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 $202x+4x^2=y^2$.
Solution: Notice that $y>2x$. Let $y=2x+k$ for some integer $k>0$. Thus, the given equation reduces to $$202x=4xk+k^2\implies x=\frac{k^2}{202-4k}\cdots (1)$$ This tells us that $202-4k|k^2,$ or that $101-2k|2k^2\implies 101-2k|101k$. However, since 101 is a prime, $\gcd(101-2k,\,101)=1\implies 101-2k|k$ or that $101-2k|2k\implies 101-2k|101\implies k=50$. Substituting in $(1)$, we get that $x$ must be $$\frac{50^2}{202-4(50)}=50\cdot 25=1250.$$ 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 $ax^2+bx+c=0$ has two distinct roots $x_1,x_2$;
- the equation $bx^2+cx+a=0$ has two distinct roots $x_2,x_3$;
- the equation $cx^2+ax+b=0$ has two distinct roots $x_3,x_1$.
Solution: This is going to be a bit computational. Here we go:
We know that $ax_1^2+bx_1+c=0$ and $cx_1^2+ax_1+b=0$. Eliminating the quadratic term, we see that $$\left(\frac{b}{a}-\frac{a}{c}\right)x_1=\frac{b}{c}-\frac{c}{a}\implies x_1=\frac{c^2-ab}{a^2-bc}.$$ Similarly, $$x_2=\frac{a^2-bc}{b^2-ca},\,x_3=\frac{b^2-ca}{c^2-ab}.$$ Further, since $x_1,$ $x_2$ are roots of $ax^2+bx+c$, their product is $c/a$. This tells us that $$\frac{c^2-ab}{b^2-ca}=\frac{c}{a}\implies c^2a-a^2b=b^2c-c^2a$$ $$\implies 2c^2a=a^2b+b^2c$$ Similarly, we must have $$2a^2b=b^2c+c^2a$$ Subtracting the two, we get $$c^2a=a^2b\implies c^2=ab.$$ By symmetry, we must also have $$a^2=bc,\,b^2=ca\implies abc=a^3=b^3=c^3.$$ $$a^3-b^3=0\implies (a-b)(a^2+ab+b^2)=0\implies (a-b)(a^2+c^2+b^2)=0$$ Since $a,b,c$ are given to be non zero reals, we must have $a-b=0\implies a=b$. Thus, it is easy to see that $a,b,c$ are all equal. However, this means that the three sets $\{x_1,x_2\},\{x_2,x_3\},$ and $\{x_3,x_1\}$ are the same. This happens if and only if $x_1=x_2=x_3$, but that cannot be possible since $ax^2+bx+c$ has two distinct roots when $a=b=c$. Phew. QED.
We know that $ax_1^2+bx_1+c=0$ and $cx_1^2+ax_1+b=0$. Eliminating the quadratic term, we see that $$\left(\frac{b}{a}-\frac{a}{c}\right)x_1=\frac{b}{c}-\frac{c}{a}\implies x_1=\frac{c^2-ab}{a^2-bc}.$$ Similarly, $$x_2=\frac{a^2-bc}{b^2-ca},\,x_3=\frac{b^2-ca}{c^2-ab}.$$ Further, since $x_1,$ $x_2$ are roots of $ax^2+bx+c$, their product is $c/a$. This tells us that $$\frac{c^2-ab}{b^2-ca}=\frac{c}{a}\implies c^2a-a^2b=b^2c-c^2a$$ $$\implies 2c^2a=a^2b+b^2c$$ Similarly, we must have $$2a^2b=b^2c+c^2a$$ Subtracting the two, we get $$c^2a=a^2b\implies c^2=ab.$$ By symmetry, we must also have $$a^2=bc,\,b^2=ca\implies abc=a^3=b^3=c^3.$$ $$a^3-b^3=0\implies (a-b)(a^2+ab+b^2)=0\implies (a-b)(a^2+c^2+b^2)=0$$ Since $a,b,c$ are given to be non zero reals, we must have $a-b=0\implies a=b$. Thus, it is easy to see that $a,b,c$ are all equal. However, this means that the three sets $\{x_1,x_2\},\{x_2,x_3\},$ and $\{x_3,x_1\}$ are the same. This happens if and only if $x_1=x_2=x_3$, but that cannot be possible since $ax^2+bx+c$ 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 $49\cdot[BDE]=25\cdot[CDF]$ 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 $A=(0,2\sqrt{3})$. Notice that $D=(-1,0)$. We compute the equation of the perpendicular bisector of $AD$ to be $$y=\frac{-1}{2\sqrt{3}}x+\frac{11}{4\sqrt{3}}$$ Intersect this with the equations of $AB,AC:$ $$y=\sqrt{3}x+2\sqrt{3},\,y=-\sqrt{3}x+2\sqrt{3}$$ to get the $y$-coordinates of $E,F$ as $15\sqrt{3}/14$ and $7\sqrt{3}/10$ respectively. Thus, $$\frac{[BDE]}{[CDF]}=\frac{1\cdot15\sqrt{3}/14}{3\cdot 7\sqrt{3}/10}=\frac{25}{49}.$$
