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RMO 2024: Discussing Solutions

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 x,y such that 202x+4x2=y2.

Solution: Notice that y>2x. Let y=2x+k for some integer k>0. Thus, the given equation reduces to 202x=4xk+k2x=k22024k(1) This tells us that 2024k|k2, or that 1012k|2k21012k|101k. However, since 101 is a prime, gcd(1012k,101)=11012k|k or that 1012k|2k1012k|101k=50. Substituting in (1), we get that x must be 5022024(50)=5025=1250. Thus, y=2x+50=2550. We conclude that the only solution is x=1250 and y=2550, which is easy to verify.

Problem 2: Show that there do not exist non-zero real numbers a,b,c such that the following statements hold simultaneously: 

  • the equation ax2+bx+c=0 has two distinct roots x1,x2
  • the equation bx2+cx+a=0 has two distinct roots x2,x3
  • the equation cx2+ax+b=0 has two distinct roots x3,x1
(Note that x1,x2,x3 may be real or complex numbers.)

Solution: This is going to be a bit computational. Here we go:
We know that ax12+bx1+c=0 and cx12+ax1+b=0. Eliminating the quadratic term, we see that (baac)x1=bccax1=c2aba2bc. Similarly, x2=a2bcb2ca,x3=b2cac2ab. Further, since x1, x2 are roots of ax2+bx+c, their product is c/a. This tells us that c2abb2ca=cac2aa2b=b2cc2a 2c2a=a2b+b2c Similarly, we must have 2a2b=b2c+c2a Subtracting the two, we get c2a=a2bc2=ab. By symmetry, we must also have a2=bc,b2=caabc=a3=b3=c3. a3b3=0(ab)(a2+ab+b2)=0(ab)(a2+c2+b2)=0 Since a,b,c are given to be non zero reals, we must have ab=0a=b. Thus, it is easy to see that a,b,c are all equal. However, this means that the three sets {x1,x2},{x2,x3}, and {x3,x1} are the same. This happens if and only if x1=x2=x3, but that cannot be possible since ax2+bx+c has two distinct roots when a=b=c. Phew. QED. 

Problem 3: Let ABC be an equilateral triangle. Suppose D is the point on BC such that BD:DC=1:3. Let the perpendicular bisector of AD intersect AB,AC at E,F respectively. Prove that 49[BDE]=25[CDF] where [XYZ] denotes the area of triangle XYZ.

Solution: Let's bash this! We define a co-ordinate plane such that point B=(2,0), C=(2,0) and A=(0,23). Notice that D=(1,0). We compute the equation of the perpendicular bisector of AD to be y=123x+1143 Intersect this with the equations of AB,AC: y=3x+23,y=3x+23 to get the y-coordinates of E,F as 153/14 and 73/10 respectively. Thus, [BDE][CDF]=1153/14373/10=2549.

Problem 4: Let n>1 be a positive integer. Call a rearrangement a1,a2,,an of 1,2,,n nice if for every k=2,3,n we have that a12+a22++ak2 is not divisible by k. Determine which positive integers n>1 have a nice rearrangement. 

Solution: Observe that 12+22++n2=n(n+1)(2n+1)/6 is divisible by n if and only if 6 and n are co-prime. Thus, we claim that we can have nice rearrangements for all n such that the gcd of n and 6 is greater than 1. We start with a1=1,a2=2. We propose the following recursive construction: if k and k+1 are both not co-prime to 6, then we can append k+1 to the nice rearrangement of k. If k+1 is not co-prime but k is, then append (k+1),k to the nice rearrangement of k1. (I've always wanted to say this) The details are left for the reader to figure out!

Problem 5: Let ABC be a triangle with ABC=20 and ACB=40. Let D be a point on BC such that BAD=DAC. Let the incircle of triangle ABC touch BC at E. Prove that BD=2CE.

Solution: If I absolutely had to pick a favorite problem from this paper, it would probably be this one. Let AB,BC,CA have lengths c,a,b respectively. We know that CE=(a+bc)/2. By the angle bisector theorem, since BD/DC=AB/AC=c/b and BD+DC=a, we compute BD=ac/(b+c). Thus, we wish to show (a+bc)(b+c)=acab+b2=c2. Let the angle bisector of C intersect AB and X. Notice that ΔACXΔABCAX/AC=AC/AB. By the angle bisector theorem, AX=cba+b. Thus, cba+b=b2cc2=b(a+b)=b2+ab. QED.

Problem 6: Let X be a set of 11 integers. Prove that one can find a non empty subset {a1,a2,,ak} of X such that 3 divides k and 9 divides the sum i=1k4iai.

Solution: The key observation in this problem is that among any 5 integers, there exist 3 such that their sum is divisible by 3 (why?). Let a1,a2,a3 be such numbers from X. Notice that s1=4a1+16a2+64a3a1+a2+a30(mod 3). If 9|s1, we are done. Thus, s1 must be 3 or 6 modulo 9. Further, from the remaining 8, there is another set of 3 numbers a4,a5,a6 such that their sum, s2 is divisible by 3. Again, if it is divisible by 9, we are done and so s2 is either 3 or 6 modulo 9. Notice that if s1+s20(mod 9), we can consider k=6 and a1,a2,a3,a4,a5,a6 to find the desired set. It must mean that s1 and s2 are both 3 or both 6 modulo 9. Now, notice that we have 5 numbers remaining in X. Again, there exist three out of these, say a7,a8,a9 such that their sum s3 is 0 mod 3. We consider three cases:
Case I: s30(mod 9). Then we are done - consider k=3 and a7,a8,a9
Case II: s33(mod 9). If s1 and s2 were both 3 modulo 9, we set k=9 and consider numbers a1 through a9, Otherwise, if s1 and s2 were both 6 modulo 9, we set k=6 and consider the numbers from s1 and s3
Case III: s36(mod 9). This case is very similar to that of Case II. Left as exercise. 

Hope this was helpful! 
Signing off,
Saee
xoxo

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