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Uhm, the REAL South African Rand does have six coins in circulation: 10c, 20c, 50c, 1R, 2R and 5R.
Would it help now?
Given three arbitrary points B, M and C on a plane. Find the locus of the points A such that angle BAM = angle CAM.
1) If B, M and C are collinear in any order, the problem is trivial (either no solution or the first family of Apollonian circles)
2) If MB=MC and B, M and C are not collinear, it is still easy: the locus includes the perpendicular bisector of BC and the segment BC itself
3) For all the other cases, I have yet to get the locus. Though I can prove that M must be equidistant from AB and AC.
Please help. Thanks.
Oh, and I have this to add:
The reason why there is only one polynomial solution (i.e. the polynomial associated with the sequence) for degrees <= n-1 if you have n numbers given is because it would then be reduced to a system of n linear equations with n unknowns.
However, in theory, there are other ways to reduce the question to such a system.
Take your sequence 1, 3, 6, 10. You can (at least in theory) interpret the sequence as 2^(f(x)), 3^(f(x)) etc., where f(x) would be a polynomial with degree <= 3 to be determined.
You would then have f(1) = log(3)1 = 0, f(2) = log(3)3 = 1, f(3) = log(3)6 = 1 + log(3)2, f(4) = log(3)10. Then remember that f(x) = ax^3 + bx^2 + cx + d and you would, in theory, be able to find the coefficients a, b, c and d.
This would be easier to show for another sequence, such as, 3, 3, 3, 9... The correct degree 3 polynomial is x^3 - 6x^2 + 11x - 3. However, the sequence could also be interpreted as f(x) = 3^g(x), where g(x) = (1/6)(x-1)(x-2)(x-3) + 1. In the first case, the next number would be 27; in the second case, the next number would be 243.
Have fun with all these sequences! Try finding as many possible functions as possible.
Perhaps a more straightforward way to present the problem is:
Suppose an integer n can be written as the sum of x+1 consecutive positive integers: a, a+1, a+2, ..., a+x.
Case 1: x is even (x+1 is odd)
Suppose x = 2k then n = a + (a+1) + ... + (a+2k) = (a+k)(2k+1) (you can prove this formula) and of course a+k >= k+1. This means that a number n can be written as the sum of an odd number of consecutive positive integers if and only if it has an odd factor (a factor of the form 2k+1) and when you divide n by that odd factor, the quotient is at least half of the odd factor plus 0.5 (i.e. >= (2k+1)/2+0.5 = k+1). That odd factor must obviously be at least 3 (since x >= 2, k >= 1)
Case 2: x is odd (x+1 is even) (x >= 3)
Suppose x = 2k - 1 (k>=2) then n = a + (a+1) + ... + (a+2k-1) = k(2a + 2k - 1) (you can prove this formula as well) and you must have 2a + 2k - 1 >= 2 + 2k - 1 = 2k + 1. In this case, n still need to have an odd factor (2a + 2k - 1), but when you divide n by that odd factor, the quotient is at most half of the odd factor minus 0.5 (k = (2k+1)/2 - 0.5 <= (2a + 2k - 1)/2 - 0.5). That odd factor is 2a + 2k - 1 >= 2k + 1 >= 5. (the case for that odd factor is 3 would be a small, trivial addition, since then a+k = 2 and the only solution would then be x = 6, which can be represented in another way: 1 + 2 + 3).
When you combine the cases above, n can be written as a sum of consecutive positive integers if and only if it has an odd factor other than 1. Therefore the only positive integers incapable of being written as such a sum are the powers of 2 (which have no odd factor other than 1).
That's right.
A set of n values always defines exactly one polynomial of degree n-1 (or less), and infinite polynomials of degrees n, or n+1, or n+2 etc.
