https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Stormstar--&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T13:47:17ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=143555User:Piphi2021-01-28T04:47:43Z<p>Stormstar--: /* User Count */</p>
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<div>{{User:Piphi/Template:Header}}<br />
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__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">448</font></center><br />
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<div style="border:2px solid black; background:#919293;-webkit-border-radius: 10px; align:center"><br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black"><font color="black"><font style="font-size: 26px;line-height: 45px;font-weight: 700;">I am currently postbanned, to talk to me, go [[User_talk:Piphi|here]].</font><br />
<br />
Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.<br><br />
<br />
Piphi started the signature trend at around May 2020.<br><br />
<br />
Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].<br><br />
<br />
Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].<br><br />
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Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.<br><br />
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Piphi has also added a lot of the info that is in the [[Reaper Archives]].<br><br />
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Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
<br />
Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br />
<br />
Piphi has a post that was made an announcement in a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br />
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Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br />
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Piphi has won 2 gold trophies at the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] and is now part of the staff.<br />
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==<font color="#f0f2f3" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px"><font color="#f0f2f3"><br />
You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br />
<br />
A User Count of 500<br />
{{User:Piphi/Template:Progress_Bar|88.8|width=100%}}<br />
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200 subpages of [[User:Piphi]]<br />
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200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|49.5|width=100%}}<br />
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Make 10,000 edits<br />
{{User:Piphi/Template:Progress_Bar|21.33|width=100%}}</font></div><br />
</div></div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_22&diff=869512005 AMC 10A Problems/Problem 222017-08-09T01:03:20Z<p>Stormstar--: </p>
<hr />
<div>==Problem==<br />
Let <math>S</math> be the [[set]] of the <math>2005</math> smallest positive multiples of <math>4</math>, and let <math>T</math> be the set of the <math>2005</math> smallest positive multiples of <math>6</math>. How many elements are common to <math>S</math> and <math>T</math>?<br />
<br />
<math> \mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001 </math><br />
<br />
==Solution==<br />
Since the [[least common multiple]] <math>\mathrm{lcm}(4,6)=12</math>, the [[element]]s that are common to <math>S</math> and <math>T</math> must be [[multiple]]s of <math>12</math>. <br />
<br />
Since <math>4\cdot2005=8020</math> and <math>6\cdot2005=12030</math>, several multiples of <math>12</math> that are in <math>T</math> won't be in <math>S</math>, but all multiples of <math>12</math> that are in <math>S</math> will be in <math>T</math>. So we just need to find the number of multiples of <math>12</math> that are in <math>S</math>. <br />
<br />
Since <math>4\cdot3=12</math> every <math>3</math>rd element of <math>S</math> will be a multiple of <math>12</math><br />
<br />
Therefore the answer is <math>\lfloor\frac{2005}{3}\rfloor=668\Longrightarrow \boxed{\mathrm{(D) 668}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2005|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=Pythagorean_Theorem&diff=86655Pythagorean Theorem2017-07-27T19:24:11Z<p>Stormstar--: </p>
<hr />
<div>The '''Pythagorean Theorem''' states that for a [[right triangle]] with legs of length <math>a</math> and <math>b</math> and [[hypotenuse]] of length <math>c</math> we have the relationship <math>{a}^{2}+{b}^{2}={c}^{2}</math>. This theorem has been know since antiquity and is a classic to prove; hundreds of proofs have been published and many can be demonstrated entirely visually(the book ''The Pythagorean Proposition'' alone consists of more than 370). The Pythagorean Theorem is one of the most frequently used theorems in [[geometry]], and is one of the many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theorem. <br />
<br />
This is generalized by the [[Geometric inequality#Pythagorean_Inequality | Pythagorean Inequality]] and the [[Law of Cosines]].<br />
<br />
== Proofs ==<br />
<br />
In these proofs, we will let <math>ABC </math> be any right triangle with a right angle at <math>{} C </math>.<br />
