https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jhawk0224&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:40:14ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_5&diff=1490752021 AIME I Problems/Problem 52021-03-11T21:38:47Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br />
<br />
==Solution==<br />
Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math>b^2=\frac{3a^2}{a-2}</math>. Simplifying further, <math>b^2=3a+6+\frac{12}{a-2}</math>. Looking at this second equation, since the right side must be an integer, <math>a-2</math> must equal <math>\pm1, 2, 3, 4, 6, 12</math>. Looking at the first equation, we see <math>a>2</math> since <math>b^2</math> is positive. This means we must test <math>a=3, 4, 5, 6, 8, 14</math>. After testing these, we see that only <math>a=5</math> and <math>a=14</math> work which give <math>b=5</math> and <math>b=7</math> respectively. Thus the answer is <math>10+21=\boxed{31}</math>.<br />
~JHawk0224<br />
<br />
===Solution 1===<br />
Let the common difference be <math> d </math> and let the middle term be <math> x </math>. Then, we have that the sequence is<br />
<cmath>x-d,~x,~x+d.</cmath><br />
This means that the sum of the sequence is <br />
<cmath> (x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2. </cmath><br />
We know that this must be equal to <math>xd^2,</math> so we can write that<br />
<cmath>3x^2+2d^2=xd^2,</cmath><br />
and it follows that<br />
<cmath>3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.</cmath><br />
<br />
Now, we can treat <math> d </math> as a constant and use the quadratic formula to get<br />
<cmath>x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.</cmath><br />
We can factor pull <math> d^2 </math> out of the square root to get<br />
<cmath>x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.</cmath><br />
Here, it is easy to test values of <math> d </math>. We find that <math> d=5 </math> and <math> d=7 </math> are the only positive integer values of <math> d </math> that make <math> \sqrt{d^2-24} </math> a positive integer. <math> d=5 </math> gives <math> x=5 </math> and <math> x=\frac{10}{3} </math>, but we can ignore the latter. <math> d=7 </math> gives <math> x=14 </math>, as well as a fraction which we can ignore. <br />
<br />
Since <math> d=5,~x=5 </math> and <math> d=7, x=14 </math> are the only two solutions and we want the sum of the third terms, our answer is <math> (5+5)+(7+14)=10+21=\boxed{031} </math>. -BorealBear<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_5&diff=1490742021 AIME I Problems/Problem 52021-03-11T21:37:03Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br />
<br />
==Solution==<br />
Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math>\displaystyle b^2=\frac{3a^2}{a-2}</math>. Simplifying further, <math>\displaystyle b^2=3a+6+\frac{12}{a-2}</math>. Looking at this second equation, since the right side must be an integer, <math>a-2</math> must equal <math>\pm1, 2, 3, 4, 6, 12</math>. Looking at the first equation, we see <math>a>2</math> since <math>b^2</math> is positive. This means we must test <math>a=3, 4, 5, 6, 8, 14</math>.<br />
<br />
===Solution 1===<br />
Let the common difference be <math> d </math> and let the middle term be <math> x </math>. Then, we have that the sequence is<br />
<cmath>x-d,~x,~x+d.</cmath><br />
This means that the sum of the sequence is <br />
<cmath> (x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2. </cmath><br />
We know that this must be equal to <math>xd^2,</math> so we can write that<br />
<cmath>3x^2+2d^2=xd^2,</cmath><br />
and it follows that<br />
<cmath>3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.</cmath><br />
<br />
Now, we can treat <math> d </math> as a constant and use the quadratic formula to get<br />
<cmath>x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.</cmath><br />
We can factor pull <math> d^2 </math> out of the square root to get<br />
<cmath>x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.</cmath><br />
Here, it is easy to test values of <math> d </math>. We find that <math> d=5 </math> and <math> d=7 </math> are the only positive integer values of <math> d </math> that make <math> \sqrt{d^2-24} </math> a positive integer. <math> d=5 </math> gives <math> x=5 </math> and <math> x=\frac{10}{3} </math>, but we can ignore the latter. <math> d=7 </math> gives <math> x=14 </math>, as well as a fraction which we can ignore. <br />
<br />
Since <math> d=5,~x=5 </math> and <math> d=7, x=14 </math> are the only two solutions and we want the sum of the third terms, our answer is <math> (5+5)+(7+14)=10+21=\boxed{031} </math>. -BorealBear<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_5&diff=1490702021 AIME I Problems/Problem 52021-03-11T21:33:29Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br />
<br />
==Solution==<br />
Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math>b^2=\frac{3a^2}{a-2}</math>. Simplifying further, <math>b^2=3a+6+\frac{12}{a-2}</math>.<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_5&diff=1490692021 AIME I Problems/Problem 52021-03-11T21:31:29Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br />
<br />
==Solution==<br />
Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math>\displaystyle b^2=\frac{3a^2}{a-2}</math>.<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490672021 AIME I Problems/Problem 62021-03-11T21:28:48Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point P have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
<cmath>(s-x)^2+y^2+z^2=(5\sqrt{10})^2</cmath><br />
<cmath>x^2+(s-y)^2+z^2=(5\sqrt{5})^2</cmath><br />
<cmath>x^2+y^2+(s-z)^2=(10\sqrt{2})^2</cmath><br />
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2</cmath><br />
These simplify into<br />
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sy=125</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sz=200</cmath><br />
<cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath><br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AP=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490662021 AIME I Problems/Problem 62021-03-11T21:25:16Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
<cmath>(s-x)^2+y^2+z^2=(5\sqrt{10})^2</cmath><br />
<cmath>x^2+(s-y)^2+z^2=(5\sqrt{5})^2</cmath><br />
<cmath>x^2+y^2+(s-z)^2=(10\sqrt{2})^2</cmath><br />
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2</cmath><br />
These simplify into<br />
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sy=125</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sz=200</cmath><br />
<cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath><br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490652021 AIME I Problems/Problem 62021-03-11T21:25:01Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
<cmath>(s-x)^2+y^2+z^2=(5\sqrt{10})^2=250</cmath><br />
<cmath>x^2+(s-y)^2+z^2=(5\sqrt{5})^2=125</cmath><br />
<cmath>x^2+y^2+(s-z)^2=(10\sqrt{2})^2=200</cmath><br />
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2=63</cmath><br />
These simplify into<br />
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sy=125</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sz=200</cmath><br />
<cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath><br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490632021 AIME I Problems/Problem 62021-03-11T21:23:22Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
<cmath>(s-x)^2+y^2+z^2=250</cmath><br />
<cmath>x^2+(s-y)^2+z^2=125</cmath><br />
<cmath>x^2+y^2+(s-z)^2=200</cmath><br />
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=63</cmath><br />
These simplify into<br />
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sy=125</cmath><br />
<cmath>s^2+x^2+y^2+z^2-2sz=200</cmath><br />
<cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath><br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.\\<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490622021 AIME I Problems/Problem 62021-03-11T21:22:59Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
<cmath>(s-x)^2+y^2+z^2=250</cmath>\\<br />
<cmath>x^2+(s-y)^2+z^2=125</cmath>\\<br />
<cmath>x^2+y^2+(s-z)^2=200</cmath>\\<br />
<cmath>(s-x)^2+(s-y)^2+(s-z)^2=63</cmath><br />
These simplify into<br />
<cmath>s^2+x^2+y^2+z^2-2sx=250</cmath>\\<br />
<cmath>s^2+x^2+y^2+z^2-2sy=125</cmath>\\<br />
<cmath>s^2+x^2+y^2+z^2-2sz=200</cmath>\\<br />
<cmath>3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath><br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490612021 AIME I Problems/Problem 62021-03-11T21:22:17Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
<cmath>(s-x)^2+y^2+z^2=250\\<br />
x^2+(s-y)^2+z^2=125\\<br />
x^2+y^2+(s-z)^2=200\\<br />
(s-x)^2+(s-y)^2+(s-z)^2=63</cmath><br />
These simplify into<br />
<cmath>s^2+x^2+y^2+z^2-2sx=250\\<br />
s^2+x^2+y^2+z^2-2sy=125\\<br />
s^2+x^2+y^2+z^2-2sz=200\\<br />
3s^2-2s(x+y+z)+x^2+y^2+z^2=63</cmath><br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490602021 AIME I Problems/Problem 62021-03-11T21:21:51Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
<cmath>(s-x)^2+y^2+z^2&=250\\<br />
x^2+(s-y)^2+z^2&=125\\<br />
x^2+y^2+(s-z)^2&=200\\<br />
(s-x)^2+(s-y)^2+(s-z)^2&=63</cmath><br />
These simplify into<br />
<cmath>s^2+x^2+y^2+z^2-2sx&=250\\<br />
s^2+x^2+y^2+z^2-2sy&=125\\<br />
s^2+x^2+y^2+z^2-2sz&=200\\<br />
3s^2-2s(x+y+z)+x^2+y^2+z^2&=63</cmath><br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490582021 AIME I Problems/Problem 62021-03-11T21:21:22Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
\[\begin{align*}<br />
(s-x)^2+y^2+z^2&=250\\<br />
x^2+(s-y)^2+z^2&=125\\<br />
x^2+y^2+(s-z)^2&=200\\<br />
(s-x)^2+(s-y)^2+(s-z)^2&=63<br />
\end{align*}\]<br />
These simplify into<br />
\[\begin{align*}<br />
s^2+x^2+y^2+z^2-2sx&=250\\<br />
s^2+x^2+y^2+z^2-2sy&=125\\<br />
s^2+x^2+y^2+z^2-2sz&=200\\<br />
3s^2-2s(x+y+z)+x^2+y^2+z^2&=63<br />
\end{align*}\]<br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_6&diff=1490572021 AIME I Problems/Problem 62021-03-11T21:20:57Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>?<br />
<br />
==Solution==<br />
First scale down the whole cube by 12. Let point M have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations<br />
\begin{align*}<br />
(s-x)^2+y^2+z^2&=250\\<br />
x^2+(s-y)^2+z^2&=125\\<br />
x^2+y^2+(s-z)^2&=200\\<br />
(s-x)^2+(s-y)^2+(s-z)^2&=63<br />
\end{align*}<br />
These simplify into<br />
\begin{align*}<br />
s^2+x^2+y^2+z^2-2sx&=250\\<br />
s^2+x^2+y^2+z^2-2sy&=125\\<br />
s^2+x^2+y^2+z^2-2sz&=200\\<br />
3s^2-2s(x+y+z)+x^2+y^2+z^2&=63<br />
\end{align*}<br />
Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>.<br />
Subtracting these, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>AM=16</math>. However, we scaled down everything by 12 so our answer is <math>16*12=\boxed{196}</math>.<br />
~JHawk0224<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_16&diff=1460542021 AMC 12B Problems/Problem 162021-02-12T16:56:29Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Let <math>g(x)</math> be a polynomial with leading coefficient <math>1,</math> whose three roots are the reciprocals of the three roots of <math>f(x)=x^3+ax^2+bx+c,</math> where <math>1<a<b<c.</math> What is <math>g(1)</math> in terms of <math>a,b,</math> and <math>c?</math><br />
<br />
<math>\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}</math><br />
<br />
==Solution==<br />
Note that <math>f(1/x)</math> has the same roots as <math>g(x)</math>, if it is multiplied by some monomial so that the leading term is <math>x^3</math> they will be equal. We have<br />
<cmath>f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c</cmath><br />
so we can see that<br />
<cmath>g(x) = \frac{x^3}{c}f(1/x)</cmath><br />
Therefore<br />
<cmath>g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}</cmath><br />
<br />
==Solution 2 (Vieta's bash)==<br />
Let the three roots of <math>f(x)</math> be <math>d</math>, <math>e</math>, and <math>f</math>. (Here e does NOT mean 2.7182818...)<br />
We know that <math>a=-(d+e+f)</math>, <math>b=de+ef+df</math>, and <math>c=-def</math>, and that <math>g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}</math> (Vieta's). This is equal to <math>\frac{def-de-df-ef+d+e+f-1}{def}</math>, which equals <math>\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</math>. -dstanz5<br />
<br />
== Video Solution by OmegaLearn (Vieta's Formula) ==<br />
https://youtu.be/afrGHNo_JcY<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_15&diff=1460522021 AMC 12B Problems/Problem 152021-02-12T16:56:09Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
The figure is constructed from <math>11</math> line segments, each of which has length <math>2</math>. The area of pentagon <math>ABCDE</math> can be written is <math>\sqrt{m} + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. What is <math>m + n ?</math><br />
<asy><br />
/* Made by samrocksnature */<br />
pair A=(-2.4638,4.10658);<br />
pair B=(-4,2.6567453480756127);<br />
pair C=(-3.47132,0.6335248637894945);<br />
pair D=(-1.464483379039766,0.6335248637894945);<br />
pair E=(-0.956630463955801,2.6567453480756127);<br />
pair F=(-2,2);<br />
pair G=(-3,2);<br />
draw(A--B--C--D--E--A);<br />
draw(A--F--A--G);<br />
draw(B--F--C);<br />
draw(E--G--D);<br />
label("A",A,N);<br />
label("B",B,W);<br />
label("C",C,S);<br />
label("D",D,S);<br />
label("E",E,dir(0));<br />
dot(A^^B^^C^^D^^E^^F^^G);<br />
</asy><br />
<br />
<math>\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24</math><br />
<br />
==Solution==<br />
<br />
Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120-30-30</math> triangles because of the equilateral triangles, <math>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}</math>. Also, <math>[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}</math> and so <math>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}</math>.<br />
<br />
~Lcz<br />
<br />
==Solution 2==<br />
<br />
Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>.<br />
<br />
== Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area) ==<br />
https://youtu.be/QtSbAKUb1VE<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=14|num-a=16}}<br />
{{AMC10 box|year=2021|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_14&diff=1460512021 AMC 12B Problems/Problem 142021-02-12T16:55:50Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Let <math>ABCD</math> be a rectangle and let <math>\overline{DM}</math> be a segment perpendicular to the plane of <math>ABCD</math>. Suppose that <math>\overline{DM}</math> has integer length, and the lengths of <math>\overline{MA},\overline{MC},</math> and <math>\overline{MB}</math> are consecutive odd positive integers (in this order). What is the volume of pyramid <math>MACD?</math><br />
<br />
<math>\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}</math><br />
<br />
==Solution==<br />
===Solution 1===<br />
This question is just about pythagorean theorem<br />
<cmath>a^2+(a+2)^2-b^2 = (a+4)^2</cmath><br />
<cmath>2a^2+4a+4-b^2 = a^2+8a+16</cmath><br />
<cmath>a^2-4a+4-b^2 = 16</cmath><br />
<cmath>(a-2+b)(a-2-b) = 16</cmath><br />
<cmath>a=3, b=7</cmath><br />
With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</math> ~Lopkiloinm<br />
===Solution 2===<br />
Let <math>\overline{AD}</math> be <math>b</math>, <math>\overline{CD}</math> be <math>a</math>, <math>\overline{MD}</math> be <math>x</math>, <br />
<math>\overline{MC}</math>, <math>\overline{MA}</math>, <math>\overline{MB}</math> be <math>t</math>, <math>t-2</math>, <math>t+2</math> respectively.<br />
