Difference between revisions of "2021 WSMO Team Round/Problem 6"
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Since the angle measure of <math>LA A_1</math> is <math>\angle{60}</math> and <math>m \angle{BA A_1} = 90</math>, we can say that the shaded area is equal to <math>4(2 \cdot [LA A_1] + [ABB_1 A_1])</math> | Since the angle measure of <math>LA A_1</math> is <math>\angle{60}</math> and <math>m \angle{BA A_1} = 90</math>, we can say that the shaded area is equal to <math>4(2 \cdot [LA A_1] + [ABB_1 A_1])</math> | ||
| − | + | <math>[LA A_1] is equal to </math>\frac{5}{2} \cdot \frac{5\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}<math> because it is a 30-60-90 triangle. | |
Next, </math>[AB B_1 A_1] = 5 \cdot \frac{5}{2} = \frac{25}{2}<math> | Next, </math>[AB B_1 A_1] = 5 \cdot \frac{5}{2} = \frac{25}{2}<math> | ||
Revision as of 13:37, 24 June 2022
Since this is a dodecagon, there are 
sides.
Thus, the angle measure for an angle on the regular polygon is:
Next, we can draw a two lines down from 
and from 
that are perpendicular to 
.
We can denote these intersection point as 
and 
, respectively.
Since the angle measure of 
is 
and 
, we can say that the shaded area is equal to ![$4(2 \cdot [LA A_1] + [ABB_1 A_1])$](http://latex.artofproblemsolving.com/9/6/c/96cba5daa70bfbc05ba1f2f0093c52108adfa66e.png)
![$[LA A_1] is equal to$](http://latex.artofproblemsolving.com/2/3/4/23435c1bc9a3fd2c695983c0c88190a8a7a9284f.png)
\frac{5}{2} \cdot \frac{5\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}$because it is a 30-60-90 triangle.
Next,$ (Error compiling LaTeX. Unknown error_msg)[AB B_1 A_1] = 5 \cdot \frac{5}{2} = \frac{25}{2}
(\frac{25\sqrt{3}}{4}) \cdot 4 + (\frac{25}{2}) \cdot 4 = 25\sqrt{3} + 50 = 50 + 25\sqrt{3}
a + b + c = 50 + 25 + 3 = 78$
