https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mc21s&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T13:07:52ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_1&diff=1893502023 AIME I Problems/Problem 12023-02-09T20:28:00Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math><br />
<br />
==Solution 1==<br />
<br />
For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are <math>14!</math> arrangements without restrictions.<br />
<br />
First, there are <math>\binom{7}{5}</math> ways to choose the man-woman diameters. Then, there are <math>10\cdot8\cdot6\cdot4\cdot2</math> ways to place the five men each in a man-woman diameter. Finally, there are <math>9!</math> ways to place the nine women without restrictions.<br />
<br />
Together, the requested probability is <cmath>\frac{\tbinom{7}{5}\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{48}{143},</cmath> from which the answer is <math>48+143 = \boxed{191}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 2==<br />
<br />
This problem is equivalent to solving for the probability that no man is standing diametrically opposite to another man. We can simply just construct this.<br />
<br />
We first place the <math>1</math>st man anywhere on the circle, now we have to place the <math>2</math>nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of <math>\frac{12}{13}</math> because there are <math>13</math> available spots, and <math>12</math> of them are not opposite to the first man.<br />
<br />
We do the same thing for the <math>3</math>rd man, finding a spot for him such that he is not opposite to the other <math>2</math> men, which would happen with a probability of <math>\frac{10}{12}</math> using similar logic. Doing this for the <math>4</math>th and <math>5</math>th men, we get probabilities of <math>\frac{8}{11}</math> and <math>\frac{6}{10}</math> respectively.<br />
<br />
Multiplying these probabilities, we get, <cmath>\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.</cmath><br />
<br />
~s214425<br />
<br />
==Solution 3==<br />
Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange <math>5</math> men in a circle and the number of ways to arrange <math>94</math> women in a circle, are constants.) The total number of ways to arrange <math>5</math> M's and <math>9</math> W's is <math>\binom{14}{5} = 2002.</math> <br />
<br />
To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly <math>2</math> of the pairs of diametrically opposite positions must be occupied by <math>2</math> W's. There are <math>\binom{7}{2} = 21</math> ways to choose these <math>2</math> pairs. For the remaining <math>5</math> pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in <math>2^{5} = 32</math> ways. Therefore, there are <math>21*32 = 672</math> valid arrangements. <br />
<br />
Therefore, the probability that an arrangement is valid is <math>\frac{672}{2002} = \frac{48}{143}</math> for an answer of <math>\boxed{191}.</math><br />
<br />
pianoboy<br />
<br />
==Solution 4==<br />
To start off, we calculate the total amount of ways to organize all <math>14</math> people irrespective of any constraints. This is simply <math>{14\choose5} = 2002</math>, because we just count how many ways we can place all <math>5</math> men in any of the <math>14</math> slots. <br />
<br />
Since men cannot be diametrically opposite with each other, because of the constraints, placing down one man in any given spot will make another spot on the opposite side of the circle unable to hold any men. This means that placing down one man will effectively take away <math>2</math> spots. <br />
<br />
There are <math>14</math> possible slots the first man can be placed. Once that man was placed, the next man only have <math>12</math> possible slots because the slot that the first man is in is taken and the diametrically opposite spot to the first man can't have any men. Similar logic applies for the third man, who has <math>10</math> possible slots. The fourth man has <math>8</math> possible slots, and the fifth man has <math>6</math> possible slots.<br />
<br />
This means the number of ways you can place all <math>5</math> men down is <math>14 \cdot 12 \cdot 10 \cdot 8 \cdot 6</math>. However, since the men are all indistinct from each other, you also have to divide that value by <math>5! = 120</math>, since there are <math>120</math> ways to arrange the <math>5</math> men in each possible positioning of the men on the circle. This means the total number of ways to arrange the men around the circle so that none of them are diametrically opposite of each other is: <math>\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6}{5!} = 672</math>. The women simply fill in the rest of the available slots in each arrangement of men.<br />
<br />
Thus, the final probability is <math>\frac{672}{2002} = \frac{48}{143}</math>, meaning the answer is <math>48 + 143 = \boxed{191}</math>.<br />
<br />
~ericshi1685<br />
<br />
==Solution 5==<br />
We will first assign seats to the men. The first man can be placed in any of the <math>14</math> slots. The second man can be placed in any of the remaining <math>13</math> seats, except for the one diametrically opposite to the first man. So, there are <math>13 - 1 = 12</math> ways to seat him. With a similar argument, the third man can be seated in <math>10</math> ways, the fourth man in <math>8</math> ways and the last man in <math>6</math> ways. <br />
<br />
So, the total number of ways to arrange the men is <math>14 \times 12 \times 10 \times 8 \times 6</math>. <br />
<br />
The women go to the remaining 9 spots. Note that since none of the seats diametrically opposite to the men is occupied, each man is opposite a woman. The number of ways to arrange the women is therefore, simply <math>9!</math>, meaning that the total number of ways to arrange the people with restrictions is <math>14 \times 12 \times 10 \times 8 \times 6 \times 9!</math> In general, there are <math>14!</math> ways to arrange the people without restrictions. So, the probability is <br />
<br />
<math>\frac{14 \times 12 \times 10 \times 8 \times 6 \times 9!}{14!} = \frac{8 \times 6}{13 \times 11} = \frac{48}{143}</math> <br />
<br />
The answer is <math>48 + 143 = \boxed{191}</math> <br />
<br />
~baassid24<br />
<br />
==Solution 6==<br />
First pin one man on one seat (to ensure no rotate situations). Then there are <math>13!</math> arrangements.<br />
Because 5 men must have women at their opposite side, we consider the 2nd man and the woman opposite as one group and name it <math>P_2</math>. There are 4 groups, <math>P_1, P_2, P_3, P_4</math> except the first man pinned on the same point. And for the rest 4 women, name them <math>P_5</math> and <math>P_6</math>. First to order <math>P_{1-6}</math>, there are <math>6!</math> ways. For the 1st man, there are 9 women to choose, 8 for the 2nd, <math>\ldots</math>, 5 for the 5th, and then for the 2 women pairs 3 and 1. Because every 2 person in the group have chance to change their position, there are <math>2^6</math> possibilities. <br />
<br />
So the possibility is <cmath>P=\frac{6!\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 3\cdot 1 \cdot 2^6}{13!}=\frac{48}{143}</cmath><br />
<br />
The answer is <math>48+143=\boxed{191}</math><br />
<br />
~PLASTA<br />
<br />
==Solution 7==<br />
We get around the condition that each man can't be opposite to another man by simply considering all <math>7</math> diagonals, and choosing <math>5</math> where there will be a single man. For each diagonal, the man can go on either side, and there are <math>\binom{14}{5}</math> ways to arrange the men and the women in total. Thus our answer is <math>\frac{\binom{7}{5}\cdot 2^5}{\binom{14}{5}} = \frac{48}{143}.</math><br />
We get <math>48 + 143 = \boxed{191}</math><br />
<br />
~AtharvNaphade<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
<br />
https://www.youtube.com/watch?v=KdKysmdgepI<br />
<br />
==See also==<br />
{{AIME box|year=2023|before=First Problem|num-a=2|n=I}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_10&diff=1893492023 AIME I Problems/Problem 102023-02-09T20:19:11Z<p>Mc21s: </p>
<hr />
<div>==Problem 10==<br />
There exists a unique positive integer <math>a</math> for which the sum <cmath>U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor</cmath> is an integer strictly between <math>-1000</math> and <math>1000</math>. For that unique <math>a</math>, find <math>a+U</math>.<br />
<br />
(Note that <math>\lfloor x\rfloor</math> denotes the greatest integer that is less than or equal to <math>x</math>.)<br />
<br />
==Solution (Bounds and Decimal Part Analysis)==<br />
<br />
Define <math>\left\{ x \right\} = x - \left\lfloor x \right\rfloor</math>.<br />
<br />
First, we bound <math>U</math>.<br />
<br />
We establish an upper bound of <math>U</math>. We have<br />
<cmath><br />
\begin{align*}<br />
U & \leq \sum_{n=1}^{2023} \frac{n^2 - na}{5} \\<br />
& = \frac{1}{5} \sum_{n=1}^{2023} n^2 - \frac{a}{5} \sum_{n=1}^{2023} n \\<br />
& = \frac{1012 \cdot 2023}{5} \left( 1349 - a \right) \\<br />
& \triangleq UB .<br />
\end{align*}<br />
</cmath><br />
<br />
We establish a lower bound of <math>U</math>. We have<br />
<cmath><br />
\begin{align*}<br />
U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\<br />
& = \sum_{n=1}^{2023} \frac{n^2 - na}{5}<br />
- \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\<br />
& = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\<br />
& \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} .<br />
\end{align*}<br />
</cmath><br />
<br />
We notice that if <math>5 | n</math>, then <math>\frac{n^2 - na}{5} \in \Bbb Z</math>.<br />
Thus,<br />
<cmath><br />
\begin{align*}<br />
U & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} \\<br />
& \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ 5 \nmid n \right\} \\<br />
& = UB - \left( 2023 - \left\lfloor \frac{2023}{5} \right\rfloor \right) \\<br />
& = UB - 1619 \\<br />
& \triangleq LB .<br />
\end{align*}<br />
</cmath><br />
<br />
Because <math>U \in \left[ - 1000, 1000 \right]</math> and <math>UB - LB = 1619 < \left( 1000 - \left( - 1000 \right) \right)</math>, we must have either <math>UB \in \left[ - 1000, 1000 \right]</math> or <math>LB \in \left[ - 1000, 1000 \right]</math>.<br />
<br />
For <math>UB \in \left[ - 1000, 1000 \right]</math>, we get a unique <math>a = 1349</math>.<br />
For <math>LB \in \left[ - 1000, 1000 \right]</math>, there is no feasible <math>a</math>.<br />
<br />
Therefore, <math>a = 1349</math>. Thus <math>UB = 0</math>.<br />
<br />
Next, we compute <math>U</math>.<br />
<br />
Let <math>n = 5 q + r</math>, where <math>r = {\rm Rem} \ \left( n, 5 \right)</math>.<br />
<br />
We have<br />
<cmath><br />
\begin{align*}<br />
\left\{ \frac{n^2 - na}{5} \right\}<br />
& = \left\{ \frac{\left( 5 q + r \right)^2 - \left( 5 q + r \right)\left( 1350 - 1 \right)}{5} \right\} \\<br />
& = \left\{ 5 q^2 + 2 q r - \left( 5 q + r \right) 270 + q + \frac{r^2 + r}{5} \right\} \\<br />
& = \left\{\frac{r^2 + r}{5} \right\} \\<br />
& = \left\{<br />
\begin{array}{ll}<br />
0 & \mbox{ if } r = 0, 4 \\<br />
\frac{2}{5} & \mbox{ if } r = 1, 3 \\<br />
\frac{1}{5} & \mbox{ if } r = 2<br />
\end{array}<br />
\right. .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore,<br />
<cmath><br />
\begin{align*}<br />
U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\<br />
& = UB<br />
- \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\<br />
& = - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\<br />
& = - \sum_{q=0}^{404} \sum_{r=0}^4 \left\{\frac{r^2 + r}{5} \right\}<br />
+ \left\{ \frac{0^2 - 0 \cdot a}{5} \right\}<br />
+ \left\{ \frac{2024^2 - 2024a}{5} \right\} \\<br />
& = - \sum_{q=0}^{404} \left( 0 + 0 + \frac{2}{5} + \frac{2}{5} + \frac{1}{5} \right)<br />
+ 0 + 0 \\<br />
& = - 405 .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefor, <math>a + U = 1349 - 405 = \boxed{\textbf{(944) }}</math>.<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 2==<br />
<br />
We define <math>U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}</math>. Since for any real number <math>x</math>, <math>\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1</math>, we have <math>U \le U' \le U + 2023</math>. Now, since <math>-1000 \le U \le 1000</math>, we have <math>-1000 \le U' \le 3023</math>.<br />
<br />
Now, we can solve for <math>U'</math> in terms of <math>a</math>. We have:<br />
<cmath><br />
\begin{align*}<br />
U' &= \sum^{2023}_{n=1} {\frac{n^2-na}{5}} \\<br />
&= \sum^{2023}_{n=1} {\frac{n^2}{5} - \frac{na}{5}} \\<br />
&= \sum^{2023}_{n=1} {\frac{n^2}{5}} - \sum^{2023}_{n=1} {\frac{na}{5}} \\<br />
&= \frac{\sum^{2023}_{n=1} {{n^2}} - \sum^{2023}_{n=1} {na}}{5} \\<br />
&= \frac{\frac{2023(2023+1)(2023 \cdot 2 + 1)}{6} - \frac{a \cdot 2023(2023+1)}{2} }{5} \\<br />
&= \frac{2023(2024)(4047-3a)}{30} \\<br />
\end{align*}<br />
</cmath><br />
So, we have <math>U' = \frac{2023(2024)(4047-3a)}{30}</math>, and <math>-1000 \le U' \le 3023</math>, so we have <math>-1000 \le \frac{2023(2024)(4047-3a)}{30} \le 3023</math>, or <math>-30000 \le 2023(2024)(4047-3a) \le 90690</math>. Now, <math>2023 \cdot 2024</math> is much bigger than <math>90690</math> or <math>30000</math>, and since <math>4047-3a</math> is an integer, to satsify the inequalities, we must have <math>4047 - 3a = 0</math>, or <math>a = 1349</math>, and <math>U' = 0</math>.<br />
<br />
Now, we can find <math>U - U'</math>. We have:<br />
<cmath><br />
\begin{align*}<br />
U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor} - \sum^{2023}_{n=1} {\frac{n^2-1349n}{5}} \\<br />
&= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}}<br />
\end{align*}.<br />
</cmath><br />
Now, if <math>n^2-1349n \equiv 0 \text{ (mod 5)}</math>, then <math>\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = 0</math>, and if <math>n^2-1349n \equiv 1 \text{ (mod 5)}</math>, then <math>\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = -\frac{1}{5}</math>, and so on. Testing with <math>n \equiv 0,1,2,3,4, \text{ (mod 5)}</math>, we get <math>n^2-1349n \equiv 0,2,1,2,0 \text{ (mod 5)}</math> respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for <math>U - U'</math>, we get:<br />
<cmath><br />
\begin{align*}<br />
U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \\<br />
&= 404 \cdot 0 - 405 \cdot \frac{2}{5} - 405 \cdot \frac{1}{5} - 405 \cdot \frac{2}{5} - 404 \cdot t0 \\<br />
&= -405(\frac{2}{5}+\frac{1}{5}+\frac{2}{5}) \\<br />
&= -405<br />
\end{align*}<br />
</cmath><br />
Since <math>U' = 0</math>, this gives <math>U = -405</math>, and we have <math>a + U = 1349-405 = \boxed{944}</math>.<br />
<br />
~ genius_007<br />
<br />
==Solution 3 (Quick)==<br />
<br />
<br />
We can view the floor function in this problem as simply subtracting the remainder of <math>n^2 - na</math> (mod <math>5</math>) from the numerator of <br />
<math>\frac{n^2-na}{5}</math>. For example, <math>\left\lfloor \frac{7}{5} \right\rfloor = \frac{7-2}{5} = 1</math>.<br />
<br />
Note that the congruence of <math>n^2 - na</math> (mod <math>5</math>) loops every time <math>n</math> increases by 5. Also, note that the congruence of <math>a</math> (mod <math>5</math>) determines the set of congruences of <math>n^2 - na</math> for each congruence of <math>n</math> (mod <math>5</math>). <br />
<br />
For example, if <math>a \equiv 1</math> (mod <math>5</math>), the set of remainders is <math>(0, 2, 1, 2, 0)</math> for <math>n \equiv 1,2,3,4,0</math> (mod <math>5</math>). Let the sum of these elements be <math>s</math>. Note that for each “loop” of the numerator (mod <math>5</math>), each element of the set will be subtracted exactly once, meaning <math>s</math> is subtracted once for each loop. The value of the numerator will loop <math>404</math> times (mod <math>5</math>) throughout the sum, as <math>5 \cdot 404=2020</math>. Then <br />
<br />
<math>U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2} -404s \right)}{5}</math><br />
<br />
Where <math>n=2023</math>. Note that since <math>5 \cdot 404=2020</math>, this is an approximation for <math>U</math> because the equation disregards the remainder (mod <math>5</math>) when <math>n=2021, 2022</math>, and <math>2023</math> so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of <math>U</math> (*). However, we will find that this is a negligible error when it comes to the inequality <math>-1000<U<1000</math>, so we can proceed with this approximation to solve for <math>a</math>. <br />
<br />
Factoring our approximation gives <math>U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}</math><br />
<br />
<br />
We set <math>a= \frac{(2n+1)}{3} = 1349</math> to make <math>\frac{(n)(n+1)(2n+1 - 3a)}{6}=0</math>, accordingly minimizing <math>|U|</math>, yielding <math>U \approx <br />
\frac{-404s}{5}</math><br />
<br />
If <math>a</math> increases or decreases by <math>1</math>, then <math>U</math> changes by <math>\frac {(n)(n+1)}{2 \cdot 5} = \frac {2023 \cdot 2024}{10}</math><br />
which clearly breaks the inequality on <math>U</math>. Therefore <math>a=1349 \equiv 4</math> (mod <math>5</math>) giving the set of remainders <math>(2,1,2,0,0)</math>, so <math>s=5</math> and our approximation yields <math>U \approx -404</math>. However, we must subtract 2, 1, and 2 (*) giving us <math>U = - 404 - \frac{(2+1+2)}{5} = - 405</math>, giving an answer of <math>1349-405= \boxed{944}</math><br />
<br />
~Spencer Danese<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
<br />
https://youtu.be/jQVbNJr0tX8<br />
<br />
==Video Solution by Steven Chen==<br />
<br />
https://youtu.be/fxsPmL6wuW4<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
<br />
https://www.youtube.com/watch?v=49XIe2jx9zg<br />
<br />
==See also==<br />
{{AIME box|year=2023|n=I|num-b=9|num-a=11}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_11&diff=1893482023 AIME I Problems/Problem 112023-02-09T20:02:32Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Find the number of subsets of <math>\{1,2,3,\ldots,10\}</math> that contain exactly one pair of consecutive integers. Examples of such subsets are <math>\{\mathbf{1},\mathbf{2},5\}</math> and <math>\{1,3,\mathbf{6},\mathbf{7},10\}.</math><br />
<br />
==Solution 1 (minimal casework)==<br />
Define <math>f(x)</math> to be the number of subsets of <math>\{1, 2, 3, 4, \ldots x\}</math> that have <math>0</math> consecutive element pairs, and <math>f'(x)</math> to be the number of subsets that have <math>1</math> consecutive pair.<br />
<br />
Using casework on where the consecutive element pair is, it is easy to see that <cmath>f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + 2f(3)^2.</cmath><br />
<br />
We see that <math>f(1) = 2</math>, <math>f(2) = 3</math>, and <math>f(n) = f(n-1) + f(n-2)</math>. This is because if the element <math>1</math> is included in our subset, then there are <math>f(n-2)</math> possibilities for the rest of the elements (because <math>2</math> cannot be used), and otherwise there are <math>f(n-1)</math> possibilities. Thus, by induction, <math>f(x)</math> is the <math>n+1</math>th Fibonacci number.<br />
<br />
This means that <math>f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = \boxed{235}</math>.<br />
<br />
~mathboy100<br />
<br />
==Solution 2==<br />
<br />
We can solve this problem using casework, with one case for each possible pair of consecutive numbers.<br />
<br />
<math>\textbf{Case 1: (1,2)}</math><br />
<br />
If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us <math>8 \choose {0}</math>. We can also pick one number, giving us <math>7 \choose {1}</math> because 3 cannot be picked. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us <math>6 \choose {2}</math>. This pattern continues for each amount of numbers, yielding <math>5 \choose {3}</math> for 3 numbers and <math>4 \choose {4}</math> for four numbers. Adding these up, we have <math>8 \choose {0}</math> + <math>7 \choose {1}</math> + <math>6 \choose {2}</math> + <math>5 \choose {3}</math> + <math>4 \choose {4}</math> = <math>\textbf{34}</math>.<br />
<br />
<br />
<br />
<math>\textbf{Case 2: (2,3)}</math><br />
<br />
If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have <math>7 \choose {0}</math> + <math>6 \choose {1}</math> + <math>5 \choose {2}</math> + <math>4 \choose {3}</math> = <math>\textbf{21}</math>.<br />
<br />
<br />
<br />
<math>\textbf{Case 3: (3,4)}</math><br />
<br />
This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say <math>2 \choose {0}</math> for choosing no numbers and <math>1 \choose {1}</math> for choosing 1, leaving us with <math>2 \choose {0}</math> + <math>1 \choose {1}</math> = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have <math>6 \choose {0}</math> + <math>5 \choose {1}</math> + <math>4 \choose {2}</math> + <math>3 \choose {3}</math> = 13. With 2 choices on one side and 13 choices on the other side, we have <math>2\cdot13</math> = <math>\textbf{26}</math> combinations in all.<br />
<br />
<br />
<br />
<math>\textbf{Case 4: (4,5)}</math><br />
<br />
Following the patterns we have already created in our previous cases, for the numbers 1-3 we have <math>3 \choose {0}</math> + <math>2 \choose {1}</math> = 3 choices (1, 2, or neither) and for the numbers 6-10 we have <math>5 \choose {0}</math> + <math>4 \choose {1}</math> + <math>3 \choose {2}</math> = 8 choices. With 3 choices on one side and 8 choices on the other side, we have <math>3\cdot8</math> = <math>\textbf{24}</math> combinations in all.<br />
<br />
<br />
<br />
<math>\textbf{Case 5: (5,6)}</math><br />
<br />
Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have <math>4 \choose {0}</math> + <math>3 \choose {1}</math> + <math>2 \choose {2}</math> = 5 choices and for the numbers 5-10 we have the same <math>4 \choose {0}</math> + <math>3 \choose {1}</math> + <math>2 \choose {2}</math> = 5 choices. <math>5\cdot5</math> = <math>\textbf{25}</math> combinations in all.<br />
<br />
<br />
<br />
<math>\textbf{Rest of the cases}</math><br />
<br />
By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1.<br />
<br />
<br />
<br />
Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 = <math>\boxed{\textbf{(235)}}</math><br />
<br />
<br />
-Algebraik<br />
<br />
==Solution 3 (double recursive equations)==<br />
<br />
Denote by <math>N_1 \left( m \right)</math> the number of subsets of a set <math>S</math> that consists of <math>m</math> consecutive integers, such that each subset contains exactly one pair of consecutive integers.<br />
<br />
Denote by <math>N_0 \left( m \right)</math> the number of subsets of a set <math>S</math> that consists of <math>m</math> consecutive integers, such that each subset does not contain any consecutive integers.<br />
<br />
Denote by <math>a</math> the smallest number in set <math>S</math>.<br />
<br />
First, we compute <math>N_1 \left( m \right)</math>.<br />
<br />
Consider <math>m \geq 3</math>.<br />
We do casework analysis.<br />
<br />
Case 1: A subset does not contain <math>a</math>.<br />
<br />
The number of subsets that has exactly one pair of consecutive integers is <math>N_1 \left( m - 1 \right)</math>.<br />
<br />
Case 2: A subset contains <math>a</math> but does not contain <math>a + 1</math>.<br />
<br />
The number of subsets that has exactly one pair of consecutive integers is <math>N_1 \left( m - 2 \right)</math>.<br />
<br />
Case 3: A subset contains <math>a</math> and <math>a + 1</math>.<br />
<br />
To have exactly one pair of consecutive integers, this subset cannot have <math>a + 2</math>, and cannot have consecutive integers in <math>\left\{ a+3, a+4, \cdots , a + m - 1 \right\}</math>.<br />
<br />
Thus, the number of subsets that has exactly one pair of consecutive integers is <math>N_0 \left( m - 3 \right)</math>.<br />
<br />
Therefore, for <math>m \geq 3</math>,<br />
<cmath><br />
N_1 \left( m \right) = N_1 \left( m - 1 \right) + N_1 \left( m - 2 \right) + N_0 \left( m - 3 \right) .<br />
</cmath><br />
<br />
For <math>m = 1</math>, we have <math>N_1 \left( 1 \right) = 0</math>.<br />
For <math>m = 2</math>, we have <math>N_1 \left( 2 \right) = 1</math>.<br />
<br />
<br />
Second, we compute <math>N_0 \left( m \right)</math>.<br />
<br />
Consider <math>m \geq 2</math>.<br />
We do casework analysis.<br />
<br />
Case 1: A subset does not contain <math>a</math>.<br />
<br />
The number of subsets that has no consecutive integers is <math>N_0 \left( m - 1 \right)</math>.<br />
<br />
Case 2: A subset contains <math>a</math>.<br />
<br />
To avoid having consecutive integers, the subset cannot have <math>a + 1</math>.<br />
<br />
Thus, the number of subsets that has no consecutive integers is <math>N_0 \left( m - 2 \right)</math>.<br />
<br />
Therefore, for <math>m \geq 2</math>,<br />
<cmath><br />
N_0 \left( m \right) = N_0 \left( m - 1 \right) + N_0 \left( m - 2 \right) .<br />
</cmath><br />
<br />
For <math>m = 0</math>, we have <math>N_0 \left( 0 \right) = 1</math>.<br />
For <math>m = 1</math>, we have <math>N_0 \left( 1 \right) = 2</math>.<br />
<br />
By solving the recursive equations above, we get <math>N_1 \left( 10 \right) = \boxed{\textbf{(235) }}</math>.<br />
<br />
~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 4==<br />
<br />
Let's consider all cases:<br />
<br />
Case 1: Our consecutive pair is (1,2).<br />
<br />
Then, let's make subcases:<br />
<br />
Case 1.1: We have 0 other numbers in our set. There is <math>1</math> way to do this.<br />
<br />
Case 1.2: We have 1 other number in our set. Let's say that this number is <math>x_1</math>. We then can say that <math>x_a = x_1 - 2</math>, with the restriction that <math>x_a \ge 2</math>. Then, we can write <math>x_a \le 8</math> (since we have a total of 8 spots for remaining numbers: 3, 4, 5, 6, 7, 8, 9, 10). To remove the restriction on x_a, we can write <math>(x_a - 2) \le 6</math>, where <math>x_a \ge 0</math>. We then can use stars and bars: <math>7 \choose {1}</math> <math> = 7</math>.<br />
<br />
Case 1.3: We have 2 numbers in our set. This is similar, so I won't go through the same process. <math>x_a = x_1 - 2, x_b = x_2 - x_1, x_{a, b} \ge 2</math> . <math>x_a + x_b \le 8, (x_a - 2) + (x_b - 2) \le 4</math>. We can use stars and bars: <math>6 \choose 2</math> <math> = 15</math>.<br />
