Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 JMPSC Sprint Problems/Problem 1"
 
Line 18: Line 18:
  
 
- kante314 -
 
- kante314 -
 +
 +
== Solution 4 ==
 +
 +
''Solution by pog''
 +
 +
Recall that, for any nonzero <math>a</math>, <math>b</math>, <math>c</math>, we have that <math>\dfrac1a+\dfrac1b+\dfrac1c = \dfrac{\sum_{\text{sym}}ab}{abc}</math>. By Vieta's, we see that <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the monic cubic polynomial <cmath>Q(x) = x^3 - px^2 + \left(\sum_{\text{sym}}ab\right) x - abc,</cmath> where <math>p</math> is an arbitrary constant equal to <math>a + b + c</math>.
 +
 +
In the given equation, we have that <math>a = 3</math>, <math>b = 6</math>, and <math>c = 9</math>, so <math>p = 3 + 6 + 9 = 18</math> and <math>abc = 3 \cdot 6 \cdot 9 = 162</math>, for <cmath>Q(x) =  x^3 - 18x^2 + \left(\sum_{\text{sym}}ab\right) x - 162.</cmath> (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 <math>3</math>, <math>6</math>, and <math>9</math>, the discriminant of our monic cubic polynomial <math>A^2B^2 + 18ABC - 4B^3 - 4A^3C - 27C^2</math> (where <math>A</math>, <math>B</math>, and <math>C</math> are the latter three coefficients of <math>Q(x)</math>, respectively--do not confuse them for the roots of our polynomial equal to <math>a=3</math>, <math>b=6</math>, and <math>c=9</math>, respectively) must not be equal to <math>0</math>, 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 <math>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</math> must not be equal to <math>0</math>. Letting <math>\sum_{\text{sym}}ab = n</math>, expanding gives that <math>324n^2 + 52488n - 4n^3 - 3779136 - 708588 = -4n^3 + 324n^2 + 52488n - 4487724</math> is not equal to <math>0</math>. Thus <math>n</math> must not be a root of our cubic, so from here we apply the cubic formula to find that <math>n</math> is not equal to (approximately) <math>-115.853202973303</math>, <math>96.4974047104675</math>, or <math>100.355798262835</math>.
 +
 +
Recall that, from earlier, our answer is equal to <math>\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}</math>. By bounding, <math>n</math> is between the aforementioned values of <math>96.4974047104675</math> and <math>100.355798262835</math>, so since it must be a multiple of <math>9</math> (all answers in the Junior Mathematicians' Problem Solving Competition are integers), it is equal to <math>99</math> and the requested answer is equal to <math>\dfrac{99}{9} = \boxed{11}</math>. This corroborates with jasperE3's strict bounding. <math>\blacksquare</math>
 +
 +
<div style='text-align: center;'>'''Figure 1:''' Final diagram from AoPS user ''awang11'':
 +
[[File:Screen Shot 2021-07-12 at 10.25.44 AM.png||400px]]</div>
  
 
==See also==
 
==See also==

Latest revision as of 09:53, 12 July 2021

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: Screen Shot 2021-07-12 at 10.25.44 AM.png

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_1&oldid=158225"