https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jskalarickal&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:54:21ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_24&diff=898642015 AMC 10B Problems/Problem 242018-01-17T19:21:57Z<p>Jskalarickal: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin <math>p_0=(0,0)</math> facing to the east and walks one unit, arriving at <math>p_1=(1,0)</math>. For <math>n=1,2,3,\dots</math>, right after arriving at the point <math>p_n</math>, if Aaron can turn <math>90^\circ</math> left and walk one unit to an unvisited point <math>p_{n+1}</math>, he does that. Otherwise, he walks one unit straight ahead to reach <math>p_{n+1}</math>. Thus the sequence of points continues <math>p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)</math>, and so on in a counterclockwise spiral pattern. What is <math>p_{2015}</math>?<br />
<br />
<math> \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) </math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
The first thing we would do is track Aaron's footsteps:<br />
<br />
He starts by taking <math>1</math> step East and <math>1</math> step North, ending at <math>(1,1)</math> after <math>2</math> steps and about to head West.<br />
<br />
Then he takes <math>2</math> steps West and <math>2</math> steps South, ending at <math>(-1,-1</math>) after <math>2+4</math> steps, and about to head East.<br />
<br />
Then he takes <math>3</math> steps East and <math>3</math> steps North, ending at <math>(2,2)</math> after <math>2+4+6</math> steps, and about to head West.<br />
<br />
Then he takes <math>4</math> steps West and <math>4</math> steps South, ending at <math>(-2,-2)</math> after <math>2+4+6+8</math> steps, and about to head East.<br />
<br />
From this pattern, we can notice that for any integer <math>k \ge 1</math> he's at <math>(-k, -k)</math> after <math>2 + 4 + 6 + ... + 4k</math> steps, and about to head East. There are <math>2k</math> terms in the sum, with an average value of <math>(2 + 4k)/2 = 2k + 1</math>, so:<br />
<br />
<cmath>2 + 4 + 6 + ... + 4k = 2k(2k + 1)</cmath><br />
<br />
If we substitute <math>k = 22</math> into the equation: <math>44(45) = 1980 < 2015</math>. So he has <math>35</math> moves to go. This makes him end up at <math>(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}</math>.<br />
<br />
===Solution 2===<br />
We are given that Aaron starts at <math>(0, 0)</math>, and we note that his net steps follow the pattern of <math>+1</math> in the <math>x</math>-direction, <math>+1</math> in the <math>y</math>-direction, <math>-2</math> in the <math>x</math>-direction, <math>-2</math> in the <math>y</math>-direction, <math>+3</math> in the <math>x</math>-direction, <math>+3</math> in the <math>y</math>-direction, and so on, where we add odd and subtract even.<br />
<br />
We want <math>2 + 4 + 6 + 8 + ... + 2n = 2015</math>, but it does not work out cleanly. Instead, we get that <math>2 + 4 + 6 + ... + 2(44) = 1980</math>, which means that there are <math>35</math> extra steps past adding <math>-44</math> in the <math>x</math>-direction (and the final number we add in the <math>y</math>-direction is <math>-44</math>).<br />
<br />
So <math>p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)</math>.<br />
<br />
We can group <math>1-2+3-4+5...-44</math> as <math>(1-2)+(3-4)+(5-6)+...+(43-44) = -22</math>.<br />
<br />
Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>.<br />
<br />
===Solution 3===<br />
Looking at his steps, we see that he walks in a spiral shape. At the 8th step, he is on the bottom right corner of the 3x3 square centered on the origin. On the 24th step, he is on the bottom right corner of the 5x5 square centered at the origin. It seems that the <math>p_{n^2-1}</math> is the bottom right corner of the <math>n</math>x<math>n</math> square. This makes sense since, after <math>n^2-1</math>, he has been on n^2 dots, including the point <math>p_{0}</math>. Also, this is only for odd <math>n</math>, because starting with the 1x1 square, we can only add one extra set of dots to each side, so we cannot get even <math>n</math>. Since <math>45^2=2025</math>, <math>p_{2024}</math> is the bottom right corner of the 45x45 square. This point is <math>\frac{45-1}{2}=22</math> over to the right, and therefore 22 down, so <math>p_{2024}=(22, -22)</math>. Since <math>p_{2024}</math> is 9 ahead of <math>p_{2015}</math>, we go back 9 spaces to <math>\boxed{\textbf{(D)}\; (13, -22)}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_24&diff=898632015 AMC 10B Problems/Problem 242018-01-17T19:21:40Z<p>Jskalarickal: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin <math>p_0=(0,0)</math> facing to the east and walks one unit, arriving at <math>p_1=(1,0)</math>. For <math>n=1,2,3,\dots</math>, right after arriving at the point <math>p_n</math>, if Aaron can turn <math>90^\circ</math> left and walk one unit to an unvisited point <math>p_{n+1}</math>, he does that. Otherwise, he walks one unit straight ahead to reach <math>p_{n+1}</math>. Thus the sequence of points continues <math>p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)</math>, and so on in a counterclockwise spiral pattern. What is <math>p_{2015}</math>?<br />
<br />
<math> \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) </math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
The first thing we would do is track Aaron's footsteps:<br />
<br />
He starts by taking <math>1</math> step East and <math>1</math> step North, ending at <math>(1,1)</math> after <math>2</math> steps and about to head West.<br />
<br />
Then he takes <math>2</math> steps West and <math>2</math> steps South, ending at <math>(-1,-1</math>) after <math>2+4</math> steps, and about to head East.<br />
<br />
Then he takes <math>3</math> steps East and <math>3</math> steps North, ending at <math>(2,2)</math> after <math>2+4+6</math> steps, and about to head West.<br />
<br />
Then he takes <math>4</math> steps West and <math>4</math> steps South, ending at <math>(-2,-2)</math> after <math>2+4+6+8</math> steps, and about to head East.<br />
<br />
From this pattern, we can notice that for any integer <math>k \ge 1</math> he's at <math>(-k, -k)</math> after <math>2 + 4 + 6 + ... + 4k</math> steps, and about to head East. There are <math>2k</math> terms in the sum, with an average value of <math>(2 + 4k)/2 = 2k + 1</math>, so:<br />
<br />
<cmath>2 + 4 + 6 + ... + 4k = 2k(2k + 1)</cmath><br />
<br />
If we substitute <math>k = 22</math> into the equation: <math>44(45) = 1980 < 2015</math>. So he has <math>35</math> moves to go. This makes him end up at <math>(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}</math>.<br />
<br />
===Solution 2===<br />
We are given that Aaron starts at <math>(0, 0)</math>, and we note that his net steps follow the pattern of <math>+1</math> in the <math>x</math>-direction, <math>+1</math> in the <math>y</math>-direction, <math>-2</math> in the <math>x</math>-direction, <math>-2</math> in the <math>y</math>-direction, <math>+3</math> in the <math>x</math>-direction, <math>+3</math> in the <math>y</math>-direction, and so on, where we add odd and subtract even.<br />
<br />
We want <math>2 + 4 + 6 + 8 + ... + 2n = 2015</math>, but it does not work out cleanly. Instead, we get that <math>2 + 4 + 6 + ... + 2(44) = 1980</math>, which means that there are <math>35</math> extra steps past adding <math>-44</math> in the <math>x</math>-direction (and the final number we add in the <math>y</math>-direction is <math>-44</math>).<br />
<br />
So <math>p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)</math>.<br />
<br />
We can group <math>1-2+3-4+5...-44</math> as <math>(1-2)+(3-4)+(5-6)+...+(43-44) = -22</math>.<br />
<br />
Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>.<br />
<br />
===Solution 3===<br />
Looking at his steps, we see that he walks in a spiral shape. At the 8th step, he is on the bottom right corner of the 3x3 square centered on the origin. On the 24th step, he is on the bottom right corner of the 5x5 square centered at the origin. It seems that the <math>p_{n^2-1}</math> is the bottom right corner of the <math>n</math>x<math>n</math> square. This makes sense since, after <math>n^2-1</math>, he has been on n^2 dots, including the point <math>p_{0}</math>. Also, this is only for odd <math>n</math>, because starting with the 1x1 square, we can only add one extra set of dots to each side, so we cannot get even <math>n</math>. Since <math>45^2=2025</math>, <math>p_{2024}</math> is the bottom right corner of the 45x45 square. This point is <math>\frac{45-1}{2}=22</math> over to the right, and therefore 22 down, so <math>p_{2024}=(22, -22). Since </math>p_{2024}<math> is 9 ahead of </math>p_{2015}<math>, we go back 9 spaces to </math>\boxed{\textbf{(D)}\; (13, -22)}$.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_24&diff=898622015 AMC 10B Problems/Problem 242018-01-17T19:20:41Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin <math>p_0=(0,0)</math> facing to the east and walks one unit, arriving at <math>p_1=(1,0)</math>. For <math>n=1,2,3,\dots</math>, right after arriving at the point <math>p_n</math>, if Aaron can turn <math>90^\circ</math> left and walk one unit to an unvisited point <math>p_{n+1}</math>, he does that. Otherwise, he walks one unit straight ahead to reach <math>p_{n+1}</math>. Thus the sequence of points continues <math>p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)</math>, and so on in a counterclockwise spiral pattern. What is <math>p_{2015}</math>?<br />
<br />
<math> \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) </math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
The first thing we would do is track Aaron's footsteps:<br />
<br />
He starts by taking <math>1</math> step East and <math>1</math> step North, ending at <math>(1,1)</math> after <math>2</math> steps and about to head West.<br />
<br />
Then he takes <math>2</math> steps West and <math>2</math> steps South, ending at <math>(-1,-1</math>) after <math>2+4</math> steps, and about to head East.<br />
<br />
Then he takes <math>3</math> steps East and <math>3</math> steps North, ending at <math>(2,2)</math> after <math>2+4+6</math> steps, and about to head West.<br />
<br />
Then he takes <math>4</math> steps West and <math>4</math> steps South, ending at <math>(-2,-2)</math> after <math>2+4+6+8</math> steps, and about to head East.<br />
<br />
From this pattern, we can notice that for any integer <math>k \ge 1</math> he's at <math>(-k, -k)</math> after <math>2 + 4 + 6 + ... + 4k</math> steps, and about to head East. There are <math>2k</math> terms in the sum, with an average value of <math>(2 + 4k)/2 = 2k + 1</math>, so:<br />
<br />
<cmath>2 + 4 + 6 + ... + 4k = 2k(2k + 1)</cmath><br />
<br />
If we substitute <math>k = 22</math> into the equation: <math>44(45) = 1980 < 2015</math>. So he has <math>35</math> moves to go. This makes him end up at <math>(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}</math>.<br />
<br />
===Solution 2===<br />
We are given that Aaron starts at <math>(0, 0)</math>, and we note that his net steps follow the pattern of <math>+1</math> in the <math>x</math>-direction, <math>+1</math> in the <math>y</math>-direction, <math>-2</math> in the <math>x</math>-direction, <math>-2</math> in the <math>y</math>-direction, <math>+3</math> in the <math>x</math>-direction, <math>+3</math> in the <math>y</math>-direction, and so on, where we add odd and subtract even.<br />