Case 1.4: We have 3 numbers in our set. <math>x_a + x_b + x_c \le 8, (x_a - 2) + (x_b - 2) + (x_c - 2) \le 2</math>. Stars and bars: <math>5 \choose 3</math> <math> = 10</math>. <br />
<br />
Case 1.4: We have 4 numbers in our set. <math>x_a + x_b + x_c + x_d \le 8, (x_a - 2) + (x_b - 2) + (x_c - 2) + (x_d - 2) \le 0</math>. Stars and bars: <math>4 \choose 4</math> <math> = 1</math>.<br />
<br />
Adding all subcases, we get <math>1 + 7 + 15 + 10 + 1 = 34</math>.<br />
<br />
Case 2: Our consecutive pair is (8,9).<br />
<br />
We use symmetry to find that there is exactly 1 case for (8,9) that matches with (1,2). This also has <math>34</math> cases.<br />
<br />
Case 3: Our consecutive pair is (2,3). The value is <math>21</math> This is left as an exercise to the reader (but I can write it if you'd like).<br />
<br />
Case 4: Our consecutive pair is (7,8). The value is <math>21</math>. This is left as an exercise to the reader (but I can write it if you'd like).<br />
<br />
*I will finish later.<br />
<br />
hia2020<br />
<br />
==Solution 5 (Similar to Solution 3)==<br />
Let <math>a_n</math> be the number of subsets of the set <math>\{1,2,3,\ldots,n\}</math> such that there exists exactly 1 pair of consecutive elements. <br />
Let <math>b_n</math> be the number of subsets of the set <math>\{1, 2, 3\ldots, n\}</math> such that there doesn't exist any pair of consecutive elements. <br />
First, lets see how we can construct <math>a_n.</math> For each subset <math>S</math> counted in <math>a_n,</math> either:<br />
1. <math>\{n-1, n\}\subseteq S,</math><br />
2. <math> n\not\in S</math>, or<br />
3. <math>n-1 \not\in S</math> and <math>n\in S.</math><br />
The first case counts <math>b_{n-3}</math> subsets (as <math>n-1</math> cannot be included and the rest cannot have any consecutive elements), The second counts <math>a_{n-1},</math> and the third counts <math>a_{n-2}.</math> Thus, <cmath>a_n = a_{n-1} + a_{n-2} + b_{n-3}.</cmath><br />
Next, Lets try to construct <math>b_n.</math> For each subset <math>T</math> counted in <math>b_n,</math> either:<br />
1.<math>n \not\in T,</math> or <br />
2.<math>n \in T.</math><br />
The first case counts <math>b_{n-1}</math> subsets and the second counts <math>b_{n-2}.</math> Thus, <cmath>b_n = b_{n-1} + b_{n-2}.</cmath><br />
Since <math>b_1 = 2</math> and <math>b_2 = 3,</math> we have that <math>b_n = F_{n+1},</math> so <math>a_n = a_{n-1} + a_{n-2} + F_{n-2}.</math> (The <math>F_i</math> is the <math>i</math>th Fibonacci number). From here, we can construct a table of the values of <math>a_n</math> until <math>n = 10.</math> By listing out possibilities, we can solve for our first 3 values. <br />
<br />
<cmath> \begin{array}{r|l} n & a_n \\ \hline 1 & 0 \\ 2 & 1\\ 3 & 2\\ 4 & 2 + 1 + F_2 = 5\\5 & 5 + 2 + F_3 = 10\\6 & 10 + 5 + F_4 = 20 \\ 7 & 20 + 10 + F_5 = 38\\ 8 & 38 + 20 + F_6 = 71 \\ 9& 71 + 38 + F_7 = 130\\ 10 & 130 + 71 + F_8 = 235 \end{array} </cmath><br />
Our answer is <math>a_{10} = \boxed{235}.</math><br />
~AtharvNaphade<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/7xqeCQiPrew<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
<br />
https://www.youtube.com/watch?v=q1ffZP63q3I<br />
<br />
==See also==<br />
{{AIME box|year=2023|num-b=10|num-a=12|n=I}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_14&diff=1762422014 AMC 10A Problems/Problem 142022-07-25T17:07:01Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
The <math>y</math>-intercepts, <math>P</math> and <math>Q</math>, of two perpendicular lines intersecting at the point <math>A(6,8)</math> have a sum of zero. What is the area of <math>\triangle APQ</math>?<br />
<br />
<math> \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1==<br />
<asy>//Needs refining (hmm I think it's fine --bestwillcui1)<br />
size(12cm);<br />
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));<br />
for(int i=-2;i<=8;i+=1)<br />
draw((i,-12)--(i,12),grey);<br />
for(int j=-12;j<=12;j+=1)<br />
draw((-2,j)--(8,j),grey);<br />
draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis<br />
draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis<br />
dot((0,0));<br />
dot((6,8));<br />
draw((-2,10.66667)--(8,7.33333),Arrows);<br />
draw((7.33333,12)--(-0.66667,-12),Arrows);<br />
draw((6,8)--(0,8));<br />
draw((6,8)--(0,0));<br />
draw(rightanglemark((0,10),(6,8),(0,-10),20));<br />
label("$A$",(6,8),NE);<br />
label("$a$", (0,5),W);<br />
label("$a$",(0,-5),W);<br />
label("$a$",(3,4),NW);<br />
label("$P$",(0,10),SW);<br />
label("$Q$",(0,-10),NW);<br />
// wanted to import graph and use xaxis/yaxis but w/e<br />
label("$x$",(9,0),E);<br />
label("$y$",(0,13),N);<br />
</asy><br />
Note that if the <math>y</math>-intercepts have a sum of <math>0</math>, the distance from the origin to each of the intercepts must be the same. Call this distance <math>a</math>. Since the <math>\angle PAQ = 90^\circ</math>, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is <math>\sqrt{6^2+8^2} = 10</math>, this means <math>a=10</math>, and the length of the hypotenuse is <math>2a = 20</math>. Since the <math>x</math>-coordinate of <math>A</math> is the same as the altitude to the hypotenuse, <math>[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can let the two lines be <cmath>y=mx+b</cmath> <cmath>y=-\frac{1}{m}x-b</cmath> This is because the lines are perpendicular, hence the <math>m</math> and <math>-\frac{1}{m}</math>, and the sum of the y-intercepts is equal to 0, hence the <math>b, -b</math>. <br />
<br />
Since both lines contain the point <math>(6,8)</math>, we can plug this into the two equations to obtain <cmath>8=6m+b</cmath> and <cmath>8=-6\frac{1}{m}-b</cmath><br />
<br />
Adding the two equations gives <cmath>16=6m+\frac{-6}{m}</cmath> Multiplying by <math>m</math> gives <cmath>16m=6m^2-6</cmath> <cmath>\implies 6m^2-16m-6=0</cmath> <cmath>\implies 3m^2-8m-3=0</cmath> Factoring gives <cmath>(3m+1)(m-3)=0</cmath> <br />
<br />
Plugging <math>m=3</math> into one of our original equations, we obtain <cmath>8=6(3)+b</cmath> <cmath>\implies b=8-6(3)=-10</cmath><br />
<br />
Since <math>\bigtriangleup APQ</math> has hypotenuse <math>2|b|=20</math> and the altitude to the hypotenuse is equal to the the x-coordinate of point <math>A</math>, or 6, the area of <math>\bigtriangleup APQ</math> is equal to <cmath>\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\ 60}</cmath><br />
<br />
==Solution 3==<br />
<br />
Like Solution 2 but solving directly for intercepts (b):<br />
<br />
1. Solve for m using: <math>8=6m+b</math><br />
<br />
<cmath>m=\frac{8-b}{6}</cmath><br />
<br />
2. Substitute into the other equation:<br />
<br />
<cmath>8=-6\cdot(\frac{1}{\frac{8-b}{6}})-b</cmath><br />
<br />
Flip the inverse:<br />
<br />
<cmath>8=-6\cdot(\frac{6}{8-b})-b</cmath><br />
<br />
Multiply <math>6</math>'s:<br />
<br />
<cmath>8=-(\frac{36}{8-b})-b</cmath><br />
<br />
<br />
3. Multiply through by <math>8-b</math> (Watch distributing minus!)<br />
<br />
<cmath>64-8b=-36-8b+b^2</cmath><br />
<br />
4. Add <math>36</math> to both sides, and cancel <math>-8b</math> by adding to both sides:<br />
<br />
<cmath>100=b^2</cmath><br />
<br />
<math>b=10</math> (or <math>-10</math>)<br />
<br />
The rest is as above.<br />
<br />
<br />
==Solution 4(Heron's Formula)==<br />
<br />
Since their sum is <math>0</math>, let the y intercepts be P<math>(0,a)</math> and Q<math>(0,-a)</math>. The slope of <math>AP</math> is <math>\frac{8-a}{6}</math>. The slope of AQ is <math>\frac{8+a}{6}</math>. Since multiplying the slopes of perpendicular lines yields a product of <math>-1</math>, we have <math>\frac{64-a^2}{36}=-1</math>, which results in <math>a^2=100</math>. We can use either the positive or negative solution because if we choose <math>10</math>, then the other y-intercept is <math>-10</math>; but if we choose <math>-10</math>, then the other y-intercept is <math>10</math>. For simplicity, we choose that <math>a=10</math> in this solution. <br />
<br />
Now we have a triangle APQ with points A<math>(6,8)</math>, P<math>(0,10)</math>, and Q<math>(0,-10)</math>. By the Pythagorean theorem, we have that <math>AP=\sqrt{6^2+2^2}=2\sqrt{10}</math>, and that <math>AQ=\sqrt{6^2+18^2}=6\sqrt{10}</math>. <math>PQ</math> is obviously <math>10-(-10)=20</math> since they have the same <math>x</math> coordinate. Now using Heron's formula, we have<br />
<math>\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{D}</math>.<br />
<br />
~smartninja2000<br />
<br />
==Solution 5 (point-slope)==<br />
Using point-slope form, the first line has the equation <cmath>y-8=m\left(x-6\right) \longrightarrow y=mx-6m+8</cmath><br />
The second line has the equation <cmath>y-8=-\frac{1}{m}\left(x-6\right) \longrightarrow y=-\frac{x}{m}+\frac{6}{m}+8</cmath><br />
At the y-intercept, the value of the x-coordinate is <math>0</math>, hence: the first equation is <math>y=-6m+8</math> and the second is <math>y=\frac{6}{m}+8</math>. Since the y-intercepts sum to <math>0</math>, they are opposites, so:<br />
<cmath>-6m+8=-\left(\frac{6}{m}+8\right)=-\frac{6}{m}-8</cmath><br />
<cmath>6m-\frac{6}{m}=16</cmath><br />
Multiply both sides by m:<br />
<math>6m^{2}-6=16m \longrightarrow 3m^{2}-8m-3=0</math>. The solution to this quadratic, using the quadratic formula, is:<br />
<math>\frac{8\pm\sqrt{64-4\left(3\right)\left(-3\right)}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}</math><br />
This yields <math>m=-\frac{1}{3}</math> and <math>m=3</math>. Plugging <math>m=3</math> into the second equation, we get <math>y=\frac{6}{3}+8=10</math>. Plugging <math>m=-\frac{1}{3}</math> into the first equation, we get <math>y=-10</math> So the base is <math>20</math> and the height is <math>6</math>, the area is <math>60 \Longrightarrow \boxed{\textbf{(D) } 60}</math>.<br />
<br />
~JH. L<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=AJdRK51xvos<br />
~Mathematical Dexterity<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_14&diff=1737102014 AIME I Problems/Problem 142022-05-02T00:18:46Z<p>Mc21s: </p>
<hr />
<div>== Problem 14 ==<br />
<br />
Let <math>m</math> be the largest real solution to the equation<br />
<br />
<math>\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4</math><br />
<br />
There are positive integers <math>a</math>, <math>b</math>, and <math>c</math> such that <math>m=a+\sqrt{b+\sqrt{c}}</math>. Find <math>a+b+c</math>.<br />
<br />
== Solution ==<br />
The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to <math>\frac{3}{x-3}</math>, then the fraction becomes of the form <math>\frac{x}{x - 3}</math>. A similar cancellation happens with the other four terms. If we assume <math>x = 0</math> is not the highest solution (which is true given the answer format) we can cancel the common factor of <math>x</math> from both sides of the equation.<br />
<br />
<math>\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11</math><br />
<br />
Then, if we make the substitution <math>y = x - 11</math>, we can further simplify.<br />
<br />
<math>\frac{1}{y + 8} + \frac{1}{y + 6} + \frac{1}{y - 6} + \frac{1}{y - 8} =y </math><br />
<br />
If we group and combine the terms of the form <math>y - n</math> and <math> y + n</math>, we get this equation:<br />
<br />
<math>\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y</math><br />
<br />
Then, we can cancel out a <math>y</math> from both sides, knowing that <math>x = 11</math> is not a possible solution given the answer format.<br />
After we do that, we can make the final substitution <math>z = y^2</math>.<br />
<br />
<math>\frac{2}{z - 64} + \frac{2}{z - 36} = 1</math><br />
<br />
<math>2z - 128 + 2z - 72 = (z - 64)(z - 36)</math><br />
<br />
<math>4z - 200 = z^2 - 100z + 64(36)</math><br />
<br />
<math>z^2 - 104z + 2504 = 0</math><br />
<br />
Using the quadratic formula, we get that the largest solution for <math>z</math> is <math>z = 52 + 10\sqrt{2}</math>. Then, repeatedly substituting backwards, we find that the largest value of <math>x</math> is <math>11 + \sqrt{52 + \sqrt{200}}</math>. The answer is thus <math>11 + 52 + 200 = \boxed{263}</math><br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtu.be/pNsmv333SE0<br />
<br />
==Video Solution by Mathematical Dexterity (Pure magic!)==<br />
https://www.youtube.com/watch?v=7b7IPOYZbrk<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_7&diff=1714402022 AIME II Problems/Problem 72022-02-18T22:58:25Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
A circle with radius <math>6</math> is externally tangent to a circle with radius <math>24</math>. Find the area of the triangular region bounded by the three common tangent lines of these two circles.<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=7NGkVu0kE08<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_4&diff=1714392022 AIME II Problems/Problem 42022-02-18T22:58:01Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
There is a positive real number <math>x</math> not equal to either <math>\tfrac{1}{20}</math> or <math>\tfrac{1}{2}</math> such that<cmath>\log_{20x} (22x)=\log_{2x} (202x).</cmath>The value <math>\log_{20x} (22x)</math> can be written as <math>\log_{10} (\tfrac{m}{n})</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2x} (202x)</math>.<br />
<br />
So that <math>(20x)^v=22x \textcircled{1}</math><br />
and <math>(2x)^v=202x \textcircled{2}</math><br />
<br />
Express <math>\textcircled{1}</math> as: <math>(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x \textcircled{3}</math><br />
<br />
Substitute <math>\textcircled{2}</math> to <math>\textcircled{3}</math>: <math>202x \cdot (10^v)=22x</math><br />
<br />
Thus, <math>v=\log_{10} (\frac{22x}{202x})= \log_{10} (\frac{11}{101})</math>, where <math>m=11</math> and <math>n=101</math>.<br />
<br />
Therefore, <math>m+n = \boxed{112}</math>.<br />
<br />
~DSAERF-CALMIT (https://binaryphi.site)<br />
<br />
==Solution 2==<br />
<br />
We have<br />
<cmath><br />
\begin{align*}<br />
\log_{20x} (22x)<br />
& = \frac{\log_k 22x}{\log_k 20x} \\<br />
& = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} .<br />
\end{align*}<br />
</cmath><br />
<br />
We have<br />
<cmath><br />
\begin{align*}<br />
\log_{2x} (202x)<br />
& = \frac{\log_k 202x}{\log_k 2x} \\<br />
& = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} .<br />
\end{align*}<br />
</cmath><br />
<br />
Because <math>\log_{20x} (22x)=\log_{2x} (202x)</math>, we get<br />
<cmath><br />
\[<br />
\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20}<br />
= \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} .<br />
\]<br />
</cmath><br />
<br />
We denote this common value as <math>\lambda</math>.<br />
<br />
By solving the equality <math>\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \lambda</math>, we get <math>\log_k x = \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}</math>.<br />
<br />
By solving the equality <math>\frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} = \lambda</math>, we get <math>\log_k x = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1}</math>.<br />
<br />
By equating these two equations, we get<br />
<cmath><br />
\[<br />
\frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}<br />
= \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1} .<br />
\]<br />
</cmath><br />
<br />
Therefore,<br />
<cmath><br />
\begin{align*}<br />
\log_{20x} (22x)<br />
& = \lambda \\<br />
& = \frac{\log_k 22 - \log_k 202}{\log_k 20 - \log_k 2} \\<br />
& = \frac{\log_k \frac{11}{101}}{\log_k 10} \\<br />
& = \log_{10} \frac{11}{101} .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, the answer is <math>11 + 101 = \boxed{\textbf{(112) }}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=4qJyvyZN630<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_3&diff=1714382022 AIME II Problems/Problem 32022-02-18T22:57:10Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
A right square pyramid with volume <math>54</math> has a base with side length <math>6.</math> The five vertices of the pyramid all lie on a sphere with radius <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius <math>\frac{6}{\sqrt{2}} = 3\sqrt{2}</math>). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: <math>l</math>. Because of the symmetrical property of the pyramid,<br />
we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.<br />
<br />
Since the volume is <math>54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h</math>, where <math>h=\frac{9}{2}</math> is the height of this pyramid, we have: <math>l^2=(\frac{9}{2}-l)^2+(3\sqrt{2})^2</math> according to pythagorean theorem.<br />
<br />
Solve this equation will give us <math>l = \frac{17}{4}</math>. Therefore, <math>m+n=\boxed{021}</math><br />
<br />
~DSAERF-CALMIT (https://binaryphi.site)<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=UJAYW6YNFVU<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_2&diff=1714372022 AIME II Problems/Problem 22022-02-18T22:56:22Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability <math>\frac23</math>. When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability <math>\frac34</math>. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=C14f91P2pYc<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_1&diff=1714362022 AIME II Problems/Problem 12022-02-18T22:56:00Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
Adults made up <math>\frac5{12}</math> of the crowd of people at a concert. After a bus carrying <math>50</math> more people arrived, adults made up <math>\frac{11}{25}</math> of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.<br />
<br />
==Solution 1==<br />
Let <math>x</math> be the number of people at the party before the bus arrives. We know that <math>x\equiv 0\pmod {12}</math>, as <math>\frac{5}{12}</math> of people at the party before the bus arrives are adults. Similarly, we know that <math>x + 50 \equiv 0 \pmod{25}</math>, as <math>\frac{11}{25}</math> of the people at the party are adults after the bus arrives. <math>x + 50 \equiv 0 \pmod{25}</math> can be reduced to <math>x \equiv 0 \pmod{25}</math>, and since we are looking for the minimum amount of people, <math>x</math> is <math>300</math>. That means there are <math>350</math> people at the party after the bus arrives, and thus there are <math>350 \cdot \frac{11}{25} = \boxed{154}</math> adults at the party.<br />
<br />
~eamo<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=gBIxZ6SUr_w<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_11&diff=1714352022 AIME I Problems/Problem 112022-02-18T22:55:37Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
Let <math>ABCD</math> be a parallelogram with <math>\angle BAD < 90^{\circ}</math>. A circle tangent to sides <math>\overline{DA}</math>, <math>\overline{AB}</math>, and <math>\overline{BC}</math> intersects diagonal <math>\overline{AC}</math> at points <math>P</math> and <math>Q</math> with <math>AP < AQ</math>, as shown. Suppose that <math>AP = 3</math>, <math>PQ = 9</math>, and <math>QC = 16</math>. Then the area of <math>ABCD</math> can be expressed in the form <math>m\sqrt n</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.<br />
<br />
<asy><br />
defaultpen(linewidth(0.6)+fontsize(11));<br />
size(8cm);<br />
pair A,B,C,D,P,Q;<br />
A=(0,0);<br />
label("$A$", A, SW);<br />
B=(6,15);<br />
label("$B$", B, NW);<br />
C=(30,15);<br />
label("$C$", C, NE);<br />
D=(24,0);<br />
label("$D$", D, SE);<br />
P=(5.2,2.6);<br />
label("$P$", (5.8,2.6), N);<br />
Q=(18.3,9.1);<br />
label("$Q$", (18.1,9.7), W);<br />
draw(A--B--C--D--cycle);<br />
draw(C--A);<br />
draw(Circle((10.95,7.45), 7.45));<br />
dot(A^^B^^C^^D^^P^^Q);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math><br />
<br />
Using POP, it is very clear that <math>PC=20,AQ=AM=6</math>, let <math>BM=BP=x,QD=14+x</math>, using LOC in <math>\triangle{ABP}</math>,<math>x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2</math>, similarly, use LOC in <math>\triangle{DQC}</math>, getting that <math>(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2</math>. We use the second equation to minus the first equation, getting that <math>28x+196-(2x+12)*14*\cos\alpha=364</math>, we can get <math>\cos\alpha=\frac{2x-12}{2x+12}</math>.<br />
<br />
Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2</math>, solving this equation to get <math>x=\frac{9}{2}</math>, then <math>\cos\alpha=-\frac{1}{7}</math>, <math>\sin\alpha=\frac{4\sqrt{3}}{7}</math>, the area is <math>\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}</math> leads to <math>\boxed{150}</math><br />
<br />
~bluesoul<br />
<br />
==Solution 2==<br />
Denote by <math>O</math> the center of the circle. Denote by <math>r</math> the radius of the circle.<br />
Denote by <math>E</math>, <math>F</math>, <math>G</math> the points that the circle meets <math>AB</math>, <math>CD</math>, <math>AD</math> at, respectively.<br />
<br />
Because the circle is tangent to <math>AD</math>, <math>CB</math>, <math>AB</math>, <math>OE = OF = OG = r</math>, <math>OE \perp AD</math>, <math>OF \perp CB</math>, <math>OG \perp AB</math>.<br />
<br />
Because <math>AD \parallel CB</math>, <math>E</math>, <math>O</math>, <math>F</math> are collinear. <br />
<br />
Following from the power of a point, <math>AG^2 = AE^2 = AP \cdot AQ</math>. Hence, <math>AG = AE = 6</math>.<br />
<br />
Following from the power of a point, <math>CF^2 = CQ \cdot CP</math>. Hence, <math>CF = 20</math>.<br />
<br />
Denote <math>BG = x</math>. Because <math>DG</math> and <math>DF</math> are tangents to the circle, <math>BF = x</math>.<br />
<br />
Because <math>AEFB</math> is a right trapezoid, <math>AB^2 = EF^2 + \left( AE - BF \right)^2</math>.<br />
Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>.<br />
This can be simplified as <br />
\[<br />
6 x = r^2 . \hspace{1cm} (1)<br />
\]<br />
<br />
In <math>\triangle ACB</math>, by applying the law of cosines, we have<br />
\begin{align*}<br />
AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\<br />
& = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\<br />
& = AB^2 + CB^2 + 2 AB \cdot CB \cdot \frac{AE - BF}{AB} \\<br />
& = AB^2 + CB^2 + 2 CB \left( AE - BF \right) \\<br />
& = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\<br />
& = 24 x + 676 .<br />
\end{align*}<br />
<br />
Because <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>.<br />
Plugging this into Equation (1), we get <math>r = 3 \sqrt{3}</math>.<br />
<br />
Therefore,<br />
\begin{align*}<br />
{\rm Area} \ ABCD & = CB \cdot EF \\<br />
& = \left( 20 + x \right) \cdot 2r \\<br />
& = 147 \sqrt{3} .<br />
\end{align*}<br />
<br />
Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Solution 3==<br />
Let <math>\omega</math> be the circle, let <math>r</math> be the radius of <math>\omega</math>, and let the points at which <math>\omega</math> is tangent to <math>AB</math>, <math>BC</math>, and <math>AD</math> be <math>X</math>, <math>Y</math>, and <math>Z</math>, respectively. Note that PoP on <math>A</math> and <math>C</math> with respect to <math>\omega</math> yields <math>AX=6</math> and <math>CY=20</math>. We can compute the area of <math>ABC</math> in two ways:<br />
<br />
1. By the half-base-height formula, <math>[ABC]=r(20+BX)</math>.<br />
<br />
2. We can drop altitudes from the center <math>O</math> of <math>\omega</math> to <math>AB</math>, <math>BC</math>, and <math>AC</math>, which have lengths <math>r</math>, <math>r</math>, and <math>\sqrt{r^2-81/4}</math>. Thus, <math>[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\sqrt{r^2-81/4}</math>.<br />
<br />
Equating the two expressions for <math>[ABC]</math> and solving for <math>r</math> yields <math>r=3\sqrt{3}</math>. Let <math>BX=BY=a</math>. By the Pythagorean Theorem, <math>(6-a)^2+(2r)^2=(a+6)^2</math>. Solving for <math>a</math> yields <math>a=9/2</math>. Thus, <math>[ABCD]=2[ABC]=2r(20+a)=147\sqrt{3}</math>, for a final answer of <math>\boxed{150}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution 2 (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=1nDKQkr9NaU<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_10&diff=1714342022 AIME I Problems/Problem 102022-02-18T22:55:08Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
Three spheres with radii <math>11</math>, <math>13</math>, and <math>19</math> are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at <math>A</math>, <math>B</math>, and <math>C</math>, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that <math>AB^2 = 560</math>. Find <math>AC^2</math>.<br />
<br />
<br />
==Solution 1==<br />
Let the distance between the center of the sphere to the center of those circular intersections as <math>a,b,c</math> separately. <math>a-11,b-13,c-19</math>. According to the problem, we have <math>a^2-11^2=b^2-13^2=c^2-19^2;(11+13)^2-(b-a)^2=560</math>. After solving we have <math>b-a=4</math>, plug this back to <math>11^2-a^2=13^2-b^2; a=4,b=8,c=16</math><br />
<br />
The desired value is <math>(11+19)^2-(16-4)^2=\boxed{756}</math><br />
<br />
~bluesoul<br />
<br />
==Solution 2==<br />
<br />
Denote by <math>r</math> the radius of three congruent circles formed by the cutting plane.<br />
Denote by <math>O_A</math>, <math>O_B</math>, <math>O_C</math> the centers of three spheres that intersect the plane to get circles centered at <math>A</math>, <math>B</math>, <math>C</math>, respectively.<br />
<br />
Because three spheres are mutually tangent, <math>O_A O_B = 11 + 13 = 24</math>, <math>O_A O_C = 11 + 19 = 32</math>.<br />
<br />
We have <math>O_A A^2 = 11^2 - r^2</math>, <math>O_B B^2 = 13^2 - r^2</math>, <math>O_C C^2 = 19^2 - r^2</math>.<br />
<br />
Because <math>O_A A</math> and <math>O_B B</math> are perpendicular to the plane, <math>O_A AB O_B</math> is a right trapezoid, with <math>\angle O_A A B = \angle O_B BA = 90^\circ</math>.<br />
<br />
Hence,<br />
<cmath><br />
\begin{align*}<br />
O_B B - O_A A & = \sqrt{O_A O_B^2 - AB^2} \\<br />
& = 4 . \hspace{1cm} (1)<br />
\end{align*}<br />
</cmath><br />
<br />
Recall that<br />
<cmath><br />
\begin{align*}<br />
O_B B^2 - O_A A^2 & = \left( 13^2 - r^2 \right) - \left( 11^2 - r^2 \right) \\<br />
& = 48 . \hspace{1cm} (2)<br />
\end{align*}<br />
