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pick dice
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tkoz0 committed Dec 16, 2024
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.b1 { border-top: 1px solid black; }
.b2 { border-left: 1px solid black; }
.td2 td { padding: 4px; text-align: center; }
.h { background: yellow; }
.dr { background: #f77; }
.db { background: lightblue; }
</style>
</head>
<body>
Expand Down Expand Up @@ -511,8 +514,223 @@ <h3 id="dec13">Dec 13</h3>

<h3 id="dec14">Dec 14</h3>

<svg version="1.1" xmlns="http://www.w3.org/2000/svg" viewBox="0 0 10 10" width="200" height="200" class="img_center">
<polygon points="2 8 4 4 6 3 8 7 7 1 3 2" stroke="black" stroke-width="0.05" fill="#7cf" />
<g stroke="black" stroke-width="0.05" fill="black">
<circle cx="2" cy="8" r="0.2" />
<circle cx="4" cy="4" r="0.2" />
<circle cx="6" cy="3" r="0.2" />
<circle cx="8" cy="7" r="0.2" />
<circle cx="7" cy="1" r="0.2" />
<circle cx="3" cy="2" r="0.2" />
</g>
<text x="4.3" y="3" fill="black" font-size="0.1em">x</text>
<g fill="black" font-size="0.05em">
<text x="1" y="9">(1,1)</text>
<text x="3" y="5">(3,5)</text>
<text x="5" y="4">(5,6)</text>
<text x="7" y="8">(7,2)</text>
<text x="7.5" y="1.5">(6,8)</text>
<text x="1" y="2.5">(2,7)</text>
</g>
</svg>

<p>
There are various methods to solve this problem. One is to split the object into
triangles and add their areas. Triangle areas can be computed by using 2 of
their sides as vectors. The determinant of 2 vectors \((a,b),(c,d)\) is
\(ad-bc\) and its absolute value is the area of the parallelogram determined by
it. Half of this parallelogram is a triangle with area \(|ad-bc|/2\). Below is
the diagram with points labeled.
</p>

<svg version="1.1" xmlns="http://www.w3.org/2000/svg" viewBox="0 0 10 10" width="200" height="200" class="img_center">
<polygon points="2 8 4 4 6 3 8 7 7 1 3 2" stroke="black" stroke-width="0.05" fill="#7cf" />
<g stroke="black" stroke-width="0.05" fill="black">
<circle cx="2" cy="8" r="0.2" />
<circle cx="4" cy="4" r="0.2" />
<circle cx="6" cy="3" r="0.2" />
<circle cx="8" cy="7" r="0.2" />
<circle cx="7" cy="1" r="0.2" />
<circle cx="3" cy="2" r="0.2" />
</g>
<text x="4.3" y="3" fill="black" font-size="0.1em">x</text>
<g fill="black" font-size="0.05em">
<text x="1" y="9">A(1,1)</text>
<text x="3" y="5">B(3,5)</text>
<text x="5" y="4">C(5,6)</text>
<text x="7" y="8">D(7,2)</text>
<text x="7.5" y="1.5">E(6,8)</text>
<text x="0.5" y="2.5">F(2,7)</text>
</g>
</svg>

<p>
The area of the "outer" quadrilateral from points \(A,D,E,F\) can be determined
from adding the following 2 triange areas:
\[{1\over2}\begin{vmatrix}F-A\\D-A\\\end{vmatrix}
={1\over2}\begin{vmatrix}1&6\\6&1\\\end{vmatrix}={35\over2}\]
\[{1\over2}\begin{vmatrix}F-E\\D-E\\\end{vmatrix}
={1\over2}\begin{vmatrix}-4&-1\\1&-6\\\end{vmatrix}={25\over2}\]
So the "outer" quadrilateral has area \(30\). For the "inner" quadrilateral
\(A,B,C,D\), we use the following 2 triangles:
\[{1\over2}\begin{vmatrix}B-A\\D-A\\\end{vmatrix}
={1\over2}\begin{vmatrix}2&4\\6&1\\\end{vmatrix}={22\over2}\]
\[{1\over2}\begin{vmatrix}B-C\\D-C\\\end{vmatrix}
={1\over2}\begin{vmatrix}-2&-1\\2&-4\\\end{vmatrix}={10\over2}\]
So the "inner" quadrilateral has area \(16\). By subtracting, the area of the
shaded part is \(30-16=14\).
</p>

<svg version="1.1" xmlns="http://www.w3.org/2000/svg" viewBox="0 0 10 10" width="200" height="200" class="img_center">
<polygon points="2 8 4 4 6 3 8 7 7 1 3 2" stroke="black" stroke-width="0.05" fill="#7cf" />
<g stroke="black" stroke-width="0.05" fill="black">
<circle cx="2" cy="8" r="0.2" />
<circle cx="4" cy="4" r="0.2" />
<circle cx="6" cy="3" r="0.2" />
<circle cx="8" cy="7" r="0.2" />
<circle cx="7" cy="1" r="0.2" />
<circle cx="3" cy="2" r="0.2" />
</g>
<text x="4.3" y="3" fill="black" font-size="0.1em">x</text>
<g fill="black" font-size="0.05em">
<text x="1" y="9">(1,1)</text>
<text x="3" y="5">(3,5)</text>
<text x="5" y="4">(5,6)</text>
<text x="7" y="8">(7,2)</text>
<text x="7.5" y="1.5">(6,8)</text>
<text x="1" y="2.5">(2,7)</text>
</g>
<g stroke="black" stroke-width="0.02" fill="none">
<line x1="0" x2="0" y1="10" y2="0" />
<line x1="1" x2="1" y1="10" y2="0" />
<line x1="2" x2="2" y1="10" y2="0" />
<line x1="3" x2="3" y1="10" y2="0" />
<line x1="4" x2="4" y1="10" y2="0" />
<line x1="5" x2="5" y1="10" y2="0" />
<line x1="6" x2="6" y1="10" y2="0" />
<line x1="7" x2="7" y1="10" y2="0" />
<line x1="8" x2="8" y1="10" y2="0" />
<line x1="9" x2="9" y1="10" y2="0" />
<line x1="10" x2="10" y1="10" y2="0" />
<line x1="0" x2="10" y1="10" y2="10" />
<line x1="0" x2="10" y1="9" y2="9" />
<line x1="0" x2="10" y1="8" y2="8" />
<line x1="0" x2="10" y1="7" y2="7" />
<line x1="0" x2="10" y1="6" y2="6" />
<line x1="0" x2="10" y1="5" y2="5" />
<line x1="0" x2="10" y1="4" y2="4" />
<line x1="0" x2="10" y1="3" y2="3" />
<line x1="0" x2="10" y1="2" y2="2" />
<line x1="0" x2="10" y1="1" y2="1" />
<line x1="0" x2="10" y1="0" y2="0" />
</g>
<g stroke="black" stroke-width="0.05" fill="red">
<circle cx="3" cy="6" r="0.15" />
<circle cx="7" cy="5" r="0.15" />
<circle cx="3" cy="4" r="0.15" />
<circle cx="5" cy="3" r="0.15" />
<circle cx="5" cy="2" r="0.15" />
<circle cx="4" cy="3" r="0.15" />
<circle cx="4" cy="2" r="0.15" />
<circle cx="6" cy="2" r="0.15" />
<circle cx="7" cy="4" r="0.15" />
<circle cx="7" cy="3" r="0.15" />
<circle cx="7" cy="2" r="0.15" />
<circle cx="3" cy="5" r="0.15" />
<circle cx="3" cy="3" r="0.15" />
</g>
</svg>

<p>
Another method is to use Pick's theorem which shows a way to calculate the area
of simple polygons with integer coordinates by counting integer coordinate
points. There are 11 interier points and 8 border points so the area is
\[i+{b\over2}-1=11+{8\over2}-1=14\]
</p>

<h3 id="dec15">Dec 15</h3>

<p>
Find the number of ways that a red and blue die can be rolled so that their
product is a multiple of 6.
</p>

<p>
This could be done by writing the multiplication table and counting, but can be
counting a bit more intelligently. One condition is that either is 6 so by
inclusion-exclusion, it is 6 for the red die, 6 for the blue die, and subtract 1
for when both are 6 so there are 11 ways. Without a 6, there are the following
pairs \((2,3),(3,2),(3,4),(4,3)\). So there is a total of 15. Below is the table
with highlighted (yellow) multiples of 6.
</p>

<table class="table_normal">
<tr>
<td></td>
<td class="dr">1</td>
<td class="dr">2</td>
<td class="dr">3</td>
<td class="dr">4</td>
<td class="dr">5</td>
<td class="dr">6</td>
</tr>
<tr class="b1">
<td class="db">1</td>
<td>1</td>
<td>2</td>
<td>3</td>
<td>4</td>
<td>5</td>
<td class="h">6</td>
</tr>
<tr>
<td class="db">2</td>
<td>2</td>
<td>4</td>
<td class="h">6</td>
<td>8</td>
<td>10</td>
<td class="h">12</td>
</tr>
<tr>
<td class="db">3</td>
<td>3</td>
<td class="h">6</td>
<td>9</td>
<td class="h">12</td>
<td>15</td>
<td class="h">18</td>
</tr>
<tr>
<td class="db">4</td>
<td>4</td>
<td>8</td>
<td class="h">12</td>
<td>16</td>
<td>20</td>
<td class="h">24</td>
</tr>
<tr>
<td class="db">5</td>
<td>5</td>
<td>10</td>
<td>15</td>
<td>20</td>
<td>25</td>
<td class="h">30</td>
</tr>
<tr>
<td class="db">6</td>
<td class="h">6</td>
<td class="h">12</td>
<td class="h">18</td>
<td class="h">24</td>
<td class="h">30</td>
<td class="h">36</td>
</tr>
</table>

<h3 id="dec16">Dec 16</h3>

<h3 id="dec17">Dec 17</h3>
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