<br />
We want <math>2 + 4 + 6 + 8 + ... + 2n = 2015</math>, but it does not work out cleanly. Instead, we get that <math>2 + 4 + 6 + ... + 2(44) = 1980</math>, which means that there are <math>35</math> extra steps past adding <math>-44</math> in the <math>x</math>-direction (and the final number we add in the <math>y</math>-direction is <math>-44</math>).<br />
<br />
So <math>p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)</math>.<br />
<br />
We can group <math>1-2+3-4+5...-44</math> as <math>(1-2)+(3-4)+(5-6)+...+(43-44) = -22</math>.<br />
<br />
Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>.<br />
<br />
===Solution 3===<br />
Looking at his steps, we see that he walks in a spiral shape. At the 8th step, he is on the bottom right corner of the 3x3 square centered on the origin. On the 24th step, he is on the bottom right corner of the 5x5 square centered at the origin. It seems that the <math>p_{n^2-1}</math> is the bottom right corner of the <math>n</math>x<math>n</math> square. This makes sense since, after <math>n^2-1</math>, he has been on n^2 dots, including the point <math>p_{0}</math>. Also, this is only for odd <math>n</math>, because starting with the 1x1 square, we can only add one extra set of dots to each side, so we cannot get even <math>n</math>. Since <math>45^2=2025</math>, <math>p_{2024}</math> is the bottom right corner of the 45x45 square. This point is <math>\frac{45-1}{2}=22</math> over to the right, and therefore 22 down, so <math>p_{2024}=(22, -22). Since </math>p_{2024} is 9 ahead of <math>p_{2015}</math>, we go back 9 spaces to <math>\boxed{\textbf{(D)}\; (13, -22)}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_24&diff=898612015 AMC 10B Problems/Problem 242018-01-17T19:20:11Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin <math>p_0=(0,0)</math> facing to the east and walks one unit, arriving at <math>p_1=(1,0)</math>. For <math>n=1,2,3,\dots</math>, right after arriving at the point <math>p_n</math>, if Aaron can turn <math>90^\circ</math> left and walk one unit to an unvisited point <math>p_{n+1}</math>, he does that. Otherwise, he walks one unit straight ahead to reach <math>p_{n+1}</math>. Thus the sequence of points continues <math>p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)</math>, and so on in a counterclockwise spiral pattern. What is <math>p_{2015}</math>?<br />
<br />
<math> \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) </math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
The first thing we would do is track Aaron's footsteps:<br />
<br />
He starts by taking <math>1</math> step East and <math>1</math> step North, ending at <math>(1,1)</math> after <math>2</math> steps and about to head West.<br />
<br />
Then he takes <math>2</math> steps West and <math>2</math> steps South, ending at <math>(-1,-1</math>) after <math>2+4</math> steps, and about to head East.<br />
<br />
Then he takes <math>3</math> steps East and <math>3</math> steps North, ending at <math>(2,2)</math> after <math>2+4+6</math> steps, and about to head West.<br />
<br />
Then he takes <math>4</math> steps West and <math>4</math> steps South, ending at <math>(-2,-2)</math> after <math>2+4+6+8</math> steps, and about to head East.<br />
<br />
From this pattern, we can notice that for any integer <math>k \ge 1</math> he's at <math>(-k, -k)</math> after <math>2 + 4 + 6 + ... + 4k</math> steps, and about to head East. There are <math>2k</math> terms in the sum, with an average value of <math>(2 + 4k)/2 = 2k + 1</math>, so:<br />
<br />
<cmath>2 + 4 + 6 + ... + 4k = 2k(2k + 1)</cmath><br />
<br />
If we substitute <math>k = 22</math> into the equation: <math>44(45) = 1980 < 2015</math>. So he has <math>35</math> moves to go. This makes him end up at <math>(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}</math>.<br />
<br />
===Solution 2===<br />
We are given that Aaron starts at <math>(0, 0)</math>, and we note that his net steps follow the pattern of <math>+1</math> in the <math>x</math>-direction, <math>+1</math> in the <math>y</math>-direction, <math>-2</math> in the <math>x</math>-direction, <math>-2</math> in the <math>y</math>-direction, <math>+3</math> in the <math>x</math>-direction, <math>+3</math> in the <math>y</math>-direction, and so on, where we add odd and subtract even.<br />
<br />
We want <math>2 + 4 + 6 + 8 + ... + 2n = 2015</math>, but it does not work out cleanly. Instead, we get that <math>2 + 4 + 6 + ... + 2(44) = 1980</math>, which means that there are <math>35</math> extra steps past adding <math>-44</math> in the <math>x</math>-direction (and the final number we add in the <math>y</math>-direction is <math>-44</math>).<br />
<br />
So <math>p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)</math>.<br />
<br />
We can group <math>1-2+3-4+5...-44</math> as <math>(1-2)+(3-4)+(5-6)+...+(43-44) = -22</math>.<br />
<br />
Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>.<br />
<br />
===Solution 2===<br />
Looking at his steps, we see that he walks in a spiral shape. At the 8th step, he is on the bottom right corner of the 3x3 square centered on the origin. On the 24th step, he is on the bottom right corner of the 5x5 square centered at the origin. It seems that the <math>p_{n^2-1}</math> is the bottom right corner of the <math>n</math>x<math>n</math> square. This makes sense since, after n^2-1, he has been on n^2 dots, including the point <math>p_{0}</math>. Also, this is only for odd <math>n</math>, because starting with the 1x1 square, we can only add one extra set of dots to each side, so we cannot get even <math>n</math>. Since <math>45^2=2025</math>, <math>p_{2024}</math> is the bottom right corner of the 45x45 square. This point is <math>\frac{45-1}{2}=22</math> over to the right, and therefore 22 down, so <math>p_{2024}=(22, -22). Since </math>p_{2024} is 9 ahead of <math>p_{2015}</math>, we go back 9 spaces to <math>\boxed{\textbf{(D)}\; (13, -22)}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_15&diff=825882010 AIME II Problems/Problem 152017-01-27T14:52:49Z<p>Jskalarickal: /* Problem 15 */</p>
<hr />
<div>== Problem 15 ==<br />
<br />
In triangle <math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>AB</math> with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <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 ==<br />
<br />
Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY}</math> since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMPN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields <math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</math>.<br />
<br />
<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. Also, <math>P</math> is the center of spiral similarity of segments <math>MD</math> and <math>NE</math>, so <math>\triangle PMD \sim \triangle PNE</math>. Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that <math>\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>.<br />
<br />
'''Note:''' Spiral similarities may sound complex, but they're really not. The fact that <math>\triangle PMD \sim \triangle PNE</math> is really just a result of simple angle chasing.<br />
<br />
Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero<br />
<br />
== Extension ==<br />
The work done in this problem leads to a nice extension of this problem:<br />
<br />
Given a <math>\triangle ABC</math> and points <math>A_1</math>, <math>A_2</math>, <math>B_1</math>, <math>B_2</math>, <math>C_1</math>, <math>C_2</math>, such that <math>A_1</math>, <math>A_2</math> <math>\in BC</math>, <math>B_1</math>, <math>B_2</math> <math>\in AC</math>, and <math>C_1</math>, <math>C_2</math> <math>\in AB</math>, then let <math>\omega_1</math> be the circumcircle of <math>\triangle AB_1C_1</math> and <math>\omega_2</math> be the circumcircle of <math>\triangle AB_2C_2</math>. Let <math>A'</math> be the intersection point of <math>\omega_1</math> and <math>\omega_2</math> distinct from <math>A</math>. Define <math>B'</math> and <math>C'</math> similarly. Then <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> concur. <br />
<br />
This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line <math>AA'</math> divides the opposite side <math>BC</math> into and similarly for the other two sides.<br />
<br />
==See Also==<br />
{{AIME box|year=2010|n=II|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823302013 AMC 10B Problems/Problem 242017-01-15T19:42:09Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the sum of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math>. <br />
<br />
If either <math>a+1</math> or <math>b+1</math> are odd, then <math>a</math> or <math>b</math> are even. Therefore, <math>a+1</math> and <math>b+1</math> are even, so <math>m</math> is a multiple of 4. The only two numbers from the range <math>2010-2019</math> that are multiples of 4 are 2012 and 2016.<br />
<br />
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.<br />
<br />
Factoring 2016, we get <math>2^5*3^2*7</math>. <math>wlog</math>, we have <math>a<b</math>.<br />
<br />
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823282013 AMC 10B Problems/Problem 242017-01-15T19:40:56Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the sum of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math>. <br />
<br />
If either <math>a+1</math> or <math>b+1</math> are odd, then <math>a</math> or <math>b</math> are even. Therefore, <math>a+1</math> and <math>b+1</math> are even, so <math>m</math> is a multiple of 4. The only two numbers from the <math>2010-2019</math> that are multiples of 4 are 2012 and 2016.<br />
<br />
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.<br />
<br />
Factoring 2016, we get <math>2^5*3^2*7</math>. <math>wlog</math>, we have <math>a<b</math>.<br />
<br />
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823272013 AMC 10B Problems/Problem 242017-01-15T19:37:39Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the the some of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math>. <br />
<br />
If either <math>a+1</math> or <math>b+1</math> are odd, then <math>a</math> or <math>b</math> are even. Therefore, <math>a+1</math> and <math>b+1</math> are even, so <math>m</math> is a multiple of 4. The only two numbers from the <math>2010-2019</math> that are multiples of 4 are 2012 and 2016.<br />
<br />
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.<br />
<br />
Factoring 2016, we get <math>2^5*3^2*7</math>. <math>wlog</math>, we have <math>a<b</math>.<br />
<br />
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823262013 AMC 10B Problems/Problem 242017-01-15T19:36:30Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the the some of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math>. <br />
<br />
If either <math>a+1</math> or <math>b+1</math> are odd, then <math>a</math> or <math>b</math> are even. Therefore, <math>a+1</math> and <math>b+1</math> are even, so <math>m</math> is a multiple of 4. The only two numbers from the <math>2010-2019</math> that are multiples of 4 are 2012 and 2016.<br />
<br />
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.<br />
<br />
Factoring 2016, we get <math>2^5*3^2*7</math>. Therefore, <math>wlog</math>, we have <math>a<b</math>.<br />
<br />
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A) \1}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823252013 AMC 10B Problems/Problem 242017-01-15T19:35:34Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the the some of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math>. <br />