</cmath><br />
<br />
Hence, taking <math>\frac{(2)}{(1)}</math>, we get<br />
<cmath><br />
\[<br />
O_B B + O_A A = 12 . \hspace{1cm} (3)<br />
\]<br />
</cmath><br />
<br />
Solving (1) and (3), we get <math>O_B B = 8</math> and <math>O_A A = 4</math>.<br />
<br />
Thus, <math>r^2 = 11^2 - O_A A^2 = 105</math>.<br />
<br />
Thus, <math>O_C C = \sqrt{19^2 - r^2} = 16</math>.<br />
<br />
Because <math>O_A A</math> and <math>O_C C</math> are perpendicular to the plane, <math>O_A AC O_C</math> is a right trapezoid, with <math>\angle O_A A C = \angle O_C CA = 90^\circ</math>.<br />
<br />
Therefore,<br />
<cmath><br />
\begin{align*}<br />
AC^2 & = O_A O_C^2 - \left( O_C C - O_A A \right)^2 \\<br />
& = \boxed{\textbf{(756) }} .<br />
\end{align*}<br />
</cmath><br />
<br />
<math>\textbf{FINAL NOTE:}</math> In our solution, we do not use the conditio that spheres <math>A</math> and <math>B</math> are externally tangent. This condition is redundant in solving this problem.<br />
<br />
~Steven Chen (www.professorcheneeu.com)<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=SqLiV2pbCpY&t=15s<br />
<br />
~Steven Chen (www.professorcheneeu.com)<br />
<br />
==Video Solution 2 (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=HbBU13YiopU<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_9&diff=1714332022 AIME I Problems/Problem 92022-02-18T22:54:35Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
Ellina has twelve blocks, two each of red <math>\left({\bf R}\right),</math> blue <math>\left({\bf B}\right),</math> yellow <math>\left({\bf Y}\right),</math> green <math>\left({\bf G}\right),</math> orange <math>\left({\bf O}\right),</math> and purple <math>\left({\bf P}\right).</math> Call an arrangement of blocks even if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement<br />
<cmath> {\text {\bf R B B Y G G Y R O P P O}}</cmath>is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
== Solution 1 ==<br />
Consider this position chart:<br />
<cmath> {\text {\bf 1 2 3 4 5 6 7 8 9 10 11 12}}</cmath><br />
Since there has to be an even number of spaces between each ball of the same color, spots <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>9</math>, and <math>11</math> contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is <math>6! \cdot 6!</math> (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of <math>\frac{12!}{(2!)^6}</math> possible arrangements, so the probability is: <cmath>\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^6}{7\cdot11\cdot12} = \frac{16}{231}</cmath><br />
, which is in simplest form. So <math>m + n = 16 + 231 = \boxed{247}</math>.<br />
<br />
-Oxymoronic15<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=dkoF7StwtrM<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_4&diff=1711462022 AIME I Problems/Problem 42022-02-17T22:38:20Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>w = \dfrac{\sqrt{3} + i}{2}</math> and <math>z = \dfrac{-1 + i\sqrt{3}}{2},</math> where <math>i = \sqrt{-1}.</math> Find the number of ordered pairs <math>(r,s)</math> of positive integers not exceeding <math>100</math> that satisfy the equation <math>i \cdot w^r = z^s.</math><br />
<br />
==Solution==<br />
<br />
We rewrite <math>w</math> and <math>z</math> in polar form:<br />
<cmath>\begin{align*}<br />
w &= e^{i\cdot\frac{\pi}{6}}, \\<br />
z &= e^{i\cdot\frac{2\pi}{3}}.<br />
\end{align*}</cmath><br />
The equation <math>i \cdot w^r = z^s</math> becomes<br />
<cmath>\begin{align*}<br />
e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\<br />
e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\<br />
\frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\<br />
3+r &= 4s+12k \\<br />
3+r &= 4(s+3k).<br />
\end{align*}</cmath><br />
for some integer <math>k.</math> <br />
<br />
Since <math>4\leq 3+r\leq 103</math> and <math>4\mid 3+r,</math> we conclude that<br />
<cmath>\begin{align*}<br />
3+r &\in \{4,8,12,\ldots,100\}, \\<br />
s+3k &\in \{1,2,3,\ldots,25\}.<br />
\end{align*}</cmath><br />
Note that the values for <math>s+3k</math> and the values for <math>r</math> have one-to-one correspondence.<br />
<br />
We apply casework to the values for <math>s+3k:</math><br />
<ol style="margin-left: 1.5em;"><br />
<li><math>s+3k\equiv0\pmod{3}</math></li><p><br />
There are <math>8</math> values for <math>s+3k,</math> so there are <math>8</math> values for <math>r.</math> It follows that <math>s\equiv0\pmod{3},</math> so there are <math>33</math> values for <math>s.</math> <p><br />
There are <math>8\cdot33=264</math> ordered pairs <math>(r,s)</math> in this case.<br />
<li><math>s+3k\equiv1\pmod{3}</math></li><p><br />
There are <math>9</math> values for <math>s+3k,</math> so there are <math>9</math> values for <math>r.</math> It follows that <math>s\equiv1\pmod{3},</math> so there are <math>34</math> values for <math>s.</math> <p><br />
There are <math>9\cdot34=306</math> ordered pairs <math>(r,s)</math> in this case.<br />
<li><math>s+3k\equiv2\pmod{3}</math></li><p><br />
There are <math>8</math> values for <math>s+3k,</math> so there are <math>8</math> values for <math>r.</math> It follows that <math>s\equiv2\pmod{3},</math> so there are <math>33</math> values for <math>s.</math> <p><br />
There are <math>8\cdot33=264</math> ordered pairs <math>(r,s)</math> in this case.<br />
</ol><br />
Together, the answer is <math>264+306+264=\boxed{834}.</math><br />
<br />
~MRENTHUSIASM ~bluesoul<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=XiEaCq5jf5s<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_3&diff=1711452022 AIME I Problems/Problem 32022-02-17T22:37:50Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
In isosceles trapezoid <math>ABCD</math>, parallel bases <math>\overline{AB}</math> and <math>\overline{CD}</math> have lengths <math>500</math> and <math>650</math>, respectively, and <math>AD=BC=333</math>. The angle bisectors of <math>\angle{A}</math> and <math>\angle{D}</math> meet at <math>P</math>, and the angle bisectors of <math>\angle{B}</math> and <math>\angle{C}</math> meet at <math>Q</math>. Find <math>PQ</math>.<br />
<br />
== Diagram ==<br />
<asy><br />
unitsize(0.016cm);<br />
pair A = (-250,324.4);<br />
pair B = (250, 324.4);<br />
pair C = (325, 0);<br />
pair D = (-325, 0);<br />
draw(A--B--C--D--cycle);<br />
pair W = (8,0);<br />
pair X = (-8, 0);<br />
pair Y = (-83,324.4);<br />
pair Z = (83,324.4);<br />
<br />
pair P = (-121, 162.2);<br />
pair Q = (121, 162.2);<br />
dot(P);<br />
dot(Q);<br />
<br />
draw(A--W, dashed);<br />
draw(B--X, dashed);<br />
draw(C--Y, dashed);<br />
draw(D--Z, dashed);<br />
label("$A$", A, N);<br />
label("$B$", B, N);<br />
label("$Y$", Y, N);<br />
label("$Z$", Z, N);<br />
label("$C$", C, S);<br />
label("$D$", D, S);<br />
label("$W$", W, SE);<br />
label("$X$", X, SW);<br />
label("$P$", P, N);<br />
label("$Q$", Q, N);<br />
</asy><br />
<br />
== Solution 1==<br />
<br />
Extend lines <math>AP</math> and <math>BQ</math> to meet line <math>DC</math> at points <math>W</math> and <math>X</math>, respectively, and extend lines <math>DP</math> and <math>CQ</math> to meet <math>AB</math> at points <math>Z</math> and <math>Y</math>, respectively.<br />
<br />
Claim: quadrilaterals <math>AZWD</math> and <math>BYXD</math> are rhombuses.<br />
<br />
Proof: Since <math>\angle DAB + \angle ADC = 180^{\circ}</math>, <math>\angle ADP + \angle PAD = 90^{\circ}</math>. Therefore, triangles <math>APD</math>, <math>APZ</math>, <math>DPW</math> and <math>PZW</math> are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, <math>\triangle PZW</math> is congruent to the other three. Therefore, <math>AD = DW = WZ = AZ</math>, so <math>AZWD</math> is a rhombus. By symmetry, <math>BYXC</math> is also a rhombus.<br />
<br />
Extend line <math>PQ</math> to meet <math>\overline{AD}</math> and <math>\overline{BC}</math> at <math>R</math> and <math>S</math>, respectively. Because of rhombus properties, <math>RP = QS = \frac{333}{2}</math>. Also, by rhombus properties, <math>R</math> and <math>S</math> are the midpoints of segments <math>AD</math> and <math>BC</math>, respectively; therefore, by trapezoid properties, <math>RS = \frac{AB + CD}{2} = 575</math>. Finally, <math>PQ = RS - RP - QS = \boxed{242}</math>.<br />
<br />
~ihatemath123<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=fNAvxXnvAxs<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_2&diff=1711442022 AIME I Problems/Problem 22022-02-17T22:37:13Z<p>Mc21s: /* Video Solution (Mathematical Dexterity */</p>
<hr />
<div>== Problem ==<br />
<br />
Find the three-digit positive integer <math>\underline{a}\,\underline{b}\,\underline{c}</math> whose representation in base nine is <math>\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are (not necessarily distinct) digits.<br />
<br />
== Solution ==<br />
<br />
We are given that <cmath>100a + 10b + c = 81b + 9c + a,</cmath> which rearranges to <cmath>99a = 71b + 8c.</cmath><br />
Taking both sides modulo <math>71,</math> we have<br />
<cmath>\begin{align*}<br />
28a &\equiv 8c \pmod{71} \\<br />
7a &\equiv 2c \pmod{71}.<br />
\end{align*}</cmath><br />
The only solution occurs at <math>(a,c)=(2,7),</math> from which <math>b=2.</math><br />
<br />
Therefore, the requested three-digit positive integer is <math>\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=z5Y4bT5rL-s<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_2&diff=1711432022 AIME I Problems/Problem 22022-02-17T22:37:02Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
<br />
Find the three-digit positive integer <math>\underline{a}\,\underline{b}\,\underline{c}</math> whose representation in base nine is <math>\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are (not necessarily distinct) digits.<br />
<br />
== Solution ==<br />
<br />
We are given that <cmath>100a + 10b + c = 81b + 9c + a,</cmath> which rearranges to <cmath>99a = 71b + 8c.</cmath><br />
Taking both sides modulo <math>71,</math> we have<br />
<cmath>\begin{align*}<br />
28a &\equiv 8c \pmod{71} \\<br />
7a &\equiv 2c \pmod{71}.<br />
\end{align*}</cmath><br />
The only solution occurs at <math>(a,c)=(2,7),</math> from which <math>b=2.</math><br />
<br />
Therefore, the requested three-digit positive integer is <math>\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Video Solution (Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=z5Y4bT5rL-s<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_1&diff=1711422022 AIME I Problems/Problem 12022-02-17T22:36:32Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
Quadratic polynomials <math>P(x)</math> and <math>Q(x)</math> have leading coefficients <math>2</math> and <math>-2,</math> respectively. The graphs of both polynomials pass through the two points <math>(16,54)</math> and <math>(20,53).</math> Find <math>P(0) + Q(0).</math><br />
<br />
==Solution 1 (Linear Polynomials)==<br />
<br />
Let <math>R(x)=P(x)+Q(x).</math> Since the <math>x^2</math>-terms of <math>P(x)</math> and <math>Q(x)</math> cancel, we conclude that <math>R(x)</math> is a linear polynomial.<br />
<br />
Note that<br />
<cmath>\begin{alignat*}{8}<br />
R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\<br />
R(20) &= P(20)+Q(20) &&= 53+53 &&= 106,<br />
\end{alignat*}</cmath><br />
so the slope of <math>R(x)</math> is <math>\frac{106-108}{20-16}=-\frac12.</math><br />
<br />
It follows that the equation of <math>R(x)</math> is <cmath>R(x)=-\frac12x+c</cmath> for some constant <math>c,</math> and we wish to find <math>R(0)=c.</math><br />
<br />
We substitute <math>x=20</math> into this equation to get <math>106=-\frac12\cdot20+c,</math> from which <math>c=\boxed{116}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 2 (Quadratic Polynomials)==<br />
<br />
Let<br />
<cmath>\begin{align*}<br />
P(x) &= 2x^2 + ax + b, \\<br />
Q(x) &= -2x^2 + cx + d,<br />
\end{align*}</cmath><br />
for some constants <math>a,b,c</math> and <math>d.</math> <br />
<br />
We are given that<br />
<cmath>\begin{alignat*}{8}<br />
P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\<br />
Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\<br />
P(20) &= &800 + 20a + b &= 53, &&(3) \\<br />
Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4)<br />
\end{alignat*}</cmath><br />
and we wish to find <cmath>P(0)+Q(0)=b+d.</cmath><br />
We need to cancel <math>a</math> and <math>c.</math> Since <math>\operatorname{lcm}(16,20)=80,</math> we subtract <math>4\cdot[(3)+(4)]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath><br />
~MRENTHUSIASM<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=sUfbEBCQ6RY<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=I|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_11&diff=1707622022 AMC 8 Problems/Problem 112022-01-30T03:20:31Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating <math>3</math> inches of pasta from the middle of one piece. In the end, he has <math>10</math> pieces of pasta whose total length is <math>17</math> inches. How long, in inches, was the piece of pasta he started with?<br />
<br />
<math>\textbf{(A) } 34\qquad\textbf{(B) } 38\qquad\textbf{(C) } 41\qquad\textbf{(D) } 44\qquad\textbf{(E) } 47</math><br />
<br />
==Solution==<br />
<br />
If there are <math>10</math> pieces of pasta, Henry took <math>10-1=9</math> bites. Each of these <math>9</math> bites took <math>3</math> inches of pasta out, and thus his bites in total took away <math>9\cdot 3 = 27</math> inches of pasta. Thus, the original piece of pasta was <math>27+17=\boxed{\textbf{(D) } 44}</math> inches long.<br />
<br />
~wamofan<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=H1K4WJ93isA<br />
<br />
~Mathematical Dexterity<br />
<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_20&diff=1707512022 AMC 8 Problems/Problem 202022-01-30T02:35:03Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
<br />
The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number <math>x</math> in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of <math>x</math>?<br />
<asy><br />
unitsize(0.5cm);<br />
draw((3,3)--(-3,3));<br />
draw((3,1)--(-3,1));<br />
draw((3,-3)--(-3,-3));<br />
draw((3,-1)--(-3,-1));<br />
draw((3,3)--(3,-3));<br />
draw((1,3)--(1,-3));<br />
draw((-3,3)--(-3,-3));<br />
draw((-1,3)--(-1,-3));<br />
label((-2,2),"$-2$");<br />
label((0,2),"$9$");<br />
label((2,2),"$5$");<br />
label((2,0),"$-1$");<br />
label((2,-2),"$8$");<br />
label((-2,-2),"$x$");<br />
</asy><br />
<math>\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad</math><br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=xnGQffaxYAA<br />
<br />
~Mathematical Dexterity<br />
<br />
== Solution 1 ==<br />
The sum of the numbers in each row is <math>12</math>. Consider the second row. In order for the sum of the numbers in this row to equal <math>12</math>, the first two numbers must add up to <math>13</math>:<br />
<asy><br />
unitsize(0.5cm);<br />
fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,lightgray);<br />
draw((3,3)--(-3,3));<br />
draw((3,1)--(-3,1));<br />
draw((3,-3)--(-3,-3));<br />
draw((3,-1)--(-3,-1));<br />
draw((3,3)--(3,-3));<br />
draw((1,3)--(1,-3));<br />
draw((-3,3)--(-3,-3));<br />
draw((-1,3)--(-1,-3));<br />
label((-2,2),"$-2$");<br />
label((0,2),"$9$");<br />
label((2,2),"$5$");<br />
label((2,0),"$-1$");<br />
label((2,-2),"$8$");<br />
label((-2,-2),"$x$");<br />
</asy><br />
If two numbers add up to <math>13</math>, one of them must be at least <math>7</math>: If both shaded numbers are no more than <math>6</math>, their sum can be at most <math>12</math>. Therefore, for <math>x</math> to be larger than the three missing numbers, <math>x</math> must be at least <math>8</math>. We can construct a working scenario where <math>x=8</math>:<br />
<asy><br />
unitsize(0.5cm);<br />
draw((3,3)--(-3,3));<br />
draw((3,1)--(-3,1));<br />
draw((3,-3)--(-3,-3));<br />
draw((3,-1)--(-3,-1));<br />
draw((3,3)--(3,-3));<br />
draw((1,3)--(1,-3));<br />
draw((-3,3)--(-3,-3));<br />
draw((-1,3)--(-1,-3));<br />
label((-2,2),"$-2$");<br />
label((0,2),"$9$");<br />
label((2,2),"$5$");<br />
label((2,0),"$-1$");<br />
label((2,-2),"$8$");<br />
label((-2,-2),"$8$");<br />
label((0,-2),"$-4$");<br />
label((-2,0),"$6$");<br />
label((0,0),"$7$");<br />
</asy><br />
So, our answer is <math>\boxed{\textbf{(D) } 8}</math>.<br />
<br />
~ihatemath123<br />
<br />
==Solution 2==<br />
<br />
The sum of the numbers in each row is <math>-2+9+5=12,</math> and the sum of the numbers in each column is <math>5+(-1)+8=12.</math><br />
<br />
Let <math>y</math> be the number in the lower middle. It follows that <math>x+y+8=12,</math> or <math>x+y=4.</math><br />
<br />
We express the other two missing numbers in terms of <math>x</math> and <math>y,</math> as shown below:<br />
<asy><br />
unitsize(0.5cm);<br />
draw((3,3)--(-3,3));<br />
draw((3,1)--(-3,1));<br />
draw((3,-3)--(-3,-3));<br />
draw((3,-1)--(-3,-1));<br />
draw((3,3)--(3,-3));<br />
draw((1,3)--(1,-3));<br />
draw((-3,3)--(-3,-3));<br />
draw((-1,3)--(-1,-3));<br />
label((-2,2),"$-2$");<br />
label((0,2),"$9$");<br />
label((2,2),"$5$");<br />
label((2,0),"$-1$");<br />
label((2,-2),"$8$");<br />
label((-2,-2),"$x$");<br />
label((0,-2),"$y$",red); label((-2,0),"$y+10$",red+fontsize(8)); label((0,0),"$x-1$",red+fontsize(8)); <br />
</asy><br />
We have <math>x>x-1, x>y+10,</math> and <math>x>y.</math> Note that the first inequality is true for all values of <math>x.</math> We only need to solve the second inequality so that the third inequality is true for all values of <math>x.</math> By substitution, we get <math>x>(4-x)+10,</math> from which <math>x>7.</math><br />
<br />
Therefore, the smallest possible value of <math>x</math> is <math>\boxed{\textbf{(D) } 8}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_21&diff=1707492022 AMC 8 Problems/Problem 212022-01-30T02:14:54Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
Steph scored <math>15</math> baskets out of <math>20</math> attempts in the first half of a game, and <math>10</math> baskets out of <math>10</math> attempts in the second half. Candace took <math>12</math> attempts in the first half and <math>18</math> attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first? <br />
<asy><br />
size(7cm);<br />
draw((-8,27)--(72,27));<br />
draw((16,0)--(16,35));<br />
draw((40,0)--(40,35));<br />
label("12", (28,3));<br />
draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle);<br />
draw((25,5.5)--(31,5.5));<br />
label("18", (56,3));<br />
draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle);<br />
draw((53,5.5)--(59,5.5));<br />
draw((53,5.5)--(59,5.5));<br />
label("20", (28,18));<br />
label("15", (28,24));<br />
draw((25,21)--(31,21));<br />
label("10", (56,18));<br />
label("10", (56,24));<br />
draw((53,21)--(59,21));<br />
label("First Half", (28,31));<br />
label("Second Half", (56,31));<br />
label("Candace", (2.35,6));<br />
label("Steph", (0,21));<br />
</asy><br />
<math>\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11</math><br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=IbsSecIq8FE<br />
<br />
~Mathematical Dexterity<br />
<br />
==Solution==<br />
<br />
Let <math>x</math> be the number of shots that Candace made in the first half, and let <math>y</math> be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have <math>x+y=10+15=25.</math> In addition, we have the following inequalities: <cmath>\frac{x}{12}<\frac{15}{20} \implies x<9,</cmath> and <cmath>\frac{y}{18}<\frac{10}{10} \implies y<18.</cmath> Pairing this up with <math>x+y=25</math> we see the <i><b>only</b></i> possible solution is <math>(x,y)=(8,17),</math> for an answer of <math>17-8 = \boxed{\textbf{(C) } 9}.</math><br />
<br />
~wamofan<br />
<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_22&diff=1707482022 AMC 8 Problems/Problem 222022-01-30T02:03:47Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
A bus takes <math>2</math> minutes to drive from one stop to the next, and waits <math>1</math> minute at each stop to let passengers board. Zia takes <math>5</math> minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus <math>3</math> stops behind. After how many minutes will Zia board the bus?<br />
<br />
[[File:2022 AMC 8 Problem 22 Diagram.png|750px|center]]<br />
<br />
<math>\textbf{(A) } 17 \qquad \textbf{(B) } 19 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 23</math><br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=SFp4HaieEBg<br />
<br />
~Mathematical Dexterity<br />
<br />
==Solution==<br />
Initially, suppose that the bus is at Stop <math>0</math> (starting point) and Zia is at Stop <math>3.</math><br />
<br />
We construct the following table of <math>5</math>-minute intervals:<br />
<cmath>\begin{array}{c||c|c}<br />
& & \\ [-2.5ex]<br />
\textbf{Time} & \textbf{Bus's Location} & \textbf{Zia's Location} \\ [0.5ex]<br />
\hline<br />
& & \\ [-2ex]<br />
\boldsymbol{5} \ \textbf{Minutes} & \text{Stop} \ 2 \ \text{(Waiting)} & \text{Stop} \ 4 \\<br />
\boldsymbol{10} \ \textbf{Minutes} & \text{Midpoint of Stops} \ 3 \ \text{and} \ 4 & \text{Stop} \ 5 \\<br />
\boldsymbol{15} \ \textbf{Minutes} & \text{Stop} \ 5 \ \text{(Leaving)} & \text{Stop} \ 6<br />
\end{array}</cmath><br />
Note that Zia will wait for the bus after <math>15</math> minutes, and the bus will arrive <math>2</math> minutes later.<br />
<br />
Therefore, the answer is <math>15+2=\boxed{\textbf{(A) } 17}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_23&diff=1707472022 AMC 8 Problems/Problem 232022-01-30T02:03:09Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
A <math>\triangle</math> or <math>\bigcirc</math> is placed in each of the nine squares in a <math>3</math>-by-<math>3</math> grid. Shown below is a sample configuration with three <math>\triangle</math>s in a line.<br />
<asy><br />
//diagram by kante314<br />
size(3.3cm);<br />
defaultpen(linewidth(1));<br />
real r = 0.37;<br />
path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle;<br />
draw((0,0)--(0,3)--(3,3)--(3,0)--cycle);<br />
draw((0,1)--(3,1)--(3,2)--(0,2)--cycle);<br />
draw((1,0)--(1,3)--(2,3)--(2,0)--cycle);<br />
draw(circle((3/2,5/2),1/3));<br />
draw(circle((5/2,1/2),1/3));<br />
draw(circle((3/2,3/2),1/3));<br />
draw(shift(0.5,0.38) * equi);<br />
draw(shift(1.5,0.38) * equi);<br />
draw(shift(0.5,1.38) * equi);<br />
draw(shift(2.5,1.38) * equi);<br />
draw(shift(0.5,2.38) * equi);<br />
draw(shift(2.5,2.38) * equi);<br />
</asy><br />
How many configurations will have three <math>\triangle</math>s in a line and three <math>\bigcirc</math>s in a line?<br />
<br />
<math>\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96</math><br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=or4pKVzQ3gI<br />
<br />
~Mathematical Dexterity<br />
<br />
==Solution 1==<br />
Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.<br />
<br />
We take casework:<br />
<br />
<i>Case 1: 3 lines</i><br />
In this case, the lines would need to be <math>2</math> of one shape and <math>1</math> of another, so there are <math>\frac{3!}{2} = 3</math> ways to arrange the lines and <math>2</math> ways to pick which shape has only one line. In total, this is <math>3\cdot 2 = 6.</math><br />
<br />
<i>Case 2: 2 lines</i><br />
In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are <math>3! = 6</math> ways to arrange the lines and <math>2^3-2 = 6</math> ways to choose the last line. In total, this is <math>6\cdot 6 = 36.</math><br />
<br />
Finally, we add and multiply: <math>2(36+6)=2(42)=\boxed{\textbf{(D) }84}</math>.<br />
<br />
~wamofan<br />
<br />
==Solution 2==<br />
We will only consider columns, but at the end our answer should be multiplied by <math>2</math>. There are <math>3</math> ways to choose a column for <math>\bigcirc</math> and <math>2</math> ways to choose a column for <math>\triangle</math>. The third column can be filled in <math>2^3=8</math> ways. Therefore, we have <math>3\cdot2\cdot8=48</math> ways. However, we overcounted the cases with <math>2</math> complete columns of with one symbol and <math>1</math> complete columns with another symbol. This happens in <math>2\cdot3=6</math> ways. <math>48-6=42</math>. However, we have to remember to double our answer giving us <math>\boxed{\textbf{(D) }84}</math>.<br />
<br />
~MathFun1000<br />
<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_24&diff=1707462022 AMC 8 Problems/Problem 242022-01-30T02:02:37Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
The figure below shows a polygon <math>ABCDEFGH</math>, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that <math>AH = EF = 8</math> and <math>GH = 14</math>. What is the volume of the prism?<br />
<br />
<asy><br />
usepackage("mathptmx");<br />
unitsize(1cm);<br />
defaultpen(linewidth(0.7)+fontsize(11));<br />
real r = 2, s = 2.5, theta = 14;<br />
pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta);<br />
pair N = (B+G)/2, J = N + s/2 * dir(180+theta);<br />
pair E = F + r * dir(- 45 - theta/2), D = I+E-F;<br />
pair H = J + r * dir(135 + theta/2), A = B+H-J;<br />
draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B));<br />
draw(J--B--G^^C--F--I,linetype ("4 4"));<br />
dot("$A$",A,N);<br />
dot("$B$",B,1.2*N);<br />
dot("$C$",C,N);<br />
dot("$D$",D,dir(0));<br />
dot("$E$",E,S);<br />
dot("$F$",F,1.5*S);<br />
dot("$G$",G,S);<br />
dot("$H$",H,W);<br />
dot("$I$",I,NE);<br />
dot("$J$",J,1.5*S);<br />
</asy><br />
<br />
<math>\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288\qquad</math><br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=2uoBPp4Kxck<br />
<br />
~Mathematical Dexterity<br />
<br />
==Solution==<br />
<br />
While imagining the folding, <math>\overline{AB}</math> goes on <math>\overline{BC},</math> <math>\overline{AH}</math> goes on <math>\overline{CI},</math> and <math>\overline{EF}</math> goes on <math>\overline{FG}.</math> So, <math>\overline{BJ}=\overline{CI}=8</math> and <math>\overline{FG}=\overline{BC}=8.</math> Also, <math>\overline{HJ}</math> becomes an edge parallel to <math>\overline{FG},</math> so that means <math>\overline{HJ}=8.</math><br />
<br />