<br />
We have three equations:<br />
<cmath>a^2 + x^2 = t^2</cmath><br />
<cmath>a^2 + b^2 + x^2 = t^2 + 4t + 4</cmath><br />
<cmath>b^2 + x^2 = t^2 - 4t + 4</cmath><br />
<br />
Subbing in the first and third equation into the second equation, we get:<br />
<cmath>t^2 - 8t - x^2 = 0</cmath><br />
<cmath>(t-4)^2 - x^2 = 16</cmath><br />
<cmath>(t-4-x)(t-4+x) = 16</cmath><br />
Therefore, <cmath>t = 9</cmath>, <cmath>x = 3</cmath><br />
Solving for other values, we get <math>b = 2\sqrt{10}</math>, <math>a = 6\sqrt{2}</math>.<br />
The volume is then <cmath>\frac{1}{3} abx = \boxed{\textbf{(A)}24\sqrt{5}}</cmath> ~jamess2022(burntTacos)<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_13&diff=1460502021 AMC 12B Problems/Problem 132021-02-12T16:54:57Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
How many values of <math>\theta</math> in the interval <math>0<\theta\le 2\pi</math> satisfy<cmath>1-3\sin\theta+5\cos3\theta = 0?</cmath><math>\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8</math><br />
<br />
==Solution==<br />
<br />
First, move terms to get <math>1+5cos3x=3sinx</math>. After graphing, we find that there are <math>\boxed{6}</math> solutions (two in each period of <math>5cos3x</math>). -dstanz5<br />
<br />
<br />
==Solution 1==<br />
We can graph two functions in this case: <math>5\cos{3x}</math> and <math>3\sin{x} -1 </math>. <cmath>\newline</cmath><br />
Using transformation of functions, we know that <math>5\cos{3x}</math> is just a cos function with<br />
amplitude 5 and frequency <math>\frac{2\pi}{3}</math>. Similarly, <math>3\sin{x} -1 </math> is just a sin function<br />
with amplitude 3 and shifted 1 unit downwards. So:<br />
<asy><br />
import graph;<br />
<br />
size(400,200,IgnoreAspect);<br />
<br />
real Sin(real t) {return 3*sin(t) - 1;}<br />
real Cos(real t) {return 5*cos(3*t);}<br />
<br />
draw(graph(Sin,0, 2pi),red,"$3\sin{x} -1 $");<br />
draw(graph(Cos,0, 2pi),blue,"$5\cos{3x}$");<br />
<br />
xaxis("$x$",BottomTop,LeftTicks);<br />
yaxis("$y$",LeftRight,RightTicks(trailingzero));<br />
<br />
<br />
<br />
add(legend(),point(E),20E,UnFill);<br />
</asy><br />
We have <math>\boxed{(A) 6}</math> solutions. ~Jamess2022 (burntTacos)<br />
<br />
== Video Solution by OmegaLearn (Using Sine and Cosine Graph) ==<br />
https://youtu.be/toBOpc6vS6s<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_12&diff=1460492021 AMC 12B Problems/Problem 122021-02-12T16:54:03Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Suppose that <math>S</math> is a finite set of positive integers. If the greatest integer in <math>S</math> is removed from <math>S</math>, then the average value (arithmetic mean) of the integers remaining is <math>32</math>. If the least integer in <math>S</math> is also removed, then the average value of the integers remaining is <math>35</math>. If the great integer is then returned to the set, the average value of the integers rises to <math>40.</math> The greatest integer in the original set <math>S</math> is <math>72</math> greater than the least integer in <math>S</math>. What is the average value of all the integers in the set <math>S?</math><br />
<br />
<math>\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37</math><br />
<br />
==Solution==<br />
===Solution 1===<br />
Let <math>x</math> be the greatest integer, <math>y</math> be the smallest, <math>z</math> be the sum of the numbers in S excluding <math>x</math> and <math>y</math>, and <math>k</math> be the number of elements in S.<br />
<br />
Then, <math>S=x+y+z</math><br />
<br />
Firstly, when the greatest integer is removed, <math>\frac{S-x}{k-1}=32</math><br />
<br />
When the smallest integer is also removed, <math>\frac{S-x-y}{k-2}=35</math><br />
<br />
When the greatest integer is added back, <math>\frac{S-y}{k-1}=40</math><br />
<br />
We are given that <math>x=y+72</math><br />
<br />
<br />
After you substitute <math>x=y+72</math>, you have 3 equations with 3 unknowns <math>S,</math>, <math>y</math> and <math>k</math>.<br />
<br />
<math>S-y-72=32k-32</math><br />
<br />
<math>S-2y-72=35k-70</math><br />
<br />
<math>S-y=40k-40</math><br />
<br />
This can be easily solved to yield <math>k=10</math>, <math>y=8</math>, <math>S=368</math>.<br />
<br />
<math>\therefore</math> average value of all integers in the set <math>=S/k = 368/10 = 36.8</math>, D)<br />
<br />
~ SoySoy4444<br />
<br />
===Solution 2===<br />
We should plug in <math>36.2</math> and assume everything is true except the <math>35</math> part. We then calculate that part and end up with <math>35.75</math>. We also see with the formulas we used with the plug in that when you increase by <math>0.2</math> the <math>35.75</math> part decreases by <math>0.25</math>. The answer is then <math>\boxed{(D) 36.8}</math>. You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm<br />
<br />
== Video Solution by OmegaLearn (System of equations) ==<br />
https://youtu.be/dRdT9gzm-Pg<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_11&diff=1460482021 AMC 12B Problems/Problem 112021-02-12T16:53:40Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Triangle <math>ABC</math> has <math>AB=13,BC=14</math> and <math>AC=15</math>. Let <math>P</math> be the point on <math>\overline{AC}</math> such that <math>PC=10</math>. There are exactly two points <math>D</math> and <math>E</math> on line <math>BP</math> such that quadrilaterals <math>ABCD</math> and <math>ABCE</math> are trapezoids. What is the distance <math>DE?</math><br />
<br />
<math>\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1 (fakesolve)===<br />
<br />
Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm<br />
<br />
===Solution 2===<br />
Using Stewart's Theorem we find <math>BP = 8\sqrt{2}</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have<br />
<cmath>DP = BP\cdot\frac{PC}{PA} = 2BP</cmath><br />
<cmath>EP = BP\cdot\frac{PA}{PC} = \frac12 BP</cmath><br />
So<br />
<cmath>DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}</cmath><br />
<br />
===Solution 3===<br />
Let <math>x</math> be the length <math>PE</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have<br />
<cmath>BP = \frac{PA}{PC}x = \frac12 x</cmath><br />
<cmath>PD = \frac{PA}{PC}BP = \frac14 x</cmath><br />
Therefore <math>BD = DE = \frac{3}{4}x</math>. Now extend line <math>CD</math> to the point <math>Z</math> on <math>AE</math>, forming parallelogram <math>ZABC</math>. As <math>BD = DE</math> we also have <math>EZ = ZC = 13</math> so <math>EC = 26</math>.<br />
<br />
We now use the Law of Cosines to find <math>x</math> (the length of <math>PE</math>):<br />
<cmath>x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)</cmath><br />
As <math>\angle PCE = \angle BAC</math>, we have (by Law of Cosines on triangle <math>BAC</math>)<br />
<cmath>\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.</cmath><br />
Therefore<br />
<cmath>\begin{align*}<br />
x^2 &= 26^2 + 10^2 - 2\cdot 26\cdot 10\cdot\frac{198}{2\cdot 13\cdot 15}\\<br />
&= 776 - 264\\<br />
&= 512<br />
\end{align*}</cmath><br />
And <math>x = 16\sqrt2</math>. The answer is then <math>\frac34x = \boxed{\textbf{(D) }12\sqrt2}</math><br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://YouTube.com/watch?v=yxt8-rUUosI&t=450s<br />
<br />
== Video Solution by OmegaLearn (Using properties of 13-14-15 triangle) ==<br />
https://youtu.be/mTcdKf5-FWg<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_10&diff=1460452021 AMC 12B Problems/Problem 102021-02-12T16:53:00Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Two distinct numbers are selected from the set <math>\{1,2,3,4,\dots,36,37\}</math> so that the sum of the remaining <math>35</math> numbers is the product of these two numbers. What is the difference of these two numbers?<br />