<br />
If either <math>a+1</math> or <math>b+1</math> are odd, then <math>a</math> or <math>b</math> are even. Therefore, <math>a+1</math> and <math>b+1</math> are even, so <math>m</math> is a multiple of 4. The only two numbers from the <math>2010-2019</math> that are multiples of 4 are 2012 and 2016.<br />
<br />
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.<br />
<br />
Factoring 2016, we get <math>2^5*3^2*7</math>. Therefore, <math>wlog</math>, we have <math>a<b</math>.<br />
<br />
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>..<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823242013 AMC 10B Problems/Problem 242017-01-15T19:34:24Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the the some of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math>. <br />
<br />
If either <math>a+1</math> or <math>b+1</math> are odd, then <math>a</math> or <math>b</math> are even. Therefore, <math>a+1</math> and <math>b+1</math> are even, so <math>m</math> is a multiple of 4. The only two numbers from the <math>2010-2019</math> that are multiples of 4 are 2012 and 2016.<br />
<br />
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.<br />
<br />
Factoring 2016, we get <math>2^5*3^2*7</math>. Therefore, <math>wlog</math>, we have <math>a<b</math>.<br />
<br />
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>\boxed{\textbf{(A) \1}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823222013 AMC 10B Problems/Problem 242017-01-15T19:33:13Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the the some of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math>. <br />
<br />
If either <math>a+1</math> or <math>b+1</math> are odd, then <math>a</math> or <math>b</math> are even. Therefore, <math>a+1</math> and <math>b+1</math> are even, so <math>m</math> is a multiple of 4. The only two numbers from the <math>2010-2019</math> that are multiples of 4 are 2012 and 2016.<br />
<br />
Factoring 2012, we get <math>2^2*503</math>. To make <math>a+1</math> and <math>b+1</math> even, <math>wlog</math>, we have <math>a+1=2</math> and <math>b+1=1006</math>. However, if <math>a</math> was 1, then <math>a</math> is not prime, so 2012 is not nice.<br />
<br />
Factoring 2016, we get <math>2^5*3^2*7</math>. Therefore, <math>wlog</math>, we have <math>a<b</math>.<br />
<br />
Testing for the lowest <math>a</math>, we get <math>a+1=4</math> and <math>b+1=504</math>. Therefore, <math>a=3</math>, and <math>b=503</math>, so <math>n=2016</math> is nice, with <math>m=1509</math>. Therefore, the answer is <math>box{A}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823212013 AMC 10B Problems/Problem 242017-01-15T19:23:36Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the the some of the divisors of <math>m</math> is <math>1+a+b+ab=(a+1)(b+1)</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=823202013 AMC 10B Problems/Problem 242017-01-15T19:22:37Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution 1==<br />
A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of <math>a*b</math>. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Easily we can see that now we can take cases again. <br />
<br />
Case 1: Either <math>a</math> or <math>b</math> is 2. <br />
<br />
If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.<br />
<br />
Case 2: Both <math>a</math> and <math>b</math> are odd primes. <br />
<br />
This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>.<br />
<math>2012=4*503</math> so either <math>(a+1)</math> or <math>(b+1)</math> both has a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>. That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work. <math>2016 = 4 * 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
If <math>m</math> has four divisors, then its divisors would be 1, <math>a</math>, <math>b</math> and <math>ab</math>, where <math>a</math> and <math>b</math> are prime. Therefore, the the some of the divisors of <math>m</math> is<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_14&diff=802891990 AIME Problems/Problem 142016-09-17T15:11:28Z<p>Jskalarickal: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
The [[rectangle]] <math>ABCD^{}_{}</math> below has dimensions <math>AB^{}_{} = 12 \sqrt{3}</math> and <math>BC^{}_{} = 13 \sqrt{3}</math>. [[Diagonal]]s <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>P^{}_{}</math>. If triangle <math>ABP^{}_{}</math> is cut out and removed, edges <math>\overline{AP}</math> and <math>\overline{BP}</math> are joined, and the figure is then creased along segments <math>\overline{CP}</math> and <math>\overline{DP}</math>, we obtain a [[triangular pyramid]], all four of whose faces are [[isosceles triangle]]s. Find the volume of this pyramid.<br />
<br />
[[Image:AIME_1990_Problem_14.png]]<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><br />
<asy><br />
import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9);<br />
currentprojection = perspective(20,-20,12);<br />
triple O=(0,0,0),A=(0,399^.5,0),D=(108^.5,0,0),C=(-108^.5,0,0);<br />
pair CENTER=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y));<br />
triple P=(CENTER.x,CENTER.y,99/133^.5); /*, Pa=(P.x,P.y,0); <br />
D(P--Pa--A);D(C--Pa--D); */<br />
D((C+D)/2--A--C--D--P--C--P--A--D);<br />
MP("A",A,NE);MP("P",P,N);MP("C",C);MP("D",D);<br />
MP("13\sqrt{3}",(A+D)/2,E,small);MP("13\sqrt{3}",(A+C)/2,N,small);MP("12\sqrt{3}",(C+D)/2,SW,small);<br />
<br />
Our triangular pyramid has base $12\sqrt{3} - 13\sqrt{3} - 13\sqrt{3} \triangle$. The area of this isosceles triangle is easy to find by $[ACD] = \frac{1}{2}bh$, where we can find $h_{ACD}$ to be $\sqrt{399}$ by the [[Pythagorean Theorem]]. Thus $A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}$.<br />
<br />
<asy><br />
<center><br />
size(280);<br />
import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9);<br />
real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12);<br />
triple O=(0,0,0),A=(0,399^.5,0),D=(108^.5,0,0),C=(-108^.5,0,0);<br />
pair CENTER=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y));<br />
triple P=(CENTER.x,CENTER.y,99/133^.5), Pa=(P.x,P.y,0);<br />
D(A--C--D--P--C--P--A--D);<br />
D(P--Pa--A);D(C--Pa--D);D(circle(Pa,h));<br />
MP("A",A,NE);MP("C",C,NW);MP("D",D);MP("P",P,N);MP("P'",Pa,SW);<br />
MP("13\sqrt{3}",(A+D)/2,E,small);MP("13\sqrt{3}",(A+C)/2,NW,small);MP("12\sqrt{3}",(C+D)/2,SW,small);<br />
MP("h",(P--Pa)/2,W);MP("\frac{\sqrt{939}}2",(C+P)/2,NW);<br />
</asy></center> <!-- Asymptote replacement for Image:1990_AIME-14c.png by azjps --><br />
<br />
To find the volume, we want to use the equation <math>\frac 13Bh = 6\sqrt{133}h</math>, so we need to find the height of the [[tetrahedron]]. By the Pythagorean Theorem, <math>AP = CP = DP = \frac{\sqrt{939}}{2}</math>. If we let <math>P</math> be the center of a [[sphere]] with radius <math>\frac{\sqrt{939}}{2}</math>, then <math>A,C,D</math> lie on the sphere. The cross section of the sphere that contains <math>A,C,D</math> is a circle, and the center of that circle is the foot of the [[perpendicular]] from the center of the sphere. Hence the foot of the height we want to find occurs at the [[circumcenter]] of <math>\triangle ACD</math>. <br />
<br />
From here we just need to perform some brutish calculations. Using the formula <math>A = 18\sqrt{133} = \frac{abc}{4R}</math> (where <math>R</math> is the [[circumradius]]), we find <math>R = \frac{12\sqrt{3} \cdot (13\sqrt{3})^2}{4\cdot 18\sqrt{133}} = \frac{13^2\sqrt{3}}{2\sqrt{133}}</math> (there are slightly [[Law of Sines|simpler ways]] to calculate <math>R</math> since we have an isosceles triangle). By the Pythagorean Theorem, <br />
<br />
<cmath><br />
\begin{align*}h^2 &= PA^2 - R^2 \\<br />
&= \left(\frac{\sqrt{939}}{2}\right)^2 - \left(\frac{13^2\sqrt{3}}{2\sqrt{133}}\right)^2\\<br />
&= \frac{939 \cdot 133 - 13^4 \cdot 3}{4 \cdot 133} = \frac{13068 \cdot 3}{4 \cdot 133} = \frac{99^2}{133}\\<br />
h &= \frac{99}{\sqrt{133}}<br />
\end{align*}<br />
</cmath><br />
<br />
Finally, we substitute <math>h</math> into the volume equation to find <math>V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}</math>.<br />
<br />
=== Solution 2 ===<br />
Let <math>\triangle{ABC}</math> (or the triangle with sides <math>12\sqrt {3}</math>, <math>13\sqrt {3}</math>, <math>13\sqrt {3}</math>) be the base of our tetrahedron. We set points <math>C</math> and <math>D</math> as <math>(6\sqrt {3}, 0, 0)</math> and <math>( - 6\sqrt {3}, 0, 0)</math>, respectively. Using Pythagoras, we find <math>A</math> as <math>(0, \sqrt {399}, 0)</math>. We know that the [[vertex]] of the tetrahedron (<math>P</math>) has to be of the form <math>(x, y, z)</math>, where <math>z</math> is the [[altitude]] of the tetrahedron. Since the distance from <math>P</math> to points <math>A</math>, <math>B</math>, and <math>C</math> is <math>\frac {\sqrt {939}}{2}</math>, we can write three equations using the [[distance formula]]:<br />
<br />
<cmath><br />
\begin{align*}<br />
x^{2} + (y - \sqrt {399})^{2} + z^{2} &= \frac {939}{4}\\<br />
(x - 6\sqrt {3})^{2} + y^{2} + z^{2} &= \frac {939}{4}\\<br />
(x + 6\sqrt {3})^{2} + y^{2} + z^{2} &= \frac {939}{4}<br />
\end{align*}<br />
</cmath><br />
<br />
Subtracting the last two equations, we get <math>x = 0</math>. Solving for <math>y,z</math> with a bit of effort, we eventually get <math>x = 0</math>, <math>y = \frac {291}{2\sqrt {399}}</math>, <math>z = \frac {99}{\sqrt {133}}</math>.<br />
Since the area of a triangle is <math>\frac {1}{2}\cdot bh</math>, we have the base area as <math>18\sqrt {133}</math>. Thus, the volume is <math>V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = 594</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1990|num-b=13|num-a=15}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems&diff=802041987 AIME Problems2016-09-05T21:03:48Z<p>Jskalarickal: /* Problem 12 */</p>
<hr />
<div>{{AIME Problems|year=1987}}<br />
<br />
== Problem 1 ==<br />
An ordered pair <math>(m,n)</math> of non-negative integers is called "simple" if the addition <math>m+n</math> in base <math>10</math> requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to <math>1492</math>. <br />
<br />
[[1987 AIME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
What is the largest possible distance between two points, one on the sphere of radius 19 with center <math>(-2,-10,5),</math> and the other on the sphere of radius 87 with center <math>(12,8,-16)</math>?<br />
<br />
[[1987 AIME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called "nice" if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?<br />
<br />
[[1987 AIME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Find the area of the region enclosed by the graph of <math>|x-60|+|y|=|x/4|.</math><br />
<br />
[[1987 AIME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Find <math>3x^2 y^2</math> if <math>x</math> and <math>y</math> are integers such that <math>y^2 + 3x^2 y^2 = 30x^2 + 517</math>.<br />
<br />
[[1987 AIME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Rectangle <math>ABCD</math> is divided into four parts of equal area by five segments as shown in the figure, where <math>XY = YB + BC + CZ = ZW = WD + DA + AX</math>, and <math>PQ</math> is parallel to <math>AB</math>. Find the length of <math>AB</math> (in cm) if <math>BC = 19</math> cm and <math>PQ = 87</math> cm.<br />
<br />
[[Image:AIME_1987_Problem_6.png]]<br />
<br />