<br />
=== Proof 1 ===<br />
<br />
We use <math>[ABC] </math> to denote the area of triangle <math>ABC </math>.<br />
<br />
Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>.<br />
<br />
<center><br />
<asy><br />
pair A, B, C, H;<br />
A = (0, 0);<br />
B = (4, 3);<br />
C = (4, 0);<br />
H = foot(C, A, B);<br />
<br />
draw(A--B--C--cycle);<br />
draw(C--H);<br />
draw(rightanglemark(A, C, B));<br />
draw(rightanglemark(C, H, B));<br />
label("$A$", A, SSW);<br />
label("$B$", B, ENE);<br />
label("$C$", C, SE);<br />
label("$H$", H, NNW);<br />
</asy><br />
</center><br />
<br />
Since <math>ABC, CBH, ACH</math> are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,<br />
<center><br />
<math> \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} </math>.<br />
</center><br />
But since triangle <math>ABC </math> is composed of triangles <math>CBH </math> and <math>ACH </math>, <math>[ABC] = [CBH] + [ACH] </math>, so <math>AB^2 = CB^2 + AC^2 </math>. {{Halmos}}<br />
<br />
=== Proof 2 ===<br />
<br />
Consider a circle <math>\omega </math> with center <math>B </math> and radius <math>BC </math>. Since <math>BC </math> and <math>AC </math> are perpendicular, <math>AC </math> is tangent to <math>\omega </math>. Let the line <math>AB </math> meet <math>\omega </math> at <math>Y </math> and <math>X </math>, as shown in the diagram:<br />
<br />
<center>[[Image:Pyth2.png]]</center><br />
<br />
Evidently, <math>AY = AB - BC </math> and <math>AX = AB + BC </math>. By considering the [[power of a point | power of point]] <math>A </math> with respect to <math>\omega </math>, we see<br />
<br />
<center><br />
<math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}}<br />
</center><br />
<br />
=== Proof 3 ===<br />
<br />
<math>ABCD</math> and <math>EFGH</math> are squares.<br />
<center><br />
<asy><br />
pair A, B,C,D;<br />
A = (-10,10);<br />
B = (10,10);<br />
C = (10,-10);<br />
D = (-10,-10);<br />
<br />
pair E,F,G,H;<br />
E = (7,10);<br />
F = (10, -7);<br />
G = (-7, -10);<br />
H = (-10, 7);<br />
<br />
draw(A--B--C--D--cycle);<br />
label("$A$", A, NNW);<br />
label("$B$", B, ENE);<br />
label("$C$", C, ESE);<br />
label("$D$", D, SSW);<br />
<br />
draw(E--F--G--H--cycle);<br />
label("$E$", E, N);<br />
label("$F$", F,SE);<br />
label("$G$", G, S);<br />
label("$H$", H, W);<br />
<br />
label("a", A--B,N);<br />
label("a", B--F,SE);<br />
label("a", C--G,S);<br />
label("a", H--D,W);<br />
label("b", E--B,N);<br />
label("b", F--C,SE);<br />
label("b", G--D,S);<br />
label("b", A--H,W);<br />
label("c", E--H,NW);<br />
label("c", E--F);<br />
label("c", F--G,SE);<br />
label("c", G--H,SW);<br />
</asy><br />
</center><br />
<math>(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2</math>. {{Halmos}}<br />
<br />
== Common Pythagorean Triples ==<br />
A [[Pythagorean Triple]] is a [[set]] of 3 [[positive integer]]s such that <math>a^{2}+b^{2}=c^{2}</math>, i.e. the 3 numbers can be the lengths of the sides of a right triangle. Among these, the [[Primitive Pythagorean Triple]]s, those in which the three numbers have no common [[divisor]], are most interesting. A few of them are:<br />
<br />
<cmath>3-4-5</cmath><br />
<cmath>5-12-13</cmath><br />
<cmath>9-12-15</cmath><br />
<cmath>7-24-25</cmath><br />
<cmath>8-15-17</cmath><br />
<cmath>9-40-41</cmath><br />
<cmath>12-35-37</cmath><br />
<cmath>20-21-29</cmath><br />
<br />
<br />
Also Pythagorean Triples can be created with the <cmath>3-4-5</cmath> triple by multiplying the lengths by any integer.<br />
For example, <br />
<cmath>6-8-10</cmath><br />
<cmath>9-12-15</cmath><br />
<cmath>12-16-20</cmath><br />
<br />
== Problems ==<br />
=== Introductory ===<br />
* [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]<br />
* [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]<br />
<br />
== External links ==<br />
*[http://www.cut-the-knot.org/pythagoras/index.shtml 118 proofs of the Pythagorean Theorem]<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]<br />
<br />
[[Category:Geometry]]</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_5&diff=866012017 AIME I Problems/Problem 52017-07-26T00:33:25Z<p>Stormstar--: </p>
<hr />
<div>==Problem 5==<br />
A rational number written in base eight is <math>\underline{ab} . \underline{cd}</math>, where all digits are nonzero. The same number in base twelve is <math>\underline{bb} . \underline{ba}</math>. Find the base-ten number <math>\underline{abc}</math>.<br />
<br />
==Solution 1==<br />
First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base twelve that have equal twelves and ones digits in base 8.<br />
<br />
<math>11_{12}=15_8</math><br />
<br />
<math>22_{12}=32_8</math><br />
<br />
<math>33_{12}=47_8</math><br />
<br />
<math>44_{12}=64_8</math><br />
<br />
<math>55_{12}=101_8</math> <br />
<br />
We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number.<br />
Compare the ones places to check if they are equal. We find that they are equal if <math>b=2</math> or <math>b=4</math>.<br />
Evaluating the places to the right side of the decimal point gives us <math>22.23_{12}</math> or <math>44.46_{12}</math>.<br />