Problem 4: Let $n>1$ be a positive integer. Call a rearrangement $a_1,a_2,\cdots,a_n$ of $1,2,\cdots,n$ $nice$ if for every $k=2,3,\cdots n$ we have that $a_1^2+a_2^2+\cdots+a_k^2$ is not divisible by $k$. Determine which positive integers $n>1$ have a nice rearrangement.
Solution: Observe that $1^2+2^2+\cdots+n^2=n(n+1)(2n+1)/6$ 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 $a_1=1,a_2=2$. 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 $k-1$. (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 $\angle ABC = 20^\circ$ and $\angle ACB=40^\circ$. Let $D$ be a point on $BC$ such that $\angle BAD=\angle DAC$. Let the incircle of triangle $ABC$ touch $BC$ at $E$. Prove that $BD=2\cdot CE$.
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 $CE=(a+b-c)/2$. 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 $$(a+b-c)(b+c)=ac\iff ab+b^2=c^2.$$ Let the angle bisector of $\angle C$ intersect $AB$ and $X$. Notice that $\Delta ACX\sim\Delta ABC\implies AX/AC=AC/AB.$ By the angle bisector theorem, $AX=\frac{cb}{a+b}$. Thus, $$\frac{cb}{a+b}=\frac{b^2}{c}\implies c^2=b(a+b)=b^2+ab.$$ QED.
Problem 6: Let $X$ be a set of $11$ integers. Prove that one can find a non empty subset $\{a_1,a_2,\cdots,a_k\}$ of $X$ such that $3$ divides $k$ and $9$ divides the sum $\sum_{i=1}^k4^ia_i$.
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 $a_1, a_2, a_3$ be such numbers from $X.$ Notice that $s_1=4a_1+16a_2+64a_3\equiv a_1+a_2+a_3\equiv 0(\text{mod }3)$. If $9|s_1$, we are done. Thus, $s_1$ must be $3$ or $6$ modulo $9$. Further, from the remaining $8$, there is another set of $3$ numbers $a_4,a_5,a_6$ such that their sum, $s_2$ is divisible by $3$. Again, if it is divisible by $9$, we are done and so $s_2$ is either $3$ or $6$ modulo $9$. Notice that if $s_1+s_2\equiv0(\text{mod }9)$, we can consider $k=6$ and $a_1,a_2,a_3,a_4,a_5,a_6$ to find the desired set. It must mean that $s_1$ and $s_2$ 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 $a_7, a_8, a_9$ such that their sum $s_3$ is $0$ mod $3$. We consider three cases:
Case I: $s_3\equiv0(\text{mod }9)$. Then we are done - consider $k=3$ and $a_7, a_8, a_9$.
Case I: $s_3\equiv0(\text{mod }9)$. Then we are done - consider $k=3$ and $a_7, a_8, a_9$.
Case II: $s_3\equiv3(\text{mod }9)$. If $s_1$ and $s_2$ were both 3 modulo 9, we set $k=9$ and consider numbers $a_1$ through $a_9$, Otherwise, if $s_1$ and $s_2$ were both 6 modulo 9, we set $k=6$ and consider the numbers from $s_1$ and $s_3$.
Case III: $s_3\equiv6(\text{mod }9)$. 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
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