<br />
<math>\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10</math><br />
<br />
==Solution==<br />
The sum of the first <math>37</math> integers is given by <math>n(n+1)/2</math>, so <math>37(37+1)/2=703</math>.<br />
<br />
Therefore, <math>703-x-y=xy</math><br />
<br />
Rearranging, <math>xy+x+y=703</math><br />
<br />
<math>(x+1)(y+1)=704</math><br />
<br />
Looking at the possible divisors of <math>704 = 2^6*11</math>, <math>22</math> and <math>32</math> are within the constraints of <math>0 < x <= y <= 37</math> so we try those:<br />
<br />
<math>(x+1)(y+1) = 22 * 32</math><br />
<br />
<math>x+1=22, y+1 = 32</math><br />
<br />
<math>x = 21, y = 31</math><br />
<br />
Therefore, the difference <math>y-x=31-21=10</math>, choice E).<br />
~ SoySoy4444<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://YouTube.com/watch?v=yxt8-rUUosI&t=292s<br />
<br />
== Video Solution by OmegaLearn (Simon's Favorite Factoring Trick) ==<br />
https://youtu.be/v8MVGAZapKU<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=p4iCAZRUESs<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_9&diff=1460422021 AMC 12B Problems/Problem 92021-02-12T16:52:10Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
What is the value of<cmath>\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?</cmath><math>\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\frac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5</math><br />
<br />
==Solution==<br />
<math>\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}</math><br />
<br />
Note that <math>\log_{40}{2}=\frac{1}{\log_{2}{40}}</math>, and similarly <math>\log_{20}{2}=\frac{1}{\log_{2}{20}}</math><br />
<br />
<math>= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot log_{2}{20}</math><br />
<br />
<math>=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}</math><br />
<br />
<math>=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}</math><br />
<br />
Expanding, <math>2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{80})^2-3\log_{2}{20}-(\log_{2}{20})^2</math><br />
<br />
All the log terms cancel, so the answer is <math>2\implies\boxed{\text{(D)}}</math>.<br />
<br />
~ SoySoy4444<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtu.be/yxt8-rUUosI&t=157s<br />
<br />
== Video Solution by OmegaLearn (Logarithmic Manipulation) ==<br />
https://youtu.be/4_dUQu0a9ZA<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=1460392021 AMC 12B Problems/Problem 82021-02-12T16:51:42Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Three equally spaced parallel lines intersect a circle, creating three chords of lengths <math>38,38,</math> and <math>34</math>. What is the distance between two adjacent parallel lines?<br />
<br />
<math>\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12</math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
unitsize(10);<br />
pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7);<br />
draw(O--A--B);<br />
draw(O--R);<br />
draw(O--L);<br />
label("$A$", A, NW);<br />
label("$B$", B, N);<br />
label("$R$", R, E);<br />
label("$L$", L, E);<br />
label("$O$", O, S);<br />
label("$d$", O--A, W);<br />
label("$2d$", A--B, W*2+0.5*N);<br />
label("$r$", O--R, S);<br />
label("$r$", O--L, S*0.5 + 1.5 * E);<br />
dot(O);<br />
dot(A);<br />
dot(B);<br />
dot(R);<br />
dot(L);<br />
<br />
draw(circle((0, 4), 4));<br />
draw((-7, 3) -- (7, 3));<br />
draw((-7, 5) -- (7, 5));<br />
draw((-7, 7) -- (7, 7));<br />
</asy><br />
<br />
<br />
<br />
Since the two chords of length <math>38</math> have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be <math>d</math>. Thus, the distance from the center of the circle to the chord of length <math>34</math> is <br />
<br />
<cmath>2d + d = 3d</cmath><br />
<br />
and the distance between each of the chords is just <math>2d</math>. Let the radius of the circle be <math>r</math>. Drawing radii to the points where the lines intersect the circle, we create two different right triangles: <br />
<br />
- One with base <math>\frac{38}{2}= 19</math>, height <math>d</math>, and hypotenuse <math>r</math><br />
<br />
- Another with base <math>\frac{34}{2} = 17</math>, height <math>3d</math>, and hypotenuse <math>r</math> <br />
<br />
By the Pythagorean theorem, we can create the following systems of equations:<br />
<br />
<cmath>19^2 + d^2 = r^2</cmath><br />
<br />
<cmath>17^2 + (3d)^2 = r^2</cmath><br />
<br />
Solving, we find <math>d = 3</math>, so <math>2d = \boxed{(B) 6}</math><br />
<br />
-Solution by Joeya and diagram by Jamess2022(burntTacos).<br />
(Someone fix the diagram if possible.)<br />
<br />
==Solution 2 (Coordinates)==<br />
<br />
Because we know that the equation of a circle is <math>(x-a)^2 + (y-b)^2 = r^2</math> where the center of the circle is <math>(a, b)</math> and the radius is <math>r</math>, we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is <math>x^2 + y^2 = r^2</math>. Now, we can set the distance between the chords as <math>2d</math> so the distance from the chord with length 38 to the diameter is <math>d</math>. <br />
<br />
Therefore, the following points are on the circle as the y-axis splits the chord in half, that is where we get our x value:<br />
<br />
<math>(19, d)</math><br />
<br />
<math>(19, -d)</math><br />
<br />
<math>(17, -3d)</math><br />
<br />
<br />
Now, we can plug one of the first two value in as well as the last one to get the following equations:<br />
<br />
<cmath>19^2 + d^2 = r^2</cmath><br />
<br />
<cmath>17^2 + (3d)^2 = r^2</cmath><br />
<br />
Subtracting these two equations, we get <math>19^2 - 17^2 = 8d^2</math> - therefore, we get <math>72 = 8d^2 \rightarrow d^2 = 9 \rightarrow d = 3</math>. We want to find <math>2d = 6</math> because that's the distance between two chords. So, our answer is <math>\boxed{B}</math>.<br />
<br />
~Tony_Li2007<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtu.be/yxt8-rUUosI<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=7|num-a=9}}<br />
{{AMC10 box|year=2021|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_7&diff=1460372021 AMC 12B Problems/Problem 72021-02-12T16:51:22Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Let <math>N = 34 \cdot 34 \cdot 63 \cdot 270</math>. What is the ratio of the sum of the odd divisors of <math>N</math> to the sum of the even divisors of <math>N</math>?<br />
<br />
<math>\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3</math><br />
==Solution 1==<br />
<br />
Prime factorize <math>N</math> to get <math>N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}</math>. For each odd divisor <math>n</math> of <math>N</math>, there exist even divisors <math>2n, 4n, 8n</math> of <math>N</math>, therefore the ratio is <math>1:(2+4+8)\rightarrow\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 2==<br />
Prime factorizing <math>N</math>, we see <math>N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}</math>. The sum of <math>N</math>'s odd divisors are the sum of the factors of <math>N</math> without <math>2</math>, and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors. The sum of odd divisors is given by <cmath>a = (1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2)</cmath> and the total sum of divisors is <cmath>(1+2+4+8)(1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2) = 15a</cmath>. Thus, our ratio is <cmath>\frac{a}{15a-a} = \frac{a}{14a} = \frac{1}{14}</cmath> <math>\boxed{C}</math><br />
<br />
~JustinLee2017<br />
<br />
== Video Solution by OmegaLearn (Prime Factorization) ==<br />
https://youtu.be/U3msAYWeMbI<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=6|num-a=8}}<br />