Since <math>\overline{GH}=14,</math> then <math>\overline{JG}=14-8=6.</math> So, the area of <math>\triangle BJG</math> is <math>\frac{8\cdot6}{2}=24.</math> If we let <math>\triangle BJG</math> be the base, then the height is <math>\overline{FG}=8.</math> So, the volume is <math>24\cdot8=\boxed{\textbf{(C) }192}.</math><br />
<br />
Solution by aops-g5-gethsemanea2<br />
<br />
==Remark==<br />
<br />
It is easy to visualize the prism when <math>\triangle BJG</math> is the base and <math>\overline{\rm AB}</math> is the height.<br />
<br />
~MathFun1000<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_25&diff=1707452022 AMC 8 Problems/Problem 252022-01-30T02:01:57Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
A cricket randomly hops between <math>4</math> leaves, on each turn hopping to one of the other <math>3</math> leaves with equal probability. After <math>4</math> hops what is the probability that the cricket has returned to the leaf where it started?<br />
<br />
[[File:2022 AMC 8 Problem 25 Picture.jpg|center|600px]]<br />
<br />
<math>\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{19}{80}\qquad\textbf{(C) }\frac{20}{81}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{7}{27}</math><br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=85A6av3oqRo<br />
<br />
~Mathematical Dexterity<br />
<br />
==Solution 1 (Casework)==<br />
<br />
Let <math>A</math> denote the leaf where the cricket starts and <math>B</math> denote one of the other <math>3</math> leaves. Note that:<br />
<br />
* If the cricket is at <math>A,</math> then the probability that it hops to <math>B</math> next is <math>1.</math><br />
<br />
* If the cricket is at <math>B,</math> then the probability that it hops to <math>A</math> next is <math>\frac13.</math><br />
<br />
* If the cricket is at <math>B,</math> then the probability that it hops to <math>B</math> next is <math>\frac23.</math><br />
<br />
We apply casework to the possible paths of the cricket:<br />
<ol style="margin-left: 1.5em;"><br />
<li><math>A \rightarrow B \rightarrow A \rightarrow B \rightarrow A</math> <p><br />
The probability for this case is <math>1\cdot\frac13\cdot1\cdot\frac13=\frac19.</math></li><p><br />
<li><math>A \rightarrow B \rightarrow B \rightarrow B \rightarrow A</math> <p><br />
The probability for this case is <math>1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.</math></li><p><br />
</ol><br />
Together, the probability that the cricket returns to <math>A</math> is <math>\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 2 (Casework)==<br />
<br />
We can label the leaves as shown below:<br />
[[File:2022 AMC 8 Problem 25 Picture 2.png|center|600px]]<br />
Carefully counting cases, we see that there are <math>7</math> ways for the cricket to return to leaf <math>A</math> after four hops if its first hop was to leaf <math>B</math>:<br />
<ol style="margin-left: 1.5em;"><br />
<li><math>A \rightarrow B \rightarrow A \rightarrow B \rightarrow A</math></li><p><br />
<li><math>A \rightarrow B \rightarrow A \rightarrow C \rightarrow A</math></li><p><br />
<li><math>A \rightarrow B \rightarrow A \rightarrow D \rightarrow A</math></li><p><br />
<li><math>A \rightarrow B \rightarrow C \rightarrow B \rightarrow A</math></li><p><br />
<li><math>A \rightarrow B \rightarrow C \rightarrow D \rightarrow A</math></li><p><br />
<li><math>A \rightarrow B \rightarrow D \rightarrow B \rightarrow A</math></li><p><br />
<li><math>A \rightarrow B \rightarrow D \rightarrow C \rightarrow A</math></li><p><br />
</ol><br />
By symmetry, we know that there are <math>7</math> ways if the cricket's first hop was to leaf <math>C</math>, and there are <math>7</math> ways if the cricket's first hop was to leaf <math>D</math>. So, there are <math>21</math> ways in total for the cricket to return to leaf <math>A</math> after four hops.<br />
<br />
Since there are <math>3^4 = 81</math> possible ways altogether for the cricket to hop to any other leaf four times, the answer is <math>\frac{21}{81} = \boxed{\textbf{(E) }\frac{7}{27}}</math>.<br />
<br />
~mahaler<br />
<br />
==Solution 3 (Recursion)==<br />
<br />
Denote <math>P_n</math> to be the probability that the cricket would return back to the first point after <math>n</math> hops. Then, we get the recursive formula <cmath>P_n = \frac13(1-P_{n-1})</cmath> because if the leaf is not on the target leaf, then there is a <math>\frac13</math> probability that it will make it back.<br />
<br />
With this formula and the fact that <math>P_0=0,</math> we have <cmath>P_1 = \frac13, P_2 = \frac29, P_3 = \frac7{27},</cmath> so our answer is <math>\boxed{\textbf{(E) }\frac{7}{27}}</math>.<br />
<br />
~wamofan<br />
<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_14&diff=1707152021 AIME II Problems/Problem 142022-01-29T18:30:44Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Let <math>\Delta ABC</math> be an acute triangle with circumcenter <math>O</math> and centroid <math>G</math>. Let <math>X</math> be the intersection of the line tangent to the circumcircle of <math>\Delta ABC</math> at <math>A</math> and the line perpendicular to <math>GO</math> at <math>G</math>. Let <math>Y</math> be the intersection of lines <math>XG</math> and <math>BC</math>. Given that the measures of <math>\angle ABC, \angle BCA, </math> and <math>\angle XOY</math> are in the ratio <math>13 : 2 : 17, </math> the degree measure of <math>\angle BAC</math> can be written as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Diagram==<br />
<asy><br />
/* Made by MRENTHUSIASM */<br />
size(375);<br />
<br />
pair A, B, C, O, G, X, Y;<br />
A = origin;<br />
B = (1,0);<br />
C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7));<br />
O = circumcenter(A,B,C);<br />
G = centroid(A,B,C);<br />
Y = intersectionpoint(G--G+(100,0),B--C);<br />
X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A));<br />
markscalefactor=3/160;<br />
draw(rightanglemark(O,G,X),red);<br />
dot("$A$",A,1.5*dir(180+585/7),linewidth(4));<br />
dot("$B$",B,1.5*dir(-585/7),linewidth(4));<br />
dot("$C$",C,1.5N,linewidth(4));<br />
dot("$O$",O,1.5N,linewidth(4));<br />
dot("$G$",G,1.5S,linewidth(4));<br />
dot("$Y$",Y,1.5E,linewidth(4));<br />
dot("$X$",X,1.5W,linewidth(4));<br />
draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C));<br />
</asy><br />
~MRENTHUSIASM<br />
<br />
==Solution 1==<br />
In this solution, all angle measures are in degrees.<br />
<br />
Let <math>M</math> be the midpoint of <math>\overline{BC}</math> so that <math>\overline{OM}\perp\overline{BC}</math> and <math>A,G,M</math> are collinear. Let <math>\angle ABC=13k,\angle BCA=2k</math> and <math>\angle XOY=17k.</math> <br />
<br />
Note that:<br />
<ol style="margin-left: 1.5em;"><br />
<li>Since <math>\angle OGX = \angle OAX = 90,</math> quadrilateral <math>OGAX</math> is cyclic by the Converse of the Inscribed Angle Theorem.<p>It follows that <math>\angle OAG = \angle OXG,</math> as they share the same intercepted arc <math>\widehat{OG}.</math></li><p><br />
<li>Since <math>\angle OGY = \angle OMY = 90,</math> quadrilateral <math>OGYM</math> is cyclic by the supplementary opposite angles.<p>It follows that <math>\angle OMG = \angle OYG,</math> as they share the same intercepted arc <math>\widehat{OG}.</math></li><p><br />
</ol><br />
Together, we conclude that <math>\triangle OAM \sim \triangle OXY</math> by AA, so <math>\angle AOM = \angle XOY = 17k.</math><br />
<br />
Next, we express <math>\angle BAC</math> in terms of <math>k.</math> By angle addition, we have<br />
<cmath>\begin{align*}<br />
\angle AOM &= \angle AOB + \angle BOM \\<br />
&= 2\angle BCA + \frac12\angle BOC \hspace{10mm} &&\text{by Inscribed Angle Theorem and Perpendicular Bisector Property} \\<br />
&= 2\angle BCA + \angle BAC. &&\text{by Inscribed Angle Theorem}<br />
\end{align*}</cmath><br />
Substituting back gives <math>17k=2(2k)+\angle BAC,</math> from which <math>\angle BAC=13k.</math><br />
<br />
For the sum of the interior angles of <math>\triangle ABC,</math> we get<br />
<cmath>\begin{align*}<br />
\angle ABC + \angle BCA + \angle BAC &= 180 \\<br />
13k+2k+13k&=180 \\<br />
28k&=180 \\<br />
k&=\frac{45}{7}.<br />
\end{align*}</cmath><br />
Finally, we obtain <math>\angle BAC=13k=\frac{585}{7},</math> from which the answer is <math>585+7=\boxed{592}.</math><br />
<br />
~Constance-variance (Fundamental Logic)<br />
<br />
~MRENTHUSIASM (Reconstruction)<br />
<br />
==Solution 2==<br />
Let <math>M</math> be the midpoint of <math>BC</math>. Because <math>\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o</math>, <math>AXOG</math> and <math>OMYG</math> are cyclic, so <math>O</math> is the center of the spiral similarity sending <math>AM</math> to <math>XY</math>, and <math>\angle{XOY}=\angle{AOM}</math>. Because <math>\angle{AOM}=2\angle{BCA}+\angle{BAC}</math>, it's easy to get <math>\frac{585}{7} \implies \boxed{592}</math> from here.<br />
<br />
~Lcz<br />
<br />
==Solution 3 (Easy and Simple)==<br />
Firstly, let <math>M</math> be the midpoint of <math>BC</math>. Then, <math>\angle OMB = 90^o</math>. Now, note that since <math>\angle OGX = \angle XAO = 90^o</math>, quadrilateral <math>AGOX</math> is cyclic. Also, because <math>\angle OMY + \angle OGY = 180^o</math>, <math>OMYG</math> is also cyclic. Now, we define some variables: let <math>\alpha</math> be the constant such that <math>\angle ABC = 13\alpha, \angle ACB = 2\alpha, </math> and <math>\angle XOY = 17\alpha</math>. Also, let <math>\beta = \angle OMG = \angle OYG</math> and <math>\theta = \angle OXG = \angle OAG</math> (due to the fact that <math>AGOX</math> and <math>OMYG</math> are cyclic). Then, <cmath>\angle XOY = 180 - \beta - \theta = 17\alpha \implies \beta + \theta = 180 - 17\alpha.</cmath> Now, because <math>AX</math> is tangent to the circumcircle at <math>A</math>, <math>\angle XAC = \angle CBA = 13\alpha</math>, and <math>\angle CAO = \angle OAX - \angle CAX = 90 - 13\alpha</math>. Finally, notice that <math>\angle AMB = \angle OMB - \angle OMG = 90 - \beta</math>. Then, <cmath>\angle BAM = 180 - \angle ABC - \angle AMB = 180 - 13\alpha - (90 - \beta) = 90 + \beta - 13\alpha.</cmath> Thus, <cmath>\angle BAC = \angle BAM + \angle MAO + \angle OAC = 90 + \beta - 13\alpha + \theta + 90 - 13\alpha = 180 - 26\alpha + (\beta + \theta),</cmath> and <cmath>180 = \angle BAC + 13\alpha + 2\alpha = 180 - 11\alpha + \beta + \theta \implies \beta + \theta = 11\alpha.</cmath> However, from before, <math>\beta+\theta = 180 - 17 \alpha</math>, so <math>11 \alpha = 180 - 17 \alpha \implies 180 = 28 \alpha \implies \alpha = \frac{180}{28}</math>. To finish the problem, we simply compute <cmath>\angle BAC = 180 - 15 \alpha = 180 \cdot \left(1 - \frac{15}{28}\right) = 180 \cdot \frac{13}{28} = \frac{585}{7},</cmath> so our final answer is <math>585+7=\boxed{592}</math>.<br />
<br />
~advanture<br />
<br />
==Solution 4 (Guessing in the Last 3 Minutes, Unreliable)==<br />
Notice that <math>\triangle ABC</math> looks isosceles, so we assume it's isosceles. Then, let <math>\angle BAC = \angle ABC = 13x</math> and <math>\angle BCA = 2x.</math> Taking the sum of the angles in the triangle gives <math>28x=180,</math> so <math>13x = \frac{13}{28} \cdot 180 = \frac{585}{7}</math> so the answer is <math>\boxed{592}.</math><br />
<br />
==Video Solution 1==<br />
https://www.youtube.com/watch?v=zFH1Z7Ydq1s<br />
~Mathematical Dexterity<br />
<br />
==Video Solution 2==<br />
https://www.youtube.com/watch?v=7Bxr2h4btWo<br />
<br />
~Osman Nal<br />
<br />
==See Also==<br />
{{AIME box|year=2021|n=II|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_21&diff=1670482021 Fall AMC 12B Problems/Problem 212021-11-29T23:50:48Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
For real numbers <math>x</math>, let <br />
<cmath>P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)</cmath><br />
where <math>i = \sqrt{-1}</math>. For how many values of <math>x</math> with <math>0\leq x<2\pi</math> does <br />
<cmath>P(x)=0?</cmath><br />
<br />
<math>\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\<br />
3 \qquad\textbf{(E)}\ 4</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>a=\cos(x)+i\sin(x)</math>. Now <math>P(a)=1+a-a^2+a^3</math>. <math>P(-1)=-2</math> and <math>P(0)=1</math> so there is a real number <math>a_1</math> between <math>-1</math> and <math>0</math>. The other <math>a</math>'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex <math>a</math>'s squared is <math>\frac{1}{a_1}</math> which is greater than <math>1</math>. If <math>x</math> is real number then <math>a</math> must have magnitude of <math>1</math>, but all the solutions for <math>a</math> do not have magnitude of <math>1</math>, so the answer is <math>\boxed{\textbf{(A)}\ 0 }</math> ~lopkiloinm<br />
<br />
== Solution 2 ==<br />
We have<br />
<cmath><br />
\begin{align*}<br />
P \left( x \right)<br />
& = 1 + e^{ix} - e^{i 2x} + e^{i 3x} .<br />
\end{align*}<br />
</cmath><br />
<br />
Denote <math>y = e^{i x}</math>. Hence, this problem asks us to find the number of <math>y</math> with <math>| y| = 1</math> that satisfy<br />
<cmath><br />
\[<br />
1 + y - y^2 + y^3 = 0 . \hspace{1cm} (1)<br />
\]<br />
</cmath><br />
<br />
Taking imaginary part of both sides, we have<br />
<cmath><br />
\begin{align*}<br />
0 & = {\rm Im} \ \left( 1 + y - y^2 + y^3 \right) \\<br />
& = \frac{1}{2i} \left( y - \bar y - y^2 + \bar y^2 + y^3 - \bar y^3 \right) \\<br />
& = \frac{y - \bar y}{2i} \left( 1 - y - \bar y + y^2 + y \bar y + \bar y^2 \right) \\<br />
& = {\rm Im} \ y \left( 1 - \left( y + \bar y \right) + \left( y + \bar y \right)^2 - y \bar y \right) \\<br />
& = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - |y|^2 \right) \\<br />
& = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - 1 \right) \\<br />
& = 2 {\rm Im} \ y \cdot {\rm Re} \ y \left( 2 {\rm Re} \ y - 1 \right) \\<br />
\end{align*}<br />
</cmath><br />
The sixth equality follows from the property that <math>|y| = 1</math>.<br />
<br />
Therefore, we have either <math>{\rm Re} \ y = 0</math> or <math>{\rm Im} \ y = 0</math> or <math>2 {\rm Re} \ y - 1 = 0</math>.<br />
<br />
Case 1: <math>{\rm Re} \ y = 0</math>.<br />
<br />
Because <math>|y| = 1</math>, <math>y = \pm i</math>.<br />
<br />
However, these solutions fail to satisfy Equation (1).<br />
<br />
Therefore, there is no solution in this case.<br />
<br />
Case 2: <math>{\rm Im} \ y = 0</math>.<br />
<br />
Because <math>|y| = 1</math>, <math>y = \pm 1</math>.<br />
<br />
However, these solutions fail to satisfy Equation (1).<br />
<br />
Therefore, there is no solution in this case.<br />
<br />
Case 3: <math>2 {\rm Re} \ y - 1 = 0</math>.<br />
<br />
Because <math>|y| = 1</math>, <math>y = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = e^{i \pm \frac{\pi}{3}}</math>.<br />
<br />
However, these solutions fail to satisfy Equation (1).<br />
<br />
Therefore, there is no solution in this case.<br />
<br />
All cases above imply that there is no solution in this problem.<br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(A) }0}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=vhAc0P09czI</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_22&diff=1670472021 Fall AMC 12B Problems/Problem 222021-11-29T23:50:13Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
Right triangle <math>ABC</math> has side lengths <math>BC=6</math>, <math>AC=8</math>, and <math>AB=10</math>. <br />
<br />
A circle centered at <math>O</math> is tangent to line <math>BC</math> at <math>B</math> and passes through <math>A</math>. A circle centered at <math>P</math> is tangent to line <math>AC</math> at <math>A</math> and passes through <math>B</math>. What is <math>OP</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{23}{8} \qquad\textbf{(B)}\ \frac{29}{10} \qquad\textbf{(C)}\ \frac{35}{12} \qquad\textbf{(D)}\<br />
\frac{73}{25} \qquad\textbf{(E)}\ 3</math><br />
<br />
==Solution 1 (Analytic Geometry) ==<br />
<br />
In a Cartesian plane, let <math>C, B,</math> and <math>A</math> be <math>(0,0),(0,6),(8,0)</math> respectively.<br />
<br />
By analyzing the behaviors of the two circles, we set <math>O</math> to be <math>(a,6)</math> and <math>P</math> be <math>(8,b)</math>.<br />
<br />
Hence derive the two equations:<br />
<br />
<math>(x-a)^2+(y-6)^2=a^2</math><br />
<br />
<math>(x-8)^2+(y-b)^2=b^2</math><br />
<br />
<br />
Considering the coordinates of <math>A</math> and <math>B</math> for the two equations respectively, we get:<br />
<br />
<math>(8-a)^2+(0-6)^2=a^2</math><br />
<br />
<math>(0-8)^2+(6-b)^2=b^2</math><br />
<br />
Solve to get <math>a=\frac{25}{4}</math> and <math>b=\frac{25}{3}</math><br />
<br />
<br />
Through using the distance formula,<br />
<br />
<math>OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}= \boxed{\textbf{(C)}\ \frac{35}{12}}</math>.<br />
<br />
<br />
~Wilhelm Z<br />
<br />
== Solution 2 ==<br />
This solution is based on this figure: [[:Image:2021_AMC_12B_(Nov)_Problem_22,_sol.png]].<br />
<br />
Because the circle with center <math>O</math> passes through points <math>A</math> and <math>B</math> and is tangent to line <math>BC</math> at point <math>B</math>, <math>O</math> is on the perpendicular bisector of segment <math>AB</math> and <math>OB \perp BC</math>.<br />
<br />
Because the circle with center <math>P</math> passes through points <math>A</math> and <math>B</math> and is tangent to line <math>AC</math> at point <math>A</math>, <math>P</math> is on the perpendicular bisector of segment <math>AB</math> and <math>OA \perp AC</math>.<br />
<br />
Let lines <math>OB</math> and <math>AP</math> intersect at point <math>D</math>.<br />
Hence, <math>ACBD</math> is a rectangle.<br />
<br />
Denote by <math>M</math> the midpoint of segment <math>AB</math>. Hence, <math>BM = \frac{AB}{2} = 5</math>.<br />
Because <math>O</math> and <math>P</math> are on the perpendicular bisector of segment <math>AB</math>, points <math>M</math>, <math>O</math>, <math>P</math> are collinear with <math>OM \perp AB</math>.<br />
<br />
We have <math>\triangle MOB \sim \triangle CBA</math>.<br />
Hence, <math>\frac{BO}{AB} = \frac{BM}{AC}</math>.<br />
Hence, <math>BO = \frac{25}{4}</math>.<br />
Hence, <math>OD = BD - BO = \frac{7}{4}</math>.<br />
<br />
We have <math>\triangle DOP \sim \triangle CBA</math>.<br />
Hence, <math>\frac{OP}{BA} = \frac{DO}{CB}</math>.<br />
Therefore, <math>OP = \frac{35}{12}</math>.<br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(C) }\frac{35}{12}}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=ctx67nltpE0<br />
<br />
{{AMC12 box|year=2021 Fall|ab=B|num-a=23|num-b=21}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_25&diff=1670462021 Fall AMC 12B Problems/Problem 252021-11-29T23:49:34Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
For <math>n</math> a positive integer, let <math>R(n)</math> be the sum of the remainders when <math>n</math> is divided by <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>9</math>, and <math>10</math>. For example, <math>R(15) = 1+0+3+0+3+1+7+6+5=26</math>. How many two-digit positive integers <math>n</math> satisfy <math>R(n) = R(n+1)\,?</math><br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4</math><br />
<br />
==Solution 1==<br />
<br />
Note that we can add <math>9</math> to <math>R(n)</math> to get <math>R(n+1)</math>, but must subtract <math>k</math> for all <math>k|n+1</math>. Hence, we see that there are four ways to do that because <math>9=7+2=6+3=5+4=4+3+2</math>. Note that only <math>7+2</math> is a plausible option, since <math>4+3+2</math> indicates <math>n+1</math> is divisible by <math>6</math>, <math>5+4</math> indicates that <math>n+1</math> is divisible by <math>2</math>, <math>6+3</math> indicates <math>n+1</math> is divisibly by <math>2</math>, and <math>9</math> itself indicates divisibility by <math>3</math>, too. So, <math>14|n+1</math> and <math>n+1</math> is not divisibly by any positive integers from <math>2</math> to <math>10</math>, inclusive, except <math>2</math> and <math>7</math>. We check and get that only <math>n+1=14 \cdot 1</math> and <math>n+1=14 \cdot 7</math> give possible solutions so our answer is <math>\boxed{\textbf{(C) }2}</math>.<br />
<br />
- kevinmathz<br />
<br />
== Solution 2 ==<br />
Denote by <math>{\rm Rem} \ \left( n, k \right)</math> the remainder of <math>n</math> divided by <math>k</math>.<br />
Define <math>\Delta \left( n, k \right) = {\rm Rem} \ \left( n + 1, k \right) - {\rm Rem} \ \left( n, k \right)</math>.<br />
<br />
Hence,<br />
<cmath><br />
\[<br />
\Delta \left( n, k \right)<br />
= \left\{<br />
\begin{array}{ll}<br />
1 & \mbox{ if } n \not\equiv -1 \pmod{k} \\<br />
- \left( k -1 \right) & \mbox{ if } n \equiv -1 \pmod{k}<br />
\end{array}<br />
\right..<br />
\]<br />
</cmath><br />
<br />
Hence, this problem asks us to find all <math>n \in \left\{ 10 , 11, \cdots , 99 \right\}</math>, such that <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>.<br />
<br />
<math>\textbf{Case 1}</math>: <math>\Delta \left( n, 10 \right) = - 9</math>.<br />
<br />
We have <math>\sum_{k = 2}^9 \Delta \left( n, k \right) \leq \sum_{k = 2}^9 1 = 8</math>.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
<math>\textbf{Case 2}</math>: <math>\Delta \left( n, 10 \right) = 1</math> and <math>\Delta \left( n, 9 \right) = -8</math>.<br />
<br />
The condition <math>\Delta \left( n, 9 \right) = -8</math> implies <math>n \equiv - 1 \pmod{9}</math>.<br />
This further implies <math>n \equiv - 1 \pmod{3}</math>.<br />
Hence, <math>\Delta \left( n, 3 \right) = -2</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) = 9</math>.<br />
<br />
However, we have <math>\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} 1 = 6</math>.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
<math>\textbf{Case 3}</math>: <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{ 9 , 10 \right\}</math> and <math>\Delta \left( n, 8 \right) = -7</math>.<br />
<br />
The condition <math>\Delta \left( n, 8 \right) = -7</math> implies <math>n \equiv - 1 \pmod{k}</math> with <math>k \in \left\{ 2, 4 \right\}</math>.<br />
Hence, <math>\Delta \left( n, 2 \right) = -1</math> and <math>\Delta \left( n, 4 \right) = -3</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) = 9</math>.<br />
<br />
However, we have <math>\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 3, 5, 6, 7 \right\}} 1 = 4</math>.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
<math>\textbf{Case 4}</math>: <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{ 8, \cdots , 10 \right\}</math> and <math>\Delta \left( n, 7 \right) = -6</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\sum_{k \in \left\{ 2, 3, 4, 5, 6 \right\}} \Delta \left( n, k \right) = 3</math>.<br />
<br />
Hence, we must have <math>\Delta \left( n, 2 \right) = -1</math> and <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{ 3 , 4 , 5 , 6 \right\}</math>.<br />
<br />
Therefore, <math>n = 13, 97</math>.<br />
<br />
<math>\textbf{Case 5}</math>: <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{ 7 , \cdots , 10 \right\}</math> and <math>\Delta \left( n, 6 \right) = -5</math>.<br />
<br />
The condition <math>\Delta \left( n, 6 \right) = -5</math> implies <math>n \equiv - 1 \pmod{k}</math> with <math>k \in \left\{ 2, 3 \right\}</math>.<br />
Hence, <math>\Delta \left( n, 2 \right) = -1</math> and <math>\Delta \left( n, 3 \right) = -2</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) = 4</math>.<br />
<br />
However, we have <math>\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 4, 5 \right\}} 1 = 2</math>.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
<br />
<math>\textbf{Case 6}</math>: <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{ 6 , \cdots , 10 \right\}</math> and <math>\Delta \left( n, 5 \right) = -4</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\sum_{k \in \left\{ 2, 3, 4 \right\}} \Delta \left( n, k \right) = -1</math>.<br />
<br />
This can be achieved if <math>\Delta \left( n, 2 \right) = 1</math>, <math>\Delta \left( n, 3 \right) = 1</math>, <math>\Delta \left( n, 4 \right) = -3</math>.<br />
<br />
However, <math>\Delta \left( n, 4 \right) = -3</math> implies <math>n \equiv - 1 \pmod{4}</math>. This implies <math>n \equiv -1 \pmod{2}</math>. Hence, <math>\Delta \left( n, 2 \right) = -1</math>.<br />
We get a contradiction.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
<math>\textbf{Case 7}</math>: <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{ 5 , \cdots , 10 \right\}</math> and <math>\Delta \left( n, 4 \right) = -3</math>.<br />
<br />
The condition <math>\Delta \left( n, 4 \right) = -3</math> implies <math>n \equiv - 1 \pmod{k}</math> with <math>k = 2</math>.<br />
Hence, <math>\Delta \left( n, 2 \right) = -1</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\Delta \left( n, 3 \right) = - 2</math>. This implies <math>n \equiv - 1 \pmod{3}</math>.<br />
<br />
Because <math>n \equiv - 1 \pmod{2}</math> and <math>n \equiv - 1 \pmod{3}</math>, we have <math>n \equiv - 1 \pmod{6}</math>.<br />
Hence, <math>\Delta \left( n, 6 \right) = - 5</math>.<br />
However, in this case, we assume <math>\Delta \left( n, 6 \right) = 1</math>.<br />
We get a contradiction.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
<math>\textbf{Case 8}</math>: <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{ 4 , \cdots , 10 \right\}</math> and <math>\Delta \left( n, 3 \right) = -2</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\Delta \left( n, 2 \right) = - 5</math>. This is infeasible.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
<math>\textbf{Case 9}</math>: <math>\Delta \left( n, k \right) = 1</math> for <math>k \in \left\{3 , \cdots , 10 \right\}</math>.<br />
<br />
To get <math>\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0</math>, we have <math>\Delta \left( n, 2 \right) = - 8</math>. This is infeasible.<br />
<br />
Therefore, there is no <math>n</math> in this case.<br />
<br />
Putting all cases together, the answer is <math>\boxed{\textbf{(C) }2}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=Fy8wU4VAzkQ<br />
<br />
