]2S(n) = \sqrt{2}^{2n + 1}(\cos \frac{3\pi}{4} + \cos \frac{-3\pi}{4} + i \sin \frac{3\pi}{4} + i \sin \frac{-3\pi}{4})[/MATH] [MATH]= -\sqrt{2}^{2n + 2} = -2^{n+1}[/MATH]. In the latter case, [MATH]2S(n) = \sqrt{2}^{2n + 1}(\cos \frac{\pi}{4} + \cos \frac{-\pi}{4} + i \sin \frac{\pi}{4} + i \sin \frac{-\pi}{4})[/MATH] [MATH]= \sqrt{2}^{2n + 2} = 2^{n+1}[/MATH]. By this we find [MATH]S(49) - 1 = -2^{49} - 1[/MATH].

I have to correct myself at some point here: If we are going to define 0^0, then it is most convenient to define 0^0 as 1, up to continuity, since lim x^x = 1, as can be seen by l'Hospitals formula: x->0+ x^x = e^(x ln x), and x ln x = (ln x)/(1/x), of the form (-infinity)/(infinity). Then lim x ln x = lim (1/x)/(-1/(x^2)) = lim (-x) = 0, and e^0 = 1. x->0+ So x = 0 cannot be said to be a solution. When it comes to the number of solutions, check for instance x^(2x) = 0.

I have some remarks to this problem: 1) This is not a quadratic equation. A quadratic equation has the form ax^2 + bx + c = 0, but the given equation has not. 2) Razorfane states that the equation has two solution. That is not correct. The equation has in fact three solutions: x = 0, x = 1/4 and a third approximated to 1,44. 3) I register that Razorfane has brought the equation into the form f(x) = f(1/4), where f(x) = x^(x - 1/2). This is a fine move, but it is not enough to state that x = 1/4 is the only solution (which it in fact isn't). First of all, Raxorfane divided by x in his first step, and therefore cancelled the solution x = 0. Secondly, f(x) = f(1/4) gives x = 1/4 only if and only if f(x) is an injective function, which it isn't. 4) To check how many solutions the equation has, we can use calculus on f(x). First, we state that the function is continuos. Then we find in which intervalls it is increasing and decreasing. Then we find which values it takes in the given intervalls. Then we (by the intermediate value theorem) find for how many intervalls (in which the function is monotone) we have some x such that f(x) = f(1/4) = 2^(1/2). 5) We easily see that there are no solution x < 0, since 2x = x^(2x) = (x^x)^2 >= 0. Therefore, we can (without loosing any solution except 0) write x = 2^v and x^(2x) = 2^(v*2^(v + 1)) = 2^(v + 1). Since the function f(x) = 2^x is one-to-one, we therefore have v * 2^(v + 1) = v + 1. It could be of some interest to work with this problem instead of the original one.