{{AMC10 box|year=2021|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=1460352021 AMC 12B Problems/Problem 62021-02-12T16:50:47Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
An inverted cone with base radius <math>12\mathrm{cm}</math> and height <math>18\mathrm{cm}</math> is full of water. The water is poured into a tall cylinder whose horizontal base has radius of <math>24\mathrm{cm}</math>. What is the height in centimeters of the water in the cylinder?<br />
<br />
<math>\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6</math><br />
==Solution==<br />
The volume of a cone is <math>\frac{1}{3} \cdot\pi \cdot r^2 \cdot h</math> where <math>r</math> is the base radius and <math>h</math> is the height. The water completely fills up the cone so the volume of the water is <math>\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi</math>.<br />
<br />
The volume of a cylinder is <math>\pi \cdot r^2 \cdot h</math> so the volume of the water in the cylinder would be <math>24\cdot24\cdot\pi\cdot h</math>.<br />
<br />
We can equate these two expressions because the water volume stays the same like this <math>24\cdot24\cdot\pi\cdot h = 6\cdot144\pi</math>. We get <math>4h = 6</math> and <math>h=\frac{6}{4}</math>.<br />
<br />
So the answer is <math>1.5 = \boxed{\textbf{(A)}}.</math><br />
<br />
<br />
--abhinavg0627<br />
<br />
== Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders) ==<br />
https://youtu.be/4JhZLAORb8c<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=5|num-a=7}}<br />
{{AMC10 box|year=2021|ab=B|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_5&diff=1460342021 AMC 12B Problems/Problem 52021-02-12T16:50:31Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
The point <math>P(a,b)</math> in the <math>xy</math>-plane is first rotated counterclockwise by <math>90\deg</math> around the point <math>(1,5)</math> and then reflected about the line <math>y = -x</math>. The image of <math>P</math> after these two transformations is at <math>(-6,3)</math>. What is <math>b - a ?</math><br />
<br />
<math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9</math><br />
==Solution==<br />
<br />
The final image of <math>P</math> is <math>(-6,3)</math>. We know the reflection rule for reflecting over <math>y=-x</math> is <math>(x,y) --> (-y, -x)</math>. So before the reflection and after rotation the point is <math>(-3,6)</math>.<br />
<br />
By definition of rotation, the slope between <math>(-3,6)</math> and <math>(1,5)</math> must be perpendicular to the slope between <math>(a,b)</math> and <math>(1,5)</math>. The first slope is <math>\frac{5-6}{1-(-3)} = \frac{-1}{4}</math>. This means the slope of <math>P</math> and <math>(1,5)</math> is <math>4</math>.<br />
<br />
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from <math>(3,-6)</math> to <math>(1,5)</math> it follows we shall only use the slope once to travel from <math>(1,5)</math> to <math>P</math>.<br />
<br />
Therefore point <math>P</math> is located at <math>(1+1, 5+4) = (2,9)</math>. The answer is <math>9-2 = 7 = \boxed{\textbf{(D)}}</math>.<br />
<br />
<br />
--abhinavg0627<br />
<br />
== Video Solution by OmegaLearn (Rotation & Reflection tricks) ==<br />
https://youtu.be/VyRWjgGIsRQ<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=4|num-a=6}}<br />
{{AMC10 box|year=2021|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_4&diff=1460332021 AMC 12B Problems/Problem 42021-02-12T16:50:15Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is <math>84</math>, and the afternoon class's mean score is <math>70</math>. The ratio of the number of students in the morning class to the number of students in the afternoon class is <math>\frac{3}{4}</math>. What is the mean of the scores of all the students?<br />
<br />
<math>\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78</math><br />
<br />
==Solution==<br />
WLOG, assume there are <math>3</math> students in the morning class and <math>4</math> in the afternoon class. Then the average is <math>\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}</math><br />
<br />
==Solution 2==<br />
Let there be <math>3x</math> students in the morning class and <math>4x</math> students in the afternoon class. The total number of students is <math>3x + 4x = 7x</math>. The average is <math>\frac{3x\cdot84 + 4x\cdot70}{7x}=76</math>. Therefore, the answer is <math>\boxed{\textbf{(C)}76}</math>.<br />
<br><br><br />
~ {TSun} ~<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
== Video Solution by OmegaLearn (Clever application of Average Formula) ==<br />
https://youtu.be/lE8v7lXT8Go<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=3|num-a=5}}<br />
{{AMC10 box|year=2021|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_3&diff=1460312021 AMC 12B Problems/Problem 32021-02-12T16:49:56Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Suppose<cmath>2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.</cmath>What is the value of <math>x?</math><br />
<br />
<math>\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}</math><br />
<br />
==Solution 1==<br />
Subtracting <math>2</math> from both sides and taking reciprocals gives <math>1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}</math>. Subtracting <math>1</math> from both sides and taking reciprocals again gives <math>2+\frac{2}{3+x}=\frac{38}{15}</math>. Subtracting <math>2</math> from both sides and taking reciprocals for the final time gives <math>\frac{x+3}{2}=\frac{15}{8}</math> or <math>x=\frac{3}{4} \implies \boxed{\text{A}}</math>.<br />
<br />
~ OlympusHero<br />
<br />
== Video Solution by OmegaLearn (Algebraic Manipulations) ==<br />
https://youtu.be/WskJI8_7Gk0<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_2&diff=1460302021 AMC 12B Problems/Problem 22021-02-12T16:49:38Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
At a math contest, <math>57</math> students are wearing blue shirts, and another <math>75</math> students are wearing yellow shirts. The 132 students are assigned into <math>66</math> pairs. In exactly <math>23</math> of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?<br />
<br />
<math>\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64</math><br />
<br />
==Solution==<br />
There are <math>46</math> students paired with a blue partner. The other <math>11</math> students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are <math>64</math> students remaining. Therefore the requested number of pairs is <math>\tfrac{64}{2}=\boxed{\textbf{(B)} ~32}</math> ~Punxsutawney Phil<br />
<br />
== Video Solution by OmegaLearn (System of Equations) ==<br />
https://youtu.be/hyYg62tT0sY<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_1&diff=1460292021 AMC 12B Problems/Problem 12021-02-12T16:49:14Z<p>Jhawk0224: </p>
<hr />
<div>How many integer values of <math>x</math> satisfy <math>|x|<3\pi</math>?<br />
<br />
<math>\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20</math><br />
<br />
==Solution==<br />
Since <math>3\pi</math> is about <math>9.42</math>, we multiply 9 by 2 and add 1 to get <math> \boxed{\textbf{(D)}\ ~19} </math>~smarty101<br />
==Solution 2==<br />
<math>|x|<3\pi</math> <math>\iff</math> <math>-3\pi<x<3\pi</math>. Since <math>\pi</math> is approximately <math>3.14</math>, <math>3\pi</math> is approximately <math>9.42</math>. We are trying to solve for <math>-9.42<x<9.42</math>, where <math>x\in\mathbb{Z}</math>. Hence, <math>-9.42<x<9.42</math> <math>\implies</math> <math>-9\leq x\leq9</math>, for <math>x\in\mathbb{Z}</math>. The number of integer values of <math>x</math> is <math>9-(-9)+1=19</math>. Therefore, the answer is <math>\boxed{\textbf{(D)}19}</math>.<br />
<br><br><br />
~ {TSun} ~<br />
<br />
== Video Solution by OmegaLearn (Basic Computation) ==<br />
