# 2021 JMPSC Sprint Problems/Problem 1

## Problem

Compute $\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)(3+6+9)$.

## Solution

Solving the right side gives $3 + 6 + 9 = 18$. Distributing into the left side gives $\frac{18}{3}+\frac{18}{6}+\frac{18}{9}$, so the answer is $6 + 3 + 2 = \boxed{11}$.

## Solution 2

$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=\frac{6}{18}+\frac{3}{18}+\frac{2}{18}=\frac{11}{18}$ and $3+6+9=18$, so the answer is $\frac{11}{18}\cdot18=\boxed{11}$.

## Solution 3

$$3 \left(\frac{1}{3}\right) + 3 \left(\frac{1}{6}\right) + 3 \left(\frac{1}{9}\right)=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$$$$6 \left(\frac{1}{3}\right) + 6 \left(\frac{1}{6}\right) + 6 \left(\frac{1}{9}\right)=2+1+\frac{2}{3}=\frac{22}{6}$$$$9 \left(\frac{1}{3}\right) + 9 \left(\frac{1}{6}\right) + 9 \left(\frac{1}{9}\right)=3+\frac{3}{2}+1=\frac{33}{6}$$ Therefore, the answer is $\frac{11}{6}+\frac{22}{6}+\frac{33}{6}=11$

- kante314 -

## Solution 4

Solution by pog

Recall that, for any nonzero $a$, $b$, $c$, we have that $\dfrac1a+\dfrac1b+\dfrac1c = \dfrac{\sum_{\text{sym}}ab}{abc}$. By Vieta's, we see that $a$, $b$, and $c$ are the roots of the monic cubic polynomial $$Q(x) = x^3 - px^2 + \left(\sum_{\text{sym}}ab\right) x - abc,$$ where $p$ is an arbitrary constant equal to $a + b + c$.

In the given equation, we have that $a = 3$, $b = 6$, and $c = 9$, so $p = 3 + 6 + 9 = 18$ and $abc = 3 \cdot 6 \cdot 9 = 162$, for $$Q(x) = x^3 - 18x^2 + \left(\sum_{\text{sym}}ab\right) x - 162.$$ (Finding the former of the latter two coefficients would lead to bashy casework, which we avoid in this extremely elegant solution.)

Since our cubic has roots $3$, $6$, and $9$, the discriminant of our monic cubic polynomial $A^2B^2 + 18ABC - 4B^3 - 4A^3C - 27C^2$ (where $A$, $B$, and $C$ are the latter three coefficients of $Q(x)$, respectively--do not confuse them for the roots of our polynomial equal to $a=3$, $b=6$, and $c=9$, respectively) must not be equal to $0$, or our cubic will have a double root, which it clearly does not (proof here is left to the reader).

Thus, substitution of the coefficients into our discriminant gives that $324 \cdot \left(\sum_{\text{sym}}ab\right)^2 + 324 \left(\sum_{\text{sym}}ab\right) \cdot 162 - 4\left(\sum_{\text{sym}}ab\right)^3 - 4 \cdot 18^3 \cdot 162 - 27 \cdot 162^2$ must not be equal to $0$. Letting $\sum_{\text{sym}}ab = n$, expanding gives that $324n^2 + 52488n - 4n^3 - 3779136 - 708588 = -4n^3 + 324n^2 + 52488n - 4487724$ is not equal to $0$. Thus $n$ must not be a root of our cubic, so from here we apply the cubic formula to find that $n$ is not equal to (approximately) $-115.853202973303$, $96.4974047104675$, or $100.355798262835$.

Recall that, from earlier, our answer is equal to $\left(\dfrac13+\dfrac16+\dfrac19\right)(3 + 6 + 9) = \left(\dfrac1a+\dfrac1b+\dfrac1c\right)(a+b+c) = \left(\dfrac{\sum_{\text{sym}}ab}{162}\right)(18) = \dfrac{n}{162} \cdot 18 = \dfrac{n}{9}$. By bounding, $n$ is between the aforementioned values of $96.4974047104675$ and $100.355798262835$, so since it must be a multiple of $9$ (all answers in the Junior Mathematicians' Problem Solving Competition are integers), it is equal to $99$ and the requested answer is equal to $\dfrac{99}{9} = \boxed{11}$. This corroborates with jasperE3's strict bounding. $\blacksquare$

Figure 1: Final diagram from AoPS user awang11: