https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mop&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T15:53:42ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_6&diff=1914192022 AIME I Problems/Problem 62023-03-27T02:55:58Z<p>Mop: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Find the number of ordered pairs of integers <math>(a, b)</math> such that the sequence<cmath>3, 4, 5, a, b, 30, 40, 50</cmath>is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.<br />
<br />
== Solution 1==<br />
Since <math>3,4,5,a</math> and <math>3,4,5,b</math> cannot be an arithmetic progression, <math>a</math> or <math>b</math> can never be <math>6</math>. Since <math>b, 30, 40, 50</math> and <math>a, 30, 40, 50</math> cannot be an arithmetic progression, <math>a</math> and <math>b</math> can never be <math>20</math>. Since <math>a < b</math>, there are <math>{24 - 2 \choose 2} = 231</math> ways to choose <math>a</math> and <math>b</math> with these two restrictions in mind.<br />
<br />
However, there are still specific invalid cases counted in these <math>231</math> pairs <math>(a,b)</math>. Since<br />
<cmath>3,5,a,b</cmath><br />
cannot form an arithmetic progression, <math>\underline{(a,b) \neq (7,9)}</math>.<br />
<cmath>a,b,30,50</cmath><br />
cannot be an arithmetic progression, so <math>(a,b) \neq (-10,10)</math>; however, since this pair was not counted in our <math>231</math>, we do not need to subtract it off.<br />
<cmath>3,a,b,30</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (12,21)}</math>.<br />
<cmath>4, a, b, 40</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (16,28)}</math>.<br />
<cmath>5, a,b, 50</cmath><br />
cannot form an arithmetic progression, <math>(a,b) \neq 20, 35</math>; however, since this pair was not counted in our <math>231</math> (since we disallowed <math>a</math> or <math>b</math> to be <math>20</math>), we do not to subtract it off.<br />
<br />
Also, the sequences <math>(3,a,b,40)</math>, <math>(3,a,b,50)</math>, <math>(4,a,b,30)</math>, <math>(4,a,b,50)</math>, <math>(5,a,b,30)</math> and <math>(5,a,b,40)</math> will never be arithmetic, since that would require <math>a</math> and <math>b</math> to be non-integers.<br />
<br />
So, we need to subtract off <math>3</math> progressions from the <math>231</math> we counted, to get our final answer of <math>\boxed{228}</math>.<br />
<br />
~ ihatemath123<br />
<br />
== Solution 2 (Rigorous) ==<br />
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.<br />
<br />
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract <math>\textbf{1}</math> from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50.<br />
(Note if we take only 1 at a time, there will have to be 3 of <math>a, b</math>, which is impossible.)<br />
<br />
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.<br />
<br />
Finally, we count the sequences that are something like (one of 3,4,5,), <math>a, b</math>, (one of 30, 40, 50). If this is to be the case, then let <math>a</math> be the starting value in the sequence. The sequence will be <math>a, a+d, a+2d, a+3d</math>; We see that if we subtract the largest term by the smallest term we have <math>3d</math>, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are <math>3,a,b,30; 4,a,b,40; 5,a,b,50</math>. Of these, only the last is invalid because it gives <math>b = 35</math>, larger than our bounds <math>6<a<b<30</math>. Therefore, we subtract <math>\textbf{2}</math> from this case.<br />
<br />
Our final answer is <math>231 - 1 - 2 = \boxed{228}</math><br />
<br />
~KingRavi<br />
<br />
==Solution 3==<br />
Denote <math>S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}</math>.<br />
<br />
Denote by <math>A</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>a</math> but not <math>b</math>.<br />
<br />
Denote by <math>B</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>b</math> but not <math>a</math>.<br />
<br />
Hence, <math>C</math> is a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes both <math>a</math> and <math>b</math>.<br />
<br />
Hence, this problem asks us to compute<br />
<cmath><br />
\[<br />
| S | - \left( | A | + | B | + | C | \right) .<br />
\]<br />
</cmath><br />
<br />
First, we compute <math>| S |</math>.<br />
<br />
We have <math>| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276</math>.<br />
<br />
Second, we compute <math>| A |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: <math>a = 6</math>.<br />
<br />
We have <math>b = 8 , \cdots , 19, 21, 22, \cdots, 29</math>.<br />
Thus, the number of solutions is 22.<br />
<br />
<math>\textbf{Case 2}</math>: <math>a = 20</math>.<br />
<br />
We have <math>b = 21, 22, \cdots , 29</math>.<br />
Thus, the number of solutions is 9.<br />
<br />
Thus, <math>| A | = 22 + 9 = 31</math>.<br />
<br />
Third, we compute <math>| B |</math>.<br />
<br />
In <math>B</math>, we have <math>b = 6, 20</math>. However, because <math>6 \leq a < b</math>, we have <math>b \geq 7</math>.<br />
Thus, <math>b = 20</math>.<br />
<br />
This implies <math>a = 7, 8, 9, 11, 12, \cdots , 19</math>. Note that <math>(a, b)=(10, 20)</math> belongs in <math>C</math>.<br />
<br />
Thus, <math>| B | = 12</math>.<br />
<br />
Fourth, we compute <math>| C |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the same side of <math>a</math> and <math>b</math>.<br />
<br />
Hence, <math>(a, b) = (6 , 7), (7, 9) , (10, 20)</math>.<br />
Therefore, the number solutions in this case is 3.<br />
<br />
<math>\textbf{Case 2}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the opposite sides of <math>a</math> and <math>b</math>.<br />
<br />
<math>\textbf{Case 2.1}</math>: The arithmetic sequence is <math>3, a, b, 30</math>.<br />
<br />
Hence, <math>(a, b) = (12, 21)</math>.<br />
<br />
<math>\textbf{Case 2.2}</math>: The arithmetic sequence is <math>4, a, b, 40</math>.<br />
<br />
Hence, <math>(a, b) = (16, 28)</math>.<br />
<br />
<math>\textbf{Case 2.3}</math>: The arithmetic sequence is <math>5, a, b, 50</math>.<br />
<br />
Hence, <math>(a, b) = (20, 35)</math>. However, the sequence <math>... 20, 35, 30, 40, 50</math> is not strictly increasing.<br />
<br />
Putting two cases together, <math>| C | = 65.</math><br />
<br />
Therefore,<br />
<cmath><br />
| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 31 + 12 + 5 \right) = \boxed{228}.<br />
</cmath><br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Solution 4==<br />
<br />
divide cases into <math>7\leq a<20; 21\leq a\leq28</math>.(Notice that <math>a</math> can't be equal to <math>6,20</math>, that's why I divide them into two parts.<br />
There are three cases that arithmetic sequence forms: <math>3,12,21,30;4,16,28,40;3,5,7,9</math>.(NOTICE that <math>5,20,35,50</math> IS NOT A VALID SEQUENCE!)<br />
So when <math>7\leq a<20</math>, there are <math>10+11+12+...+22-3-13=192</math> possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)<br />
<br />
When <math>21\leq a \leq 28</math>, there are <math>1+2+\cdots+8=36</math> ways.<br />
<br />
In all, there are <math>192+36=\boxed{228}</math> possible sequences.<br />
<br />
~bluesoul<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_6&diff=1914182022 AIME I Problems/Problem 62023-03-27T02:55:16Z<p>Mop: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Find the number of ordered pairs of integers <math>(a, b)</math> such that the sequence<cmath>3, 4, 5, a, b, 30, 40, 50</cmath>is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.<br />
<br />
== Solution 1==<br />
Since <math>3,4,5,a</math> and <math>3,4,5,b</math> cannot be an arithmetic progression, <math>a</math> or <math>b</math> can never be <math>6</math>. Since <math>b, 30, 40, 50</math> and <math>a, 30, 40, 50</math> cannot be an arithmetic progression, <math>a</math> and <math>b</math> can never be <math>20</math>. Since <math>a < b</math>, there are <math>{24 - 2 \choose 2} = 231</math> ways to choose <math>a</math> and <math>b</math> with these two restrictions in mind.<br />
<br />
However, there are still specific invalid cases counted in these <math>231</math> pairs <math>(a,b)</math>. Since<br />
<cmath>3,5,a,b</cmath><br />
cannot form an arithmetic progression, <math>\underline{(a,b) \neq (7,9)}</math>.<br />
<cmath>a,b,30,50</cmath><br />
cannot be an arithmetic progression, so <math>(a,b) \neq (-10,10)</math>; however, since this pair was not counted in our <math>231</math>, we do not need to subtract it off.<br />
<cmath>3,a,b,30</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (12,21)}</math>.<br />
<cmath>4, a, b, 40</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (16,28)}</math>.<br />
<cmath>5, a,b, 50</cmath><br />
cannot form an arithmetic progression, <math>(a,b) \neq 20, 35</math>; however, since this pair was not counted in our <math>231</math> (since we disallowed <math>a</math> or <math>b</math> to be <math>20</math>), we do not to subtract it off.<br />
<br />
Also, the sequences <math>(3,a,b,40)</math>, <math>(3,a,b,50)</math>, <math>(4,a,b,30)</math>, <math>(4,a,b,50)</math>, <math>(5,a,b,30)</math> and <math>(5,a,b,40)</math> will never be arithmetic, since that would require <math>a</math> and <math>b</math> to be non-integers.<br />
<br />
So, we need to subtract off <math>3</math> progressions from the <math>231</math> we counted, to get our final answer of <math>\boxed{228}</math>.<br />
<br />
~ ihatemath123<br />
<br />
== Solution 2 (Rigorous) ==<br />
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.<br />
<br />
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract <math>\textbf{1}</math> from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50.<br />
(Note if we take only 1 at a time, there will have to be 3 of <math>a, b</math>, which is impossible.)<br />
<br />
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.<br />
<br />
Finally, we count the sequences that are something like (one of 3,4,5,), <math>a, b</math>, (one of 30, 40, 50). If this is to be the case, then let <math>a</math> be the starting value in the sequence. The sequence will be <math>a, a+d, a+2d, a+3d</math>; We see that if we subtract the largest term by the smallest term we have <math>3d</math>, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are <math>3,a,b,30; 4,a,b,40; 5,a,b,50</math>. Of these, only the last is invalid because it gives <math>b = 35</math>, larger than our bounds <math>6<a<b<30</math>. Therefore, we subtract <math>\textbf{2}</math> from this case.<br />
<br />
Our final answer is <math>231 - 1 - 2 = \boxed{228}</math><br />
<br />
~KingRavi<br />
<br />
==Solution 3==<br />
Denote <math>S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}</math>.<br />
<br />
Denote by <math>A</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>a</math> but not <math>b</math>.<br />
<br />
Denote by <math>B</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>b</math> but not <math>a</math>.<br />
<br />
Hence, <math>C</math> is a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes both <math>a</math> and <math>b</math>.<br />
<br />
Hence, this problem asks us to compute<br />
<cmath><br />
\[<br />
| S | - \left( | A | + | B | + | C | \right) .<br />
\]<br />
</cmath><br />
<br />
First, we compute <math>| S |</math>.<br />
<br />
We have <math>| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276</math>.<br />
<br />
Second, we compute <math>| A |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: <math>a = 6</math>.<br />
<br />
We have <math>b = 8 , \cdots , 19, 21, 22, \cdots, 29</math>.<br />
Thus, the number of solutions is 22.<br />
<br />
<math>\textbf{Case 2}</math>: <math>a = 20</math>.<br />
<br />
We have <math>b = 21, 22, \cdots , 29</math>.<br />
Thus, the number of solutions is 9.<br />
<br />
Thus, <math>| A | = 22 + 9 = 31</math>.<br />
<br />
Third, we compute <math>| B |</math>.<br />
<br />
In <math>B</math>, we have <math>b = 6, 20</math>. However, because <math>6 \leq a < b</math>, we have <math>b \geq 7</math>.<br />
Thus, <math>b = 20</math>.<br />
<br />
This implies <math>a = 7, 8, 9, 11, 12, \cdots , 19</math>. Note that <math>(a, b)=(10, 20)</math> belongs in <math>C</math>.<br />
Thus, <math>| B | = 12</math>.<br />
<br />
Fourth, we compute <math>| C |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the same side of <math>a</math> and <math>b</math>.<br />
<br />
Hence, <math>(a, b) = (6 , 7), (7, 9) , (10, 20)</math>.<br />
Therefore, the number solutions in this case is 3.<br />
<br />
<math>\textbf{Case 2}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the opposite sides of <math>a</math> and <math>b</math>.<br />
<br />
<math>\textbf{Case 2.1}</math>: The arithmetic sequence is <math>3, a, b, 30</math>.<br />
<br />
Hence, <math>(a, b) = (12, 21)</math>.<br />
<br />
<math>\textbf{Case 2.2}</math>: The arithmetic sequence is <math>4, a, b, 40</math>.<br />
<br />
Hence, <math>(a, b) = (16, 28)</math>.<br />
<br />
<math>\textbf{Case 2.3}</math>: The arithmetic sequence is <math>5, a, b, 50</math>.<br />
<br />
Hence, <math>(a, b) = (20, 35)</math>. However, the sequence <math>... 20, 35, 30, 40, 50</math> is not strictly increasing.<br />
<br />
Putting two cases together, $| C | = 65.<br />
<br />
Therefore,<br />
<cmath><br />
| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 31 + 12 + 5 \right) = \boxed{228}.<br />
</cmath><br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Solution 4==<br />
<br />
divide cases into <math>7\leq a<20; 21\leq a\leq28</math>.(Notice that <math>a</math> can't be equal to <math>6,20</math>, that's why I divide them into two parts.<br />
There are three cases that arithmetic sequence forms: <math>3,12,21,30;4,16,28,40;3,5,7,9</math>.(NOTICE that <math>5,20,35,50</math> IS NOT A VALID SEQUENCE!)<br />
So when <math>7\leq a<20</math>, there are <math>10+11+12+...+22-3-13=192</math> possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)<br />
<br />
When <math>21\leq a \leq 28</math>, there are <math>1+2+\cdots+8=36</math> ways.<br />
<br />
In all, there are <math>192+36=\boxed{228}</math> possible sequences.<br />
<br />
~bluesoul<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_6&diff=1914172022 AIME I Problems/Problem 62023-03-27T02:53:28Z<p>Mop: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Find the number of ordered pairs of integers <math>(a, b)</math> such that the sequence<cmath>3, 4, 5, a, b, 30, 40, 50</cmath>is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.<br />
<br />
== Solution 1==<br />
Since <math>3,4,5,a</math> and <math>3,4,5,b</math> cannot be an arithmetic progression, <math>a</math> or <math>b</math> can never be <math>6</math>. Since <math>b, 30, 40, 50</math> and <math>a, 30, 40, 50</math> cannot be an arithmetic progression, <math>a</math> and <math>b</math> can never be <math>20</math>. Since <math>a < b</math>, there are <math>{24 - 2 \choose 2} = 231</math> ways to choose <math>a</math> and <math>b</math> with these two restrictions in mind.<br />
<br />
However, there are still specific invalid cases counted in these <math>231</math> pairs <math>(a,b)</math>. Since<br />
<cmath>3,5,a,b</cmath><br />
cannot form an arithmetic progression, <math>\underline{(a,b) \neq (7,9)}</math>.<br />
<cmath>a,b,30,50</cmath><br />
cannot be an arithmetic progression, so <math>(a,b) \neq (-10,10)</math>; however, since this pair was not counted in our <math>231</math>, we do not need to subtract it off.<br />
<cmath>3,a,b,30</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (12,21)}</math>.<br />
<cmath>4, a, b, 40</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (16,28)}</math>.<br />
<cmath>5, a,b, 50</cmath><br />
cannot form an arithmetic progression, <math>(a,b) \neq 20, 35</math>; however, since this pair was not counted in our <math>231</math> (since we disallowed <math>a</math> or <math>b</math> to be <math>20</math>), we do not to subtract it off.<br />
<br />
Also, the sequences <math>(3,a,b,40)</math>, <math>(3,a,b,50)</math>, <math>(4,a,b,30)</math>, <math>(4,a,b,50)</math>, <math>(5,a,b,30)</math> and <math>(5,a,b,40)</math> will never be arithmetic, since that would require <math>a</math> and <math>b</math> to be non-integers.<br />
<br />
So, we need to subtract off <math>3</math> progressions from the <math>231</math> we counted, to get our final answer of <math>\boxed{228}</math>.<br />
<br />
~ ihatemath123<br />
<br />
== Solution 2 (Rigorous) ==<br />
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.<br />
<br />
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract <math>\textbf{1}</math> from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50.<br />
(Note if we take only 1 at a time, there will have to be 3 of <math>a, b</math>, which is impossible.)<br />
<br />
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.<br />
<br />
Finally, we count the sequences that are something like (one of 3,4,5,), <math>a, b</math>, (one of 30, 40, 50). If this is to be the case, then let <math>a</math> be the starting value in the sequence. The sequence will be <math>a, a+d, a+2d, a+3d</math>; We see that if we subtract the largest term by the smallest term we have <math>3d</math>, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are <math>3,a,b,30; 4,a,b,40; 5,a,b,50</math>. Of these, only the last is invalid because it gives <math>b = 35</math>, larger than our bounds <math>6<a<b<30</math>. Therefore, we subtract <math>\textbf{2}</math> from this case.<br />
<br />
Our final answer is <math>231 - 1 - 2 = \boxed{228}</math><br />
<br />
~KingRavi<br />
<br />
==Solution 3==<br />
Denote <math>S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}</math>.<br />
<br />
Denote by <math>A</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>a</math> but not <math>b</math>.<br />
<br />
Denote by <math>B</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>b</math> but not <math>a</math>.<br />
<br />
Hence, <math>C</math> is a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes both <math>a</math> and <math>b</math>.<br />
<br />
Hence, this problem asks us to compute<br />
<cmath><br />
\[<br />
| S | - \left( | A | + | B | + | C | \right) .<br />
\]<br />
</cmath><br />
<br />
First, we compute <math>| S |</math>.<br />
<br />
We have <math>| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276</math>.<br />
<br />
Second, we compute <math>| A |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: <math>a = 6</math>.<br />
<br />
We have <math>b = 8 , \cdots , 19, 21, 22, \cdots, 29</math>.<br />
Thus, the number of solutions is 22.<br />
<br />
<math>\textbf{Case 2}</math>: <math>a = 20</math>.<br />
<br />
We have <math>b = 21, 22, \cdots , 29</math>.<br />
Thus, the number of solutions is 9.<br />
<br />
Thus, <math>| A | = 22 + 9 = 31</math>.<br />
<br />
Third, we compute <math>| B |</math>.<br />
<br />
In <math>B</math>, we have <math>b = 6, 20</math>. However, because <math>6 \leq a < b</math>, we have <math>b \geq 7</math>.<br />
Thus, <math>b = 20</math>.<br />
<br />
This implies <math>a = 7, 9, 11, 12, \cdots , 19</math>. <br />
Thus, <math>| B | = 12</math>.<br />
<br />
Fourth, we compute <math>| C |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the same side of <math>a</math> and <math>b</math>.<br />
<br />
Hence, <math>(a, b) = (6 , 7), (7, 9) , (10, 20)</math>.<br />
Therefore, the number solutions in this case is 3.<br />
<br />
<math>\textbf{Case 2}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the opposite sides of <math>a</math> and <math>b</math>.<br />
<br />
<math>\textbf{Case 2.1}</math>: The arithmetic sequence is <math>3, a, b, 30</math>.<br />
<br />
Hence, <math>(a, b) = (12, 21)</math>.<br />
<br />
<math>\textbf{Case 2.2}</math>: The arithmetic sequence is <math>4, a, b, 40</math>.<br />
<br />
Hence, <math>(a, b) = (16, 28)</math>.<br />
<br />
<math>\textbf{Case 2.3}</math>: The arithmetic sequence is <math>5, a, b, 50</math>.<br />
<br />
Hence, <math>(a, b) = (20, 35)</math>. However, the sequence <math>... 20, 35, 30, 40, 50</math> is not strictly increasing.<br />
<br />
Putting two cases together, $| C | = 65.<br />
<br />
Therefore,<br />
<cmath><br />
| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 31 + 12 + 5 \right) = \boxed{228}.<br />
</cmath><br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Solution 4==<br />
<br />
divide cases into <math>7\leq a<20; 21\leq a\leq28</math>.(Notice that <math>a</math> can't be equal to <math>6,20</math>, that's why I divide them into two parts.<br />
There are three cases that arithmetic sequence forms: <math>3,12,21,30;4,16,28,40;3,5,7,9</math>.(NOTICE that <math>5,20,35,50</math> IS NOT A VALID SEQUENCE!)<br />
So when <math>7\leq a<20</math>, there are <math>10+11+12+...+22-3-13=192</math> possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)<br />
<br />
When <math>21\leq a \leq 28</math>, there are <math>1+2+\cdots+8=36</math> ways.<br />
<br />
In all, there are <math>192+36=\boxed{228}</math> possible sequences.<br />
<br />
~bluesoul<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_6&diff=1914162022 AIME I Problems/Problem 62023-03-27T02:49:30Z<p>Mop: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Find the number of ordered pairs of integers <math>(a, b)</math> such that the sequence<cmath>3, 4, 5, a, b, 30, 40, 50</cmath>is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.<br />
<br />
== Solution 1==<br />
Since <math>3,4,5,a</math> and <math>3,4,5,b</math> cannot be an arithmetic progression, <math>a</math> or <math>b</math> can never be <math>6</math>. Since <math>b, 30, 40, 50</math> and <math>a, 30, 40, 50</math> cannot be an arithmetic progression, <math>a</math> and <math>b</math> can never be <math>20</math>. Since <math>a < b</math>, there are <math>{24 - 2 \choose 2} = 231</math> ways to choose <math>a</math> and <math>b</math> with these two restrictions in mind.<br />
<br />
However, there are still specific invalid cases counted in these <math>231</math> pairs <math>(a,b)</math>. Since<br />
<cmath>3,5,a,b</cmath><br />
cannot form an arithmetic progression, <math>\underline{(a,b) \neq (7,9)}</math>.<br />
<cmath>a,b,30,50</cmath><br />
cannot be an arithmetic progression, so <math>(a,b) \neq (-10,10)</math>; however, since this pair was not counted in our <math>231</math>, we do not need to subtract it off.<br />
<cmath>3,a,b,30</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (12,21)}</math>.<br />
<cmath>4, a, b, 40</cmath><br />
cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (16,28)}</math>.<br />
<cmath>5, a,b, 50</cmath><br />
cannot form an arithmetic progression, <math>(a,b) \neq 20, 35</math>; however, since this pair was not counted in our <math>231</math> (since we disallowed <math>a</math> or <math>b</math> to be <math>20</math>), we do not to subtract it off.<br />
<br />
Also, the sequences <math>(3,a,b,40)</math>, <math>(3,a,b,50)</math>, <math>(4,a,b,30)</math>, <math>(4,a,b,50)</math>, <math>(5,a,b,30)</math> and <math>(5,a,b,40)</math> will never be arithmetic, since that would require <math>a</math> and <math>b</math> to be non-integers.<br />
<br />
So, we need to subtract off <math>3</math> progressions from the <math>231</math> we counted, to get our final answer of <math>\boxed{228}</math>.<br />
<br />
~ ihatemath123<br />
<br />
== Solution 2 (Rigorous) ==<br />
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.<br />
<br />
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract <math>\textbf{1}</math> from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50.<br />
(Note if we take only 1 at a time, there will have to be 3 of <math>a, b</math>, which is impossible.)<br />
<br />
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.<br />
<br />
Finally, we count the sequences that are something like (one of 3,4,5,), <math>a, b</math>, (one of 30, 40, 50). If this is to be the case, then let <math>a</math> be the starting value in the sequence. The sequence will be <math>a, a+d, a+2d, a+3d</math>; We see that if we subtract the largest term by the smallest term we have <math>3d</math>, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are <math>3,a,b,30; 4,a,b,40; 5,a,b,50</math>. Of these, only the last is invalid because it gives <math>b = 35</math>, larger than our bounds <math>6<a<b<30</math>. Therefore, we subtract <math>\textbf{2}</math> from this case.<br />
<br />
Our final answer is <math>231 - 1 - 2 = \boxed{228}</math><br />
<br />
~KingRavi<br />
<br />
==Solution 3==<br />
Denote <math>S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}</math>.<br />
<br />
Denote by <math>A</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>a</math> but not <math>b</math>.<br />
<br />
Denote by <math>B</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>b</math> but not <math>a</math>.<br />
<br />
Hence, <math>C</math> is a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes both <math>a</math> and <math>b</math>.<br />
<br />
Hence, this problem asks us to compute<br />
<cmath><br />
\[<br />
| S | - \left( | A | + | B | + | C | \right) .<br />
\]<br />
</cmath><br />
<br />
First, we compute <math>| S |</math>.<br />
<br />
We have <math>| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276</math>.<br />
<br />
Second, we compute <math>| A |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: <math>a = 6</math>.<br />
<br />
We have <math>b = 8 , \cdots , 19, 21, 22, \cdots, 29</math>.<br />
Thus, the number of solutions is 22.<br />
<br />
<math>\textbf{Case 2}</math>: <math>a = 20</math>.<br />
<br />
We have <math>b = 21, 22, \cdots , 29</math>.<br />
Thus, the number of solutions is 9.<br />
<br />
Thus, <math>| A | = 22 + 9 = 31</math>.<br />
<br />
Third, we compute <math>| B |</math>.<br />
<br />
In <math>B</math>, we have <math>b = 6, 20</math>. However, because <math>6 \leq a < b</math>, we have <math>b \geq 7</math>.<br />
Thus, <math>b = 20</math>.<br />
<br />
This implies <math>a = 7, 8, 9, 11, 12, \cdots , 19</math>.<br />
Thus, <math>| B | = 12</math>.<br />
<br />
Fourth, we compute <math>| C |</math>.<br />
<br />
<math>\textbf{Case 1}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the same side of <math>a</math> and <math>b</math>.<br />
<br />
Hence, <math>(a, b) = (6 , 7), (7, 9) , (10, 20)</math>.<br />
Therefore, the number solutions in this case is 3.<br />
<br />
<math>\textbf{Case 2}</math>: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the opposite sides of <math>a</math> and <math>b</math>.<br />
<br />
<math>\textbf{Case 2.1}</math>: The arithmetic sequence is <math>3, a, b, 30</math>.<br />
<br />
Hence, <math>(a, b) = (12, 21)</math>.<br />
<br />
<math>\textbf{Case 2.2}</math>: The arithmetic sequence is <math>4, a, b, 40</math>.<br />
<br />
Hence, <math>(a, b) = (16, 28)</math>.<br />
<br />
<math>\textbf{Case 2.3}</math>: The arithmetic sequence is <math>5, a, b, 50</math>.<br />
<br />
Hence, <math>(a, b) = (20, 35)</math>. However, the sequence <math>... 20, 35, 30, 40, 50</math> is not strictly increasing.<br />
<br />
Putting two cases together, $| C | = 65.<br />
<br />
Therefore,<br />
<cmath><br />
| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 31 + 12 + 5 \right) = \boxed{228}.<br />
</cmath><br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
fixed by mop<br />
<br />
==Solution 4==<br />
<br />
divide cases into <math>7\leq a<20; 21\leq a\leq28</math>.(Notice that <math>a</math> can't be equal to <math>6,20</math>, that's why I divide them into two parts.<br />
There are three cases that arithmetic sequence forms: <math>3,12,21,30;4,16,28,40;3,5,7,9</math>.(NOTICE that <math>5,20,35,50</math> IS NOT A VALID SEQUENCE!)<br />
So when <math>7\leq a<20</math>, there are <math>10+11+12+...+22-3-13=192</math> possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)<br />
<br />
When <math>21\leq a \leq 28</math>, there are <math>1+2+\cdots+8=36</math> ways.<br />
<br />
In all, there are <math>192+36=\boxed{228}</math> possible sequences.<br />
<br />
~bluesoul<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_20&diff=1703182008 AMC 10A Problems/Problem 202022-01-27T06:53:24Z<p>Mop: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
[[Trapezoid]] <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K.</math> Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24.</math> What is the area of trapezoid <math>ABCD</math>?<br />
<br />
<math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math><br />
<br />
==Solution==<br />
==Solution 1==<br />
<center><asy><br />
pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */<br />
pen sm = fontsize(10); /* small font pen */<br />
pair D=(0,0),C=(12,0), K=(7,16/3); /* note that K.x is arbitrary, as generator for A,B */<br />
pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;<br />
D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D);D(A--MP("K",K)--D--cycle,linewidth(0.7));<br />
MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E);<br />
</asy></center><br />
Since <math>\overline{AB} \parallel \overline{DC}</math> it follows that <math>\triangle ABK \sim \triangle CDK</math>. Thus <math>\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}</math>. <br />
<br />
We now introduce the concept of [[area ratios]]: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since <math>\triangle AKB, \triangle AKD</math> share a common [[altitude]] to <math>\overline{BD}</math>, it follows that (we let <math>[\triangle \ldots]</math> denote the area of the triangle) <math>\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}</math>, so <math>[\triangle AKB] = \frac{3}{4}(24) = 18</math>. Similarly, we find <math>[\triangle DKC] = \frac{4}{3}(24) = 32</math> and <math>[\triangle BKC] = 24</math>.<br />
<br />
Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>.<br />
==Solution 2==<br />
We denote <math>KA</math> with length <math>x</math> and <math>KD</math> with length <math>\frac{4x}{3}</math> (which follows from similar triangles), and we denote <math>\angle{AKD}=\theta</math>. Note that <math>\frac{4x^2}{3}\cdot \sin\theta=48\implies 4x^2\cdot \sin\theta=36</math>. The areas of triangles <math>ABK</math> and <math>CDK</math> combined are <math>\frac{x^2\cdot\sin\theta+\frac{16x^2}{9}\cdot\sin\theta}{2}=\frac{25x^2}{18}\cdot\sin\theta=36\cdot\frac{25}{18}=50</math>. Thus, <math>[ABCD]=[ABK]+[BCK]+[CDK]+[ADK]=48+50=98\ \mathrm{(D)}</math>, as desired.<br />
-mop<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_20&diff=1703172008 AMC 10A Problems/Problem 202022-01-27T06:52:47Z<p>Mop: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
[[Trapezoid]] <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K.</math> Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24.</math> What is the area of trapezoid <math>ABCD</math>?<br />
<br />
<math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math><br />
<br />
==Solution==<br />
==Solution 1==<br />
<center><asy><br />
pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */<br />
pen sm = fontsize(10); /* small font pen */<br />
pair D=(0,0),C=(12,0), K=(7,16/3); /* note that K.x is arbitrary, as generator for A,B */<br />
pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;<br />
D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D);D(A--MP("K",K)--D--cycle,linewidth(0.7));<br />
MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E);<br />
</asy></center><br />
Since <math>\overline{AB} \parallel \overline{DC}</math> it follows that <math>\triangle ABK \sim \triangle CDK</math>. Thus <math>\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}</math>. <br />
<br />
We now introduce the concept of [[area ratios]]: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since <math>\triangle AKB, \triangle AKD</math> share a common [[altitude]] to <math>\overline{BD}</math>, it follows that (we let <math>[\triangle \ldots]</math> denote the area of the triangle) <math>\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}</math>, so <math>[\triangle AKB] = \frac{3}{4}(24) = 18</math>. Similarly, we find <math>[\triangle DKC] = \frac{4}{3}(24) = 32</math> and <math>[\triangle BKC] = 24</math>.<br />
<br />
Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>.<br />
==Solution 2==<br />
We denote <math>KA</math> with length <math>x</math> and <math>KD</math> with length <math>\frac{4x}{3}</math> (which follows from similar triangles), and we denote <math>\angle{AKD}=\theta</math>. Note that <math>\frac{4x^2}{3}\cdot \sin\theta=48\implies 4x^2\cdot \sin\theta=36</math>. The areas of triangles <math>ABK</math> and <math>CDK</math> combined are <math>\frac{x^2\cdot\sin\theta+\frac{16x^2}{9}\cdot\sin\theta}{2}=\frac{25x^2}{18}\cdot\sin\theta=36\cdot\frac{25}{18}=50</math>. Thus, <math>[ABCD]=[ABK]+[BCK]+[CDK]+[ADK]=48+50=98</math>, as desired.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_20&diff=1703162008 AMC 10A Problems/Problem 202022-01-27T06:49:33Z<p>Mop: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
[[Trapezoid]] <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K.</math> Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24.</math> What is the area of trapezoid <math>ABCD</math>?<br />
<br />
<math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math><br />
<br />
==Solution==<br />
==Solution 1==<br />
<center><asy><br />
pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */<br />
pen sm = fontsize(10); /* small font pen */<br />
pair D=(0,0),C=(12,0), K=(7,16/3); /* note that K.x is arbitrary, as generator for A,B */<br />
pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;<br />
D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D);D(A--MP("K",K)--D--cycle,linewidth(0.7));<br />
MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E);<br />
</asy></center><br />
Since <math>\overline{AB} \parallel \overline{DC}</math> it follows that <math>\triangle ABK \sim \triangle CDK</math>. Thus <math>\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}</math>. <br />
<br />
