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 "1962 AHSME Problems/Problem 31"

(→‎Solution)
m (→‎Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
{{solution}}
 
  
 
The formula for the measure of the interior angle of a regular polygon with <math>n</math>-sides is <math>180 - \frac{360}{n}</math>. Letting our two polygons have side length <math>r</math> and <math>k</math>, we have that the ratio of the interior angles is <math>\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}</math>.  Cross multiplying both sides, we have <math>2rk-4k = 3kr-6r \Rightarrow -rk-4k+6r = 0</math>. Using Simon's Favorite Factoring Trick, we have <math>-k(r+4)+6r+24 = 24 \Rightarrow (r+4)(6-k)=24</math>. Because <math>k</math> and <math>r</math> are both more than <math>2</math>, we know that <math>6-k < r+4</math>. Now, we just set these factors equal to the factors of 24. We can set <math>6-k</math> to <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math> and <math>r+4</math> to <math>24</math>, <math>12</math>, <math>8</math>, or <math>6</math> respectively to get the following pairs for <math>(k, r)</math>: <math>(5, 20)</math>, <math>(4, 8)</math>, <math>(3, 4)</math>, and <math>(2, 2)</math>. However, we have to take out the solution with <math>k = 2</math>, because <math>k</math> and <math>r</math> are both more than <math>2</math>, leaving us with <math>\boxed{\textbf{(C)}\ 3}</math> as the correct answer.
 
The formula for the measure of the interior angle of a regular polygon with <math>n</math>-sides is <math>180 - \frac{360}{n}</math>. Letting our two polygons have side length <math>r</math> and <math>k</math>, we have that the ratio of the interior angles is <math>\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}</math>.  Cross multiplying both sides, we have <math>2rk-4k = 3kr-6r \Rightarrow -rk-4k+6r = 0</math>. Using Simon's Favorite Factoring Trick, we have <math>-k(r+4)+6r+24 = 24 \Rightarrow (r+4)(6-k)=24</math>. Because <math>k</math> and <math>r</math> are both more than <math>2</math>, we know that <math>6-k < r+4</math>. Now, we just set these factors equal to the factors of 24. We can set <math>6-k</math> to <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math> and <math>r+4</math> to <math>24</math>, <math>12</math>, <math>8</math>, or <math>6</math> respectively to get the following pairs for <math>(k, r)</math>: <math>(5, 20)</math>, <math>(4, 8)</math>, <math>(3, 4)</math>, and <math>(2, 2)</math>. However, we have to take out the solution with <math>k = 2</math>, because <math>k</math> and <math>r</math> are both more than <math>2</math>, leaving us with <math>\boxed{\textbf{(C)}\ 3}</math> as the correct answer.

Revision as of 14:52, 30 October 2018

Problem

The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$. How many such pairs are there?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$

Solution

The formula for the measure of the interior angle of a regular polygon with $n$-sides is $180 - \frac{360}{n}$. Letting our two polygons have side length $r$ and $k$, we have that the ratio of the interior angles is $\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}$. Cross multiplying both sides, we have $2rk-4k = 3kr-6r \Rightarrow -rk-4k+6r = 0$. Using Simon's Favorite Factoring Trick, we have $-k(r+4)+6r+24 = 24 \Rightarrow (r+4)(6-k)=24$. Because $k$ and $r$ are both more than $2$, we know that $6-k < r+4$. Now, we just set these factors equal to the factors of 24. We can set $6-k$ to $1$, $2$, $3$, or $4$ and $r+4$ to $24$, $12$, $8$, or $6$ respectively to get the following pairs for $(k, r)$: $(5, 20)$, $(4, 8)$, $(3, 4)$, and $(2, 2)$. However, we have to take out the solution with $k = 2$, because $k$ and $r$ are both more than $2$, leaving us with $\boxed{\textbf{(C)}\ 3}$ as the correct answer.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1962_AHSME_Problems/Problem_31&oldid=98422"