Hello everyone!
Congratulations to everyone who attempted the RMO 2024. As you might know, we had an amazing livesolve of the paper with Archit, Adhitya, Abel and Kanav which you can check out here. We also have question wise video solutions to all the problems, thanks to Nanda, Om and Shreya!
We had a lot of people interested in solutions for the KV/JNV paper, which is what this blog post will be about. Without further ado, let's get started!
Problem 1: Find all positive integers such that .
Solution: Notice that . Let for some integer . Thus, the given equation reduces to This tells us that or that . However, since 101 is a prime, or that . Substituting in , we get that must be Thus, We conclude that the only solution is and , which is easy to verify.
Problem 2: Show that there do not exist non-zero real numbers such that the following statements hold simultaneously:
- the equation has two distinct roots ;
- the equation has two distinct roots ;
- the equation has two distinct roots .
(Note that may be real or complex numbers.)
Solution: This is going to be a bit computational. Here we go:
We know that and . Eliminating the quadratic term, we see that Similarly, Further, since are roots of , their product is . This tells us that Similarly, we must have Subtracting the two, we get By symmetry, we must also have Since are given to be non zero reals, we must have . Thus, it is easy to see that are all equal. However, this means that the three sets and are the same. This happens if and only if , but that cannot be possible since has two distinct roots when . Phew. QED.
Problem 3: Let be an equilateral triangle. Suppose is the point on such that . Let the perpendicular bisector of intersect at respectively. Prove that where denotes the area of triangle
Solution: Let's bash this! We define a co-ordinate plane such that point , and . Notice that . We compute the equation of the perpendicular bisector of to be Intersect this with the equations of to get the -coordinates of as and respectively. Thus,
Problem 4: Let be a positive integer. Call a rearrangement of if for every we have that is not divisible by . Determine which positive integers have a nice rearrangement.
Solution: Observe that is divisible by if and only if and are co-prime. Thus, we claim that we can have nice rearrangements for all such that the gcd of and is greater than . We start with . We propose the following recursive construction: if and are both not co-prime to , then we can append to the nice rearrangement of . If is not co-prime but is, then append to the nice rearrangement of . (I've always wanted to say this) The details are left for the reader to figure out!
Problem 5: Let be a triangle with and . Let be a point on such that . Let the incircle of triangle touch at . Prove that .
Solution: If I absolutely had to pick a favorite problem from this paper, it would probably be this one. Let have lengths respectively. We know that . By the angle bisector theorem, since and , we compute . Thus, we wish to show Let the angle bisector of intersect and . Notice that By the angle bisector theorem, . Thus, QED.
Problem 6: Let be a set of integers. Prove that one can find a non empty subset of such that divides and divides the sum .
Solution: The key observation in this problem is that among any integers, there exist such that their sum is divisible by 3 (why?). Let be such numbers from Notice that . If , we are done. Thus, must be or modulo . Further, from the remaining , there is another set of numbers such that their sum, is divisible by . Again, if it is divisible by , we are done and so is either or modulo . Notice that if , we can consider and to find the desired set. It must mean that and are both or both modulo . Now, notice that we have numbers remaining in . Again, there exist three out of these, say such that their sum is mod . We consider three cases:
Case I: . Then we are done - consider and .
Case II: . If and were both 3 modulo 9, we set and consider numbers through , Otherwise, if and were both 6 modulo 9, we set and consider the numbers from and .
Case III: . This case is very similar to that of Case II. Left as exercise.
Hope this was helpful!
Signing off,
Saee
xoxo
Isn't writing solutions a form of art :)
ReplyDeleteSo true. Saee didi orz
Delete