[[1987 AIME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Let <math>[r,s]</math> denote the least common multiple of positive integers <math>r</math> and <math>s</math>. Find the number of ordered triples <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>.<br />
<br />
[[1987 AIME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
What is the largest positive integer <math>n</math> for which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>?<br />
<br />
[[1987 AIME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Triangle <math>ABC</math> has right angle at <math>B</math>, and contains a point <math>P</math> for which <math>PA = 10</math>, <math>PB = 6</math>, and <math>\angle APB = \angle BPC = \angle CPA</math>. Find <math>PC</math>.<br />
<br />
[[Image:AIME_1987_Problem_9.png]]<br />
<br />
[[1987 AIME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) <br />
<br />
[[1987 AIME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Find the largest possible value of <math>k</math> for which <math>3^{11}</math> is expressible as the sum of <math>k</math> consecutive positive integers.<br />
<br />
[[1987 AIME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math>m</math> be the smallest integer whose cube root is of the form <math>n+r</math>, where n is a positive integer and <math>r</math> is a positive real number less than <math>1/1000</math>. Find <math>n</math>.<br />
<br />
[[1987 AIME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
A given sequence <math>r_1, r_2, \dots, r_n</math> of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, <math>r_n</math>, with its current predecessor and exchanging them if and only if the last term is smaller. <br />
<br />
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.<br />
<center><math>\underline{1 \quad 9} \quad 8 \quad 7</math></center><br />
<center><math>1 \quad {}\underline{9 \quad 8} \quad 7</math></center><br />
<center><math>1 \quad 8 \quad \underline{9 \quad 7}</math></center><br />
<center><math>1 \quad 8 \quad 7 \quad 9</math></center><br />
Suppose that <math>n = 40</math>, and that the terms of the initial sequence <math>r_1, r_2, \dots, r_{40}</math> are distinct from one another and are in random order. Let <math>p/q</math>, in lowest terms, be the probability that the number that begins as <math>r_{20}</math> will end up, after one bubble pass, in the <math>30^{\mbox{th}}</math> place. Find <math>p + q</math>.<br />
<br />
[[1987 AIME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Compute<br />
<center><math>\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}</math></center>.<br />
<br />
[[1987 AIME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Squares <math>S_1</math> and <math>S_2</math> are inscribed in right triangle <math>ABC</math>, as shown in the figures below. Find <math>AC + CB</math> if area <math>(S_1) = 441</math> and area <math>(S_2) = 440</math>.<br />
<br />
[[Image:AIME_1987_Problem_15.png]]<br />
<br />
[[1987 AIME Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:AIME Problems|1987]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems&diff=802031987 AIME Problems2016-09-05T21:02:48Z<p>Jskalarickal: /* Problem 12 */</p>
<hr />
<div>{{AIME Problems|year=1987}}<br />
<br />
== Problem 1 ==<br />
An ordered pair <math>(m,n)</math> of non-negative integers is called "simple" if the addition <math>m+n</math> in base <math>10</math> requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to <math>1492</math>. <br />
<br />
[[1987 AIME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
What is the largest possible distance between two points, one on the sphere of radius 19 with center <math>(-2,-10,5),</math> and the other on the sphere of radius 87 with center <math>(12,8,-16)</math>?<br />
<br />
[[1987 AIME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called "nice" if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?<br />
<br />
[[1987 AIME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Find the area of the region enclosed by the graph of <math>|x-60|+|y|=|x/4|.</math><br />
<br />
[[1987 AIME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Find <math>3x^2 y^2</math> if <math>x</math> and <math>y</math> are integers such that <math>y^2 + 3x^2 y^2 = 30x^2 + 517</math>.<br />
<br />
[[1987 AIME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Rectangle <math>ABCD</math> is divided into four parts of equal area by five segments as shown in the figure, where <math>XY = YB + BC + CZ = ZW = WD + DA + AX</math>, and <math>PQ</math> is parallel to <math>AB</math>. Find the length of <math>AB</math> (in cm) if <math>BC = 19</math> cm and <math>PQ = 87</math> cm.<br />
<br />
[[Image:AIME_1987_Problem_6.png]]<br />
<br />
[[1987 AIME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Let <math>[r,s]</math> denote the least common multiple of positive integers <math>r</math> and <math>s</math>. Find the number of ordered triples <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>.<br />
<br />
[[1987 AIME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
What is the largest positive integer <math>n</math> for which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>?<br />
<br />
[[1987 AIME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Triangle <math>ABC</math> has right angle at <math>B</math>, and contains a point <math>P</math> for which <math>PA = 10</math>, <math>PB = 6</math>, and <math>\angle APB = \angle BPC = \angle CPA</math>. Find <math>PC</math>.<br />
<br />
[[Image:AIME_1987_Problem_9.png]]<br />
<br />
[[1987 AIME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) <br />
<br />
[[1987 AIME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Find the largest possible value of <math>k</math> for which <math>3^{11}</math> is expressible as the sum of <math>k</math> consecutive positive integers.<br />
<br />
[[1987 AIME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let m be the smallest integer whose cube root is of the form n+r, where n is a positive integer and r is a positive real number less than <math>1/1000</math>. Find n.<br />
<br />
[[1987 AIME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
A given sequence <math>r_1, r_2, \dots, r_n</math> of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, <math>r_n</math>, with its current predecessor and exchanging them if and only if the last term is smaller. <br />
<br />
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.<br />
<center><math>\underline{1 \quad 9} \quad 8 \quad 7</math></center><br />
<center><math>1 \quad {}\underline{9 \quad 8} \quad 7</math></center><br />
<center><math>1 \quad 8 \quad \underline{9 \quad 7}</math></center><br />
<center><math>1 \quad 8 \quad 7 \quad 9</math></center><br />
Suppose that <math>n = 40</math>, and that the terms of the initial sequence <math>r_1, r_2, \dots, r_{40}</math> are distinct from one another and are in random order. Let <math>p/q</math>, in lowest terms, be the probability that the number that begins as <math>r_{20}</math> will end up, after one bubble pass, in the <math>30^{\mbox{th}}</math> place. Find <math>p + q</math>.<br />
<br />
[[1987 AIME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Compute<br />
<center><math>\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}</math></center>.<br />
<br />
[[1987 AIME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Squares <math>S_1</math> and <math>S_2</math> are inscribed in right triangle <math>ABC</math>, as shown in the figures below. Find <math>AC + CB</math> if area <math>(S_1) = 441</math> and area <math>(S_2) = 440</math>.<br />
<br />
[[Image:AIME_1987_Problem_15.png]]<br />
<br />
[[1987 AIME Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:AIME Problems|1987]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_15&diff=801732000 AIME II Problems/Problem 152016-09-01T21:24:29Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center><br />
<br />
== Solution 1 ==<br />
We apply the identity <br />
<br />
<cmath>\begin{align*}<br />
\frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}</cmath><br />
<br />
The motivation for this identity arises from the need to decompose those fractions, possibly into [[telescoping]]. <br />
<br />
Thus our summation becomes <br />
<br />
<cmath>\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).</cmath><br />
<br />
Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>.<br />
<br />
== Solution 2 ==<br />
We can make an approximation by observing the following points:<br />
<br />
The average term is around the 60's which gives 4/3.<br />
<br />
There are 45 terms so the approximate sum is 60.<br />
<br />
Therefore the entire thing equals approximately <math>\frac{1}{60}</math>.<br />
<br />
Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 in degrees is about sin<math>\frac{1}{57}</math> in radians, or is about <math>\frac{1}{57}</math> because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of sin1=<math>\frac{1}{60}</math> in degrees, convert to radians and use the small angle approximation sinx=x.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_15&diff=801722000 AIME II Problems/Problem 152016-09-01T21:24:18Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center><br />
<br />
== Solution 1 ==<br />
We apply the identity <br />
<br />
<cmath>\begin{align*}<br />
\frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}</cmath><br />
<br />
The motivation for this identity arises from the need to decompose those fractions, possibly into [[telescoping]]. <br />
<br />
Thus our summation becomes <br />
<br />
<cmath>\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).</cmath><br />
<br />
Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>.<br />
<br />
======= Solution 2 =======<br />
We can make an approximation by observing the following points:<br />
<br />
The average term is around the 60's which gives 4/3.<br />
<br />
There are 45 terms so the approximate sum is 60.<br />
<br />
Therefore the entire thing equals approximately <math>\frac{1}{60}</math>.<br />
<br />
Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 in degrees is about sin<math>\frac{1}{57}</math> in radians, or is about <math>\frac{1}{57}</math> because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of sin1=<math>\frac{1}{60}</math> in degrees, convert to radians and use the small angle approximation sinx=x.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_15&diff=801712000 AIME II Problems/Problem 152016-09-01T21:24:05Z<p>Jskalarickal: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center><br />
<br />
== Solution 1 ==<br />
We apply the identity <br />
<br />
<cmath>\begin{align*}<br />
\frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}</cmath><br />
<br />
The motivation for this identity arises from the need to decompose those fractions, possibly into [[telescoping]]. <br />
<br />
Thus our summation becomes <br />
<br />
<cmath>\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).</cmath><br />
<br />
Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>.<br />
<br />
=== Solution 2 ===<br />
We can make an approximation by observing the following points:<br />
<br />
The average term is around the 60's which gives 4/3.<br />
<br />
There are 45 terms so the approximate sum is 60.<br />
<br />
Therefore the entire thing equals approximately <math>\frac{1}{60}</math>.<br />
<br />
Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 in degrees is about sin<math>\frac{1}{57}</math> in radians, or is about <math>\frac{1}{57}</math> because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of sin1=<math>\frac{1}{60}</math> in degrees, convert to radians and use the small angle approximation sinx=x.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_15&diff=801702000 AIME II Problems/Problem 152016-09-01T21:23:45Z<p>Jskalarickal: /* = Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center><br />
<br />
== Solution 1 ==<br />
We apply the identity <br />
<br />
<cmath>\begin{align*}<br />
\frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}</cmath><br />
<br />
The motivation for this identity arises from the need to decompose those fractions, possibly into [[telescoping]]. <br />
<br />
Thus our summation becomes <br />
<br />
<cmath>\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).</cmath><br />
<br />
Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>.<br />
<br />
Solution 2 =<br />
We can make an approximation by observing the following points:<br />
<br />
The average term is around the 60's which gives 4/3.<br />
<br />
There are 45 terms so the approximate sum is 60.<br />
<br />
Therefore the entire thing equals approximately <math>\frac{1}{60}</math>.<br />
<br />
Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 in degrees is about sin<math>\frac{1}{57}</math> in radians, or is about <math>\frac{1}{57}</math> because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of sin1=<math>\frac{1}{60}</math> in degrees, convert to radians and use the small angle approximation sinx=x.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_15&diff=801692000 AIME II Problems/Problem 152016-09-01T21:23:28Z<p>Jskalarickal: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center><br />
<br />
== Solution 1 ==<br />
We apply the identity <br />
<br />
<cmath>\begin{align*}<br />
\frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}</cmath><br />
<br />
The motivation for this identity arises from the need to decompose those fractions, possibly into [[telescoping]]. <br />
<br />
Thus our summation becomes <br />
<br />
<cmath>\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).</cmath><br />
<br />
Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>.<br />
<br />
== Solution 2 =<br />
We can make an approximation by observing the following points:<br />
<br />
The average term is around the 60's which gives 4/3.<br />
<br />
There are 45 terms so the approximate sum is 60.<br />
<br />
Therefore the entire thing equals approximately <math>\frac{1}{60}</math>.<br />
<br />
Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 in degrees is about sin<math>\frac{1}{57}</math> in radians, or is about <math>\frac{1}{57}</math> because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of sin1=<math>\frac{1}{60}</math> in degrees, convert to radians and use the small angle approximation sinx=x.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_13&diff=800632014 AIME I Problems/Problem 132016-08-22T20:12:53Z<p>Jskalarickal: /* Strategy */</p>
<hr />
<div>==Problem 13==<br />
On square <math>ABCD</math>, points <math>E,F,G</math>, and <math>H</math> lie on sides <math>\overline{AB},\overline{BC},\overline{CD},</math> and <math>\overline{DA},</math> respectively, so that <math>\overline{EG} \perp \overline{FH}</math> and <math>EG=FH = 34</math>. Segments <math>\overline{EG}</math> and <math>\overline{FH}</math> intersect at a point <math>P</math>, and the areas of the quadrilaterals <math>AEPH, BFPE, CGPF,</math> and <math>DHPG</math> are in the ratio <math>269:275:405:411.</math> Find the area of square <math>ABCD</math>.<br />
<br />
<asy><br />
pair A = (0,sqrt(850));<br />
pair B = (0,0);<br />
pair C = (sqrt(850),0);<br />
pair D = (sqrt(850),sqrt(850));<br />
draw(A--B--C--D--cycle);<br />
dotfactor = 3;<br />
dot("$A$",A,dir(135));<br />
dot("$B$",B,dir(215));<br />
dot("$C$",C,dir(305));<br />
dot("$D$",D,dir(45));<br />
pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));<br />
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);<br />
dot("$H$",H,dir(90));<br />
dot("$F$",F,dir(270));<br />
draw(H--F);<br />
pair E = (0,(sqrt(850)-6)/2);<br />
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);<br />
dot("$E$",E,dir(180));<br />
dot("$G$",G,dir(0));<br />
draw(E--G);<br />
pair P = extension(H,F,E,G);<br />
dot("$P$",P,dir(60));<br />
label("$w$", intersectionpoint( A--P, E--H ));<br />
label("$x$", intersectionpoint( B--P, E--F ));<br />
label("$y$", intersectionpoint( C--P, G--F ));<br />
label("$z$", intersectionpoint( D--P, G--H ));</asy><br />
<br />
== Solution ==<br />
<br />
Notice that <math>269+411=275+405</math>. This means <math>\overline{EG}</math> passes through the centre of the square. <br />
<br />
Draw <math>\overline{IJ} \parallel \overline{HF}</math> with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the centre of the square <math>O</math>.<br />
<br />
Let the area of the square be <math>1360a</math>. Then the area of <math>HPOI=71a</math> and the area of <math>FPOJ=65a</math>. This is because <math>\overline{HF}</math> is perpendicular to <math>\overline{EG}</math> (given in the problem), so <math>\overline{IJ}</math> is also perpendicular to <math>\overline{EG}</math>. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.<br />
<br />
Let the side length of the square be <math>d=\sqrt{1360a}</math>. <br />
<br />
Draw <math>\overline{OK}\parallel \overline{HI}</math> and intersects <math>\overline{HF}</math> at <math>K</math>. <math>OK=d\cdot\frac{HFJI}{ABCD}=\frac{d}{10}</math>. <br />
<br />
The area of <math>HKOI=\frac12\cdot HFJI=68a</math>, so the area of <math>POK=3a</math>.<br />
<br />
Let <math>\overline{PO}=h</math>. Then <math>KP=\frac{6a}{h}</math><br />
<br />
Consider the area of <math>PFJO</math>. <br />
<cmath>\frac12(PF+OJ)(PO)=65a</cmath><br />
<cmath>(17-\frac{3a}{h})h=65a</cmath><br />
<cmath>h=4a</cmath><br />
<br />
Thus, <math>KP=1.5</math>.<br />
<br />
Solving <math>(4a)^2+1.5^2=(\frac{d}{10})^2=13.6a</math>, we get <math>a=\frac58</math>.<br />
<br />
Therefore, the area of <math>ABCD=1360a=\boxed{850}</math><br />
<br />
==Lazy Solution==<br />
<math>269+275+405+411=1360</math>, a multiple of <math>17</math>. In addition, <math>EG=FH=34</math>, which is <math>17\cdot 2</math>.<br />
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to <math>EG</math> and <math>FH</math> must be a multiple of <math>17</math>. All of these triples are primitive:<br />
<br />
<cmath>17=1^2+4^2</cmath><br />
<cmath>34=3^2+5^2</cmath><br />
<cmath>51=\emptyset</cmath><br />
<cmath>68=\emptyset\text{ others}</cmath><br />
<cmath>85=2^2+9^2=6^2+7^2</cmath><br />
<cmath>102=\emptyset</cmath><br />
<cmath>119=\emptyset \dots</cmath><br />
<br />
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting <math>EG=FH=34</math>:<br />
<cmath>\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}</cmath><br />
<cmath>\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850</cmath><br />
<cmath>\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}</cmath><br />
<br />
Thus, <math>\boxed{850}</math> is the only valid answer.<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_7&diff=800622014 AIME I Problems/Problem 72016-08-22T20:12:12Z<p>Jskalarickal: /* Solution 2 (No calculus) */</p>
<hr />
<div>== Problem 7 ==<br />
Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for <math>w \neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane)<br />
<br />
== Solution ==<br />
Let <math>w = \mathrm{cis}{(\alpha)}</math> and <math>z = 10\mathrm{cis}{(\beta)}</math>. Then, <math>\dfrac{w - z}{z} = \dfrac{\mathrm{cis}{(\alpha)} - 10\mathrm{cis}{(\beta)}}{10\mathrm{cis}{\beta}}</math>.<br />
<br />
Multiplying both the numerator and denominator of this fraction by <math>\mathrm{cis}{(-\beta)}</math> gives us:<br />
<br />
<math>\dfrac{w - z}{z} = \dfrac{1}{10}\mathrm{cis}{(\alpha - \beta)} - 1 = \dfrac{1}{10}\mathrm{cos}{(\alpha - \beta)} + \dfrac{1}{10}i\mathrm{sin}{(\alpha - \beta)} - 1</math>.<br />
<br />
We know that <math>\mathrm{tan}{\theta}</math> is equal to the imaginary part of the above expression divided by the real part. Let <math>x = \alpha - \beta</math>. Then, we have that:<br />
<br />
<math>\mathrm{tan}{\theta} = \dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10}.</math><br />
<br />
We need to find a maximum of this expression, so we take the derivative:<br />
<br />
<math>\dfrac{d}{dx} \left (\dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10} \right) = \dfrac{1 - 10\mathrm{cos}{x}}{(\mathrm{cos}{x} - 10)^2}</math><br />
<br />
Thus, we see that the maximum occurs when <math>\mathrm{cos}{x} = \dfrac{1}{10}</math>. Therefore, <math>\mathrm{sin}{x} = \pm\dfrac{\sqrt{99}}{10}</math>, and <math>\mathrm{tan}{\theta} = \pm\dfrac{\sqrt{99}}{99}</math>. Thus, the maximum value of <math>\mathrm{tan^2}{\theta}</math> is <math>\dfrac{99}{99^2}</math>, or <math>\dfrac{1}{99}</math>, and our answer is <math>1 + 99 = \boxed{100}</math>.<br />
<br />
== Solution 2 ==<br />
Without the loss of generality one can let <math>z</math> lie on the positive x axis and since <math>arg(\theta)</math> is a measure of the angle if <math>z=10</math> then <math>arg(\dfrac{w-z}{z})=arg(w-z)</math> and we can see that the question is equivelent to having a triangle <math>OAB</math> with sides <math>OA =10</math> <math>AB=1</math> and <math>OB=t</math> and trying to maximize the angle <math>BOA</math><br />
<asy><br />
pair O = (0,0);<br />
pair A = (100,0);<br />
pair B = (80,30);<br />
pair D = (sqrt(850),sqrt(850));<br />
draw(A--B--O--cycle);<br />
dotfactor = 3;<br />
dot("$A$",A,dir(45));<br />
dot("$B$",B,dir(45));<br />
dot("$O$",O,dir(135));<br />
dot("$ \theta$",O,(7,1.2));<br />
<br />
label("$1$", ( A--B ));<br />
label("$10$",(O--A));<br />
label("$t$",(O--B));<br />
</asy><br />
<br />
using the law of cosines we get:<br />
<math>1^2=10^2+t^2-t*10*2\cos\theta</math><br />
rearranging:<br />
<cmath>20t\cos\theta=t^2+99</cmath><br />
solving for <math>\cos\theta</math> we get:<br />
<br />
<cmath>\frac{99}{20t}+\frac{t}{20}=\cos\theta</cmath><br />
if we want to maximize <math>\theta</math> we need to minimize <math>\cos\theta</math><br />
, using AM-GM inequality we get that the minimum value for <math>\cos\theta= 2(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}})=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}</math><br />
hence using the identity <math>\tan^2\theta=\sec^2\theta-1</math><br />
we get <math>\tan^2\theta=\frac{1}{99}</math>and our answer is <math>1 + 99 = \boxed{100}</math>.<br />
<br />
== Solution 3 ==<br />
<br />
Note that <math>\frac{w-z}{z}=\frac{w}{z}-1</math>, and that <math>\left|\frac{w}{z}\right|=\frac{1}{10}</math>. Thus <math>\frac{w}{z}-1</math> is a complex number on the circle with radius <math>\frac{1}{10}</math> and centered at <math>-1</math> on the complex plane. Let <math>\omega</math> denote this circle. <br />
<br />
Let <math>A</math> and <math>C</math> be the points that represent <math>\frac{w}{z}-1</math> and <math>-1</math> respectively on the complex plane. Let <math>O</math> be the origin. In order to maximize <math>\tan^2(\theta)</math>, we need to maximize <math>\angle{AOC}</math>. This angle is maximized when <math>AO</math> is tangent to <math>\omega</math>. Using the Pythagorean Theorem, we get <br />
<br />
<cmath>AO^2=1^2-\left(\frac{1}{10}\right)^2=\frac{99}{100}</cmath><br />
<br />
Thus<br />
<br />
<cmath>\tan^2(\theta)=\frac{AC^2}{AO^2}=\frac{1/100}{99/100}=\frac{1}{99}</cmath><br />
<br />
And the answer is <math>1+99=\boxed{100}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_15&diff=800612014 AIME I Problems/Problem 152016-08-22T20:11:48Z<p>Jskalarickal: /* Solutions */</p>
