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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.
+
<math>[LA A_1]</math> 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>
  
−
So the whole shaded area is equal to </math>(\frac{25\sqrt{3}}{4}) \cdot 4 + (\frac{25}{2}) \cdot 4 = 25\sqrt{3} + 50 = 50 + 25\sqrt{3}<math>
+
So the whole shaded area is equal to <math>(\frac{25\sqrt{3}}{4}) \cdot 4 + (\frac{25}{2}) \cdot 4 = 25\sqrt{3} + 50 = 50 + 25\sqrt{3}</math>
  
−
Thus, </math>a + b + c = 50 + 25 + 3 = 78$
+
Thus, <math>a + b + c = 50 + 25 + 3 = 78</math>
 +
 
 +
~akliu

Latest revision as of 13:37, 24 June 2022

Since this is a dodecagon, there are $12$ sides.

Thus, the angle measure for an angle on the regular polygon is:

$\frac{10 \cdot 180}{12} = 150$

Next, we can draw a two lines down from $A$ and from $B$ that are perpendicular to $LC$.

We can denote these intersection point as $A_1$ and $B_1$, respectively.

Since the angle measure of $LA A_1$ is $\angle{60}$ and $m \angle{BA A_1} = 90$, we can say that the shaded area is equal to $4(2 \cdot [LA A_1] + [ABB_1 A_1])$

$[LA A_1]$ is equal to $\frac{5}{2} \cdot \frac{5\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}$ because it is a 30-60-90 triangle.

Next, $[AB B_1 A_1] = 5 \cdot \frac{5}{2} = \frac{25}{2}$

So the whole shaded area is equal to $(\frac{25\sqrt{3}}{4}) \cdot 4 + (\frac{25}{2}) \cdot 4 = 25\sqrt{3} + 50 = 50 + 25\sqrt{3}$

Thus, $a + b + c = 50 + 25 + 3 = 78$

~akliu

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