Difference between revisions of "2024 SSMO Accuracy Round Problems/Problem 2"
(Created page with "==Problem== Equilateral triangle <math>N</math> is inscribed within circle <math>O</math>. A smaller equilateral triangle <math>P</math> is inscribed within <math>N</math> su...") |
|||
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
| + | WLOG, let the radius of <math>O</math> be 1. Then, the side length of <math>N</math> is <math>\sqrt{3},</math> meaning the side length of <math>P</math> is <math>\frac{\sqrt{3}}{2}.</math> Thus, the answer is <cmath>\frac{\pi}{\left(\frac{\sqrt{3}}{2}\right)^2\cdot\left(\frac{\sqrt{3}}{4}\right)} = \frac{\pi}{\frac{3\sqrt{3}}{16}} = \frac{16\sqrt{3}\pi}{9}\implies 16+3+9 = \boxed{28}.</cmath> | ||
| + | |||
| + | ~SMO_Team | ||
Latest revision as of 14:31, 10 September 2025
Problem
Equilateral triangle 
is inscribed within circle 
. A smaller equilateral triangle 
is inscribed within such that the vertices of lie on the midpoints of . The ratio of the areas between and can be expressed as 
for relatively prime positive integers 
and squarefree 
Find 
.
Solution
WLOG, let the radius of be 1. Then, the side length of is 
meaning the side length of is 
Thus, the answer is ![\[\frac{\pi}{\left(\frac{\sqrt{3}}{2}\right)^2\cdot\left(\frac{\sqrt{3}}{4}\right)} = \frac{\pi}{\frac{3\sqrt{3}}{16}} = \frac{16\sqrt{3}\pi}{9}\implies 16+3+9 = \boxed{28}.\]](http://latex.artofproblemsolving.com/6/0/c/60c490800303370848cecbb86cb100c1358381f9.png)
~SMO_Team