So in general, if you are given something like
1, 3, 5, 7...
you'll find one matching <4 polynomial, and infinite matching degree n (n>3) polynomials for each n. To find that degree 3 polynomial, simply plug in the values x = 1, 2, 3, 4 into P(x) = ax^3 + bx^2 + cx + d (you'll get a system of 4 degree 1 equations with 4 unknowns, so it will have exactly one solution). Of course there will be special cases, but in general, if you have n given numbers in the sequence, there would be infinite formulas that fit unless you limit the function to be a polynomial function of degree not exceeding n-1.
Something that goes with my previous post: http://mth.bz/fit/polyfit/findcurv.htm
I do not recognise the code, but I'll assume its for omega^n.
Since omega is a cube root of unity, omega ^ 3 = 1. Therefore omega^n takes up to 3 values: omega, omega^2 (or 1/omega) and omega^3=1.
Since omega is not 1, omega, omega^2 and omega^3 have three distinct values, so there are exactly 3 values.
- Since Albert does not know Cheryl's birthday on the first time, the month have at least 2 days in contention (doesn't really help).
- Albert knows that Bernard doesn't know it, so all of the days in that month (which Albert knows) have alternatives. This must be July or August.
- Now Bernard knows the birthday, just from knowing that it is July OR August, and from the day. Therefore the birthday cannot be July/August 14, or Bernard would never have figured out.
- If Albert has got August initially, at this point he wouldn't be able to tell whether if it was August 15 or 17. Therefore Albert must have received July at first, and so the birthday was July 16.
Tangram, you forgot that the two As can be in different places.
I would count it this way:
625 ways
4^4 = 256 ways with 0 A
4 * 4^3 = 256 ways with 1 A
So I get 1 - (256 + 256)/625 = 113/625
Bob, you should be using combinatorics here. Since 972 = 2^2 × 3^5, each factor can only have prime factors 2 or 3.
Suppose factor n = 2^i × 5^j. We have 3 choices for i (0, 1, 2) and 6 choices for j (0, 1, 2, 3, 4, 5) and so 3×6=18 factors
My solution doesn't use 20 or 25 degrees, but only uses angle PAQ = 45
That perimeter will always be 2 given that angle PAQ = 45.
Let E be a point on the ray opposite of ray DC such that DE = BQ. Then triangle ABQ = triangle ADE so AQ=AE and also angle QAP equals angle PAE equals 45. Therefore triangle AQP equals triangle AEP (s.a.s.). So QP = EP = ED + DP = QB + DP. Therefore the perimeter of triangle CPQ would be CP + PQ + QC = CP + PD + BQ + QC = CD + CB = 2.
Ignore the D and consider only the isosceles right triangle ABC. Draw isosceles right triangle APE (isosceles at P) so that E lies on the other side of AC.
We can see that angle BAP = 45o - angle PAC = angle CAE. Also AB/AC=AP/AE or AB/AP = AC/AE. Thus triangles BAP and CAE are similar, so CE/AE = BP/AP = 2 or CE = 2 * sqrt2. This means PE^2 + CE^2 = CP^2 and so angle PEC = 90. Also angle AEP = 45 because APE is a right isosceles triangle. So angle AEC = 135 and angle APB =135 (remember the similar triangles).
Okay. Is there a better way to count them?
I mean, there should be a way to compute the number of permutations without listing all those permutations out, right?
Explain?
Yeah, I guessed it.
Have you programmed it and get an answer?
P(n!) = |P(n)|! for all positive integers n.
Order the integers from 1 to 37 in a sequence such that the sum of the first n numbers is divisible by the (n+1)th number.
a) If the first integer is 37 and the second integer is 1, how many possible sequences are there?
b) How many possible sequences are there? (in general)
If there is less than 50 sequences for a, please list them for me.
Thanks.
You got to use vectors, but not trigonometry yet?
That's a weird way of ordering the textbook.
So a + b = 10.
Thus (a+b)^2 = 100
2(a^2 + b^2) - (a-b)^2 = 100
2(a^2 + b^2) = 100 + (a-b)^2 >= 100
(a^2 + b^2) >= 50.
Equality holds if (a-b)^2 = 0 or a = b = 5.
It doesn't matter if the numbers are real or integral or whatever.