{{AMC12 box|year=2021 Fall|ab=B|num-b=24|after=Last problem}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_19&diff=1670452021 Fall AMC 12A Problems/Problem 192021-11-29T23:48:31Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Let <math>x</math> be the least real number greater than <math>1</math> such that <math>\sin(x)= \sin(x^2)</math>, where the arguments are in degrees. What is <math>x</math> rounded up to the closest integer?<br />
<br />
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20</math><br />
<br />
==Solution 1==<br />
The smallest <math>x</math> to make <math>\sin(x) = \sin(x^2)</math> would require <math>x=x^2</math>, but since <math>x</math> needs to be greater than <math>1</math>, these solutions are not valid.<br />
<br />
The next smallest <math>x</math> would require <math>x=180-x^2</math>, or <math>x^2+x=180</math>. <br />
<br />
After a bit of guessing and checking, we find that <math>12^2+12=156</math>, and <math>13^2+13=182</math>, so the solution lies between <math>12{ }</math> and <math>13</math>, making our answer <math>\boxed{\textbf{(B) } 13}.</math><br />
<br />
Note: One can also solve the quadratic and estimate the radical.<br />
<br />
~kingofpineapplz<br />
<br />
== Solution 2 ==<br />
For choice <math>\textbf{(A)},</math> we have<br />
<cmath><br />
\begin{align*}<br />
\left| \sin x - \sin \left( x^2 \right) \right|<br />
& = \left| \sin 10^\circ - \sin \left( \left( 10^2 \right)^\circ \right) \right| \\<br />
& = \left| \sin 10^\circ - \sin 100^\circ \right| \\<br />
& = \left| \sin 10^\circ - \sin 80^\circ \right| \\<br />
& = \sin 80^\circ - \sin 10^\circ .<br />
\end{align*}<br />
</cmath><br />
For choice <math>\textbf{(B)},</math> we have<br />
<cmath><br />
\begin{align*}<br />
\left| \sin x - \sin \left( x^2 \right) \right|<br />
& = \left| \sin 13^\circ - \sin \left( \left( 13^2 \right)^\circ \right) \right| \\<br />
& = \left| \sin 13^\circ - \sin 169^\circ \right| \\<br />
& = \left| \sin 10^\circ - \sin 11^\circ \right| \\<br />
& = \sin 11^\circ - \sin 10^\circ .<br />
\end{align*}<br />
</cmath><br />
For choice <math>\textbf{(C)},</math> we have<br />
<cmath><br />
\begin{align*}<br />
\left| \sin x - \sin \left( x^2 \right) \right|<br />
& = \left| \sin 14^\circ - \sin \left( \left( 14^2 \right)^\circ \right) \right| \\<br />
& = \left| \sin 14^\circ - \sin 196^\circ \right| \\<br />
& = \left| \sin 14^\circ + \sin 16^\circ \right| \\<br />
& = \sin 14^\circ + \sin 16^\circ .<br />
\end{align*}<br />
</cmath><br />
For choice <math>\textbf{(D)},</math> we have<br />
<cmath><br />
\begin{align*}<br />
\left| \sin x - \sin \left( x^2 \right) \right|<br />
& = \left| \sin 19^\circ - \sin \left( \left( 19^2 \right)^\circ \right) \right| \\<br />
& = \left| \sin 19^\circ - \sin 361^\circ \right| \\<br />
& = \left| \sin 19^\circ - \sin 1^\circ \right| \\<br />
& = \sin 19^\circ - \sin 1^\circ .<br />
\end{align*}<br />
</cmath><br />
For choice <math>\textbf{(E)},</math> we have<br />
<cmath><br />
\begin{align*}<br />
\left| \sin x - \sin \left( x^2 \right) \right|<br />
& = \left| \sin 20^\circ - \sin \left( \left( 20^2 \right)^\circ \right) \right| \\<br />
& = \left| \sin 20^\circ - \sin 400^\circ \right| \\<br />
& = \left| \sin 20^\circ - \sin 40^\circ \right| \\<br />
& = \sin 40^\circ - \sin 20^\circ .<br />
\end{align*}<br />
</cmath><br />
Therefore, the answer is <math>\boxed{\textbf{(B) }13},</math> as <math>\sin 11^\circ - \sin 10^\circ</math> is the closest to <math>0.</math><br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=H0pNJFbV4jE<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021 Fall|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_21&diff=1670442021 Fall AMC 12A Problems/Problem 212021-11-29T23:47:54Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCD</math> be an isosceles trapezoid with <math>\overline{BC}\parallel \overline{AD}</math> and <math>AB=CD</math>. Points <math>X</math> and <math>Y</math> lie on diagonal <math>\overline{AC}</math> with <math>X</math> between <math>A</math> and <math>Y</math>, as shown in the figure. Suppose <math>\angle AXD = \angle BYC = 90^\circ</math>, <math>AX = 3</math>, <math>XY = 1</math>, and <math>YC = 2</math>. What is the area of <math>ABCD?</math><br />
<br />
<asy><br />
size(10cm);<br />
usepackage("mathptmx");<br />
import geometry;<br />
void perp(picture pic=currentpicture,<br />
pair O, pair M, pair B, real size=5,<br />
pen p=currentpen, filltype filltype = NoFill){<br />
perpendicularmark(pic, M,unit(unit(O-M)+unit(B-M)),size,p,filltype);<br />
}<br />
pen p=black+linewidth(1),q=black+linewidth(5);<br />
pair C=(0,0),Y=(2,0),X=(3,0),A=(6,0),B=(2,sqrt(5.6)),D=(3,-sqrt(12.6));<br />
draw(A--B--C--D--cycle,p);<br />
draw(A--C,p);<br />
draw(B--Y,p);<br />
draw(D--X,p);<br />
dot(A,q);<br />
dot(B,q);<br />
dot(C,q);<br />
dot(D,q);<br />
dot(X,q);<br />
dot(Y,q);<br />
label("2",C--Y,S);<br />
label("1",Y--X,S);<br />
label("3",X--A,S);<br />
label("$A$",A,2*E);<br />
label("$B$",B,2*N);<br />
label("$C$",C,2*W);<br />
label("$D$",D,2*S);<br />
label("$Y$",Y,2*sqrt(2)*NE);<br />
label("$X$",X,2*N);<br />
perp(B,Y,C,8,p);<br />
perp(A,X,D,8,p);<br />
</asy><br />
<math>\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}</math><br />
<br />
== Solution 1 ==<br />
First realize that <math>\triangle BCY \sim \triangle DAX.</math> Thus, because <math>CY: XA = 2:3,</math> we can say that <math>BY = 2s</math> and <math>DX = 3s.</math> From the Pythagorean Theorem, we have <math>AB =(2s)^2 + 4^2 = 4s^2 + 16</math> and <math>CD = (3s)^2 + 3^2 = 9s^2 + 9.</math> Because <math>AB = CD,</math> from the problem statement, we have that <cmath>4s^2 + 16 = 9s^2 + 9.</cmath> Solving, gives <math>s = \frac{\sqrt{7}}{\sqrt{5}}.</math> To find the area of the trapezoid, we can compute the area of <math>\triangle ABC</math> and add it to the area of <math>\triangle ACD.</math> Thus, the area of the trapezoid is <math>\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \left(\frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 \right) = \frac{15\sqrt{7}}{\sqrt{5}} = 3\sqrt{35}.</math> Thus, the answer is <math>\boxed{\textbf {(C)} \: 3\sqrt{35}}.</math><br />
<br />
~NH14<br />
<br />
== Solution 2 ==<br />
We put the graph to a coordinate plane. We put <math>X</math> to the origin, <math>AC</math> to the <math>x</math>-axis, and <math>DX</math> to the <math>y</math>-axis.<br />
<br />
We have the coordinates of the following points: <math>X = \left( 0 , 0 \right)</math>, <math>A = \left( 3 , 0 \right)</math>, <math>Y = \left( - 1 , 0 \right)</math>, <math>C = \left( - 3 , 0 \right)</math>.<br />
<br />
Denote <math>BY = u</math>, <math>DX = v</math>. Hence, <math>B = \left( - 1 , u \right)</math>, <math>D = \left( 0 , - v \right)</math>.<br />
<br />
Hence, the slope of <math>BC</math> is <math>m_{BC} = \frac{u}{2}</math>.<br />
The slope of <math>AD</math> is <math>m_{AD} = \frac{v}{3}</math>.<br />
<br />
Because <math>BC \parallel AD</math>, <math>m_{BC} = m_{AD}</math>. Hence,<br />
<cmath><br />
\[<br />
\frac{u}{2} = \frac{v}{3} . \hspace{1cm} (1)<br />
\]<br />
</cmath><br />
<br />
In <math>\triangle AYB</math>, following from the Pythagorean theorem, we have <math>AB^2 = AY^2 + BY^2 = 4^2 + u^2</math>.<br />
<br />
In <math>\triangle CXD</math>, following from the Pythagorean theorem, we have <math>CD^2 = CX^2 + XD^2 = 3^2 + v^2</math>.<br />
<br />
Because <math>AB = CD</math>,<br />
<cmath><br />
\[<br />
4^2 + u^2 = 3^2 + v^2 . \hspace{1cm} (2)<br />
\]<br />
</cmath><br />
<br />
Solving Equations (1) and (2), we get <math>u = \frac{2 \sqrt{35}}{5}</math>, <math>V = \frac{3 \sqrt{35}}{5}</math>.<br />
<br />
Therefore,<br />
<cmath><br />
\begin{align*}<br />
{\rm Area} \ ABCD<br />
& = {\rm Area} \ \triangle ABC + {\rm Area} \ \triangle ADC \\<br />
& = \frac{1}{2} AC \cdot BY + \frac{1}{2} AC \cdot DX \\<br />
& = \frac{1}{2} AC \left( BY + DY \right) \\<br />
& = 3 \sqrt{35} .<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(C) }3 \sqrt{35}}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution by The Power of Logic==<br />
https://youtu.be/R1cyesL2t8A<br />
<br />
~math2718281828459<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=TCeYkekkjrU<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2021 Fall|ab=A|num-b=20|num-a=22}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_22&diff=1670432021 Fall AMC 12A Problems/Problem 222021-11-29T23:47:04Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Azar and Carl play a game of tic-tac-toe. Azar places an in <math>X</math> one of the boxes in a <math>3</math>-by-<math>3</math> array of boxes, then Carl places an <math>O</math> in one of the remaining boxes. After that, Azar places an <math>X</math> in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third <math>O</math>. How many ways can the board look after the game is over?<br />
<br />
<math>\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160</math><br />
<br />
== Solution ==<br />
We need to find out the number of configurations with 3 <math>O</math> and 3 <math>X</math> with 3 <math>O</math> in a row, and 3 <math>X</math> not in a row.<br />
<br />
<math>\textbf{Case 1}</math>: 3 <math>O</math> are in a horizontal row or a vertical row.<br />
<br />
Step 1: We determine the row that 3 <math>O</math> occupy.<br />
<br />
The number of ways is 6.<br />
<br />
Step 2: We determine the configuration of 3 <math>X</math>.<br />
<br />
The number of ways is <math>\binom{6}{3} - 2 = 18</math>.<br />
<br />
In this case, following from the rule of product, the number of ways is <math>6 \cdot 18 = 108</math>.<br />
<br />
<math>\textbf{Case 2}</math>: 3 <math>O</math> are in a diagonal row.<br />
<br />
Step 1: We determine the row that 3 <math>O</math> occupy.<br />
<br />
The number of ways is 2.<br />
<br />
Step 2: We determine the configuration of 3 <math>X</math>.<br />
<br />
The number of ways is <math>\binom{6}{3} = 20</math>.<br />
<br />
In this case, following from the rule of product, the number of ways is <math>2 \cdot 20 = 40</math>.<br />
<br />
Putting all cases together, the total number of ways is <math>108 + 40 = 148</math>.<br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(D) }148}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=OpRk-iposj8<br />
<br />
{{AMC12 box|year=2021 Fall|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_23&diff=1654012021 Fall AMC 10A Problems/Problem 232021-11-23T14:00:45Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
<br />
For each positive integer <math>n</math>, let <math>f_1(n)</math> be twice the number of positive integer divisors of <math>n</math>, and for <math>j \ge 2</math>, let <math>f_j(n) = f_1(f_{j-1}(n))</math>. For how many values of <math>n \le 50</math> is <math>f_{50}(n) = 12?</math><br />
<br />
<math>\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11</math><br />
<br />
==Solution==<br />
First, we can test values that would make <math>f(x)=12</math> true. For this to happen <math>x</math> must have <math>6</math> divisors, which means its prime factorization is in the form <math>pq^2</math> or <math>p^5</math>, where <math>p</math> and <math>q</math> are prime numbers. Listing out values less than <math>50</math> which have these prime factorizations, we find <math>12,20,28,44,18,45,50</math> for <math>pq^2</math>, and just <math>32</math> for <math>p^5</math>. Here <math>12</math> especially catches our eyes, as this means if one of <math>f_i(n)=12</math>, each of <math>f_{i+1}(n), f_{i+2}(n), ...</math> will all be <math>12</math>. This is because <math>f_{i+1}(n)=f(f_i(n))</math> (as given in the problem statement), so were <math>f_i(n)=12</math>, plugging this in we get <math>f_{i+1}(n)=f(12)=12</math>, and thus the pattern repeats. Hence, as long as for a <math>i</math>, such that <math>i\leq 50</math> and <math>f_{i}(n)=12</math>, <math>f_{50}(n)=12</math> must be true, which also immediately makes all our previously listed numbers, where <math>f(x)=12</math>, possible values of <math>n</math>.<br />
<br />
<br />
We also know that if <math>f(x)</math> were to be any of these numbers, <math>x</math> would satisfy <math>f_{50}(n)</math> as well. Looking through each of the possibilities aside from <math>12</math>, we see that <math>f(x)</math> could only possibly be equal to <math>20</math> and <math>18</math>, and still have <math>x</math> less than or equal to <math>50</math>. This would mean <math>x</math> must have <math>10</math>, or <math>9</math> divisors, and testing out, we see that <math>x</math> will then be of the form <math>p^4q</math>, or <math>p^2q^2</math>. The only two values less than or equal to <math>50</math> would be <math>48</math> and <math>36</math> respectively. From here there are no more possible values, so tallying our possibilities we count<math>\boxed{D:10}</math> values (Namely <math>12,20,28,44,18,45,50,32,36,48</math>).<br />
<br />
~Ericsz<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=WQQVjCdoqWI</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_23&diff=1654002021 Fall AMC 10A Problems/Problem 232021-11-23T14:00:39Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
<br />
For each positive integer <math>n</math>, let <math>f_1(n)</math> be twice the number of positive integer divisors of <math>n</math>, and for <math>j \ge 2</math>, let <math>f_j(n) = f_1(f_{j-1}(n))</math>. For how many values of <math>n \le 50</math> is <math>f_{50}(n) = 12?</math><br />
<br />
<math>\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11</math><br />
<br />
==Solution==<br />
First, we can test values that would make <math>f(x)=12</math> true. For this to happen <math>x</math> must have <math>6</math> divisors, which means its prime factorization is in the form <math>pq^2</math> or <math>p^5</math>, where <math>p</math> and <math>q</math> are prime numbers. Listing out values less than <math>50</math> which have these prime factorizations, we find <math>12,20,28,44,18,45,50</math> for <math>pq^2</math>, and just <math>32</math> for <math>p^5</math>. Here <math>12</math> especially catches our eyes, as this means if one of <math>f_i(n)=12</math>, each of <math>f_{i+1}(n), f_{i+2}(n), ...</math> will all be <math>12</math>. This is because <math>f_{i+1}(n)=f(f_i(n))</math> (as given in the problem statement), so were <math>f_i(n)=12</math>, plugging this in we get <math>f_{i+1}(n)=f(12)=12</math>, and thus the pattern repeats. Hence, as long as for a <math>i</math>, such that <math>i\leq 50</math> and <math>f_{i}(n)=12</math>, <math>f_{50}(n)=12</math> must be true, which also immediately makes all our previously listed numbers, where <math>f(x)=12</math>, possible values of <math>n</math>.<br />
<br />
<br />
We also know that if <math>f(x)</math> were to be any of these numbers, <math>x</math> would satisfy <math>f_{50}(n)</math> as well. Looking through each of the possibilities aside from <math>12</math>, we see that <math>f(x)</math> could only possibly be equal to <math>20</math> and <math>18</math>, and still have <math>x</math> less than or equal to <math>50</math>. This would mean <math>x</math> must have <math>10</math>, or <math>9</math> divisors, and testing out, we see that <math>x</math> will then be of the form <math>p^4q</math>, or <math>p^2q^2</math>. The only two values less than or equal to <math>50</math> would be <math>48</math> and <math>36</math> respectively. From here there are no more possible values, so tallying our possibilities we count<math>\boxed{D:10}</math> values (Namely <math>12,20,28,44,18,45,50,32,36,48</math>).<br />
<br />
~Ericsz<br />
<br />
==Video Solution by Mathematical Dexterity==</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_20&diff=1653992021 Fall AMC 10A Problems/Problem 202021-11-23T13:59:58Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
<br />
How many ordered pairs of positive integers <math>(b,c)</math> exist where both <math>x^2+bx+c=0</math> and <math>x^2+cx+b=0</math> do not have distinct, real solutions?<br />
<br />
<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad</math><br />
<br />
== Solution 1 (Casework) ==<br />
A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:<br />
<ol style="margin-left: 1.5em;"><br />
<li>Since <math>x^2+bx+c=0</math> does not have real solutions, we have <math>b^2\leq 4c.</math></li><p><br />
<li>Since <math>x^2+cx+b=0</math> does not have real solutions, we have <math>c^2\leq 4b.</math></li><p><br />
</ol><br />
Squaring the first inequality, we get <math>b^4\leq 16c^2.</math> Multiplying the second inequality by <math>16,</math> we get <math>16c^2\leq 64b.</math> Combining these results, we get <cmath>b^4\leq 16c^2\leq 64b.</cmath><br />
We apply casework to the value of <math>b:</math><br />
<br />
* If <math>b=1,</math> then <math>1\leq 16c^2\leq 64,</math> from which <math>c=1,2.</math><br />
<br />
* If <math>b=2,</math> then <math>16\leq 16c^2\leq 128,</math> from which <math>c=1,2.</math><br />
<br />
* If <math>b=3,</math> then <math>81\leq 16c^2\leq 192,</math> from which <math>c=3.</math><br />
<br />
* If <math>b=4,</math> then <math>256\leq 16c^2\leq 256,</math> from which <math>c=4.</math><br />
<br />
Together, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs <math>(b,c),</math> namely <math>(1,1),(1,2),(2,1),(2,2),(3,3),</math> and <math>(4,4).</math><br />
<br />
~MRENTHUSIASM<br />
<br />
== Solution 2 (Graphing) ==<br />
Similar to Solution 1, use the discriminant to get <math>b^2\leq 4c</math> and <math>c^2\leq 4b</math>. These can be rearranged to <math>c\geq \frac{1}{4}b^2</math> and <math>b\geq \frac{1}{4}c^2</math>. Now, we can roughly graph these two inequalities, letting one of them be the <math>x</math> axis and the other be <math>y</math>. The graph of solutions should look like this:<br />
<asy><br />
unitsize(2);<br />
Label f; <br />
f.p=fontsize(6); <br />
xaxis("$x$",0,5,Ticks(f, 1.0)); <br />
yaxis("$y$",0,5,Ticks(f, 1.0)); <br />
real f(real x) <br />
{ <br />
return 0.25x^2; <br />
} <br />
real g(real x) <br />
{ <br />
return 2*sqrt(x); <br />
} <br />
dot((1,1));<br />
dot((2,1));<br />
dot((1,2));<br />
dot((2,2));<br />
dot((3,3));<br />
dot((4,4));<br />
draw(graph(f,0,sqrt(20)));<br />
draw(graph(g,0,5));<br />
</asy><br />
We are looking for lattice points (since <math>b</math> and <math>c</math> are positive integers), of which we can count <math>\boxed{\textbf{(B) } 6}</math>.<br />
<br />
~aop2014<br />
<br />
==Solution 3 (Oversimplified but Risky)==<br />
A quadratic equation <math>Ax^2+Bx+C=0</math> has one real solution if and only if <math>\sqrt{B^2-4AC}=0.</math> Similarly, it has imaginary solutions if and only if <math>\sqrt{B^2-4AC}<0.</math> We proceed as following:<br />
<br />
We want both <math>x^2+bx+c</math> to be <math>1</math> value or imaginary and <math>x^2+cx+b</math> to be <math>1</math> value or imaginary. <math>x^2+4x+4</math> is one such case since <math>\sqrt {b^2-4ac}</math> is <math>0.</math> Also, <math>x^2+3x+3, x^2+2x+2, x^2+x+1</math> are always imaginary for both <math>b</math> and <math>c.</math> We also have <math>x^2+x+2</math> along with <math>x^2+2x+1</math> since the latter has one solution, while the first one is imaginary. Therefore, we have <math>6</math> total ordered pairs of integers, which is <math>\boxed{\textbf{(B) } 6}.</math><br />
<br />
~Arcticturn<br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=EkaKfkQgFbI<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=A|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_21&diff=1653982021 Fall AMC 10A Problems/Problem 212021-11-23T13:59:22Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Each of the <math>20</math> balls is tossed independently and at random into one of the <math>5</math> bins. Let <math>p</math> be the probability that some bin ends up with <math>3</math> balls, another with <math>5</math> balls, and the other three with <math>4</math> balls each. Let <math>q</math> be the probability that every bin ends up with <math>4</math> balls. What is <math>\frac{p}{q}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\<br />
12 \qquad\textbf{(E)}\ 16</math><br />
<br />
==Solution 1 (Multinomial Coefficients)==<br />
For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable.<br />
<br />
Let <math>d</math> be the number of ways to distribute <math>20</math> balls into <math>5</math> bins. We have<br />
<cmath>p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.</cmath> Therefore, the answer is <cmath>\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.</cmath><br />
<br />
<u><b>Remark</b></u><br />
<br />
By the stars and bars argument, we get <math>d=\binom{20+5-1}{5-1}=\binom{24}{4}.</math> <br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 2 (Simple) ==<br />
Since both of the boxes will have <math>3</math> boxes with <math>4</math> balls in them, we can leave those out. There are <math>\binom {6}{3}</math> = <math>20</math> ways to choose where to place the <math>3</math> and the <math>5</math>. After that, there are <math>\binom {8}{3} = 56</math> ways to put the <math>3</math> and <math>5</math> balls being put into the boxes. For the <math>4,4,4,4,4</math> case, after we canceled the <math>4,4,4</math> out, we have <math>\binom {8}{4}</math> = <math>70</math> ways to put the <math>4</math> balls inside the boxes. Therefore, we have <math>\frac {56\cdot 20}{70}</math> which is equal to <math>8 \cdot 2</math> = <math>\boxed {(E)16}</math><br />
<br />
~Arcticturn<br />
<br />
==Solution 3 (30-second set-theoretic solution) ==<br />
Construct the set <math>A</math> consisting of all possible <math>3-5-4-4-4</math> bin configurations, and construct set <math>B</math> consisting of all possible <math>4-4-4-4-4</math> configurations. If we let <math>N</math> be the total number of configurations possible, it's clear we want to solve for <math>\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}</math>. <br />
<br />
Consider drawing an edge between an element in <math>A</math> and an element in <math>B</math> if it is possible to reach one configuration from the other by moving a single ball (note this process is reversible). Let us consider the total number of edges drawn.<br />
<br />
From any element in <math>A</math>, we may take one of the <math>5</math> balls in the 5-bin and move it to the 3-bin to get a valid element in <math>B</math>. This implies the number of edges is <math>5|A|</math>. <br />
<br />
On the other hand for any element in <math>B</math>, we may choose one of the <math>20</math> balls and move it to one of the other <math>4</math> bins to get a valid element in <math>A</math>. This implies the number of edges is <math>80|B|</math>.<br />
<br />
Since they must be equal, then <math>5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed {(E)16}</math><br />
<br />
==Video Solution by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=Lu6eSvY6RHE<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://YouTube.com/watch?v=bvd2VjMxiZ4<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=A|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_15&diff=1601692021 AIME II Problems/Problem 152021-08-13T16:08:01Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Let <math>f(n)</math> and <math>g(n)</math> be functions satisfying<br />
<cmath>f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\<br />
1 + f(n+1) & \text{ otherwise}<br />
\end{cases}</cmath>and<br />
<cmath>g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\<br />
2 + g(n+2) & \text{ otherwise}<br />
\end{cases}</cmath>for positive integers <math>n</math>. Find the least positive integer <math>n</math> such that <math>\tfrac{f(n)}{g(n)} = \tfrac{4}{7}</math>.<br />
<br />
==Solution 1==<br />
<br />
Consider what happens when we try to calculate <math>f(n)</math> where n is not a square. If <math>k^2<n<(k+1)^2</math> for (positive) integer k, recursively calculating the value of the function gives us <math>f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n</math>. Note that this formula also returns the correct value when <math>n=(k+1)^2</math>, but not when <math>n=k^2</math>. Thus <math>f(n)=k^2+3k+2-n</math> for <math>k^2<n \leq (k+1)^2</math>.<br />
<br />
If <math>2 \mid (k+1)^2-n</math>, <math>g(n)</math> returns the same value as <math>f(n)</math>. This is because the recursion once again stops at <math>(k+1)^2</math>. We seek a case in which <math>f(n)<g(n)</math>, so obviously this is not what we want. We want <math>(k+1)^2,n</math> to have a different parity, or <math>n, k</math> have the same parity. When this is the case, <math>g(n)</math> instead returns <math>(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n</math>.<br />
<br />
Write <math>7f(n)=4g(n)</math>, which simplifies to <math>3k^2+k-10=3n</math>. Notice that we want the <math>LHS</math> expression to be divisible by 3; as a result, <math>k \equiv 1 \pmod{3}</math>. We also want n to be strictly greater than <math>k^2</math>, so <math>k-10>0, k>10</math>. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is <math>k=16</math>, giving <math>n=258</math>.<br />
<br />
Indeed - if we check our answer, it works. Therefore, the answer is <math>\boxed{258}</math>.<br />
<br />
-Ross Gao<br />
<br />
==Solution 2 (Four Variables)==<br />
We consider <math>f(n)</math> and <math>g(n)</math> separately:<br />
<ul style="list-style-type:square;"><br />
<li><math>\boldsymbol{f(n)}</math><p><br />
We restrict <math>n</math> in which <math>k^2<n\leq(k+1)^2</math> for some positive integer <math>k,</math> or <cmath>n=(k+1)^2-p\hspace{40mm}(1)</cmath> for some integer <math>p</math> such that <math>0\leq p<2k+1.</math> By recursion, we get<br />
<cmath>\begin{align*}<br />
f\left((k+1)^2\right)&=k+1, \\<br />
f\left((k+1)^2-1\right)&=k+2, \\<br />
f\left((k+1)^2-2\right)&=k+3, \\<br />
&\cdots \\<br />
f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{19mm}(2) \\<br />
\end{align*}</cmath></li><p><br />
<li><math>\boldsymbol{g(n)}</math><p><br />
If <math>n</math> and <math>(k+1)^2</math> have the same parity, then we get <math>g(n)=f(n)</math> by a similar process from <math>g\left((k+1)^2\right)=k+1.</math> This contradicts the precondition <math>\frac{f(n)}{g(n)} = \frac{4}{7}.</math> Therefore, <math>n</math> and <math>(k+1)^2</math> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. <p><br />
