Difference between revisions of "2024 SSMO Team Round Problems/Problem 9"
(Created page with "==Problem== Let <math>ABCDEFGH</math> be an equiangular octagon such that <math>AB=6, BC=8, CD=10, DE=12, EF=6, FG=8, GH=10,</math> and <math>AH=12</math>. The radius of the...") |
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==Solution== | ==Solution== | ||
| + | Note that <math>AB \parallel EF, BC \parallel FG, CD \parallel GH,</math> and <math>DE \parallel HA.</math> Thus, the diameter of the circle is the least distance between any pair of these opposite parallel sides. Note that a side of length <math>x</math> oriented at a <math>45</math>-degree angle relative to a pair of opposite parallel sides contributes <math>\frac{x}{\sqrt2}</math> to the distance and a side of length <math>x</math> oriented perpendicularly contributes <math>x</math> to the distance. Obviously, we only have to add up the contributions of one run of three sides; for example for <math>AB</math> and <math>EF</math> we only have to consider sides from <math>B</math> to <math>E</math> or <math>F</math> to <math>A.</math> However, it doesn't matter which run we pick since opposite sides are congruent. Let the distances be <math>a,b,c,</math> and <math>d</math> respectively, with the order given in the first sentence. We have \begin{align*} | ||
| + | a &= 10 + \frac{8}{\sqrt2} + \frac{12}{\sqrt2} = 10 + 10\sqrt2, \\ | ||
| + | b &= 12 + \frac{10}{\sqrt2} + \frac{6}{\sqrt2} = 12 + 8\sqrt2, \\ | ||
| + | c &= 6 + \frac{12}{\sqrt2} + \frac{8}{\sqrt2} = 6 + 10\sqrt2, \text{ and} \\ | ||
| + | d &= 8 + \frac{6}{\sqrt2} + \frac{10}{\sqrt2} = 8 + 8\sqrt2. \end{align*} Obviously, <math>10 + 10\sqrt2 \ge 8+8\sqrt2</math> and <math>12+8\sqrt2 \ge 8+8\sqrt2.</math> We also have <math>6+10\sqrt2 \ge 8+8\sqrt2</math> since <math>\sqrt2 \ge 1.</math> So, the least distance between pairs of opposite parallel sides is <math>8+8\sqrt2</math> and the radius of the largest circle that fits inside the octagon is <math>4+4\sqrt2.</math> Thus, the answer is <math>4+4+2=\boxed{10}.</math> | ||
| + | |||
| + | ~SMO_Team | ||
Latest revision as of 14:38, 10 September 2025
Problem
Let 
be an equiangular octagon such that 
and 
. The radius of the largest circle that fits inside the octagon can be expressed as 
where \(a,b,\) and \(c\) are integers and \(c\) is squarefree. Find 
Solution
Note that 
and 
Thus, the diameter of the circle is the least distance between any pair of these opposite parallel sides. Note that a side of length 
oriented at a 
-degree angle relative to a pair of opposite parallel sides contributes 
to the distance and a side of length oriented perpendicularly contributes to the distance. Obviously, we only have to add up the contributions of one run of three sides; for example for 
and 
we only have to consider sides from 
to 
or 
to 
However, it doesn't matter which run we pick since opposite sides are congruent. Let the distances be 
and 
respectively, with the order given in the first sentence. We have \begin{align*}
a &= 10 + \frac{8}{\sqrt2} + \frac{12}{\sqrt2} = 10 + 10\sqrt2, \\
b &= 12 + \frac{10}{\sqrt2} + \frac{6}{\sqrt2} = 12 + 8\sqrt2, \\
c &= 6 + \frac{12}{\sqrt2} + \frac{8}{\sqrt2} = 6 + 10\sqrt2, \text{ and} \\
d &= 8 + \frac{6}{\sqrt2} + \frac{10}{\sqrt2} = 8 + 8\sqrt2. \end{align*} Obviously, 
and 
We also have 
since 
So, the least distance between pairs of opposite parallel sides is 
and the radius of the largest circle that fits inside the octagon is 
Thus, the answer is 
~SMO_Team