https://youtu.be/_C4ceJn6Iaw<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=VzwxbsuSQ80<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_8&diff=1452092021 AMC 12A Problems/Problem 82021-02-11T21:00:29Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
A sequence of numbers is defined by <math>D_0=0,D_0=1,D_2=1</math> and <math>D_n=D_{n-1}+D_{n-3}</math> for <math>n\ge 3</math>. What are the parities (evenness or oddness) of the triple of numbers <math>(D_{2021},D_{2022},D_{2023})</math>, where <math>E</math> denotes even and <math>O</math> denotes odd?<br />
<br />
<math>\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)</math><br />
<br />
==Solution==<br />
Making a small chart, we have<br />
<br />
<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}<br />
D0&D1&D2&D3&D4&D5&D6&D7&D8&D9\\\hline<br />
0&0&1&1&1&2&3&4&6&9\\\hline<br />
E&E&O&O&O&E&O&E&E&O<br />
\end{tabular}</math><br />
<br />
This starts repeating every 7 terms, so <math>D_{2021}=D_5=E</math>, <math>D_{2022}=D_6=O</math>, and <math>D_{2023}=D_7=E</math>. Thus, the answer is <math>\boxed{\textbf{(C) }(E, O, E)}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_8&diff=1452072021 AMC 12A Problems/Problem 82021-02-11T20:59:45Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
A sequence of numbers is defined by <math>D_0=0,D_0=1,D_2=1</math> and <math>D_n=D_{n-1}+D_{n-3}</math> for <math>n\ge 3</math>. What are the parities (evenness or oddness) of the triple of numbers <math>(D_{2021},D_{2022},D_{2023})</math>, where <math>E</math> denotes even and <math>O</math> denotes odd?<br />
<br />
<math>\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)</math><br />
<br />
==Solution==<br />
Making a small chart, we have<br />
<br />
<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}<br />
D\textsubscript{0}&D\textsubscript{1}&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline<br />
0&0&1&1&1&2&3&4&6&9\\\hline<br />
E&E&O&O&O&E&O&E&E&O<br />
\end{tabular}</math><br />
<br />
This starts repeating every 7 terms, so <math>D_{2021}=D_5=E</math>, <math>D_{2022}=D_6=O</math>, and <math>D_{2023}=D_7=E</math>. Thus, the answer is <math>\boxed{\textbf{(C) }(E, O, E)}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_8&diff=1452042021 AMC 12A Problems/Problem 82021-02-11T20:58:52Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
A sequence of numbers is defined by <math>D_0=0,D_0=1,D_2=1</math> and <math>D_n=D_{n-1}+D_{n-3}</math> for <math>n\ge 3</math>. What are the parities (evenness or oddness) of the triple of numbers <math>(D_{2021},D_{2022},D_{2023})</math>, where <math>E</math> denotes even and <math>O</math> denotes odd?<br />
<br />
<math>\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)</math><br />
<br />
==Solution==<br />
Making a small chart, we have<br />
<br />
<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}<br />
</math>D_0<math>&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline<br />
0&0&1&1&1&2&3&4&6&9\\\hline<br />
E&E&O&O&O&E&O&E&E&O<br />
\end{tabular}</math><br />
<br />
This starts repeating every 7 terms, so <math>D_{2021}=D_5=E</math>, <math>D_{2022}=D_6=O</math>, and <math>D_{2023}=D_7=E</math>. Thus, the answer is <math>\boxed{\textbf{(C) }(E, O, E)}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_8&diff=1452022021 AMC 12A Problems/Problem 82021-02-11T20:58:17Z<p>Jhawk0224: /* Solution */</p>
<hr />
<div>==Problem==<br />
A sequence of numbers is defined by <math>D_0=0,D_0=1,D_2=1</math> and <math>D_n=D_{n-1}+D_{n-3}</math> for <math>n\ge 3</math>. What are the parities (evenness or oddness) of the triple of numbers <math>(D_{2021},D_{2022},D_{2023})</math>, where <math>E</math> denotes even and <math>O</math> denotes odd?<br />
<br />
<math>\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)</math><br />
<br />
==Solution==<br />
Making a small chart, we have<br />
<br />
<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}<br />
D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline<br />
0&0&1&1&1&2&3&4&6&9\\\hline<br />
E&E&O&O&O&E&O&E&E&O<br />
\end{tabular}</math><br />
<br />
This starts repeating every 7 terms, so <math>D_{2021}=D_5=E</math>, <math>D_{2022}=D_6=O</math>, and <math>D_{2023}=D_7=E</math>. Thus, the answer is <math>\boxed{\textbf{(C) }(E, O, E)}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_18&diff=1451982021 AMC 12A Problems/Problem 182021-02-11T20:56:31Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b) = f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Furthermore, suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x) < 0</math>?<br />
<br />
<math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math><br />
<br />
==Solution 1==<br />
Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>.<br />
<br />
-Lemonie<br />
<br />
==Solution 2==<br />
We know that <math>f(2)=2</math>. Adding <math>f(1)</math> to both sides, we get <cmath>\begin{align*}<br />
f(2)+f(1)&=2+f(1)\\<br />
f(2)&=2+f(1)\\<br />
2&=2+f(1)\\<br />
f(1)&=0<br />
\end{align*}</cmath><br />
Also<br />
<cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath><br />
<cmath>f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3</cmath><br />
<cmath>f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11</cmath><br />
In <math>\textbf{(A)}</math> we have <math>f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7</math>.<br />
<br />
In <math>\textbf{(B)}</math> we have <math>f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3</math>.<br />
<br />
In <math>\textbf{(C)}</math> we have <math>f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1</math>.<br />
<br />
In <math>\textbf{(D)}</math> we have <math>f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2</math>.<br />
<br />
In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>.<br />
<br />
Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=dvlTA8Ncp58<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtu.be/8gGcj95rlWY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_16&diff=1451962021 AMC 12A Problems/Problem 162021-02-11T20:55:45Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
In the following list of numbers, the integer <math>n</math> appears <math>n</math> times in the list for <math>1 \leq n \leq 200</math>.<cmath>1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \cdot, 200, 200, \cdot , 200</cmath>What is the median of the numbers in this list?<br />
<br />
==Solution==<br />
===Solution 1===<br />
There are <math>1+2+..+199+200=\frac{(200)(201)}{2}=20100</math> numbers in total. Let the median be <math>k</math>. We want to find the median <math>k</math> such that<br />
<cmath> \frac{k(k+1)}{2}=20100/2,</cmath><br />
or<br />
<cmath> k(k+1)=20100.</cmath><br />
Note that <math>\sqrt{20100} \approx 142</math>. Plugging this value in as <math>k</math> gives<br />
<cmath>\frac{1}{2}(142)(143)=10153.</cmath><br />
<math>10153-142<10050</math>, so <math>142</math> is the <math>152</math>nd and <math>153</math>rd numbers, and hence, our desired answer. <math>\fbox{(C) 142}.</math>.<br />
===Solution 2===<br />
The <math>x</math>th number of this sequence is obviously <math>\frac{-1\pm\sqrt{1+8x}}{2}</math> via the quadratic formula. We can see that if we halve <math>x</math> we end up getting <math>\frac{-1\pm\sqrt{1+4x}}{2}</math>. This is approximately the number divided by <math>\sqrt{2}</math>. <math>\frac{200}{\sqrt{2}} = 141.4</math> and since <math>142</math> looks like the only number close to it, it is answer <math>\boxed{(C) 142}</math> ~Lopkiloinm<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=AjQARBvdZ20<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_15&diff=1451952021 AMC 12A Problems/Problem 152021-02-11T20:55:32Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