It follows that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{38.25mm}(3)</cmath> for some integer <math>q</math> such that <math>0<q<2k+2.</math> By recursion, we get<br />
<cmath>\begin{align*}<br />
g\left((k+2)^2\right)&=k+2, \\<br />
g\left((k+2)^2-2\right)&=k+4, \\<br />
g\left((k+2)^2-4\right)&=k+6, \\<br />
&\cdots \\<br />
g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\<br />
\end{align*}</cmath></li><p><br />
<li><b>Answer</b><p><br />
By <math>(2)</math> and <math>(4),</math> we have <cmath>\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)</cmath><br />
From <math>(1)</math> and <math>(3),</math> equating the expressions for <math>n</math> gives <math>(k+1)^2-p=(k+2)^2-2q.</math> Solving for <math>k</math> produces <cmath>k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)</cmath><br />
We substitute <math>(6)</math> into <math>(5),</math> then simplify, cross-multiply, and rearrange:<br />
<cmath>\begin{align*}<br />
\frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\<br />
\frac{p+2q-1}{-p+6q+1}&=\frac{4}{7} \\<br />
7p+14q-7&=-4p+24q+4 \\<br />
11p-11&=10q \\<br />
11(p-1)&=10q. \hspace{29mm}(7)<br />
\end{align*}</cmath><br />
Since <math>\gcd(11,10)=1,</math> we know that <math>p-1</math> must be divisible by <math>10,</math> and <math>q</math> must be divisible by <math>11.</math> <p> Recall that the restrictions on <math>(1)</math> and <math>(2)</math> are <math>0\leq p<2k+1</math> and <math>0<q<2k+2,</math> respectively. Substituting <math>(6)</math> into either inequality gives <math>p+1<q.</math> Combining all these results produces <cmath>0<p+1<q<2k+2. \hspace{28mm}(8)</cmath><br />
<br />
To minimize <math>n</math> in either <math>(1)</math> or <math>(3),</math> we minimize <math>k,</math> so we minimize <math>p</math> and <math>q</math> in <math>(8).</math> From <math>(6)</math> and <math>(7),</math> we construct the following table:<br />
<cmath>\begin{array}{c|c|c|c}<br />
& & & \\ [-2.5ex]<br />
\boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{(8)?} \\ [0.5ex]<br />
\hline<br />
& & & \\ [-2ex]<br />
11 & 11 & 4 & \\<br />
21 & 22 & 10 & \\<br />
31 & 33 & 16 & \checkmark \\<br />
\geq41 & \geq44 & \geq22 & \checkmark \\<br />
\end{array}</cmath><br />
Finally, we have <math>(p,q,k)=(31,33,16).</math> Substituting this result into either <math>(1)</math> or <math>(3)</math> generates <math>n=\boxed{258}.</math></li><p><br />
<li><b>Remark</b><p><br />
We can verify that <cmath>\frac{f(258)}{g(258)}=\frac{1\cdot31+f(258+1\cdot31)}{2\cdot33+g(258+2\cdot33)}=\frac{31+\overbrace{f(289)}^{17}}{66+\underbrace{g(324)}_{18}}=\frac{48}{84}=\frac47.</cmath></li><p><br />
</ul><br />
~MRENTHUSIASM<br />
<br />
==Solution 3==<br />
Since <math>n</math> isn't a perfect square, let <math>n=m^2+k</math> with <math>0<k<2m+1</math>. If <math>m</math> is odd, then <math>f(n)=g(n)</math>. If <math>m</math> is even, then<br />
<cmath>\begin{align*}<br />
f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\<br />
g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k,<br />
\end{align*}</cmath><br />
from which<br />
<cmath>\begin{align*}<br />
7(3m+2-k)&=4(5m+6-k) \\<br />
m&=3k+10.<br />
\end{align*}</cmath><br />
Since <math>m</math> is even, <math>k</math> is even. Since <math>k\neq 0</math>, the smallest <math>k</math> is <math>2</math> which produces the smallest <math>n</math>: <cmath>k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.</cmath><br />
~Afo<br />
<br />
==Solution 4 (Quadratics With Two Variables)==<br />
To begin, note that if <math>n</math> is a perfect square, <math>f(n)=g(n)</math>, so <math>f(n)/g(n)=1</math>, so we must look at values of <math>n</math> that are not perfect squares (what a surprise). First, let the distance between <math>n</math> and the first perfect square greater than or equal to it be <math>k</math>, making the values of <math>f(n+k)</math> and <math>g(n+k)</math> integers. Using this notation, we see than <math>f(n)=k+f(n+k)</math>, giving us a formula for the numerator of our ratio. However, since the function of <math>g(n)</math> does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of <math>\sqrt{n+k}</math> in <math>g(n)</math> unless <math>k</math> is an even number. However, this is impossible, since if <math>k</math> was an even number, <math>f(n)=g(n)</math>, giving a ratio of one. Thus, <math>k</math> must be an odd number.<br />
<br />
Thus, since <math>k</math> must be an odd number, regardless of whether <math>n</math> is even or odd, to get an integral value in <math>g(n)</math>, we must get to the next perfect square after <math>n+k</math>. To make matters easier, let <math>z^2=n+k</math>. Thus, in <math>g(n)</math>, we want to achieve <math>(z+1)^2</math>. <br />
<br />
Expanding <math>(z+1)^2</math> and substituting in the fact that <math>z=\sqrt{n+k}</math> yields:<br />
<br />
<cmath>(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1</cmath><br />
<br />
Thus, we must add the quantity <math>k+2z+1</math> to <math>n</math> to achieve a integral value in the function <math>g(n)</math>. Thus.<br />
<br />
<cmath>g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}</cmath><br />
<br />
However, note that the quantity within the square root is just <math>(z+1)^2</math>, and so:<br />
<br />
<cmath>g(n)=k+3z+2</cmath><br />
<br />
Thus,<br />
<cmath>\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}</cmath><br />
<br />
Since we want this quantity to equal <math>\frac{4}{7}</math>, we can set the above equation equal to this number and collect all the variables to one side to achieve<br />
<br />
<cmath>3k-5z=8</cmath><br />
<br />
Substituting back in that <math>z=\sqrt{n+k}</math>, and then separating variables and squaring yields that<br />
<br />
<cmath>9k^2-73k+64=25n</cmath><br />
<br />
Now, if we treat <math>n</math> as a constant, we can use the quadratic formula in respect to <math>k</math> to get an equation for <math>k</math> in terms of <math>n</math> (without all the squares). Doing so yields<br />
<br />
<cmath>\frac{73\pm\sqrt{3025+900n}}{18}=k</cmath><br />
<br />
Now, since <math>n</math> and <math>k</math> are integers, we want the quantity within the square root to be a perfect square. Note that <math>55^2=3025</math>. Thus, assume that the quantity within the root is equal to the perfect square, <math>m^2</math>. Thus, after using a difference of squares, we have<br />
<cmath>(m-55)(m+55)=900n</cmath><br />
Since we want <math>n</math> to be an integer, we know that the <math>LHS</math> should be divisible by five, so, let's assume that we should have <math>m</math> divisible by five. If so, the quantity <math>18k-73</math> must be divisible by five, meaning that <math>k</math> leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!). <br />
<br />
Thus, we see that to achieve integers <math>n</math> and <math>k</math> that could potentially satisfy the problem statement, we must try the values of <math>k</math> congruent to one modulo five. However, if we recall a statement made earlier in the problem, we see that we can skip all even values of <math>k</math> produced by this modulo argument.<br />
<br />
Also, note that <math>k=1,6</math> won't work, as they are too small, and will give an erroneous value for <math>n</math>. After trying <math>k=11,21,31</math>, we see that <math>k=31</math> will give a value of <math>m=485</math>, which yields <math>n=\boxed{258}</math>, which, if plugged in to for our equations of <math>f(n)</math> and <math>g(n)</math>, will yield the desired ratio, and we're done.<br />
<br />
Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with "needs proof". If you can prove it, please add a lemma to the solution doing so :)<br />
<br />
-mathislife52<br />
<br />
==Solution 5 (Basic substitution)==<br />
<br />
First of all, if <math>n</math> is a perfect square, <math>f(n)=g(n)=\sqrt{n}</math> and their quotient is <math>1.</math> So for the rest of this solution, assume <math>n</math> is not a perfect square.<br />
<br />
Let <math>a^2</math> be the smallest perfect square greater than <math>n</math> and let <math>b^2</math> be the smallest perfect square greater than <math>n</math> with the same parity as <math>n,</math> and note that either <math>b=a</math> or <math>b=a+1.</math> Notice that <math>(a-1)^2 < n < a^2.</math><br />
<br />
With a bit of inspection, it becomes clear that <math>f(n) = a+(a^2-n)</math> and <math>g(n) = b+(b^2-n).</math><br />
<br />
If <math>a</math> and <math>n</math> have the same parity, we get <math>a=b</math> so <math>f(n) = g(n)</math> and their quotient is <math>1.</math> So for the rest of this solution, we let <math>a</math> and <math>n</math> have opposite parity. We have two cases to consider.<br />
<br />
Case 1: <math>n</math> is odd and <math>a</math> is even<br />
<br />
Here, we get <math>a=2k</math> for some positive integer <math>k.</math> Then, <math>b = 2k+1.</math> We let <math>n = a^2-(2m+1)</math> for some positive integer <math>m</math> so <math>f(n) = 2k+2m+1</math> and <math>g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.</math><br />
<br />
We set <math>\frac{2k+2m+1}{6k+2m+3}=\frac{4}{7},</math> cross multiply, and rearrange to get <math>6m-10k=5.</math> Since <math>k</math> and <math>m</math> are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions.<br />
<br />
Case 2: <math>n</math> is even and <math>a</math> is odd<br />
<br />
Here, we get <math>a=2k+1</math> for some positive ineger <math>k.</math> Then, <math>b=2k+2.</math> We let <math>n = a^2-(2m+1)</math> for some positive integer <math>m</math> so <math>f(n)=2k+1+2m+1=2k+2m+2</math> and <math>g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.</math><br />
<br />
We set <math>\frac{2k+2m+2}{6k+2m+6} = \frac{4}{7},</math> cross multiply, and rearrange to get <math>5k=3m-5,</math> or <math>k=\frac{3}{5}m-1.</math> Since <math>k</math> and <math>m</math> are integers, <math>m</math> must be a multiple of <math>5.</math> Some possible solutions for <math>(k,m)</math> with the least <math>k</math> and <math>m</math> are <math>(2,5), (5,10), (8,15),</math> and <math>(11,20).</math><br />
<br />
We wish to minimize <math>k</math> since <math>a=2k+1.</math> One thing to keep in mind is the initial assumption <math>(a-1)^2 < n < a^2.</math><br />
<br />
The pair <math>(2,5)</math> gives <math>a=2(2)+1=5</math> and <math>n=5^2-(2(5)+1)=14.</math> But <math>4^2<14<5^2</math> is clearly false, so we discard this case.<br />
<br />
The pair <math>(5,10)</math> gives <math>a=2(5)+1=11</math> and <math>n=11^2-(2(10)+1)=100</math> which is a perfect square and therefore can be discarded.<br />
<br />
The pair <math>(8,15)</math> gives <math>a=2(8)+1=17</math> and <math>n=17^2-(2(15)+1)=258</math> which is between <math>16^2</math> and <math>17^2</math> so it is our smallest solution.<br />
<br />
So <math>\boxed{258}</math> is the correct answer.<br />
<br />
- mc21s<br />
<br />
==Video Solution==<br />
https://youtu.be/tRVe2bKwIY8<br />
~Mathematical Dexterity<br />
<br />
==See Also==<br />
{{AIME box|year=2021|n=II|num-b=14|after=Last Question}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_15&diff=1601682021 AIME II Problems/Problem 152021-08-13T16:07:33Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Let <math>f(n)</math> and <math>g(n)</math> be functions satisfying<br />
<cmath>f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\<br />
1 + f(n+1) & \text{ otherwise}<br />
\end{cases}</cmath>and<br />
<cmath>g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\<br />
2 + g(n+2) & \text{ otherwise}<br />
\end{cases}</cmath>for positive integers <math>n</math>. Find the least positive integer <math>n</math> such that <math>\tfrac{f(n)}{g(n)} = \tfrac{4}{7}</math>.<br />
<br />
==Solution 1==<br />
<br />
Consider what happens when we try to calculate <math>f(n)</math> where n is not a square. If <math>k^2<n<(k+1)^2</math> for (positive) integer k, recursively calculating the value of the function gives us <math>f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n</math>. Note that this formula also returns the correct value when <math>n=(k+1)^2</math>, but not when <math>n=k^2</math>. Thus <math>f(n)=k^2+3k+2-n</math> for <math>k^2<n \leq (k+1)^2</math>.<br />
<br />
If <math>2 \mid (k+1)^2-n</math>, <math>g(n)</math> returns the same value as <math>f(n)</math>. This is because the recursion once again stops at <math>(k+1)^2</math>. We seek a case in which <math>f(n)<g(n)</math>, so obviously this is not what we want. We want <math>(k+1)^2,n</math> to have a different parity, or <math>n, k</math> have the same parity. When this is the case, <math>g(n)</math> instead returns <math>(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n</math>.<br />
<br />
Write <math>7f(n)=4g(n)</math>, which simplifies to <math>3k^2+k-10=3n</math>. Notice that we want the <math>LHS</math> expression to be divisible by 3; as a result, <math>k \equiv 1 \pmod{3}</math>. We also want n to be strictly greater than <math>k^2</math>, so <math>k-10>0, k>10</math>. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is <math>k=16</math>, giving <math>n=258</math>.<br />
<br />
Indeed - if we check our answer, it works. Therefore, the answer is <math>\boxed{258}</math>.<br />
<br />
-Ross Gao<br />
<br />
==Solution 2 (Four Variables)==<br />
We consider <math>f(n)</math> and <math>g(n)</math> separately:<br />
<ul style="list-style-type:square;"><br />
<li><math>\boldsymbol{f(n)}</math><p><br />
We restrict <math>n</math> in which <math>k^2<n\leq(k+1)^2</math> for some positive integer <math>k,</math> or <cmath>n=(k+1)^2-p\hspace{40mm}(1)</cmath> for some integer <math>p</math> such that <math>0\leq p<2k+1.</math> By recursion, we get<br />
<cmath>\begin{align*}<br />
f\left((k+1)^2\right)&=k+1, \\<br />
f\left((k+1)^2-1\right)&=k+2, \\<br />
f\left((k+1)^2-2\right)&=k+3, \\<br />
&\cdots \\<br />
f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{19mm}(2) \\<br />
\end{align*}</cmath></li><p><br />
<li><math>\boldsymbol{g(n)}</math><p><br />
If <math>n</math> and <math>(k+1)^2</math> have the same parity, then we get <math>g(n)=f(n)</math> by a similar process from <math>g\left((k+1)^2\right)=k+1.</math> This contradicts the precondition <math>\frac{f(n)}{g(n)} = \frac{4}{7}.</math> Therefore, <math>n</math> and <math>(k+1)^2</math> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. <p><br />
It follows that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{38.25mm}(3)</cmath> for some integer <math>q</math> such that <math>0<q<2k+2.</math> By recursion, we get<br />
<cmath>\begin{align*}<br />
g\left((k+2)^2\right)&=k+2, \\<br />
g\left((k+2)^2-2\right)&=k+4, \\<br />
g\left((k+2)^2-4\right)&=k+6, \\<br />
&\cdots \\<br />
g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\<br />
\end{align*}</cmath></li><p><br />
<li><b>Answer</b><p><br />
By <math>(2)</math> and <math>(4),</math> we have <cmath>\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)</cmath><br />
From <math>(1)</math> and <math>(3),</math> equating the expressions for <math>n</math> gives <math>(k+1)^2-p=(k+2)^2-2q.</math> Solving for <math>k</math> produces <cmath>k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)</cmath><br />
We substitute <math>(6)</math> into <math>(5),</math> then simplify, cross-multiply, and rearrange:<br />
<cmath>\begin{align*}<br />
\frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\<br />
\frac{p+2q-1}{-p+6q+1}&=\frac{4}{7} \\<br />
7p+14q-7&=-4p+24q+4 \\<br />
11p-11&=10q \\<br />
11(p-1)&=10q. \hspace{29mm}(7)<br />
\end{align*}</cmath><br />
Since <math>\gcd(11,10)=1,</math> we know that <math>p-1</math> must be divisible by <math>10,</math> and <math>q</math> must be divisible by <math>11.</math> <p> Recall that the restrictions on <math>(1)</math> and <math>(2)</math> are <math>0\leq p<2k+1</math> and <math>0<q<2k+2,</math> respectively. Substituting <math>(6)</math> into either inequality gives <math>p+1<q.</math> Combining all these results produces <cmath>0<p+1<q<2k+2. \hspace{28mm}(8)</cmath><br />
<br />
To minimize <math>n</math> in either <math>(1)</math> or <math>(3),</math> we minimize <math>k,</math> so we minimize <math>p</math> and <math>q</math> in <math>(8).</math> From <math>(6)</math> and <math>(7),</math> we construct the following table:<br />
<cmath>\begin{array}{c|c|c|c}<br />
& & & \\ [-2.5ex]<br />
\boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{(8)?} \\ [0.5ex]<br />
\hline<br />
& & & \\ [-2ex]<br />
11 & 11 & 4 & \\<br />
21 & 22 & 10 & \\<br />
31 & 33 & 16 & \checkmark \\<br />
\geq41 & \geq44 & \geq22 & \checkmark \\<br />
\end{array}</cmath><br />
Finally, we have <math>(p,q,k)=(31,33,16).</math> Substituting this result into either <math>(1)</math> or <math>(3)</math> generates <math>n=\boxed{258}.</math></li><p><br />
<li><b>Remark</b><p><br />
We can verify that <cmath>\frac{f(258)}{g(258)}=\frac{1\cdot31+f(258+1\cdot31)}{2\cdot33+g(258+2\cdot33)}=\frac{31+\overbrace{f(289)}^{17}}{66+\underbrace{g(324)}_{18}}=\frac{48}{84}=\frac47.</cmath></li><p><br />
</ul><br />
~MRENTHUSIASM<br />
<br />
==Solution 3==<br />
Since <math>n</math> isn't a perfect square, let <math>n=m^2+k</math> with <math>0<k<2m+1</math>. If <math>m</math> is odd, then <math>f(n)=g(n)</math>. If <math>m</math> is even, then<br />
<cmath>\begin{align*}<br />
f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\<br />
g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k,<br />
\end{align*}</cmath><br />
from which<br />
<cmath>\begin{align*}<br />
7(3m+2-k)&=4(5m+6-k) \\<br />
m&=3k+10.<br />
\end{align*}</cmath><br />
Since <math>m</math> is even, <math>k</math> is even. Since <math>k\neq 0</math>, the smallest <math>k</math> is <math>2</math> which produces the smallest <math>n</math>: <cmath>k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.</cmath><br />
~Afo<br />
<br />
==Solution 4 (Quadratics With Two Variables)==<br />
To begin, note that if <math>n</math> is a perfect square, <math>f(n)=g(n)</math>, so <math>f(n)/g(n)=1</math>, so we must look at values of <math>n</math> that are not perfect squares (what a surprise). First, let the distance between <math>n</math> and the first perfect square greater than or equal to it be <math>k</math>, making the values of <math>f(n+k)</math> and <math>g(n+k)</math> integers. Using this notation, we see than <math>f(n)=k+f(n+k)</math>, giving us a formula for the numerator of our ratio. However, since the function of <math>g(n)</math> does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of <math>\sqrt{n+k}</math> in <math>g(n)</math> unless <math>k</math> is an even number. However, this is impossible, since if <math>k</math> was an even number, <math>f(n)=g(n)</math>, giving a ratio of one. Thus, <math>k</math> must be an odd number.<br />
<br />
Thus, since <math>k</math> must be an odd number, regardless of whether <math>n</math> is even or odd, to get an integral value in <math>g(n)</math>, we must get to the next perfect square after <math>n+k</math>. To make matters easier, let <math>z^2=n+k</math>. Thus, in <math>g(n)</math>, we want to achieve <math>(z+1)^2</math>. <br />
<br />
Expanding <math>(z+1)^2</math> and substituting in the fact that <math>z=\sqrt{n+k}</math> yields:<br />
<br />
<cmath>(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1</cmath><br />
<br />
Thus, we must add the quantity <math>k+2z+1</math> to <math>n</math> to achieve a integral value in the function <math>g(n)</math>. Thus.<br />
<br />
<cmath>g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}</cmath><br />
<br />
However, note that the quantity within the square root is just <math>(z+1)^2</math>, and so:<br />
<br />
<cmath>g(n)=k+3z+2</cmath><br />
<br />
Thus,<br />
<cmath>\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}</cmath><br />
<br />
Since we want this quantity to equal <math>\frac{4}{7}</math>, we can set the above equation equal to this number and collect all the variables to one side to achieve<br />
<br />
<cmath>3k-5z=8</cmath><br />
<br />
Substituting back in that <math>z=\sqrt{n+k}</math>, and then separating variables and squaring yields that<br />
<br />
<cmath>9k^2-73k+64=25n</cmath><br />
<br />
Now, if we treat <math>n</math> as a constant, we can use the quadratic formula in respect to <math>k</math> to get an equation for <math>k</math> in terms of <math>n</math> (without all the squares). Doing so yields<br />
<br />
<cmath>\frac{73\pm\sqrt{3025+900n}}{18}=k</cmath><br />
<br />
Now, since <math>n</math> and <math>k</math> are integers, we want the quantity within the square root to be a perfect square. Note that <math>55^2=3025</math>. Thus, assume that the quantity within the root is equal to the perfect square, <math>m^2</math>. Thus, after using a difference of squares, we have<br />
<cmath>(m-55)(m+55)=900n</cmath><br />
Since we want <math>n</math> to be an integer, we know that the <math>LHS</math> should be divisible by five, so, let's assume that we should have <math>m</math> divisible by five. If so, the quantity <math>18k-73</math> must be divisible by five, meaning that <math>k</math> leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!). <br />
<br />
Thus, we see that to achieve integers <math>n</math> and <math>k</math> that could potentially satisfy the problem statement, we must try the values of <math>k</math> congruent to one modulo five. However, if we recall a statement made earlier in the problem, we see that we can skip all even values of <math>k</math> produced by this modulo argument.<br />
<br />
Also, note that <math>k=1,6</math> won't work, as they are too small, and will give an erroneous value for <math>n</math>. After trying <math>k=11,21,31</math>, we see that <math>k=31</math> will give a value of <math>m=485</math>, which yields <math>n=\boxed{258}</math>, which, if plugged in to for our equations of <math>f(n)</math> and <math>g(n)</math>, will yield the desired ratio, and we're done.<br />
<br />
Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with "needs proof". If you can prove it, please add a lemma to the solution doing so :)<br />
<br />
-mathislife52<br />
<br />
==Solution 6==<br />
<br />
First of all, if <math>n</math> is a perfect square, <math>f(n)=g(n)=\sqrt{n}</math> and their quotient is <math>1.</math> So for the rest of this solution, assume <math>n</math> is not a perfect square.<br />
<br />
Let <math>a^2</math> be the smallest perfect square greater than <math>n</math> and let <math>b^2</math> be the smallest perfect square greater than <math>n</math> with the same parity as <math>n,</math> and note that either <math>b=a</math> or <math>b=a+1.</math> Notice that <math>(a-1)^2 < n < a^2.</math><br />
<br />
With a bit of inspection, it becomes clear that <math>f(n) = a+(a^2-n)</math> and <math>g(n) = b+(b^2-n).</math><br />
<br />
If <math>a</math> and <math>n</math> have the same parity, we get <math>a=b</math> so <math>f(n) = g(n)</math> and their quotient is <math>1.</math> So for the rest of this solution, we let <math>a</math> and <math>n</math> have opposite parity. We have two cases to consider.<br />
<br />
Case 1: <math>n</math> is odd and <math>a</math> is even<br />
<br />
Here, we get <math>a=2k</math> for some positive integer <math>k.</math> Then, <math>b = 2k+1.</math> We let <math>n = a^2-(2m+1)</math> for some positive integer <math>m</math> so <math>f(n) = 2k+2m+1</math> and <math>g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.</math><br />
<br />
We set <math>\frac{2k+2m+1}{6k+2m+3}=\frac{4}{7},</math> cross multiply, and rearrange to get <math>6m-10k=5.</math> Since <math>k</math> and <math>m</math> are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions.<br />
<br />
Case 2: <math>n</math> is even and <math>a</math> is odd<br />
<br />
Here, we get <math>a=2k+1</math> for some positive ineger <math>k.</math> Then, <math>b=2k+2.</math> We let <math>n = a^2-(2m+1)</math> for some positive integer <math>m</math> so <math>f(n)=2k+1+2m+1=2k+2m+2</math> and <math>g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.</math><br />
<br />
We set <math>\frac{2k+2m+2}{6k+2m+6} = \frac{4}{7},</math> cross multiply, and rearrange to get <math>5k=3m-5,</math> or <math>k=\frac{3}{5}m-1.</math> Since <math>k</math> and <math>m</math> are integers, <math>m</math> must be a multiple of <math>5.</math> Some possible solutions for <math>(k,m)</math> with the least <math>k</math> and <math>m</math> are <math>(2,5), (5,10), (8,15),</math> and <math>(11,20).</math><br />
<br />
We wish to minimize <math>k</math> since <math>a=2k+1.</math> One thing to keep in mind is the initial assumption <math>(a-1)^2 < n < a^2.</math><br />
<br />
The pair <math>(2,5)</math> gives <math>a=2(2)+1=5</math> and <math>n=5^2-(2(5)+1)=14.</math> But <math>4^2<14<5^2</math> is clearly false, so we discard this case.<br />
<br />
The pair <math>(5,10)</math> gives <math>a=2(5)+1=11</math> and <math>n=11^2-(2(10)+1)=100</math> which is a perfect square and therefore can be discarded.<br />
<br />
The pair <math>(8,15)</math> gives <math>a=2(8)+1=17</math> and <math>n=17^2-(2(15)+1)=258</math> which is between <math>16^2</math> and <math>17^2</math> so it is our smallest solution.<br />
<br />
So <math>\boxed{258}</math> is the correct answer.<br />
<br />
- mc21s<br />
<br />
==Video Solution==<br />
https://youtu.be/tRVe2bKwIY8<br />
~Mathematical Dexterity<br />
<br />
==See Also==<br />
{{AIME box|year=2021|n=II|num-b=14|after=Last Question}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=1601062011 AIME I Problems/Problem 42021-08-12T23:46:30Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>.<br />
<br />
== Solution 1 == <br />
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. <br />
Since <math>{BM}</math> is the angle bisector of angle <math>B</math>, and <math>{CM}</math> is perpendicular to <math>{BM}</math>, so <math>BP=BC=120</math>, and <math>M</math> is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>, and <math>N</math> is the midpoint of <math>{CQ}</math>.<br />
Hence <math>MN=\frac{PQ}{2}</math>. <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>IM \perp MC</math> and <math>IN \perp NC</math>, we have <math>CMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = CI</math>. Since <math>\sin \angle MIN = \sin (90 - \frac{\angle BAC}{2}) = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math>. Thus, <math>MN=\frac{117+120-125}{2}=\boxed{056}</math><br />