We now introduce the concept of [[area ratios]]: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since <math>\triangle AKB, \triangle AKD</math> share a common [[altitude]] to <math>\overline{BD}</math>, it follows that (we let <math>[\triangle \ldots]</math> denote the area of the triangle) <math>\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}</math>, so <math>[\triangle AKB] = \frac{3}{4}(24) = 18</math>. Similarly, we find <math>[\triangle DKC] = \frac{4}{3}(24) = 32</math> and <math>[\triangle BKC] = 24</math>.<br />
<br />
Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>.<br />
==Solution 2==<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_20&diff=1703152008 AMC 10A Problems/Problem 202022-01-27T06:48:56Z<p>Mop: /* Solution */</p>
<hr />
<div>==Problem==<br />
[[Trapezoid]] <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K.</math> Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24.</math> What is the area of trapezoid <math>ABCD</math>?<br />
<br />
<math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math><br />
<br />
==Solution==<br />
==Solution 1==<br />
<center><asy><br />
pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */<br />
pen sm = fontsize(10); /* small font pen */<br />
pair D=(0,0),C=(12,0), K=(7,16/3); /* note that K.x is arbitrary, as generator for A,B */<br />
pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;<br />
D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D);D(A--MP("K",K)--D--cycle,linewidth(0.7));<br />
MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E);<br />
</asy></center><br />
Since <math>\overline{AB} \parallel \overline{DC}</math> it follows that <math>\triangle ABK \sim \triangle CDK</math>. Thus <math>\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}</math>. <br />
<br />
We now introduce the concept of [[area ratios]]: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since <math>\triangle AKB, \triangle AKD</math> share a common [[altitude]] to <math>\overline{BD}</math>, it follows that (we let <math>[\triangle \ldots]</math> denote the area of the triangle) <math>\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}</math>, so <math>[\triangle AKB] = \frac{3}{4}(24) = 18</math>. Similarly, we find <math>[\triangle DKC] = \frac{4}{3}(24) = 32</math> and <math>[\triangle BKC] = 24</math>.<br />
<br />
Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_19&diff=1652972021 Fall AMC 10A Problems/Problem 192021-11-23T05:43:25Z<p>Mop: Created page with "<math>19</math>. A disk of radius <math>1</math> rolls all the way around in the inside of a square of side length <math>s>4</math> and sweeps out a region of area <math>A</ma..."</p>
<hr />
<div><math>19</math>. A disk of radius <math>1</math> rolls all the way around in the inside of a square of side length <math>s>4</math> and sweeps out a region of area <math>A</math>. A second disk of radius <math>1</math> rolls all the way around the outside of the same square and sweeps out a region of area <math>2A</math>. The value of <math>s</math> can be written as <math>a + \dfrac{b\pi}{c}</math>, where <math>a,b,</math> and <math>c</math> are positive integers and <math>b</math> and <math>c</math> are relatively prime. What is <math>a+b+c?</math><br />
<br />
<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math></div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_16&diff=1652952021 Fall AMC 10A Problems/Problem 162021-11-23T05:42:26Z<p>Mop: Created page with "The graph of <math>f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|</math> is symmetric about which of the following? (Here <math>\lfloor x \rfloor</math> is the greatest..."</p>
<hr />
<div>The graph of <math>f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|</math> is symmetric about which of the following? (Here <math>\lfloor x \rfloor</math> is the greatest integer not exceeding <math>x</math>.)<br />
<br />
<math>\textbf{(A) }</math> the <math>y</math>-axis <math>\qquad \textbf{(B) }</math> the line <math>x = 1</math> <math>\qquad \textbf{(C) }</math> the origin <math>\qquad<br />
\textbf{(D) }</math> the point <math>\left(\dfrac12, 0\right)</math> <math>\qquad \textbf{(E) }</math> the point <math>(1,0)</math></div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_14&diff=1652932021 Fall AMC 10A Problems/Problem 142021-11-23T05:40:18Z<p>Mop: Blanked the page</p>
<hr />
<div></div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_14&diff=1652922021 Fall AMC 10A Problems/Problem 142021-11-23T05:39:24Z<p>Mop: Created page with "<math>14</math>. How many ordered pairs <math>(x,y)</math> of real numbers satisfy the following system of equations? \begin{align*} x^2+3y &= 9\\ (|x|+|y|-4)^2 &=1 \end{align..."</p>
<hr />
<div><math>14</math>. How many ordered pairs <math>(x,y)</math> of real numbers satisfy the following system of equations?<br />
\begin{align*}<br />
x^2+3y &= 9\\<br />
(|x|+|y|-4)^2 &=1<br />
\end{align*}<br />
<br />
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad\textbf{(E) }7</math></div>Mophttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_20&diff=1567082015 AMC 10B Problems/Problem 202021-06-23T23:24:31Z<p>Mop: /* Solution 2 (3D Geometry) */</p>
<hr />
<div>==Problem==<br />
Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?<br />
<br />
<math>\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}</math><br />
<br />
==Solution 1 (Coloring)==<br />
<asy>import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br />
label("2",(0,0,0),S);<br />
label("A",(1,0,0),W);<br />
label("B",(0,0,1),N);<br />
label("1",(1,0,1),NW);<br />
label("3",(1,1,0),S);<br />
label("C",(0,1,0),E);<br />
label("D",(1,1,1),SE);<br />
label("4",(0,1,1),NE);<br />
</asy><br />
<br />
We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled <math>A</math>.<br />
<br />
<br />
