Vi har och . Om så . .
Det är ett skönt problem. Tack så mycket, Kurre.
Yes that is the trigonometric approach
Have you been studying swedish? I think I remember that you wrote that you are interested in languages. Nicely done!
Anyway here is my method:
Reflect AB in the horizontal line and draw line BC as in 2. Now reflect the triangle ABC in AC to make a quadrilateral as in 3. Now, all angles in the quadrilateral ABCD are equal, and all sides are equal, thus it is a square, and since AC is a diagonal,must be half of a right angle, ie
This is an easy problem if you use the standard approach with trigonometry (the problem is for ~17 years old students). But I found a much nicer geometrical method, without using any trigonometry.
So my challenge for you is to find it.
(or solve it any other way without trigonometry is fine too )
Il give you some tips. First prove that the first number (4-digit minus reverse) is divisible by 9. We know (or you can prove that too) that a number is divisible by 9 iff its sum of digits is divisible by 9. So when we first add the digits it must be a number divisible by 9. What possible numbers are there?
We first use the cosine theorem on each of the squared sides:
What if one angle is obtuse?
Then we can argue as follows. WLOG we assume
I guess he means that, because we assumed that p,p+2 where the largest pair there does not exist any numbers with factors p+2k,p+2(k+1).
As I posted before, this statement is false. (p+2k)*(p+2(k+1)) has precisely those numbers as factors.
hm I meant prime factors, but maybe he does not use that they are prime factors, idk, im too lazy to read through the oproof atm.
I'm having a lot of trouble understanding what it is you're trying to go for.
If we suppose that there exist a larger pair of the form p+2k and p+2(k+1) Than the set of numbers that have as factors p+2k and p+2(k+1) must be equal to 0 (since they do not exist).
We are supposing that there exists a larger pair where p+2k and p+2(k+1) are both prime? This doesn't make sense, we've already supposed that p and p+2 were the largest pair with this property.
And you say that the number of integers that have p+2k and p+2(k+1) as factors must be zero. This is false, just look at the number:
This number has both those as a factor.
I guess he means that, because we assumed that p,p+2 where the largest pair there does not exist any numbers with factors p+2k,p+2(k+1). IM not sure though and I have not tried to understand the proof either.
Thanks for the tips!
Actually I made this thread to see what kinds of programs that existed, since I got the idea to create such a program myself. I have tried Geometers Sketchpad and Cabri now, and Cabri was really nice but was not exactly the type of program I had in mind, altough I guess I will not be able to make a program close as good. But Il maybe give it a try anyway
Are there any good programs for drawing geometrical pictures? I mean that has functions for lets say creating circumcircles, marking centroids/circumcentres/medians etc, basically a program that is designed for creating proofs/problems/solutions in euclidean geometry??
Write it as C(n,0)*C(n,n)+C(n,1)*C(n,n-1)+...+C(n,k)*C(n,n-k)+...+C(n,n)*C(n,0).
Then assume you have a set of 2n persons, and divide it into two sets A and B of n persons in each. Now you want to choose a comittee of n persons. Then for each choice there will be say k persons in set A, and n-k persons in set B, and summing will yield exactly the sum above.