<br />
== Solution 3 (Bash) == <br />
Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines (LoC) formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>.<br />
<br />
--lucasxia01<br />
<br />
== Solution 4==<br />
<br />
Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. <br />
<br />
Ptolemy on CMIN:<br />
<br />
<math>CN*MI+CM*IN=CI*MN</math><br />
<br />
<math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN</math><br />
<br />
<math>MN = CI \sin \angle MCN</math> by angle addition formula.<br />
<br />
<math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>. <br />
<br />
Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>.<br />
<math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>.<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=yIUBhWiJ4Dk<br />
~Mathematical Dexterity<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=vkniYGN45F4<br />
<br />
~Shreyas S<br />
<br />
Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&t=12s<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_7&diff=1596942021 AIME I Problems/Problem 72021-08-06T22:57:35Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Find the number of pairs <math>(m,n)</math> of positive integers with <math>1\le m<n\le 30</math> such that there exists a real number <math>x</math> satisfying <cmath>\sin(mx)+\sin(nx)=2.</cmath><br />
<br />
==Solution 1==<br />
The maximum value of <math>\sin \theta</math> is <math>1</math>, which is achieved at <math>\theta = \frac{\pi}{2}+2k\pi</math> for some integer <math>k</math>. This is left as an exercise to the reader.<br />
<br />
This implies that <math>\sin(mx) = \sin(nx) = 1</math>, and that <math>mx = \frac{\pi}{2}+2a\pi</math> and <math>nx = \frac{\pi}{2}+2b\pi</math>, for integers <math>a, b</math>.<br />
<br />
Taking their ratio, we have <cmath>\frac{mx}{nx} = \frac{\frac{\pi}{2}+2a\pi}{\frac{\pi}{2}+2b\pi} \implies \frac{m}{n} = \frac{4a + 1}{4b + 1} \implies \frac{m}{4a + 1} = \frac{n}{4b + 1} = k.</cmath><br />
It remains to find all <math>m, n</math> that satisfy this equation.<br />
<br />
If <math>k = 1</math>, then <math>m \equiv n \equiv 1 \pmod 4</math>. This corresponds to choosing two elements from the set <math>\{1, 5, 9, 13, 17, 21, 25, 29\}</math>. There are <math>\binom 82</math> ways to do so.<br />
<br />
If <math>k < 1</math>, by multiplying <math>m</math> and <math>n</math> by the same constant <math>c = \frac{1}{k}</math>, we have that <math>mc \equiv nc \equiv 1 \pmod 4</math>. Then either <math>m \equiv n \equiv 1 \pmod 4</math>, or <math>m \equiv n \equiv 3 \pmod 4</math>. But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set <math>\{3, 7, 11, 15, 19, 23, 27\}</math>. There are <math>\binom 72</math> ways here.<br />
<br />
Finally, if <math>k > 1</math>, note that <math>k</math> must be an integer. This means that <math>m, n</math> belong to the set <math>\{k, 5k, 9k, \dots\}</math>, or <math>\{3k, 7k, 11k, \dots\}</math>. Taking casework on <math>k</math>, we get the sets <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}</math>. Some sets have been omitted; this is because they were counted in the other cases already. This sums to <math>\binom 42 + \binom 42 + \binom 22 + \binom 22</math>.<br />
<br />
In total, there are <math>\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{63}</math> pairs of <math>(m, n)</math>.<br />
<br />
This solution was brought to you by ~Leonard_my_dude~<br />
<br />
==Solution 2==<br />
In order for <math>\sin(mx) + \sin(nx) = 2</math>, <math>\sin(mx) = \sin(nx) = 1</math>. <br />
<br />
This happens when <br />
<math>mx \equiv nx \equiv \frac{\pi}{2} (</math>mod <math>2\pi).</math><br />
<br />
This means that <math>mx = \frac{\pi}{2} + 2\pi\alpha</math> and <math>nx = \frac{\pi}{2} + 2\pi\beta</math> for any integers <math>\alpha</math> and <math>\beta</math>. <br />
<br />
As in Solution 1, take the ratio of the two equations:<br />
<cmath>\frac{mx}{nx} = \frac{\frac{\pi}{2}+2\pi\alpha}{\frac{\pi}{2}+2\pi\beta} \implies \frac{m}{n} = \frac{\frac{1}{2}+2\alpha}{\frac{1}{2}+2\beta} \implies \frac{m}{n} = \frac{4\alpha+1}{4\beta+1}</cmath><br />
<br />
Now notice that the numerator and denominator of <math>\frac{4\alpha+1}{4\beta+1}</math> are both odd, which means that <math>m</math> and <math>n</math> have the same power of two (the powers of 2 cancel out).<br />
<br />
Let the common power be <math>p</math>: then <math>m = 2^p\cdot a</math>, and <math>n = 2^p\cdot b</math> where <math>a</math> and <math>b</math> are integers between 1 and 30.<br />
<br />
We can now rewrite the equation: <br />
<cmath>\frac{2^p\cdot a}{2^p\cdot b} = \frac{4\alpha+1}{4\beta+1} \implies \frac{a}{b} = \frac{4\alpha+1}{4\beta+1}</cmath><br />
<br />
Now it is easy to tell that <math>a \equiv 1 (</math>mod <math>4)</math> and <math>b \equiv 1 (</math>mod <math>4)</math>. However, there is another case: that <br />
<br />
<math>a \equiv 3 (</math>mod <math>4)</math> and <math>b \equiv 3 (</math>mod <math>4)</math>. This is because multiplying both <math>4\alpha+1</math> and <math>4\beta+1</math> by <math>-1</math> will not change the fraction, but each congruence will be changed to <math>-1 (</math>mod <math>4) \equiv 3 (</math>mod <math>4)</math>.<br />
<br />
From the first set of congruences, we find that <math>a</math> and <math>b</math> can be two of <br />
<math>\{1, 5, 9, \cdots, 29\}</math>.<br />
<br />
From the second set of congruences, we find that <math>a</math> and <math>b</math> can be two of <br />
<math>\{3, 7, 11, \cdots, 27\}</math>.<br />
<br />
Now all we have to do is multiply by <math>2^p</math> to get back to <math>m</math> and <math>n</math>. <br />
Let’s organize the solutions in order of increasing values of <math>p</math>, keeping in mind that <math>m</math> and <math>n</math> are bounded between 1 and 30.<br />
<br />
For <math>p = 0</math> we get <math>\{1, 5, 9, \cdots, 29\}, \{3, 7, 11, \cdots, 27\}</math>.<br />
<br />
For <math>p = 1</math> we get <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}</math><br />
<br />
For <math>p = 2</math> we get <math>\{4, 20\}, \{12, 28\}</math><br />
<br />
If we increase the value of <math>p</math> more, there will be less than two integers in our sets, so we are done there.<br />
<br />
There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.<br />
<br />
In each of these sets we can choose 2 numbers to be <math>m</math> and <math>n</math> and then assign them in increasing order. Thus there are:<br />
<br />
<cmath>\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2} = 28+21+6+6+1+1 = \boxed{63}</cmath> possible pairs <math>(m,n)</math> that satisfy the conditions.<br />
<br />
-KingRavi<br />
<br />
==Solution 3==<br />
We know that the range of <math>sin</math> is between <math>-1</math> and <math>1</math>.<br />
<br />
Thus, the only way for the sum to be <math>2</math> is for <math>sin</math> of <math>mx</math> and <math>nx</math> to both be <math>1</math>.<br />
<br />
The <math>sin</math> of <math>(90+360k)</math> is equal to 1.<br />
<br />
Assuming <math>mx</math> and <math>nm</math> are both positive, m and n could be <math>1,5,9,13,17,21,25,29</math>. There are <math>8</math> ways, so <math>\dbinom{8}{2}</math>.<br />
<br />
If both are negative, m and n could be <math>3,7,11,15,19,23,27</math>. There are <math>7</math> ways, so <math>\dbinom{7}{2}</math>.<br />
<br />
However, the pair <math>(1,5)</math> could also be <math>(2, 10)</math> and so on. The same goes for some other pairs.<br />
<br />
In total there are <math>14</math> of these extra pairs.<br />
<br />
The answer is <math>28+21+14 = \boxed{063}</math><br />
<br />
==Remark==<br />
The graphs of <math>r\leq\sin(m\theta)+\sin(n\theta)</math> and <math>r=2</math> are shown here in Desmos: https://www.desmos.com/calculator/busxadywja<br />
<br />
Move the sliders around for <math>1\leq m \leq 29</math> and <math>2\leq m+1\leq n\leq30</math> to observe the geometric representation generated by each pair <math>(m,n).</math><br />
<br />
~MRENTHUSIASM (inspired by TheAMCHub)<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=LUkQ7R1DqKo<br />
~Mathematical Dexterity<br />
<br />
==See Also==<br />
{{AIME box|year=2021|n=I|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_15&diff=1596932021 AIME II Problems/Problem 152021-08-06T22:57:10Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
Let <math>f(n)</math> and <math>g(n)</math> be functions satisfying<br />
<cmath>f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\<br />
1 + f(n+1) & \text{ otherwise}<br />
\end{cases}</cmath>and<br />
<cmath>g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\<br />
2 + g(n+2) & \text{ otherwise}<br />
\end{cases}</cmath>for positive integers <math>n</math>. Find the least positive integer <math>n</math> such that <math>\tfrac{f(n)}{g(n)} = \tfrac{4}{7}</math>.<br />
<br />
==Solution 1==<br />
<br />
Consider what happens when we try to calculate <math>f(n)</math> where n is not a square. If <math>k^2<n<(k+1)^2</math> for (positive) integer k, recursively calculating the value of the function gives us <math>f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n</math>. Note that this formula also returns the correct value when <math>n=(k+1)^2</math>, but not when <math>n=k^2</math>. Thus <math>f(n)=k^2+3k+2-n</math> for <math>k^2<n \leq (k+1)^2</math>.<br />
<br />
If <math>2 \mid (k+1)^2-n</math>, <math>g(n)</math> returns the same value as <math>f(n)</math>. This is because the recursion once again stops at <math>(k+1)^2</math>. We seek a case in which <math>f(n)<g(n)</math>, so obviously this is not what we want. We want <math>(k+1)^2,n</math> to have a different parity, or <math>n, k</math> have the same parity. When this is the case, <math>g(n)</math> instead returns <math>(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n</math>.<br />
<br />
Write <math>7f(n)=4g(n)</math>, which simplifies to <math>3k^2+k-10=3n</math>. Notice that we want the <math>LHS</math> expression to be divisible by 3; as a result, <math>k \equiv 1 \pmod{3}</math>. We also want n to be strictly greater than <math>k^2</math>, so <math>k-10>0, k>10</math>. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is <math>k=16</math>, giving <math>n=258</math>.<br />
<br />
Indeed - if we check our answer, it works. Therefore, the answer is <math>\boxed{258}</math>.<br />
<br />
-Ross Gao<br />
<br />
==Solution 2 (Four Variables)==<br />
We consider <math>f(n)</math> and <math>g(n)</math> separately:<br />
<ul style="list-style-type:square;"><br />
<li><math>\boldsymbol{f(n)}</math><p><br />
We restrict <math>n</math> in which <math>k^2<n\leq(k+1)^2</math> for some positive integer <math>k,</math> or <cmath>n=(k+1)^2-p\hspace{40mm}(1)</cmath> for some integer <math>p</math> such that <math>0\leq p<2k+1.</math> By recursion, we get<br />
<cmath>\begin{align*}<br />
f\left((k+1)^2\right)&=k+1, \\<br />
f\left((k+1)^2-1\right)&=k+2, \\<br />
f\left((k+1)^2-2\right)&=k+3, \\<br />
&\cdots \\<br />
f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{19mm}(2) \\<br />
\end{align*}</cmath></li><p><br />
<li><math>\boldsymbol{g(n)}</math><p><br />
If <math>n</math> and <math>(k+1)^2</math> have the same parity, then we get <math>g(n)=f(n)</math> by a similar process from <math>g\left((k+1)^2\right)=k+1.</math> This contradicts the precondition <math>\frac{f(n)}{g(n)} = \frac{4}{7}.</math> Therefore, <math>n</math> and <math>(k+1)^2</math> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. <p><br />
It follows that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{38.25mm}(3)</cmath> for some integer <math>q</math> such that <math>0<q<2k+2.</math> By recursion, we get<br />
<cmath>\begin{align*}<br />
g\left((k+2)^2\right)&=k+2, \\<br />
g\left((k+2)^2-2\right)&=k+4, \\<br />
g\left((k+2)^2-4\right)&=k+6, \\<br />
&\cdots \\<br />
g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\<br />
\end{align*}</cmath></li><p><br />
<li><b>Answer</b><p><br />
By <math>(2)</math> and <math>(4),</math> we have <cmath>\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)</cmath><br />
From <math>(1)</math> and <math>(3),</math> equating the expressions for <math>n</math> gives <math>(k+1)^2-p=(k+2)^2-2q.</math> Solving for <math>k</math> produces <cmath>k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)</cmath><br />
We substitute <math>(6)</math> into <math>(5),</math> then simplify, cross-multiply, and rearrange:<br />
<cmath>\begin{align*}<br />
\frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\<br />
\frac{p+2q-1}{-p+6q+1}&=\frac{4}{7} \\<br />
7p+14q-7&=-4p+24q+4 \\<br />
11p-11&=10q \\<br />
11(p-1)&=10q. \hspace{29mm}(7)<br />
\end{align*}</cmath><br />
Since <math>\gcd(11,10)=1,</math> we know that <math>p-1</math> must be divisible by <math>10,</math> and <math>q</math> must be divisible by <math>11.</math> <p> Recall that the restrictions on <math>(1)</math> and <math>(2)</math> are <math>0\leq p<2k+1</math> and <math>0<q<2k+2,</math> respectively. Substituting <math>(6)</math> into either inequality gives <math>p+1<q.</math> Combining all these results produces <cmath>0<p+1<q<2k+2. \hspace{28mm}(8)</cmath><br />
<br />
To minimize <math>n</math> in either <math>(1)</math> or <math>(3),</math> we minimize <math>k,</math> so we minimize <math>p</math> and <math>q</math> in <math>(8).</math> From <math>(6)</math> and <math>(7),</math> we construct the following table:<br />
<cmath>\begin{array}{c|c|c|c}<br />
& & & \\ [-2.5ex]<br />
\boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{(8)?} \\ [0.5ex]<br />
\hline<br />
& & & \\ [-2ex]<br />
11 & 11 & 4 & \\<br />
21 & 22 & 10 & \\<br />
31 & 33 & 16 & \checkmark \\<br />
\geq41 & \geq44 & \geq22 & \checkmark \\<br />
\end{array}</cmath><br />
Finally, we have <math>(p,q,k)=(31,33,16).</math> Substituting this result into either <math>(1)</math> or <math>(3)</math> generates <math>n=\boxed{258}.</math></li><p><br />
<li><b>Remark</b><p><br />
We can verify that <cmath>\frac{f(258)}{g(258)}=\frac{1\cdot31+f(258+1\cdot31)}{2\cdot33+g(258+2\cdot33)}=\frac{31+\overbrace{f(289)}^{17}}{66+\underbrace{g(324)}_{18}}=\frac{48}{84}=\frac47.</cmath></li><p><br />
</ul><br />
~MRENTHUSIASM<br />
<br />
==Solution 3==<br />
Since <math>n</math> isn't a perfect square, let <math>n=m^2+k</math> with <math>0<k<2m+1</math>. If <math>m</math> is odd, then <math>f(n)=g(n)</math>. If <math>m</math> is even, then<br />
<cmath>\begin{align*}<br />
f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\<br />
g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k,<br />
\end{align*}</cmath><br />
from which<br />
<cmath>\begin{align*}<br />
7(3m+2-k)&=4(5m+6-k) \\<br />
m&=3k+10.<br />
\end{align*}</cmath><br />
Since <math>m</math> is even, <math>k</math> is even. Since <math>k\neq 0</math>, the smallest <math>k</math> is <math>2</math> which produces the smallest <math>n</math>: <cmath>k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.</cmath><br />
~Afo<br />
<br />
==Solution 4 (Quadratics With Two Variables)==<br />
To begin, note that if <math>n</math> is a perfect square, <math>f(n)=g(n)</math>, so <math>f(n)/g(n)=1</math>, so we must look at values of <math>n</math> that are not perfect squares (what a surprise). First, let the distance between <math>n</math> and the first perfect square greater than or equal to it be <math>k</math>, making the values of <math>f(n+k)</math> and <math>g(n+k)</math> integers. Using this notation, we see than <math>f(n)=k+f(n+k)</math>, giving us a formula for the numerator of our ratio. However, since the function of <math>g(n)</math> does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of <math>\sqrt{n+k}</math> in <math>g(n)</math> unless <math>k</math> is an even number. However, this is impossible, since if <math>k</math> was an even number, <math>f(n)=g(n)</math>, giving a ratio of one. Thus, <math>k</math> must be an odd number.<br />
<br />
Thus, since <math>k</math> must be an odd number, regardless of whether <math>n</math> is even or odd, to get an integral value in <math>g(n)</math>, we must get to the next perfect square after <math>n+k</math>. To make matters easier, let <math>z^2=n+k</math>. Thus, in <math>g(n)</math>, we want to achieve <math>(z+1)^2</math>. <br />
<br />
Expanding <math>(z+1)^2</math> and substituting in the fact that <math>z=\sqrt{n+k}</math> yields:<br />
<br />
<cmath>(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1</cmath><br />
<br />
Thus, we must add the quantity <math>k+2z+1</math> to <math>n</math> to achieve a integral value in the function <math>g(n)</math>. Thus.<br />
<br />
<cmath>g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}</cmath><br />
<br />
However, note that the quantity within the square root is just <math>(z+1)^2</math>, and so:<br />
<br />
<cmath>g(n)=k+3z+2</cmath><br />
<br />
Thus,<br />
<cmath>\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}</cmath><br />
<br />
Since we want this quantity to equal <math>\frac{4}{7}</math>, we can set the above equation equal to this number and collect all the variables to one side to achieve<br />
<br />
<cmath>3k-5z=8</cmath><br />
<br />
Substituting back in that <math>z=\sqrt{n+k}</math>, and then separating variables and squaring yields that<br />
<br />
<cmath>9k^2-73k+64=25n</cmath><br />
<br />
Now, if we treat <math>n</math> as a constant, we can use the quadratic formula in respect to <math>k</math> to get an equation for <math>k</math> in terms of <math>n</math> (without all the squares). Doing so yields<br />
<br />
<cmath>\frac{73\pm\sqrt{3025+900n}}{18}=k</cmath><br />
<br />
Now, since <math>n</math> and <math>k</math> are integers, we want the quantity within the square root to be a perfect square. Note that <math>55^2=3025</math>. Thus, assume that the quantity within the root is equal to the perfect square, <math>m^2</math>. Thus, after using a difference of squares, we have<br />
<cmath>(m-55)(m+55)=900n</cmath><br />
Since we want <math>n</math> to be an integer, we know that the <math>LHS</math> should be divisible by five, so, let's assume that we should have <math>m</math> divisible by five. If so, the quantity <math>18k-73</math> must be divisible by five, meaning that <math>k</math> leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!). <br />
<br />
Thus, we see that to achieve integers <math>n</math> and <math>k</math> that could potentially satisfy the problem statement, we must try the values of <math>k</math> congruent to one modulo five. However, if we recall a statement made earlier in the problem, we see that we can skip all even values of <math>k</math> produced by this modulo argument.<br />
<br />
Also, note that <math>k=1,6</math> won't work, as they are too small, and will give an erroneous value for <math>n</math>. After trying <math>k=11,21,31</math>, we see that <math>k=31</math> will give a value of <math>m=485</math>, which yields <math>n=\boxed{258}</math>, which, if plugged in to for our equations of <math>f(n)</math> and <math>g(n)</math>, will yield the desired ratio, and we're done.<br />
<br />
Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with "needs proof". If you can prove it, please add a lemma to the solution doing so :)<br />
<br />
-mathislife52<br />
<br />
==Video Solution==<br />
https://youtu.be/tRVe2bKwIY8<br />
~Mathematical Dexterity<br />
<br />
==See Also==<br />
{{AIME box|year=2021|n=II|num-b=14|after=Last Question}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=1583172020 AMC 10A Problems/Problem 142021-07-13T16:32:17Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
Real numbers <math>x</math> and <math>y</math> satisfy <math>x + y = 4</math> and <math>x \cdot y = -2</math>. What is the value of <br />
<br />
<cmath>x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?</cmath><br />
<br />
<math>\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480</math><br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
<cmath>x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}</cmath><br />
<br />
Continuing to combine <cmath>\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}</cmath><br />
From the givens, it can be concluded that <math>x^2y^2=4</math>. Also, <cmath>(x+y)^2=x^2+2xy+y^2=16</cmath> This means that <math>x^2+y^2=20</math>. Substituting this information into <math>\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}</math>, we have <math>\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}</math>. ~PCChess<br />
<br />
=== Solution 2 ===<br />
As above, we need to calculate <math>\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}</math>. Note that <math>x,y,</math> are the roots of <math>x^2-4x-2</math> and so <math>x^3=4x^2+2x</math> and <math>y^3=4y^2+2y</math>. Thus <math>x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88</math> where <math>x^2+y^2=20</math> and <math>x^2y^2=4</math> as in the previous solution. Thus the answer is <math>\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}</math>. Note(<math>x^2+y^2=(x+y)^2-2xy=20</math>, and <math>x^2y^2 = (xy)^2 = 4</math>)<br />
<br />
=== Solution 3 ===<br />
Note that <math>( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.</math> Now, we only need to find the values of <math>x^3 + y^3</math> and <math>\frac{1}{y^2} + \frac{1}{x^2}.</math> <br><br />
<br><br />
Recall that <math>x^3 + y^3 = (x + y) (x^2 - xy + y^2),</math> and that <math>x^2 - xy + y^2 = (x + y)^2 - 3xy.</math> We are able to solve the second equation, and doing so gets us <math>4^2 - 3(-2) = 22.</math> Plugging this into the first equation, we get <math>x^3 + y^3 = 4(22) = 88.</math> <br><br />
<br><br />
In order to find the value of <math>\frac{1}{y^2} + \frac{1}{x^2},</math> we find a common denominator so that we can add them together. This gets us <math>\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.</math> Recalling that <math>x^2 + y^2 = (x+y)^2 - 2xy</math> and solving this equation, we get <math>4^2 - 2(-2) = 20.</math> Plugging this into the first equation, we get <math>\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.</math> <br><br />
<br><br />
Solving the original equation, we get <math>x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.</math> ~[[User:emerald_block|emerald_block]]<br />
<br />
=== Solution 4 (Bashing) ===<br />
This is basically bashing using Vieta's formulas to find <math> x </math> and <math> y </math> (which I highly do not recommend, I only wrote this solution for fun).<br />
<br />
We use Vieta's to find a quadratic relating <math> x </math> and <math> y </math>. We set <math> x </math> and <math> y </math> to be the roots of the quadratic <math> Q ( n ) = n^2 - 4n - 2 </math> (because <math> x + y = 4 </math>, and <math> xy = -2 </math>). We can solve the quadratic to get the roots <math> 2 + \sqrt{6} </math> and <math> 2 - \sqrt{6} </math>. <math> x </math> and <math> y </math> are "interchangeable", meaning that it doesn't matter which solution <math> x </math> or <math> y </math> is, because it'll return the same result when plugged in. So we plug in <math> 2 + \sqrt{6} </math> for <math> x </math> and <math> 2 - \sqrt{6} </math> and get <math> \boxed{\textbf{(D)}\ 440} </math> as our answer.<br />
<br />
~Baolan<br />
<br />
=== Solution 5 (Bashing Part 2) ===<br />
This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br />
<br />
We first change the original expression to <math>4 + \frac{x^5 + y^5}{x^2 y^2}</math>, because <math>x + y = 4</math>. This is equal to <math>4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8</math>. We can factor and reduce <math>x^4 + y^4</math> to <math>(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392</math>. Now our expression is just <math>400 - (x^3 y + x y^3)</math>. We factor <math>x^3 y + x y^3</math> to get <math>(xy)(x^2 + y^2) = -40</math>. So the answer would be <math>400 - (-40)<br />
= \boxed{\textbf{(D)} 440} </math>.<br />
<br />
=== Solution 6 (Complete Binomial Theorem) ===<br />
<br />
We first simplify the expression to <cmath>x + y + \frac{x^5 + y^5}{x^2y^2}.</cmath><br />
Then, we can solve for <math>x</math> and <math>y</math> given the system of equations in the problem.<br />
Since <math>xy = -2,</math> we can substitute <math>\frac{-2}{x}</math> for <math>y</math>. <br />
Thus, this becomes the equation <cmath>x - \frac{2}{x} = 4.</cmath><br />
Multiplying both sides by <math>x</math>, we obtain <math>x^2 - 2 = 4x,</math> or <br />
<cmath>x^2 - 4x - 2 = 0.</cmath><br />
By the quadratic formula we obtain <math>x = 2 \pm \sqrt{6}</math>. <br />
We also easily find that given <math>x = 2 \pm \sqrt{6}</math>, <math>y</math> equals the conjugate of <math>x</math>. <br />
Thus, plugging our values in for <math>x</math> and <math>y</math>, our expression equals<br />
<cmath>4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}</cmath><br />
By the binomial theorem, we observe that every second terms of the expansions <math>x^5</math> and <math>y^5</math> will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of <math>x^5 + y^5</math>.<br />
Thus, our expression equals<br />
<cmath>4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.</cmath><br />
which equals<br />
<cmath>4 + \frac{2(872)}{4}</cmath><br />
which equals <math>\boxed{\textbf{(D)} 440}</math>.<br />
<br />
~ fidgetboss_4000<br />
<br />
=== Solution 7 ===<br />
As before, simplify the expression to <cmath>x + y + \frac{x^5 + y^5}{x^2y^2}.</cmath><br />
Since <math>x + y = 4</math> and <math>x^2y^2 = 4</math>, we substitute that in to obtain <cmath> 4 + \frac{x^5 + y^5}{4}.</cmath><br />
Now, we must solve for <math>x^5 + y^5</math>. Start by squaring <math>x + y</math>, to obtain <cmath>x^2 + 2xy + y^2 = 16</cmath><br />
Simplifying, <math>x^2 + y^2 = 20</math>. Squaring once more, we obtain <cmath>x^4 + y^4 + 2x^2y^2 = 400</cmath><br />
Once again simplifying, <math>x^4 + y^4 = 392</math>. Now, to obtain the fifth powers of <math>x</math> and <math>y</math>, we multiply both sides by <math>x + y</math>.<br />
We now have <br />
<cmath>x^5 + x^4y + xy^4 + y^5 = 1568</cmath>, or <br />