I have found a full solution not involving circumcircle, similar triangles or side lengths. However it involves the intercept theorem.
First of all mark D, E and K as I have described in post #6.
Triangles CDK and CEK are halves of the equilateral triangle CDE, so CK = CD/2 = CA. Therefore triangle CKA is isosceles at C and thus angle CAK = (180o - 120o) / 2 = 30o = angle CDK, so KA = KD.
Angle KAB = angle CAB - angle CAK = 45o - 30o = 15o = angle KBA. Thus KA = KB.
Triangle KBD has KB = KD and angle BKD = 90o, so it is an isosceles right triangle. Thus angle KBD = 45o, so angle MBD = angle MBK + angle KBD = 15o + 45o = 60o.
The last part uses the intercept theorem (although I believe there should be a simple way to overcome this obstacle and turn it into an even easier solution). Since AM/AB = 1/3 = AC/AD, MC and BD are parallel. Therefore angle BMC = 180o - angle MBD = 180o - 60o = 120o. (a pair of interior angles on the same side of the transversal).
That is QED. So far this is the simplest solution I could find.
The key to the solution is that if you draw the points exactly, you should see that MC and BD are parallel. Although I managed to get the D before, I failed to see this because I didn't try to draw accurately enough.
So lesson is: try to draw it as accurately as possible. Also, remember some key triangles that will come to the questions: 20-80-80, 50-50-80, 20-20-140, 15-45-120 etc. and remember the approach(es) to each triangles. There is usually only a few for each triangle.
Nikolasekulov,
Have you learnt the Pythagorean theorem? If not, I suppose getting an answer to this problem will be much more difficult. If yes, you probably must have worked with irrational numbers. (think about a right triangle with legs 1 and 2. what is the hypotenuse?)
The 45-45-90 triangle's side lengths is in a ratio of 1-1-sqrt 2 because it is isosceles and thus the two legs are equal.
For the 30-60-90 triangle, it is half an equilateral triangle. Thus the hypotenuse is twice the length of the leg opposite the 30 degrees angle. Using the Pythagorean theorem will help you compute the last side.
First of all you can prove that BED = ACB = 120 degrees.
Draw an altitude BJ on AC. Let AJ = BJ = x then BC = x*2/(sqrt(3)), CJ = x/(sqrt(3)), AC = AJ - CJ = x*(3-sqrt(3))/3 etc.
Basically by using 45-45-90 (sides 1-1-sqrt2) and 30-60-90 (sides 1-sqrt3-2) triangles we can compute the sides and prove that their ratios are equal. In my country we learn these two special right triangles two years before trig.
No, I don't have the expression.
I can prove S(2) = 19.
First of all, if 16 is in a square with 3 or 4 adjacent squares then one of its neighbours will be at least 3. Thus there will be a sum of at least 19.
Otherwise, 16 is in a corner. We prove that then the maximum sum will still be at least 19. If 3, 4, 5 etc is adjacent to 16 then QED. Suppose 16's neighbours are 1 and 2. Then wherever 15 is, it must have at least two neighbours other than 1 or 2. Thus 15 must have a neighbour at least 4 in value, and the sum would then be 4.
However, I did try to use a kind of checkerboard pattern like this:
16 - 02 - 13 - 06
01 - 14 - 05 - 10
15 - 04 - 11 - 08
03 - 12 - 07 - 09
When I apply this kind of checkerboard pattern to n from 3 to 5 I get S(3) = 40, S(4) = 69 and S(5) = 106. (ofc I'm not sure if this pattern yields the smallest S). Therefore I'm proposing that:
I used to have multiple problems on this 15-45-120 triangle. The solutions have always relied on the point D, which lies on ray AC and such that CD = 2 CA. If you also take E on BC such that CE = CD = 2CA then triangles ABC and BDE would be similar.
If you also construct a perpendicular DK from D to BC then AK = DK = BK.
I hope those can be starters for you.
That is right.
For example, min S(1) = 6 and min S(2) = 19.