Difference between revisions of "1962 AHSME Problems/Problem 31"
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− | 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 17:19, 25 March 2021
Problem
The ratio of the interior angles of two regular polygons with sides of unit length is . How many such pairs are there?
Solution
The formula for the measure of the interior angle of a regular polygon with -sides is . Letting our two polygons have side length and , we have that the ratio of the interior angles is . Cross multiplying both sides, we have . Using [Simon's Favorite Factoring Trick], we have . Because and are both more than , we know that . Now, we just set these factors equal to the factors of 24. We can set to , , , or and to , , , or respectively to get the following pairs for : , , , and . However, we have to take out the solution with , because and are both more than , leaving us with as the correct answer.