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Difference between revisions of "2013 AMC 12A Problems/Problem 24"

(Created page with "Suppose <math>p</math> is the answer. We calculate <math>1-p</math>. Assume that the circumradius of the 12-gon is <math>1</math>, and the 6 different lengths are <math>a_1</mat...")
 
Line 3: Line 3:
 
Assume that the circumradius of the 12-gon is <math>1</math>, and the 6 different lengths are <math>a_1</math>, <math>a_2</math>, <math>\cdot</math>, <math>a_6</math>, in increasing order. Then
 
Assume that the circumradius of the 12-gon is <math>1</math>, and the 6 different lengths are <math>a_1</math>, <math>a_2</math>, <math>\cdot</math>, <math>a_6</math>, in increasing order. Then
  
−
<math>a_k = 2\sin ( \frac{k\pi}{12} )</math>. So <math>a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5</math>, <math>a_2=1</math>, <math>a_3=\sqrt{2}\approx 1.4</math>, <math>a_4=\sqrt{3}\approx 1.7</math>, <math>a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3  </math>, <math>a_6 = 2</math>.
+
<math>a_k = 2\sin ( \frac{k\pi}{12} )</math>.  
  
−
Critical triples to consider:
+
So <math>a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5</math>,
 +
 
 +
<math>a_2=1</math>,
 +
 
 +
<math>a_3=\sqrt{2}\approx 1.4</math>,
 +
 
 +
<math>a_4=\sqrt{3}\approx 1.7</math>,
 +
 
 +
<math>a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3  </math>,
 +
 
 +
<math>a_6 = 2</math>.
 +
 
 +
Note that there are <math>12</math> segments of each length of <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_5</math>, respectively, and <math>6</math> segments of length <math>a_6</math>. There are <math>66</math> segments in total.
 +
 
 +
Now, Consider the following inequalities:
  
 
-  <math>a_1 + a_1 > a_2</math>: Since <math>6 > (1+\sqrt{2})^2</math>  
 
-  <math>a_1 + a_1 > a_2</math>: Since <math>6 > (1+\sqrt{2})^2</math>  
 +
 
-  <math>a_1 + a_1 < a_3</math>.
 
-  <math>a_1 + a_1 < a_3</math>.
 +
 
-  <math>a_1 + a_2</math> is greater than <math>a_3</math> but less than <math>a_4</math>.
 
-  <math>a_1 + a_2</math> is greater than <math>a_3</math> but less than <math>a_4</math>.
 +
 
-  <math>a_1 + a_3</math> is greater than <math>a_4</math> but equal to <math>a_5</math>.
 
-  <math>a_1 + a_3</math> is greater than <math>a_4</math> but equal to <math>a_5</math>.
 +
 
-  <math>a_1 + a_4</math> is greater than <math>a_6</math>.
 
-  <math>a_1 + a_4</math> is greater than <math>a_6</math>.
  
−
- <math>a_2+a_2 = 2 = a_6</math>.
+
- <math>a_2+a_2 = 2 = a_6</math>. Then obviously any two segments with at least one them longer than <math>a_2</math> have a sum greater than <math>a_6</math>.
  
−
Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following: (Note: x-y-z means <math>(a_x, a_y, a_z)</math>)
+
Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means <math>(a_x, a_y, a_z)</math>:
  
−
1-1-3, 1-1-4, 1-1-5, 1-1-6,
+
1-1-3, 1-1-4, 1-1-5, 1-1-6,
 +
1-2-4, 1-2-5, 1-2-6,
 +
1-3-5, 1-3-6,
 +
2-2-6
  
−
1-2-4, 1-2-5, 1-2-6,
+
In the above list there are <math>3</math> triples of the type a-a-b without ''6'', <math>2</math> triples of a-a-6 where a is not ''6'', <math>3</math> triples of a-b-c without ''6'', and <math>2</math> triples of a-b-6 where a, b are not ''6''. So,
−
 
 
−
1-3-5, 1-3-6,
 
−
 
 
−
2-2-6
 
−
 
 
−
Note that there are <math>12</math> segments of length <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_5</math>, respectively, and <math>6</math> segments of length <math>a_6</math>. Also in the above list there are <math>3</math> triples of the type a-a-b without ''6'', <math>2</math> triples of a-a-6 where a is not ''6'', <math>3</math> triples of a-b-c without ''6'', and <math>2</math> triples of a-b-6 where a, b are not ''6''. So,
 
  
 
<cmath>1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)</cmath>
 
<cmath>1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)</cmath>
  
−
<cmath> = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) </cmath>
+
<cmath> = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} </cmath>
−
 
 
−
<cmath> = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} </cmath>
 
  
 
So <math>p = 223/286</math>.
 
So <math>p = 223/286</math>.

Revision as of 15:00, 11 February 2013

Suppose $p$ is the answer. We calculate $1-p$.

Assume that the circumradius of the 12-gon is $1$, and the 6 different lengths are $a_1$, $a_2$, $\cdot$, $a_6$, in increasing order. Then

$a_k = 2\sin ( \frac{k\pi}{12} )$.

So $a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5$,

$a_2=1$,

$a_3=\sqrt{2}\approx 1.4$,

$a_4=\sqrt{3}\approx 1.7$,

$a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3$,

$a_6 = 2$.

Note that there are $12$ segments of each length of $a_1$, $a_2$, $\cdots$, $a_5$, respectively, and $6$ segments of length $a_6$. There are $66$ segments in total.

Now, Consider the following inequalities:

- $a_1 + a_1 > a_2$: Since $6 > (1+\sqrt{2})^2$

- $a_1 + a_1 < a_3$.

- $a_1 + a_2$ is greater than $a_3$ but less than $a_4$.

- $a_1 + a_3$ is greater than $a_4$ but equal to $a_5$.

- $a_1 + a_4$ is greater than $a_6$.

- $a_2+a_2 = 2 = a_6$. Then obviously any two segments with at least one them longer than $a_2$ have a sum greater than $a_6$.

Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means $(a_x, a_y, a_z)$: 

1-1-3, 1-1-4, 1-1-5, 1-1-6,
1-2-4, 1-2-5, 1-2-6,
1-3-5, 1-3-6,
2-2-6

In the above list there are $3$ triples of the type a-a-b without 6, $2$ triples of a-a-6 where a is not 6, $3$ triples of a-b-c without 6, and $2$ triples of a-b-6 where a, b are not 6. So,

\[1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)\]

\[= \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286}\]

So $p = 223/286$.

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