<hr />
<div>== Problem 15 ==<br />
<br />
In <math>\triangle ABC</math>, <math>AB = 3</math>, <math>BC = 4</math>, and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and <math>D</math>, and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4}</math>, length <math>DE=\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>.<br />
<br />
== Solution 1 ==<br />
<br />
Since <math>\angle DBE = 90^\circ</math>, <math>DE</math> is the diameter of <math>\omega</math>. Then <math>\angle DFE=\angle DGE=90^\circ</math>. But <math>DF=FE</math>, so <math>\triangle DEF</math> is a 45-45-90 triangle. Letting <math>DG=3x</math>, we have that <math>EG=4x</math>, <math>DE=5x</math>, and <math>DF=EF=\frac{5x}{\sqrt{2}}</math>. <br />
<br />
Note that <math>\triangle DGE \sim \triangle ABC</math> by SAS similarity, so <math>\angle BAC = \angle GDE</math> and <math>\angle ACB = \angle DEG</math>. Since <math>DEFG</math> is a cyclic quadrilateral, <math>\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE</math> and <math>\angle ACB = \angle DEG = \angle GFD</math>, implying that <math>\triangle AFE</math> and <math>\triangle CDF</math> are isosceles. As a result, <math>AE=CD=\frac{5x}{\sqrt{2}}</math>, so <math>BE=3-\frac{5x}{\sqrt{2}}</math> and <math>BD =4-\frac{5x}{\sqrt{2}}</math>. <br />
<br />
Finally, using the Pythagorean Theorem on <math>\triangle BDE</math>, <br />
<cmath> \left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2</cmath><br />
Solving for <math>x</math>, we get that <math>x=\frac{5\sqrt{2}}{14}</math>, so <math>DE=5x=\frac{25\sqrt{2}}{14}</math>. Thus, the answer is <math>25+2+14=\boxed{041}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
<asy><br />
pair A = (0,3);<br />
pair B = (0,0);<br />
pair C = (4,0);<br />
draw(A--B--C--cycle);<br />
dotfactor = 3;<br />
dot("$A$",A,dir(135));<br />
dot("$B$",B,dir(215));<br />
dot("$C$",C,dir(305));<br />
pair D = (2.21, 0);<br />
pair E = (0, 1.21);<br />
pair F = (1.71, 1.71);<br />
pair G = (2, 1.5);<br />
dot("$D$",D,dir(270));<br />
dot("$E$",E,dir(180));<br />
dot("$F$",F,dir(90));<br />
dot("$G$",G,dir(0));<br />
draw(Circle((1.109, 0.609), 1.28));<br />
draw(D--E);<br />
draw(E--F);<br />
draw(D--F);<br />
draw(E--G);<br />
draw(D--G);<br />
draw(B--F);<br />
draw(B--G);<br />
</asy><br />
<br />
First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>\overline{DE}</math> the same as the diameter of <math>\omega</math>. We also note that <math>\triangle DGE \sim \triangle ABC</math> since <math>\angle EGD</math> is a right angle and the ratios of the sides are <math>3:4:5</math>. <br />
<br />
From congruent arc intersections, we know that <math>\angle GED \cong \angle GBC</math>, and that from similar triangles <math>\angle GED</math> is also congruent to <math>\angle GCB</math>. Thus, <math>\triangle BGC</math> is an isosceles triangle with <math>BG = GC</math>, so <math>G</math> is the midpoint of <math>\overline{AC}</math> and <math>AG = GC = 5/2</math>. Similarly, we can find from angle chasing that <math>\angle ABF = \angle EDF = \frac{\pi}4</math>. Therefore, <math>\overline{BF}</math> is the angle bisector of <math>\angle B</math>. From the angle bisector theorem, we have <math>\frac{AF}{AB} = \frac{CF}{CB}</math>, so <math>AF = 15/7</math> and <math>CF = 20/7</math>. <br />
<br />
Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math><br />
<br />
<br />
Also: <math>FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}</math>. We can also use Ptolemy's Theorem on quadrilateral <math>DEFG</math> to figure what <math>FG</math> is in terms of <math>d</math>:<br />
<cmath>DE\cdot FG+DG\cdot EF=DF\cdot EG</cmath><br />
<cmath>d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}</cmath><br />
<cmath>d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}</cmath><br />
Thus <math>\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}</math>. <math>a+b+c=25+2+14= \boxed{041}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=14|after=Last Question}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1975_Canadian_MO_Problems&diff=799491975 Canadian MO Problems2016-08-11T16:17:55Z<p>Jskalarickal: /* Problem 6 */</p>
<hr />
<div>== Problem 1 ==<br />
Simplify<br />
<cmath>\left(\frac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{1/3}</cmath>.<br />
<br />
[[1975 Canadian MO Problems/Problem 1 | Solution]]<br />
<br />
== Problem 2 ==<br />
A sequence of numbers <math>a_1, a_2, a_3, \dots</math> satisfies<br />
<div class=ol><br />
<div class=li><span class=num>(i)</span> <math>a_1 = \frac{1}{2}</math></div><br />
<div class=li><span class=num>(ii)</span> <math>a_1+a_2+\cdots+a_n=n^2a_n\quad(n\ge1).</math></div><br />
</div><br />
Determine the value of <math>a_n\quad(n\ge1).</math><br />
<br />
[[1975 Canadian MO Problems/Problem 2 | Solution]]<br />
<br />
== Problem 3 ==<br />
For each real number <math>r</math>, <math>[r]</math> denotes the largest integer less than or equal to <math>r</math>, <math>e.g.,</math> <math>[6] = 6, [\pi] = 3, [-1.5] = -2.</math> Indicate on the <math>(x,y)</math>-plane the set of all points <math>(x,y)</math> for which <math>[x]^2+[y]^2 = 4</math>.<br />
<br />
[[1975 Canadian MO Problems/Problem 3 | Solution]]<br />
== Problem 4 ==<br />
For a positive number such as <math>3.27</math>, <math>3</math> is referred to as the integral part of the number and <math>.27</math> as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression.<br />
<br />
[[1975 Canadian MO Problems/Problem 4 | Solution]]<br />
== Problem 5 ==<br />
<math>A, B, C, D</math> are four "consecutive" points on the circumference of a circle and <math>P, Q, R, S</math> are points on the circumference which are respectively the midpoints of the arcs <math>AB, BC, CD, DA.</math> Prove that <math>PR</math> is perpendicular to <math>QS</math>.<br />
<br />
[[1975 Canadian MO Problems/Problem 5 | Solution]]<br />
== Problem 6 ==<br />
<div class=ol><br />
<div class=li><span class=num>(i)</span><math>15</math> chairs are equally place around a circular table on which are name cards for <math>15</math> guests. The guests fail to notice these cards until after they have sat down, and it turns out that no one is sitting in the correct seat. Prove that the table can be rotated so that at least two of the guests are simultaneously correctly seated. </div><br />
<div class=li><span class=num>(ii)</span> Give an example of an arrangement in which just one of the 15 quests is correctly seated and for which no rotation correctly places more than one person.</div><br />
</div><br />
<br />
[[1975 Canadian MO Problems/Problem 6 | Solution]]<br />
<br />
== Problem 7 ==<br />
A function <math>f(x)</math> is <math>\textit{periodic}</math> if there is a positive integer such that <math>f(x+p) = f(x)</math> for all <math>x</math>. For example, <math>\sin x</math> is periodic with period <math>2\pi</math>. Is the function <math>\sin(x^2)</math> periodic? Prove your assertion.<br />
<br />
[[1975 Canadian MO Problems/Problem 7 | Solution]]<br />
== Problem 8 ==<br />
Let <math>k</math> be a positive integer. Find all polynomials<br />
<cmath>P(x) = a_0+a_1x+\cdots+a_nx^n</cmath><br />
where the <math>a_i</math> are real, which satisfy the equation<br />
<cmath>P(P(x)) = \{P(x)\}^k</cmath>.<br />
<br />
[[1975 Canadian MO Problems/Problem 8 | Solution]]</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_7&diff=699912001 AMC 12 Problems/Problem 72015-04-20T16:37:22Z<p>Jskalarickal: /* Problem */</p>
<hr />
<div>{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #7]] and [[2001 AMC 10 Problems|2001 AMC 10A #14]]}}<br />
<br />
== Problem ==<br />
<br />
A charity sells <math>140</math> benefit tickets for a total of <math></math>2001<math>. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?<br />
<br />
</math>\text{(A) }<math> <dollar/></math>782\qquad \text{(B) }<math> <dollar/></math>986\qquad \text{(C) }<math> <dollar/></math>1158\qquad \text{(D) }<math> <dollar/></math>1219\qquad \text{(E) }<math> <dollar/></math>1449$<br />
<br />
== Solution ==<br />
<br />
Let's multiply ticket costs by <math>2</math>, then the half price becomes an integer, and the charity sold <math>140</math> tickets worth a total of <math>4002</math> dollars.<br />
<br />
Let <math>h</math> be the number of half price tickets, we then have <math>140-h</math> full price tickets. The cost of <math>140-h</math> full price tickets is equal to the cost of <math>280-2h</math> half price tickets. <br />
<br />
Hence we know that <math>h+(280-2h) = 280-h</math> half price tickets cost <math>4002</math> dollars. Then a single half price ticket costs <math>\frac{4002}{280-h}</math> dollars, and this must be an integer. Thus <math>280-h</math> must be a divisor of <math>4002</math>. Keeping in mind that <math>0\leq h\leq 140</math>, we are looking for a divisor between <math>140</math> and <math>280</math>, inclusive.<br />
<br />
The prime factorization of <math>4002</math> is <math>4002=2\cdot 3\cdot 23\cdot 29</math>. We can easily find out that the only divisor of <math>4002</math> within the given range is <math>2\cdot 3\cdot 29 = 174</math>. <br />
<br />
This gives us <math>280-h=174</math>, hence there were <math>h=106</math> half price tickets and <math>140-h = 34</math> full price tickets.<br />
<br />
In our modified setting (with prices multiplied by <math>2</math>) the price of a half price ticket is <math>\frac{4002}{174} = 23</math>. In the original setting this is the price of a full price ticket. Hence <math>23\cdot 34 = \boxed{(\text{A})782}</math> dollars are raised by the full price tickets.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2001|num-b=6|num-a=8}}<br />
{{AMC10 box|year=2001|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_31&diff=693991950 AHSME Problems/Problem 312015-03-22T18:46:44Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math> 50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:<br />
<br />
<math>\textbf{(A)}\ 4:1 \qquad<br />
\textbf{(B)}\ 2:1 \qquad<br />
\textbf{(C)}\ 1:4 \qquad<br />
\textbf{(D)}\ 1:2 \qquad<br />
\textbf{(E)}\ 1:8</math><br />
<br />
==Solution==<br />
Let the number of blue socks be represented as <math>b</math>. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of blue socks is <math>\$1</math>. That means the price of one pair of black socks is <math>\$2</math>.<br />
<br />
Now from the third and fourth sentence, we see that <math>1.5(2(4)+1(b))=1(4)+2(b)</math>. Simplifying gives <math>b=16</math>. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is <math>\boxed{\textbf{(C)}\ 1:4}</math><br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1950|num-b=30|num-a=32}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_31&diff=693981950 AHSME Problems/Problem 312015-03-22T18:45:18Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math> 50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:<br />
<br />
<math>\textbf{(A)}\ 4:1 \qquad<br />
\textbf{(B)}\ 2:1 \qquad<br />
\textbf{(C)}\ 1:4 \qquad<br />
\textbf{(D)}\ 1:2 \qquad<br />
\textbf{(E)}\ 1:8</math><br />
<br />
==Solution==<br />
Let the number of blue socks be represented as <math>b</math>. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of blue socks is <math>\$1</math>. That means the price of one pair of black socks is <math>\$2</math>.<br />
<br />