A choir direction must select a group of singers from among his <math>6</math> tenors and <math>8</math> basses. The only requirements are that the difference between the number of tenors and basses must be a multiple of <math>4</math>, and the group must have at least one singer. Let <math>N</math> be the number of different groups that could be selected. What is the remainder when <math>N</math> is divided by <math>100</math>?<br />
<br />
<math>\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad</math><br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtube.com/watch?v=FD9BE7hpRvg&t=533s<br />
<br />
==Solution 2==<br />
We know the choose function and we know the pair multiplication <math>MN</math> so we do the multiplications and additions.<br />
<math>\binom{6}{0}(\binom{8}{4}+\binom{8}{8})+\binom{6}{1}(\binom{8}{1}+\binom{8}{5})+\binom{6}{2}(\binom{8}{2}+\binom{8}{6})+\binom{6}{3}(\binom{8}{3}+\binom{8}{7})+\binom{6}{4}(\binom{8}{0}+\binom{8}{4}+\binom{8}{8})+\binom{6}{5}(\binom{8}{1}+\binom{8}{5})+\binom{6}{6}(\binom{8}{2}+\binom{8}{6}) = \boxed{(D) 4095}</math><br />
~Lopkiloinm<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=AjQARBvdZ20<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=14|num-a=16}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_14&diff=1451932021 AMC 12A Problems/Problem 142021-02-11T20:55:14Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
What is the value of<cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?</cmath><math>\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000</math><br />
==Solution==<br />
This equals<br />
<cmath>\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}</cmath><br />
~JHawk0224<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtube.com/watch?v=FD9BE7hpRvg&t=322s<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=AjQARBvdZ20<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_13&diff=1451922021 AMC 12A Problems/Problem 132021-02-11T20:54:59Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Of the following complex numbers <math>z</math>, which one has the property that <math>z^5</math> has the greatest real part?<br />
<br />
<math>\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i</math><br />
<br />
==Solution==<br />
First, <math>\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)</math><math>, \textbf{(D)} =2\text{cis}(120)</math>.<br />
<br />
Taking the real part of the 5th power of each we have:<br />
<br />
<math>\textbf{(A): }(-2)^5=-32</math>,<br />
<br />
<math>\textbf{(B): }32\cos(650)=32\cos(30)=16\sqrt{3}</math><br />
<br />
<math>\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}</math><br />
<br />
<math>\textbf{(D): }32\cos(600)=32\cos(240)</math> which is negative<br />
<br />
<math>\textbf{(E): }(2i)^5</math> which is imaginary<br />
<br />
Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.<br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=AjQARBvdZ20<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_12&diff=1451902021 AMC 12A Problems/Problem 122021-02-11T20:54:49Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
All the roots of the polynomial <math>z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16</math> are positive integers, possibly repeated. What is the value of <math>B</math>?<br />
<br />
<math>\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40</math><br />
<br />
==Solution==<br />
By Vieta's formulae, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, so the function is <math>(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)</math>. Therefore, <math>B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=AjQARBvdZ20<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_11&diff=1451892021 AMC 12A Problems/Problem 112021-02-11T20:54:38Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
A laser is placed at the point <math>(3,5)</math>. The laser bean travels in a straight line. Larry wants the beam to hit and bounce off the <math>y</math>-axis, then hit and bounce off the <math>x</math>-axis, then hit the point <math>(7,5)</math>. What is the total distance the beam will travel along this path?<br />
<br />
<math>\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5</math><br />
<br />
==Solution==<br />
Every time the laser bounces off a wall, instead we can imagine it going straight by reflecting it about the wall. Thus, the laser starts at <math>(3, 5)</math> and ends at <math>(-7, -5)</math>, so the path's length is <math>\sqrt{10^2+10^2}=\boxed{\textbf{(C)} 10\sqrt{2}}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=AjQARBvdZ20<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_10&diff=1451882021 AMC 12A Problems/Problem 102021-02-11T20:54:25Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are <math>3</math> cm and <math>6</math> cm. Into each cone is dropped a spherical marble of radius <math>1</math> cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?<br />
<br />
<asy><br />
size(350);<br />
defaultpen(linewidth(0.8));<br />
real h1 = 10, r = 3.1, s=0.75;<br />
pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q;<br />
path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9);<br />
draw(ellipse(origin,r*(s-0.1),0.8));<br />
fill(ep,gray(0.8));<br />
fill(origin--Pp--Qp--cycle,gray(0.8));<br />
draw((-r,h1)--(0,0)--(r,h1)^^e);<br />
draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4"));<br />
draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br />
draw(Qp--(0,Qp.y),Arrows(size=8));<br />
draw(origin--(0,12),linetype("4 4"));<br />
draw(origin--(r*(s-0.1),0));<br />
label("$3$",(-0.9,h1*s),N,fontsize(10));<br />
<br />
real h2 = 7.5, r = 6, s=0.6, d = 14;<br />
pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0);<br />
path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1);<br />
draw(ellipse((d,0),r*(s-0.1),0.8));<br />
fill(ep,gray(0.8));<br />
fill((d,0)--Pp--Qp--cycle,gray(0.8));<br />
draw(P--(d,0)--Q^^e);<br />
draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4"));<br />
draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br />
draw(Qp--(d,Qp.y),Arrows(size=8));<br />
draw((d,0)--(d,10),linetype("4 4"));<br />
draw((d,0)--(d+r*(s-0.1),0));<br />
label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10));<br />
</asy><br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) }\frac{47}{43} \qquad \textbf{(C) }2 \qquad \textbf{(D) }\frac{40}{13} \qquad \textbf{(E) }4</math><br />
<br />
==Solution==<br />
The answer is <math>\boxed{\textbf{(E) } 4}</math><br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=AjQARBvdZ20<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=9|num-a=11}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_9&diff=1451872021 AMC 12A Problems/Problem 92021-02-11T20:53:45Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Which of the following is equivalent to<br />
<cmath>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</cmath><br />
<math>\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}</math><br />
<br />
==Solution 1==<br />
<br />
All you need to do is multiply the entire equation by <math>(3-2)</math>. Then all the terms will easily simplify by difference of squares and you will get <math>3^{128}-2^{128}</math> or <math>\boxed{C}</math> as your final answer. Notice you don't need to worry about <math>3-2</math> because that's equal to <math>1</math>.<br />
<br />
-Lemonie<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_8&diff=1451862021 AMC 12A Problems/Problem 82021-02-11T20:53:32Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
A sequence of numbers is defined by <math>D_0=0,D_0=1,D_2=1</math> and <math>D_n=D_{n-1}+D_{n-3}</math> for <math>n\ge 3</math>. What are the parities (evenness or oddness) of the triple of numbers <math>(D_{2021},D_{2022},D_{2023})</math>, where <math>E</math> denotes even and <math>O</math> denotes odd?<br />