<cmath>x^5 + y^5 + xy(x^3 + y^3) = 1568</cmath><br />
We now solve for <math>x^3 + y^3</math>. <math>(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64</math>, so <math>x^3 + y^3 = 88</math>. <br />
Plugging this back into<math>x^5 + x^4y + xy^4 + y^5 = 1568</math>, we find that <math>x^5 + y^5 = 1744</math>, so we have <cmath> 4 + \frac{1744}{4}.</cmath>. This equals 440, so our answer is <math>\boxed{\textbf{(D)} 440}</math>.<br />
<br />
~Binderclips1<br />
<br />
=== Solution 8 ===<br />
We can use Newton Sums to solve this problem.<br />
We start by noticing that we can rewrite the equation as <math>\frac{x^3}{y^2} + \frac{y^3}{x^2} + x + y.</math><br />
Then, we know that <math>x + y = 4,</math> so we have <math>\frac{x^3}{y^2} + \frac{y^3}{x^2} + 4.</math><br />
We can use the equation <math>x \cdot y = -2</math> to write <math>x = \frac{-2}{y}</math> and <math>y = \frac{-2}{x}.</math><br />
Next, we can plug in these values of <math>x</math> and <math>y</math> to get <math>\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5}{4} + \frac{y^5}{4},</math> which is the same as <cmath>\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4}.</cmath><br />
Then, we use Newton sums where <math>S_n</math> is the elementary symmetric sum of the sequence and <math>P_n</math> is the power sum (<math>x^n + y^n</math>). Using this, we can make the following Newton sums:<br />
<cmath>P_1 = S_1</cmath><br />
<cmath>P_2 = P_1 S_1 - 2S_2</cmath><br />
<cmath>P_3 = P_2 S_1 - P_1 S_2</cmath><br />
<cmath>P_4 = P_3 S_1 - P_2 S_2</cmath><br />
<cmath>P_5 = P_4 S_1 - P_3 S_2.</cmath><br />
We also know that <math>S_1</math> is 4 because <math>x + y</math> is four, and we know that <math>S_2</math> is <math>-2</math> because <math>x \cdot y</math> is <math>-2</math> as well.<br />
Then, we can plug in values! We have<br />
<cmath>P_1 = S_1 = 4</cmath><br />
<cmath>P_2 = P_1 S_1 - 2S_2 = 16 - (-4) = 20</cmath><br />
<cmath>P_3 = P_2 S_1 - P_1 S_2 = 80 - (-8) = 88</cmath><br />
<cmath>P_4 = P_3 S_1 - P_2 S_2 = 88 \cdot 4 - (-40) = 392</cmath><br />
<cmath>P_5 = P_4 S_1 - P_3 S_2 = 392 \cdot 4 - (-2) \cdot 88 = 1744.</cmath><br />
We earlier noted that <math>\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4},</math> so we have that this equals <math>\frac{1744}{4},</math> or <math>436.</math> Then, plugging this back into the original equation, this is <math>436 + 4</math> or <math>440,</math> so our answer is <math>\boxed{\textbf{(D)}\ 440}.</math><br />
<br />
~Coolpeep<br />
<br />
=== Solution 9 ===<br />
As in the first solution, we get the expression to be <math>\frac{x^3+y^3}{x^2} + \frac{x^3+y^3}{y^2}.</math><br />
<br />
Then, since the numerators are the same, we can put the two fractions as a common denominator and multiply the numerator by <math>x^2y^2.</math> This gets us <math>\frac{(x^2 + y^2)(x^3 + y^3)}{x^2y^2}.</math><br />
<br />
Now, since we know <math>x+y=4</math> and <math>xy=-2,</math> instead of solving for <math>x</math> and <math>y,</math> we will try to manipulate the above expression them into a manner that we can substitute the sum and product that we know. Also, another form of <math>x^3+y^3</math> is <math>(x+y)(x^2-xy+y^2).</math><br />
<br />
Thus, we can convert the current expression to <math>\frac{(x^2 + y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}.</math><br />
<br />
Doing some algebraic multiplications, we get <math>\frac{((x+y)^2 - 2xy)(x+y)((x+y)^2 - 3xy)}{(xy)^2}.</math><br />
<br />
Since we know <math>x+y=4</math> and <math>xy=-2,</math> we have <math>\frac{(16-(-4))(4)(16-(-6))}{4} = \frac{20 \cdot 4 \cdot 22}{4} = 20 \cdot 22 = 440.</math><br />
<br />
Therefore the answer is <math>\boxed{\textbf{(D)} 440}.</math><br />
<br />
~mathboy282<br />
<br />
<br />
<br />
=== Video Solution ===<br />
<br />
Education, The Study of Everything<br />
<br />
https://youtu.be/PNkRlUKWCzg<br />
<br />
https://www.youtube.com/watch?v=jlRmDrL_jmk<br />
~Mathematical Dexterity (Don't Worry, Be Hoppy!)<br />
<br />
https://youtu.be/ZGwAasE32Y4<br />
<br />
~IceMatrix<br />
<br />
https://youtu.be/XEtzvxfFEJk<br />
<br />
~savannahsolver<br />
<br />
https://youtu.be/ba6w1OhXqOQ?t=3551<br />
<br />
~ pi_is_3.14<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_15&diff=1581742021 AMC 12B Problems/Problem 152021-07-12T02:21:48Z<p>Mc21s: </p>
<hr />
<div>{{duplicate|[[2021 AMC 10B Problems#Problem 20|2021 AMC 10B #20]] and [[2021 AMC 12B Problems#Problem 15|2021 AMC 12B #15]]}}<br />
<br />
==Problem==<br />
The figure is constructed from <math>11</math> line segments, each of which has length <math>2</math>. The area of pentagon <math>ABCDE</math> can be written as <math>\sqrt{m} + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. What is <math>m + n ?</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(-2.4638,4.10658);<br />
pair B=(-4,2.6567453480756127);<br />
pair C=(-3.47132,0.6335248637894945);<br />
pair D=(-1.464483379039766,0.6335248637894945);<br />
pair E=(-0.956630463955801,2.6567453480756127);<br />
pair F=(-2,2);<br />
pair G=(-3,2);<br />
draw(A--B--C--D--E--A);<br />
draw(A--F--A--G);<br />
draw(B--F--C);<br />
draw(E--G--D);<br />
label("A",A,N);<br />
label("B",B,W);<br />
label("C",C,S);<br />
label("D",D,S);<br />
label("E",E,dir(0));<br />
dot(A^^B^^C^^D^^E^^F^^G);<br />
</asy><br />
<br />
<math>\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120-30-30</math> triangles because of the equilateral triangles, <cmath>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}.</cmath> Also, <math>[AED]=2\cdot2\cdot\frac{1}{2}\cdot\sin{120^o}=\sqrt{3}</math> and so <cmath>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}.</cmath><br />
<br />
==Solution 2==<br />
<br />
<asy><br />
/* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again xD */<br />
pair A=(-2.4638,4.10658);<br />
pair B=(-4,2.6567453480756127);<br />
pair C=(-3.47132,0.6335248637894945);<br />
pair D=(-1.464483379039766,0.6335248637894945);<br />
pair E=(-0.956630463955801,2.6567453480756127);<br />
pair F=(-1.85,2);<br />
pair G=(-3.1,2);<br />
draw(A--B--C--D--E--A);<br />
draw(A--C--A--D);<br />
label("A",A,N);<br />
label("B",B,W);<br />
label("C",C,S);<br />
label("D",D,S);<br />
label("E",E,dir(0));<br />
dot(A^^B^^C^^D^^E);<br />
dot(F^^G, gray);<br />
draw(A--G--A--F, gray);<br />
draw(B--F--C, gray);<br />
draw(E--G--D, gray);<br />
</asy><br />
<br />
Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>.<br />
<br />
== Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area) ==<br />
https://youtu.be/QtSbAKUb1VE<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==Video Solution by Mathematical Dexterity (Basic Area Formulas)==<br />
https://www.youtube.com/watch?v=7kDTlVixsD0<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/FV9AnyERgJQ?t=1226<br />
<br />
~IceMatrix<br />
==Video Solution by Interstigation (Ignore Useless Segments)==<br />
https://youtu.be/9eChInz03UQ<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=14|num-a=16}}<br />
{{AMC10 box|year=2021|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_19&diff=1574842019 AMC 10A Problems/Problem 192021-07-07T13:52:15Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
<br />
What is the least possible value of<br />
<cmath>(x+1)(x+2)(x+3)(x+4)+2019</cmath>where <math>x</math> is a real number?<br />
<br />
<math>\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021</math><br />
<br />
==Solution 1==<br />
<br />
Grouping the first and last terms and two middle terms gives <math>(x^2+5x+4)(x^2+5x+6)+2019</math>, which can be simplified to <math>(x^2+5x+5)^2-1+2019</math>. Noting that squares are nonnegative, and verifying that <math>x^2+5x+5=0</math> for some real <math>x</math>, the answer is <math>\boxed{\textbf{(B) } 2018}</math>.<br />
<br />
==Solution 2==<br />
<br />
Let <math>a=x+\tfrac{5}{2}</math>. Then the expression <math>(x+1)(x+2)(x+3)(x+4)</math> becomes <math>\left(a-\tfrac{3}{2}\right)\left(a-\tfrac{1}{2}\right)\left(a+\tfrac{1}{2}\right)\left(a+\tfrac{3}{2}\right)</math>. <br />
<br />
We can now use the difference of two squares to get <math>\left(a^2-\tfrac{9}{4}\right)\left(a^2-\tfrac{1}{4}\right)</math>, and expand this to get <math>a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}</math>.<br />
<br />
Refactor this by completing the square to get <math>\left(a^2-\tfrac{5}{4}\right)^2-1</math>, which has a minimum value of <math>-1</math>. The answer is thus <math>2019-1=\boxed{\textbf{(B) }2018}</math>.<br />
<br />
==Solution 3 (calculus)==<br />
<br />
Similar to Solution 1, grouping the first and last terms and the middle terms, we get <math>(x^2+5x+4)(x^2+5x+6)+2019</math>. <br />
<br />
Letting <math>y=x^2+5x</math>, we get the expression <math>(y+4)(y+6)+2019</math>. Now, we can find the critical points of <math>(y+4)(y+6)</math> to minimize the function:<br />
<br />
<math>\frac{d}{dx}(y^2+10y+24)=0</math><br />
<br />
<math>2y+10=0</math><br />
<br />
<math>2y(y+5)=0</math><br />
<br />
<math>y=-5,0</math><br />
<br />
To minimize the result, we use <math>y=-5</math>. Hence, the minimum is <math>(-5+4)(-5+6)=-1</math>, so <math>-1+2019 = \boxed{\textbf{(B) }2018}</math>.<br />
<br />
''Note'': We could also have used the result that minimum/maximum point of a parabola <math>y = ax^2 + bx + c</math> occurs at <math>x=-\frac{b}{2a}</math>.<br />
<br />
==Solution 4==<br />
<br />
The expression is negative when an odd number of the factors are negative. This happens when <math>-2 < x < -1</math> or <math>-4 < x < -3</math>. Plugging in <math>x = -\frac32</math> or <math>x = -\frac72</math> yields <math>-\frac{15}{16}</math>, which is very close to <math>-1</math>. Thus the answer is <math>-1 + 2019 = \boxed{\textbf{(B) }2018}</math>.<br />
<br />
==Solution 5 (using the answer choices) ==<br />
<br />
Answer choices <math>C</math>, <math>D</math>, and <math>E</math> are impossible, since <math>(x+1)(x+2)(x+3)(x+4)</math> can be negative (as seen when e.g. <math>x = -\frac{3}{2}</math>). Plug in <math>x = -\frac{3}{2}</math> to see that it becomes <math>2019 - \frac{15}{16}</math>, so round this to <math>\boxed{\textbf{(B) }2018}</math>.<br />
<br />
We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.<br />
<br />
<br />
==Video Solutions==<br />
https://www.youtube.com/watch?v=Lis8yKT9WXc (less than 2 minutes)<br />
*https://youtu.be/NRa3VnjNVbw - Education, the Study of Everything<br />
*https://www.youtube.com/watch?v=Mfa7j2BoNjI<br />
*https://youtu.be/tIzJtgJbHGc - savannahsolver<br />
*https://youtu.be/3dfbWzOfJAI?t=3319 - pi_is_3.14<br />
*https://youtu.be/GmUWIXXf_uk?t=1134 ~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=1571262021 AMC 10B Problems/Problem 22021-06-30T23:50:13Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
What is the value of <math>\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}</math>?<br />
<br />
<math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6</math><br />
<br />
==Solution==<br />
Note that the square root of any square is always the absolute value of the squared number because the square root function will only return a positive number. By squaring both <math>3</math> and <math>2\sqrt{3}</math>, we see that <math>2\sqrt{3}>3</math>, thus <math>3-2\sqrt{3}</math> is negative, so we must take the absolute value of <math>3-2\sqrt{3}</math>, which is just <math>2\sqrt{3}-3</math>. Knowing this, the first term in the expression equals <math>2\sqrt{3}-3</math> and the second term is <math>3+2\sqrt3</math>, and summing the two gives <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math>.<br />
<br />
~bjc, abhinavg0627 and JackBocresion<br />
<br />
==Solution 2==<br />
Let <math>x = \sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}</math>, then <math>x^2 = (3-2\sqrt{3})^2+2\sqrt{(-3)^2}+(3+2\sqrt3)^2</math>. The <math>2\sqrt{(-3)^2}</math> term is there due to difference of squares. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel<br />
<br />
==Video Solution==<br />
https://youtu.be/HHVdPTLQsLc<br />
~Math Python<br />
<br />
==Video Solution (50 seconds) by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=ScZ5VK7QTpY<br />
<br />
== Video Solution by OmegaLearn ==<br />
https://youtu.be/Df3AIGD78xM<br />
<br />
==Video Solution 4==<br />
https://youtu.be/v71C6cFbErQ<br />
<br />
~savannahsolver<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/gLahuINjRzU?t=154<br />
<br />
~IceMatrix<br />
<br />
==Video Solution by Interstigation==<br />
https://youtu.be/DvpN56Ob6Zw?t=101<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=1571252021 AMC 10B Problems/Problem 22021-06-30T23:50:05Z<p>Mc21s: </p>
<hr />
<div>==Problem==<br />
What is the value of <math>\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}</math>?<br />
<br />
<math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6</math><br />
<br />
==Solution==<br />
Note that the square root of any square is always the absolute value of the squared number because the square root function will only return a positive number. By squaring both <math>3</math> and <math>2\sqrt{3}</math>, we see that <math>2\sqrt{3}>3</math>, thus <math>3-2\sqrt{3}</math> is negative, so we must take the absolute value of <math>3-2\sqrt{3}</math>, which is just <math>2\sqrt{3}-3</math>. Knowing this, the first term in the expression equals <math>2\sqrt{3}-3</math> and the second term is <math>3+2\sqrt3</math>, and summing the two gives <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math>.<br />
<br />
~bjc, abhinavg0627 and JackBocresion<br />
<br />
==Solution 2==<br />
Let <math>x = \sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}</math>, then <math>x^2 = (3-2\sqrt{3})^2+2\sqrt{(-3)^2}+(3+2\sqrt3)^2</math>. The <math>2\sqrt{(-3)^2}</math> term is there due to difference of squares. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel<br />
<br />
==Video Solution==<br />
https://youtu.be/HHVdPTLQsLc<br />
~Math Python<br />
<br />
==Video Solution (50 seconds) by Mathematical Dexterity==<br />
https://www.youtube.com/watch?v=ScZ5VK7QTpY<br />
<br />
== Video Solution by OmegaLearn ==<br />
https://youtu.be/Df3AIGD78xM<br />
<br />
==Video Solution 3==<br />
https://youtu.be/v71C6cFbErQ<br />
<br />
~savannahsolver<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/gLahuINjRzU?t=154<br />
<br />
~IceMatrix<br />
<br />
==Video Solution by Interstigation==<br />
https://youtu.be/DvpN56Ob6Zw?t=101<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_17&diff=1571082021 AMC 12A Problems/Problem 172021-06-30T20:34:47Z<p>Mc21s: </p>
<hr />
<div>{{duplicate|[[2021 AMC 10A Problems#Problem 17|2021 AMC 10A #17]] and [[2021 AMC 12A Problems#Problem 17|2021 AMC 12A #17]]}}<br />
<br />
==Problem==<br />
Trapezoid <math>ABCD</math> has <math>\overline{AB}\parallel\overline{CD},BC=CD=43</math>, and <math>\overline{AD}\perp\overline{BD}</math>. Let <math>O</math> be the intersection of the diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>, and let <math>P</math> be the midpoint of <math>\overline{BD}</math>. Given that <math>OP=11</math>, the length of <math>AD</math> can be written in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215</math><br />
<br />
==Diagram==<br />
[[File:2021 AMC 12A Problem 17 (Revised).png|center|600px]]<br />
~MRENTHUSIASM (by Geometry Expressions)<br />
<br />
== Solution 1 ==<br />
<br />
Angle chasing reveals that <math>\triangle BPC\sim\triangle BDA</math>, therefore <br />
<cmath>2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43}</cmath><br />
<cmath>AB=86</cmath><br />
Additional angle chasing shows that <math>\triangle ABO \sim\triangle CDO</math>, therefore<br />
<cmath>2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11}</cmath><br />
<cmath>BP=33 \Rightarrow BD=66</cmath><br />
Since <math>\triangle ADB</math> is right, the Pythagorean theorem implies that<br />
<cmath>AD=\sqrt{86^2-66^2}</cmath><br />
<cmath>AD=4\sqrt{190}</cmath><br />
<math>4\sqrt{190}\implies 4 + 190 = \boxed{\textbf{D) } 194}</math><br />
<br />
~mn28407 (minor edits by eagleye)<br />
<br />
==Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)==<br />
Since <math>\triangle BCD</math> is isosceles with base <math>\overline{BD},</math> it follows that median <math>\overline{CP}</math> is also an altitude. Let <math>OD=x</math> and <math>CP=h,</math> so <math>PB=x+11.</math><br />
<br />
Since <math>\angle AOD=\angle COP</math> by vertical angles, we conclude that <math>\triangle AOD\sim\triangle COP</math> by AA, from which <math>\frac{AD}{CP}=\frac{OD}{OP},</math> or <cmath>AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.</cmath> Let the brackets denote areas. Notice that <math>[AOD]=[BOC]</math> (By the same base and height, <math>[ACD]=[BDC].</math> Subtracting <math>[OCD]</math> from both sides gives <math>[AOD]=[BOC].</math>). Doubling both sides produces<br />
<cmath>\begin{align*}<br />
2[AOD]&=2[BOC] \\<br />
OD\cdot AD&=OB\cdot CP \\<br />
x\left(\frac{hx}{11}\right)&=(x+22)h \\<br />
x^2&=11(x+22).<br />
\end{align*}</cmath><br />
Rearranging and factoring result in <math>(x-22)(x+11)=0,</math> from which <math>x=22.</math><br />
<br />
Applying the Pythagorean Theorem to right <math>\triangle CPB,</math> we have <cmath>h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.</cmath> Finally, we get <cmath>AD=h\cdot\frac{x}{11}=4\sqrt{190},</cmath> so the answer is <math>4+190=\boxed{\textbf{(D) }194}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3 (Short)==<br />
Let <math>CP = y</math> and <math>CP</math> is perpendicular bisector of <math>DB.</math> Let <math>DO = x,</math> so <math>DP = PB = 11+x.</math><br />
<br />
(1) <math>\triangle CPO \sim \triangle ADO,</math> so we get <math>\frac{AD}{x} = \frac{y}{11},</math> or <math>AD = \frac{xy}{11}.</math><br />
<br />
(2) pythag on <math>\triangle CDP</math> gives <math>(11+x)^2 + y^2 = 43^2.</math><br />
<br />
(3) <math>\triangle BPC \sim \triangle BDA</math> with ratio <math>1:2,</math> so <math>AD = 2y.</math> (remember that <math>P</math> is the midpoint of <math>BD</math>)<br />
<br />
Thus, <math>xy/11 = 2y,</math> or <math>x = 22.</math> And <math>y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},</math> so <math>AD = 4 \sqrt{190}</math> and the answer is <math>\boxed{194}.</math><br />
<br />
~ ccx09<br />
<br />
==Solution 4 - Extending the line==<br />
Observe that <math>\triangle BPC</math> is congruent to <math>\triangle DPC</math>; both are similar to <math>\triangle BDA</math>. Let's extend <math>\overline{AD}</math> and <math>\overline{BC}</math> past points <math>D</math> and <math>C</math> respectively, such that they intersect at a point <math>E</math>. Observe that <math>\angle BDE</math> is <math>90</math> degrees, and that <math>\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA</math>. Thus, by ASA, we know that <math>\triangle ABD \cong \triangle EBD</math>, thus, <math>AD = ED</math>, meaning <math>D</math> is the midpoint of <math>AE</math>.<br />
Let <math>M</math> be the midpoint of <math>\overline{DE}</math>. Note that <math>\triangle CME</math> is congruent to <math>\triangle BPC</math>, thus <math>BC = CE</math>, meaning <math>C</math> is the midpoint of <math>\overline{BE}.</math><br />
<br />
Therefore, <math>\overline{AC}</math> and <math>\overline{BD}</math> are both medians of <math>\triangle ABE</math>. This means that <math>O</math> is the centroid of <math>\triangle ABE</math>; therefore, because the centroid divides the median in a 2:1 ratio, <math>\frac{BO}{2} = DO = \frac{BD}{3}</math>. Recall that <math>P</math> is the midpoint of <math>BD</math>; <math>DP = \frac{BD}{2}</math>. The question tells us that <math>OP = 11</math>; <math>DP-DO=11</math>; we can write this in terms of <math>DB</math>; <math>\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66</math>.<br />
<br />
<br />
We are almost finished. Each side length of <math>\triangle ABD</math> is twice as long as the corresponding side length <math>\triangle CBP</math> or <math>\triangle CPD</math>, since those triangles are similar; this means that <math>AB = 2 \cdot 43 = 86</math>. Now, by Pythagorean theorem on <math>\triangle ABD</math>, <math>AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}</math>. <math>4+190 = \boxed{194, \textbf{D}}</math><br />
<br />
~ ihatemath123<br />
<br />
== Video Solution by Mathematical Dexterity ==<br />
https://www.youtube.com/watch?v=QzAVdsgBBqg<br />
<br />
== Video Solution (Using Similar Triangles, Pythagorean Theorem) ==<br />
https://youtu.be/gjeSGJy_ld4<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtube.com/watch?v=rtdovluzgQs <br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=16|num-a=18}}<br />
{{AMC12 box|year=2021|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_9&diff=1564452018 AIME II Problems/Problem 92021-06-20T14:48:37Z<p>Mc21s: </p>
<hr />
<div>__TOC__<br />
==Problem==<br />
Octagon <math>ABCDEFGH</math> with side lengths <math>AB = CD = EF = GH = 10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <math>23</math> <math>\times</math> <math>27</math> rectangle with side <math>\overline{AH}</math> on a short side of the rectangle, as shown. Let <math>J</math> be the midpoint of <math>\overline{AH}</math>, and partition the octagon into 7 triangles by drawing segments <math>\overline{JB}</math>, <math>\overline{JC}</math>, <math>\overline{JD}</math>, <math>\overline{JE}</math>, <math>\overline{JF}</math>, and <math>\overline{JG}</math>. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.<br />
<br />
==Solution 1 (Massive Shoelace)==<br />
We represent octagon <math>ABCDEFGH</math> in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that <math>A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})</math>. Recall that the centroid is <math>\frac{1}{3}</math> way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point <math>J</math>. Furthermore, we can take advantage of the reflective symmetry across the line parallel to <math>BC</math> going through <math>J</math> by dealing with less coordinates and ommiting the <math>\frac{1}{2}</math> in the shoelace formula. <br />
<br />
By doing some basic algebra, we find that the coordinates of the centroids of <math>\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE</math> are <math>\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),</math> and <math>\left(18, -\frac{23}{2}\right)</math>, respectively. We'll have to throw in the projection of the centroid of <math>\bigtriangleup JAB</math> to the line of reflection to apply shoelace, and that point is <math>\left( \frac{8}{3}, -\frac{23}{2}\right)</math><br />
<br />
Finally, applying Shoelace, we get: <br />
<math>\left|\left(\frac{8}{3}\cdot (-\frac{23}{6})+9\cdot (-\frac{35}{6})+\frac{46}{3}\cdot (-\frac{23}{2})+18\cdot (\frac{-23}{2})+\frac{8}{3}\cdot (-\frac{35}{6})\right) - \left((-\frac{35}{6}\cdot 9) +\\(-\frac{23}{6}\cdot \frac{46}{3})+ (-\frac{35}{6}\cdot 18)+(\frac{-23}{2}\cdot \frac{8}{3})+(-\frac{23}{2}\cdot \frac{8}{3})\right)\right|</math><br />
<math>=\left|\left(-\frac{92}{9}-\frac{105}{2}-\frac{529}{3}-207-\frac{140}{9}\right)-\left(-\frac{105}{2}-\frac{529}{9}-105-\frac{92}{3}-\frac{92}{3}\right)\right|</math><br />
<math>=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|</math><br />
<math>=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}</math><br />
<br />
Solution by ktong<br />
<br />
==Solution 2 (Homothety)==<br />
Draw the heptagon whose vertices are the midpoints of octagon <math>ABCDEFGH</math> except <math>J</math>.<br />
We have a homothety since:<br />
<br />
1. <math>J</math> passes through corresponding vertices of the two heptagons.<br />
<br />
2. By centroid properties, our ratio between the sidelengths is <math>\frac{2}{3},</math> and their area ratio is hence <math>\frac{4}{9}.</math> <br />
<br />
Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is <cmath>= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.</cmath> The area of each triangle is <cmath>=<br />
\frac{1}{2}\cdot 17\cdot 4=34.</cmath><br />
<br />
Hence, the area of the large heptagon is <cmath>2\cdot 190+34=414.</cmath> Then, from our homothety, the area of the required heptagon is <cmath>\frac{4}{9}\cdot 414=\boxed{184}.</cmath><br />
~novus677<br />
<br />
==Video Solution (Mathematical Dexterity)==<br />
https://www.youtube.com/watch?v=HUwJqixBLUI<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=II|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=How_many_prime_numbers_are_between_30_and_40%3F&diff=156007How many prime numbers are between 30 and 40?2021-06-15T17:02:13Z<p>Mc21s: </p>
<hr />
<div>The answer is <math>2</math>: <math>31</math> and <math>37</math>.</div>Mc21shttps://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_3&diff=1558941994 AIME Problems/Problem 32021-06-13T18:58:19Z<p>Mc21s: </p>
<hr />
<div>== Problem ==<br />
The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math><br />
<center><math>f(x)+f(x-1) = x^2.\,</math></center><br />
If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>?<br />
<br />
== Solution 1 ==<br />
<cmath>\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\<br />
&= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94 \\<br />
&= 4561 \end{align*}</cmath><br />
<br />
So, the remainder is <math>\boxed{561}</math>.<br />
<br />
== Solution 2 ==<br />
Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is,<br />
<cmath>T_{n-1} + T_n = n^2,</cmath><br />
where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number.<br />
<br />
Using this, as well as using the fact that the value of <math>f(x)</math> directly determines the value of <math>f(x+1)</math> and <math>f(x-1),</math> we conclude that <math>f(n) = T_n + K</math> for all odd <math>n</math> and <math>f(n) = T_n - K</math> for all even <math>n,</math> where <math>K</math> is a constant real number.<br />
<br />
Since <math>f(19) = 94</math> and <math>T_{19} = 190,</math> we see that <math>K = -96.</math> It follows that <math>f(94) = T_{94} - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,</math> so the answer is <math>561.</math><br />
<br />
== See also ==<br />
{{AIME box|year=1994|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mc21s