Now from the third and fourth sentence, we see that <math>1.5(1(4)+2(b))=1(b)+2(4)</math>. Simplifying gives <math>b=1</math>. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is <math>\boxed{\textbf{(C)}\ 1:4}</math><br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1950|num-b=30|num-a=32}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_31&diff=693971950 AHSME Problems/Problem 312015-03-22T18:44:37Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math> 50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:<br />
<br />
<math>\textbf{(A)}\ 4:1 \qquad<br />
\textbf{(B)}\ 2:1 \qquad<br />
\textbf{(C)}\ 1:4 \qquad<br />
\textbf{(D)}\ 1:2 \qquad<br />
\textbf{(E)}\ 1:8</math><br />
<br />
==Solution==<br />
Let the number of blue socks be represented as <math>b</math>. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of black socks is <math>\$1</math>. That means the price of one pair of blue socks is <math>\$2</math>.<br />
<br />
Now from the third and fourth sentence, we see that <math>1.5(1(4)+2(b))=1(b)+2(4)</math>. Simplifying gives <math>b=1</math>. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is <math>\boxed{\textbf{(C)}\ 1:4}</math><br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1950|num-b=30|num-a=32}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_31&diff=693961950 AHSME Problems/Problem 312015-03-22T18:43:42Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math> 50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:<br />
<br />
<math>\textbf{(A)}\ 4:1 \qquad<br />
\textbf{(B)}\ 2:1 \qquad<br />
\textbf{(C)}\ 1:4 \qquad<br />
\textbf{(D)}\ 1:2 \qquad<br />
\textbf{(E)}\ 1:8</math><br />
<br />
==Solution==<br />
Let the number of blue socks be represented as <math>b</math>. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of black socks is <math>\$1</math>. That means the price of one pair of blue socks is <math>\$2</math>.<br />
<br />
Now from the third and fourth sentence, we see that <math>1.5(1(4)+2(b))=1(b)+2(4)</math>. Simplifying gives <math>b=1</math>. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is <math>\boxed{\textbf{(A)}\ 4:1}</math><br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1950|num-b=30|num-a=32}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_3&diff=673422015 AMC 10A Problems/Problem 32015-02-04T21:25:30Z<p>Jskalarickal: Created page with "i"</p>
<hr />
<div>i</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_1&diff=673402015 AMC 10A Problems/Problem 12015-02-04T21:24:57Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math><br />
<br />
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math></div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems&diff=673392015 AMC 10A Problems2015-02-04T21:21:05Z<p>Jskalarickal: /* Problem 11 */</p>
<hr />
<div>==Problem 1==<br />
<br />
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math><br />
<br />
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2015 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
A box contains a collection of triangular and square tiles. There are <math>25</math> tiles in the box, containing <math>84</math> edges total. How many square tiles are there in the box?<br />
<br />
<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}}\ 9\qquad\textbf{(E)}\ 11</math><br />
<br />
[[2015 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
Ann made a 3-step staircase using 18 toothpicks. How many toothpicks does she need to add to complete a 5-step staircase?<br />
<br />
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}}\ 22\qquad\textbf{(E)}\ 24</math><br />
<br />
[[2015 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2015 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Mr. Patrick teaches math to <math> 15 </math> students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was <math> 80 </math>. After he graded Payton's test, the test average became <math> 81 </math>. What was Payton's score on the test?<br />
<br />
<math> \textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}}\ 94\qquad\textbf{(E)}\ 95 </math><br />
<br />
[[2015 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
The sum of two positive numbers is <math> 5 </math> times their difference. What is the ratio of the larger number to the smaller number?<br />
<br />
<math> \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}}\ 2 \qquad\textbf{(E)}\ \frac{5}{2} </math><br />
<br />
[[2015 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
How many terms are there in the arithmetic sequence <math>13</math>, <math>16</math>, <math>19</math>, . . ., <math>70</math>, <math>73</math>?<br />
<br />
<math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 60\qquad\textbf{(E)}\ 61 </math><br />
<br />
[[2015 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be <math>2</math> : <math>1</math>?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}}\ 6\qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2015 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
Two right circular cylinders have the same volume. The radius of the second cylinder is <math>10\%</math> more than the radius of the first. What is the relationship between the heights of the two cyclinders?<br />
<br />
<math>\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}</math><br />
<br />
[[2015 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
How many rearrangements of <math>abcd</math> are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either <math>ab</math> or <math>ba</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 />
[[2015 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
The ratio of the length to the width of a rectangle is 4:3. If the rectangle has diagonal of length <math>d</math>, then the area may be expressed as <math>kd^2</math> for some constant <math>k</math>. What is <math>k</math>?<br />
<br />
==Problem 12==<br />
<br />
==Problem 13==<br />
<br />
==Problem 14==<br />
<br />
==Problem 15==<br />
<br />
==Problem 16==<br />
<br />
==Problem 17==<br />
<br />
==Problem 18==<br />
<br />
==Problem 19==<br />
<br />
==Problem 20==<br />
<br />
A rectangle has area <math>A</math> <math>\text{cm}^2</math> and perimeter <math>P</math> <math>\text{cm}</math>, where <math>A</math> and <math>P</math> are positive integers. Which of the following numbers cannot equal <math>A+P</math>?<br />
<br />
<math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math><br />
<br />
[[2015 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
==Problem 22==<br />
<br />
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br />
<br />
<math>\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} </math><br />
<br />
[[2015 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
The zeros of the function <math>f(x)=x^2-ax+2a</math> are integers. What is the sum of the possible values of <math>a</math>?<br />
<br />
<math> \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18</math><br />
<br />
[[2015 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
==Problem 25==<br />
<br />
== See also ==<br />
* [[AMC Problems and Solutions]]<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_5&diff=663822000 AMC 10 Problems/Problem 52014-11-29T00:02:11Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Points <math>M</math> and <math>N</math> are the midpoints of sides <math>PA</math> and <math>PB</math> of <math>\triangle PAB</math>. As <math>P</math> moves along a line that is parallel to side <math>AB</math>, how many of the four quantities listed below change?<br />
<br />
(a) the length of the segment <math>MN</math><br />
<br />
(b) the perimeter of <math>\triangle PAB</math><br />
<br />
(c) the area of <math>\triangle PAB</math><br />
<br />
(d) the area of trapezoid <math>ABNM</math><br />
<br />
<asy><br />
draw((2,0)--(8,0)--(6,4)--cycle);<br />
draw((4,2)--(7,2));<br />
draw((1,4)--(9,4),Arrows);<br />
label("$A$",(2,0),SW);<br />
label("$B$",(8,0),SE);<br />
label("$M$",(4,2),W);<br />
label("$N$",(7,2),E);<br />
label("$P$",(6,4),N);<br />
</asy><br />
<br />
<math>\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 3 \qquad\mathrm{(E)}\ 4</math><br />
<br />
==Solution==<br />
<br />
(a) Clearly <math>AB</math> does not change, and <math>MN=\frac{1}{2}AB</math>, so <math>MN</math> doesn't change either.<br />
<br />
(b) Obviously, the perimeter changes.<br />
<br />
(c) The area clearly doesn't change, as both the base <math>AB</math> and its corresponding height remain the same.<br />
<br />
(d) The bases <math>AB</math> and <math>MN</math> do not change, and neither does the height, so the area of the trapezoid remains the same.<br />
<br />
Only <math>1</math> quantity changes, so the correct answer is <math>\boxed{\text{B}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2000|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_20&diff=660322011 AMC 10B Problems/Problem 202014-11-25T21:15:07Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>{{duplicate|[[2011 AMC 12B Problems|2011 AMC 12B #16]] and [[2011 AMC 10B Problems|2011 AMC 10B #20]]}}<br />
<br />
== Problem==<br />
<br />
Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
== Solution ==<br />
<br />
Suppose that <math>P</math> is a point in the rhombus <math>ABCD</math> and let <math>\ell_{BC}</math> be the [[perpendicular bisector]] of <math>\overline{BC}</math>. Then <math>PB < PC</math> if and only if <math>P</math> is on the same side of <math>\ell_{BC}</math> as <math>B</math>. The line <math>\ell_{BC}</math> divides the plane into two half-planes; let <math>S_{BC}</math> be the half-plane containing <math>B</math>. Let us define similarly <math>\ell_{BD},S_{BD}</math> and <math>\ell_{BA},S_{BA}</math>. Then <math>R</math> is equal to <math>ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}</math>. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:<br />
<br />
<asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(0.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2;<br />
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,lightgray);<br />
draw(A--B--C--D--cycle);<br />
draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G);<br />
<br />
label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);<br />
label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE);<br />
label("$2$",(D--C),SW);<br />
</asy><br />
Since <math>\triangle BCD</math> and <math>\triangle BAD</math> are equilateral, <math>\ell_{BC}</math> contains <math>D</math>, <math>\ell_{BD}</math> contains <math>A</math> and <math>C</math>, and <math>\ell_{BA}</math> contains <math>D</math>. Then <math>\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH</math> with <math>BE = 1</math> and <math>EF = \frac{1}{\sqrt{3}}</math> so <math>[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}</math> and <math>R</math> has area <math>\boxed{(C)\frac{2\sqrt{3}}{3}}</math>.<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=19|num-a=21}}<br />
<br />
{{AMC12 box|year=2011|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_20&diff=660312011 AMC 10B Problems/Problem 202014-11-25T21:14:34Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>{{duplicate|[[2011 AMC 12B Problems|2011 AMC 12B #16]] and [[2011 AMC 10B Problems|2011 AMC 10B #20]]}}<br />
<br />
== Problem==<br />
<br />
Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
== Solution ==<br />
<br />