<br />
<math>\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)</math><br />
<br />
==Solution==<br />
Making a small chart, we have<br />
<cmath>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}<br />
D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline<br />
0&0&1&1&1&2&3&4&6&9\\\hline<br />
E&E&O&O&O&E&O&E&E&O<br />
\end{tabular}</cmath><br />
This starts repeating every 7 terms, so <math>D_{2021}=D_5=E</math>, <math>D_{2022}=D_6=O</math>, and <math>D_{2023}=D_7=E</math>. Thus, the answer is <math>\boxed{\textbf{(C) }(E, O, E)}</math><br />
~JHawk0224<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_7&diff=1451852021 AMC 12A Problems/Problem 72021-02-11T20:53:17Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
What is the least possible value of <math>(xy-1)^2+(x+y)^2</math> for real numbers <math>x</math> and <math>y</math>?<br />
<br />
<math>\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2</math><br />
<br />
==Solution==<br />
Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the trivial inequality(all squares are nonnegative) the minimum value for this is <math>\boxed{(D) 1}</math>, which can be achieved at <math>x=y=0</math>. ~aop2014<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_6&diff=1451832021 AMC 12A Problems/Problem 62021-02-11T20:52:58Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is <math>\frac13</math>. When <math>4</math> black cards are added to the deck, the probability of choosing red becomes <math>\frac14</math>. How many cards were in the deck originally?<br />
<br />
<br />
<math>\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18</math><br />
<br />
==Solution==<br />
If the probability of choosing a red card is <math>\frac{1}{3}</math>, the red and black cards are in ratio <math>1:2</math>. This means at the beginning there are <math>x</math> red cards and <math>2x</math> black cards.<br />
<br />
After <math>4</math> black cards are added, there are <math>2x+4</math> black cards. This time, the probability of choosing a red card is <math>\frac{1}{4}</math> so the ratio of red to black cards is <math>1:3</math>. This means in the new deck the number of black cards is also <math>3x</math> for the same <math>x</math> red cards.<br />
<br />
So, <math>3x = 2x + 4</math> and <math>x=4</math> meaning there are <math>4</math> red cards in the deck at the start and <math>2(4) = 8</math> black cards. <br />
<br />
So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>.<br />
<br />
<br />
--abhinavg0627<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_5&diff=1451822021 AMC 12A Problems/Problem 52021-02-11T20:52:43Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
When a student multiplied the number <math>66</math> by the repeating decimal <cmath>\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},</cmath> where <math>a</math> and <math>b</math> are digits, he did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit number <math>\underline{a}\underline{b}?</math><br />
<br />
<math>\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75</math><br />
<br />
==Solution==<br />
It is known that <math>0.\overline{ab}=\frac{ab}{99}</math> and <math>0.ab=\frac{ab}{100}</math>. Let <math>\overline {ab} = x</math>. We have that <math>66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})</math>. Solving gives that <math>100x-75=99x</math> so <math>x=\boxed{(E) 75}</math>. ~aop2014<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtube.com/watch?v=MUHja8TpKGw&t=359s<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_4&diff=1451812021 AMC 12A Problems/Problem 42021-02-11T20:52:03Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Tom has a collection of <math>13</math> snakes, <math>4</math> of which are purple and <math>5</math> of which are happy. He observes that all of his happy snakes can add, none of his purple snakes can subtract, and all of his snakes that can't subtract also can't add. Which of these conclusions can be drawn about Tom's snakes?<br />
<br />
<math>\textbf{(A) }</math> Purple snakes can add.<br />
<br />
<math>\textbf{(B) }</math> Purple snakes are happy.<br />
<br />
<math>\textbf{(C) }</math> Snakes that can add are purple.<br />
<br />
<math>\textbf{(D) }</math> Happy snakes are not purple.<br />
<br />
<math>\textbf{(E) }</math> Happy snakes can't subtract.<br />
<br />
<br />
==Solution 1==<br />
<br />
We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
--abhinavg0627<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtube.com/watch?v=MUHja8TpKGw&t=259s (Note that there's a slight error in the video I corrected in the description)<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=3|num-a=5}}<br />
{{AMC10 box|year=2021|ab=A|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&diff=1451802021 AMC 12A Problems/Problem 32021-02-11T20:51:34Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br />
<br />
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math><br />
<br />
==Solution==<br />
<br />
The units digit of a multiple of <math>10</math> will always be <math>0</math>. We add a <math>0</math> whenever we multiply by <math>10</math>. So, removing the units digit is equal to dividing by <math>10</math>.<br />
<br />
Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>.<br />
<br />
We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = 14238 = \boxed{\textbf{(D)}}</math>.<br />
<br />
<br />
--abhinavg0627<br />
<br />
==Video Solution by Punxsutawney Phil==<br />
https://youtube.com/watch?v=MUHja8TpKGw&t=143s<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==Video Solution==<br />
https://youtu.be/d2musztzDjw<br />
-pi_is_3.14<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_2&diff=1451782021 AMC 12A Problems/Problem 22021-02-11T20:50:57Z<p>Jhawk0224: </p>
<hr />
<div>==Problem==<br />
Under what conditions does <math>\sqrt{a^2+b^2}=a+b</math> hold, where <math>a</math> and <math>b</math> are real numbers?<br />
<br />
<math>\textbf{(A) }</math> It is never true.<br />
<br />
<math>\textbf{(B) }</math> It is true if and only if <math>ab=0</math>.<br />
<br />
<math>\textbf{(C) }</math> It is true if and only if <math>a+b\ge 0</math>.<br />
<br />
<math>\textbf{(D) }</math> It is true if and only if <math>ab=0</math> and <math>a+b\ge 0</math>.<br />
<br />
<math>\textbf{(E) }</math> It is always true.<br />
<br />
==Solution==<br />
Square both sides to get <math>a^{2}+b^{2}=a^{2}+2ab+b^{2}</math>. Then, <math>0=2ab\rightarrow ab=0</math>. Then, the answer is <math>\boxed{\textbf{(B)}}</math>. Consider a right triangle with legs <math>a</math> and <math>b</math> and hypotenuse <math>\sqrt{a^{2}+b^{2}}</math>. Then one of the legs must be equal to <math>0</math>, but they are also nonnegative as they are lengths. Therefore, both <math>\textbf{(B)}</math> and <math>\textbf{(D)}</math> are correct.<br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Jhawk0224https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_1&diff=1451762021 AMC 12A Problems/Problem 12021-02-11T20:50:09Z<p>Jhawk0224: /* Video Solution */</p>
<hr />
<div>==Problem==<br />
What is the value of<cmath>2^{1+2+3}-(2^1+2^2+2^3)?</cmath><math>\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57</math><br />
<br />
==Solution==<br />
We evaluate the given expression to get that <cmath>2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{\text{(B)}}</cmath><br />
<br />
==Video Solution by Hawk Math==<br />
https://www.youtube.com/watch?v=P5al76DxyHY<br />
<br />
==See also==<br />
{{AMC12 box|year=2021|ab=A|before=First problem|num-a=2}}<br />
{{MAA Notice}}</div>Jhawk0224