If we define a "move" as each time Erin crawls along a single edge from one vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be one unit away from <math>A</math> (since Erin cannot crawl back to <math>A</math>), this vertex must be <math>4</math>.<br />
<br />
<br />
Therefore, we now just need to count the number of paths from <math>A</math> to <math>4</math>. To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to <math>4</math>, and 2 choices for which vertex Erin was at 2 moves before <math>4</math>. All of Erin's previous moves were forced, so the total number of legal paths from <math>A</math> to <math>4</math> is <math>3 \cdot 2 = \boxed{\textbf{(A)}\; 6}</math>.<br />
<br />
==Solution 2 (3D Geometry)==<br />
Lets say that this cube is an unit cube and the given corner is <math>(0,0,0)</math>. Because Erin can not return back to its starting point, she can not be on <math>(0,0,1)</math>, <math>(0,1,0)</math>, or <math>(1,0,0)</math>. He can not be on <math>(1,1,0)</math> , <math>(1,0,1)</math>, or <math>(0,1,1,)</math> because after <math>7</math> moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at <math>(1,1,1)</math>. Now, we draw a net and see that there are <math>3</math> choices for the first move, <math>2</math> for the second, and the rest are forced. Thus the answer is <math>3*2 = \boxed{\textbf{(A)}\; 6}</math>.<br />
<br />
-Lcz<br />
<br />
== Solution 3 (3D Geo & Logic) ==<br />
Let's suppose the given corner on the cube is <math>(0,0,0)</math>. Erin has 3 identical ways to proceed. Suppose she goes to <math>(0,1,0)</math>. She now has two more identical ways to go. Let's say she goes to <math>(0,1,1)</math>. She has to go to <math>(0,0,1)</math>, otherwise, she will end up on that point after 7 moves. This is because once if she chooses the other path to <math>(1,1,1)</math>, the endpoint is certain to be <math>(0,0,1)</math>, which is directly connected to <math>(0,0,0)</math>. After she goes to <math>(0,0,1)</math>, the only option she has is to go to <math>(1,0,1)</math>, then <math>(1,0,0)</math>, after that <math>(1,1,0)</math>, and finally <math>(1,1,1)</math>. She is forced to go this way because she cannot end up on <math>(1,0,0)</math>. At the start, she had 3 ways to choose, and after that 2 ways to choose, so <math>3*2=\boxed{\textbf{(A)}\; 6}</math>.<br />
<br />
~HelloWorld21<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=Vieta%27s_Formulas&diff=142344Vieta's Formulas2021-01-17T01:12:10Z<p>Mop: </p>
<hr />
<div>'''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.<br />
<br />
== Introduction ==<br />
<br />
Vieta's Formulas were discovered by the French mathematician [[François Viète]]. <br />
Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as:<br />
<br />
<center><math>x^2+ax+b=(x-p)(x-q)</math></center><br />
<br />
(Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we now have<br />
<br />
<center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center> <br />
<br />
Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form <math>ax^2 + bx +c</math> with roots <math>r_1</math> and <math>r_2.</math> They state that: <br />
<br />
<cmath>r_1 + r_2 = -\frac{b}{a}</cmath> and <cmath>r_1 \cdot r_2 = \frac{c}{a}.</cmath><br />
<br />
We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.<br />
<br />
A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.<br />
<br />
We can state Vieta's formulas more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,<br />
where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that<br />
<br />
<center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center><br />
<br />
Expanding out the right-hand side gives us<br />
<br />
<center><math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math></center><br />
<br />
The coefficient of <math> x^k </math> in this expression will be the <math>(n-k)</math>-th [[elementary symmetric sum]] of the <math>r_i</math>. <br />
<br />
We now have two different expressions for <math>P(x)</math>. These must be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that<br />
<br />
<center><math>a_n = a_n</math></center><br />
<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center><br />
<center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center><br />
<center><math>\vdots</math></center><br />
<center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center><br />
<br />
More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).<br />
<br />
If we denote <math>\sigma_k</math> as the <math>k</math>-th elementary symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.<br />
Also, <math>-b/a = p + q, c/a = p \cdot q</math>.<br />
<br />
==Proving Vieta's Formula==<br />
Basic proof:<br />
This has already been proved earlier, but I will explain it more.<br />
If we have <br />
<math>x^2+ax+b=(x-p)(x-q)</math>, the roots are <math>p</math> and <math>q</math>.<br />
Now expanding the left side, we get: <math>x^2+ax+b=x^2-qx-px+pq</math>. <br />
Factor out an <math>x</math> on the right hand side and we get: <math>x^2+ax+b=x^2-x(p+q)+pq</math><br />
Looking at the two sides, we can quickly see that the coefficient <math>a</math> is equal to <math>-(p+q)</math>. <math>p+q</math> is the actual sum of roots, however. Therefore, it makes sense that <math>p+q= \frac{-b}{a}</math>. The same proof can be given for <math>pq=\frac{c}{a}</math>. <br />
<br />
Note: If you do not understand why we must divide by a, try rewriting the original equation as <math>ax^2+bx+c=(x-p)(x-q)</math><br />
SuperJJ</div>Mophttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_forums&diff=121771AoPS Wiki:AoPS forums2020-04-28T04:39:19Z<p>Mop: /* Bot */</p>
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