Suppose that <math>P</math> is a point in the rhombus <math>ABCD</math> and let <math>\ell_{BC}</math> be the [[perpendicular bisector]] of <math>\overline{BC}</math>. Then <math>PB < PC</math> if and only if <math>P</math> is on the same side of <math>\ell_{BC}</math> as <math>B</math>. The line <math>\ell_{BC}</math> divides the plane into two half-planes; let <math>S_{BC}</math> be the half-plane containing <math>B</math>. Let us define similarly <math>\ell_{BD},S_{BD}</math> and <math>\ell_{BA},S_{BA}</math>. Then <math>R</math> is equal to <math>ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}</math>. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:<br />
<br />
<asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(0.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2;<br />
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,lightgray);<br />
draw(A--B--C--D--cycle);<br />
draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G);<br />
<br />
label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);<br />
label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE);<br />
label("$2$",(D--C),SW);<br />
</asy><br />
Since <math>\triangle BCD</math> and <math>\triangle BAD</math> are equilateral, <math>\ell_{BC}</math> contains <math>D</math>, <math>\ell_{BD}</math> contains <math>A</math> and <math>C</math>, and <math>\ell_{BA}</math> contains <math>D</math>. Then <math>\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH</math> with <math>BE = 1</math> and <math>EF = \frac{1}{\sqrt{3}}</math> so <math>[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}</math> and <math>R</math> has area <math>\boxed(C){\frac{2\sqrt{3}}{3}}</math>.<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=19|num-a=21}}<br />
<br />
{{AMC12 box|year=2011|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_25&diff=660192013 AMC 10A Problems/Problem 252014-11-24T22:14:29Z<p>Jskalarickal: /* Solution 2 (elimination) */</p>
<hr />
<div>==Problem==<br />
<br />
All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br />
of the octagon (not on the boundary) do two or more diagonals intersect?<br />
<br />
<math> \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1 (drawing)==<br />
<br />
If you draw a good diagram like the one below, it is easy to see that there are <math>\boxed{\textbf{(A) }49}</math>, points.<br />
<br />
<asy><br />
size(14cm);<br />
pathpen = white + 1.337;<br />
// Initialize octagon<br />
pair[] A;<br />
for (int i=0; i<8; ++i) {<br />
A[i] = dir(45*i);<br />
}<br />
D(CR( (0,0), 1));<br />
// Draw diagonals<br />
// choose pen colors<br />
pen[] colors;<br />
colors[1] = orange + 1.337;<br />
colors[2] = blue;<br />
colors[3] = green;<br />
colors[4] = black;<br />
for (int d=1; d<=4; ++d) {<br />
pathpen = colors[d];<br />
for (int j=0; j<8; ++j) {<br />
D(A[j]--A[(j+d) % 8]);<br />
}<br />
}<br />
pathpen = blue + 2;<br />
// Draw all the intersections<br />
pointpen = red + 7;<br />
for (int x1=0; x1<8; ++x1) {<br />
for (int x2=x1+1; x2<8; ++x2) {<br />
for (int x3=x2+1; x3<8; ++x3) {<br />
for (int x4=x3+1; x4<8; ++x4) {<br />
D(IP(A[x1]--A[x2], A[x3]--A[x4]));<br />
D(IP(A[x1]--A[x3], A[x4]--A[x2]));<br />
D(IP(A[x1]--A[x4], A[x2]--A[x3]));<br />
}<br />
}<br />
}<br />
}</asy><br />
<br />
==Solution 2 (elimination)==<br />
<br />
Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every 4 points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract <math>\dbinom{4}{2} -1 = 5</math> from this count. <math>70-5 = 65</math>. Note that diagonals like AD, CG, and BE all intersect at the same point. There are <math>8</math> of this type with three diagonals intersecting at the same point, so we need to subtract <math>2</math> of the <math>\dbinom{3}{2}</math> (one is kept as the actual intersection). In the end, we obtain <math>65 - 16 = \boxed{\textbf{(A) }49}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2013|ab=A|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_6&diff=660092014 AMC 10A Problems/Problem 62014-11-24T01:29:18Z<p>Jskalarickal: /* Solution 1 */</p>
<hr />
<div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #4]] and [[2014 AMC 10A Problems|2014 AMC 10A #6]]}}<br />
<br />
==Problem==<br />
<br />
Suppose that <math>a</math> cows give <math>b</math> gallons of milk in <math>c</math> days. At this rate, how many gallons of milk will <math>d</math> cows give in <math>e</math> days?<br />
<br />
<math> \textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}</math><br />
<br />
==Solution 1==<br />
<br />
We need to multiply <math>b</math> by <math>\frac{d}{a}</math> for the new cows and <math>\frac{e}{c}</math> for the new time, so the answer is <math>b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}</math>, or <math>\boxed{\textbf{(A)}}</math>.<br />
<br />
==Solution 2==<br />
We see that the the amount of cows is inversely proportional to the amount of days and directly proportional to the gallons of milk. So our constant is <math>\dfrac{ac}{b}</math>.<br />
<br />
Let <math>g</math> be the answer to the question. We have <math>\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=5|num-a=7}}<br />
{{AMC12 box|year=2014|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_14&diff=660082014 AMC 10A Problems/Problem 142014-11-24T01:06:50Z<p>Jskalarickal: /* Solution 1 */</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 />
// 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 />
We can just let <math>m=3</math>, since the two values of <math>m</math> do not affect our solution - one is the slope of one line and the other is the slope of the other line. <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 altutude 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 />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_16&diff=660062014 AMC 10A Problems/Problem 162014-11-24T00:58:01Z<p>Jskalarickal: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=1</math>, <math>BC=2</math>, and points <math>E</math>, <math>F</math>, and <math>G</math> are midpoints of <math>\overline{BC}</math>, <math>\overline{CD}</math>, and <math>\overline{AD}</math>, respectively. Point <math>H</math> is the midpoint of <math>\overline{GE}</math>. What is the area of the shaded region?<br />
<br />
<asy><br />
import graph;<br />
size(9cm);<br />
pen dps = fontsize(10); defaultpen(dps);<br />
pair D = (0,0);<br />
pair F = (1/2,0);<br />
pair C = (1,0);<br />
pair G = (0,1);<br />
pair E = (1,1);<br />
pair A = (0,2);<br />
pair B = (1,2);<br />
pair H = (1/2,1);<br />
<br />
// do not look<br />
pair X = (1/3,2/3);<br />
pair Y = (2/3,2/3);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(G--E);<br />
draw(A--F--B);<br />
draw(D--H--C);<br />
filldraw(H--X--F--Y--cycle,grey);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E,E);<br />
label("$F$",F,S);<br />
label("$G$",G,W);<br />
label("$H$",H,N);<br />
<br />
label("$\frac12$",(0.25,0),S);<br />
label("$\frac12$",(0.75,0),S);<br />
label("$1$",(1,0.5),E);<br />
label("$1$",(1,1.5),E);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1==<br />
Denote <math>D=(0,0)</math>. Then <math>A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)</math>. Let the intersection of <math>AF</math> and <math>DH</math> be <math>X</math>, and the intersection of <math>BF</math> and <math>CH</math> be <math>Y</math>. Then we want to find the coordinates of <math>X</math> so we can find <math>XY</math>. From our points, the slope of <math>AF</math> is <math>\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4</math>, and its <math>y</math>-intercept is just <math>2</math>. Thus the equation for <math>AF</math> is <math>y = -4x + 2</math>. We can also quickly find that the equation of <math>DH</math> is <math>y = 2x</math>. Setting the equations equal, we have <math>2x = -4x +2 \implies x = \frac13</math>. Because of symmetry, we can see that the distance from <math>Y</math> to <math>BC</math> is also <math>\frac13</math>, so <math>XY = 1 - 2 \cdot \frac13 = \frac13</math>. Now the area of the kite is simply the product of the two diagonals over <math>2</math>. Since the length <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.<br />
<br />
<asy><br />
import graph;<br />
size(9cm);<br />
pen dps = fontsize(10); defaultpen(dps);<br />
pair D = (0,0);<br />
pair F = (1/2,0);<br />
pair C = (1,0);<br />
pair G = (0,1);<br />
pair E = (1,1);<br />
pair A = (0,2);<br />
pair B = (1,2);<br />
pair H = (1/2,1);<br />
<br />
// do not look<br />
pair X = (1/3,2/3);<br />
pair Y = (2/3,2/3);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(G--E);<br />
draw(A--F--B);<br />
draw(D--H--C);<br />
filldraw(H--X--F--Y--cycle,grey);<br />
draw(X--Y,dashed);<br />
<br />
<br />
<br />
label("$A\: (0,2)$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D \: (0,0)$",D,SW);<br />
label("$E$",E,E);<br />
label("$F\: (\frac12,0)$",F,S);<br />
label("$G$",G,W);<br />
label("$H \: (\frac12,1)$",H,N);<br />
label("$Y$",Y,E);<br />
label("$X$",X,W);<br />
<br />
<br />
label("$\frac12$",(0.25,0),S);<br />
label("$\frac12$",(0.75,0),S);<br />
label("$1$",(1,0.5),E);<br />
label("$1$",(1,1.5),E);<br />
</asy><br />
<br />
==Solution 2==<br />
<br />
Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2}* IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = \boxed{\textbf{(E)} \: \frac{1}{6}}</math><br />
<br />
==Solution 3==<br />
<br />
From the diagram in Solution 1, let <math>e</math> be the height of <math>XHY</math> and <math>f</math> be the height of <math>XFY</math>. It is clear that their sum is <math>1</math> as they are parallel to <math>GD</math>. Let <math>k</math> be the ratio of the sides of the similar triangles <math>XFY</math> and <math>AFB</math>, which are similar because <math>XY</math> is parallel to <math>AB</math> and the triangles share angle <math>F</math>. Then <math>k = f/2</math>, as 2 is the height of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height of <math>XHY</math> will be equal to the base, like in <math>DHC</math>, making <math>XY = e</math>. However, <math>XY</math> is also the base of <math>XFY</math>, so <math>k = e / AB</math> where <math>AB = 1</math> so <math>k = e</math>. Subbing into <math>k = f/2</math> gives a system of linear equations, <math>e + f = 1</math> and <math>e = f/2</math>. Solving yields <math>e = XY = 1/3</math> and <math>f = 2/3</math>, and since the area of the kite is simply the product of the two diagonals over <math>2</math> and <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Jskalarickalhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_21&diff=660052014 AMC 10A Problems/Problem 212014-11-24T00:34:42Z<p>Jskalarickal: /* Solution */</p>
<hr />
<div>==Problem==<br />
Positive integers <math>a</math> and <math>b</math> are such that the graphs of <math>y=ax+5</math> and <math>y=3x+b</math> intersect the <math>x</math>-axis at the same point. What is the sum of all possible <math>x</math>-coordinates of these points of intersection?<br />
<br />
<math> \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} </math><br />
<br />
==Solution==<br />
Note that when <math>y=0</math>, the <math>x</math> values of the equations should be equal by the problem statement. We have that<br />
<cmath>0 = ax + 5 \implies x = -\dfrac{5}{a}</cmath> <cmath>0 = 3x+b \implies x= -\dfrac{b}{3}</cmath><br />
Which means that <cmath>-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15</cmath><br />
The only possible pairs <math>(a,b)</math> then are <math>(a,b) = (1,15), (3,5), (5,3), (15, 1)</math>. These pairs give respective <math>x</math>-values of <math>-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}</math> which have a sum of <math>\boxed{\textbf{(E)} \: -8}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Jskalarickal