When the numbers are converted into base 8, we get <math>32.14_8</math> and <math>64.30_8</math>. Since <math>d\neq0</math>, the first value is correct. Compiling the necessary digits leaves us a final answer of <math>\boxed{321} QED \blacksquare</math><br />
<br />
<br />
==Solution 2==<br />
The parts before and after the decimal points must be equal. Therefore <math>8a + b = 12b + b</math> and <math>c/8 + d/64 = b/12 + a/144</math>. Simplifying the first equation gives <math>a = (3/2)b</math>. Plugging this into the second equation gives <math>3b/32 = c/8 + d/64</math>. Multiplying both sides by 64 gives <math>6b = 8c + d</math>. <math>a</math> and <math>b</math> are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using <math>a = 3/2b</math>, <math>(a,b) = (3,2)</math> or <math>(6,4)</math>. Testing these gives that <math>(6,4)</math> doesn't work, and <math>(3,2)</math> gives <math>a = 3, b = 2, c = 1</math>, and <math>d = 4</math>. Therefore <math>abc = \boxed{321}</math><br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=866002017 AIME I Problems/Problem 42017-07-26T00:33:08Z<p>Stormstar--: </p>
<hr />
<div>==Problem 4==<br />
A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. We find that the altitude to side <math>\overline {AB}</math> is <math>16</math>, so the area of <math>\triangle ABC</math> is <math>(24*16)/2 = 192</math>.<br />
<br />
Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C</math>. Then,<br />
<br />
<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath><br />
<br />
Let <math>\overline {OM} = d</math>.<br />
Equation <math>(1)</math>:<br />
<cmath>d + \sqrt {d^2 + 144} = 16</cmath><br />
<br />
Squaring both sides, we have<br />
<br />
<cmath>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</cmath><br />
<br />
<cmath>2d^2 + 2d\sqrt {d^2+144} = 112</cmath><br />
<br />
<cmath>2d(d + \sqrt {d^2+144}) = 112</cmath><br />
<br />
Substituting with equation <math>(1)</math>:<br />
<br />
<cmath>2d(16) = 112</cmath><br />
<br />
<cmath>d = 7/2</cmath><br />
<br />
We now find that <math>\sqrt{d^2 + 144} = 25/2</math>.<br />
<br />
Let the distance <math>\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by SSS):<br />
<br />
<cmath>25^2 = h^2 + (25/2)^2</cmath><br />
<br />
<cmath>625 = h^2 + 625/4</cmath><br />
<br />
<cmath>1875/4 = h^2</cmath><br />
<br />
<cmath>25\sqrt {3} / 2 = h</cmath><br />
<br />
<br />
Finally, by the formula for volume of a pyramid,<br />
<br />
<cmath>V = Bh/3</cmath><br />
<br />
<cmath>V = (192)(25\sqrt{3}/2)/3</cmath><br />
This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.<br />
<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_3&diff=865992017 AIME I Problems/Problem 32017-07-26T00:32:18Z<p>Stormstar--: </p>
<hr />
<div>==Problem 3==<br />
For a positive integer <math>n</math>, let <math>d_n</math> be the units digit of <math>1 + 2 + \dots + n</math>. Find the remainder when<br />
<cmath>\sum_{n=1}^{2017} d_n</cmath>is divided by <math>1000</math>.<br />
<br />
==Solution==<br />
We see that <math>d(n)</math> appears in cycles of <math>20</math>, adding a total of <math>70</math> each cycle.<br />
Since <math>\left\lfloor\frac{2017}{20}\right\rfloor=100</math>, we know that by <math>2017</math>, there have been <math>100</math> cycles, or <math>7000</math> has been added. This can be discarded, as we're just looking for the last three digits.<br />
Adding up the first <math>17</math> of the cycle of <math>20</math>, we get that the answer is <math>\boxed{69}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&diff=865982017 AIME I Problems/Problem 22017-07-26T00:31:52Z<p>Stormstar--: </p>
<hr />
<div>==Problem 2==<br />
When each of <math>702</math>, <math>787</math>, and <math>855</math> is divided by the positive integer <math>m</math>, the remainder is always the positive integer <math>r</math>. When each of <math>412</math>, <math>722</math>, and <math>815</math> is divided by the positive integer <math>n</math>, the remainder is always the positive integer <math>s \neq r</math>. Find <math>m+n+r+s</math>.<br />
<br />
==Solution==<br />
Let's tackle the first part of the problem first. We can safely assume: <cmath>702 = xm + r</cmath> <cmath>787 = ym + r</cmath> <cmath>855 = zm + r</cmath><br />
Now, if we subtract two values: <cmath>787-702 = 85 = 17\cdot5</cmath><br />
which also equals <cmath>(ym+r)-(xm+r) = m\cdot(y-x)</cmath><br />
Similarly, <cmath>855-787 = 68 = 17\cdot4; (zm+r)-(ym+r) = m\cdot(z-y)</cmath><br />
Since <math>17</math> is the only common factor, we can assume that <math>m=17</math>, and through simple division, that <math>r=5</math>.<br />
<br />
Using the same method on the second half: <cmath>412 = an + s</cmath> <cmath>722 = bn + s</cmath> <cmath>815 = cn + s</cmath><br />
Then. <cmath>722-412 = 310 = 31\cdot10; (bn+s)-(an+s) = n\cdot(b-a)</cmath> <cmath>815-722 = 93 = 31\cdot3; (cn+s)-(bn+s) = n\cdot(c-b)</cmath><br />
The common factor is <math>31</math>, so <math>n=31</math> and through division, <math>s=9</math>.<br />
<br />
The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{62}</math><br />
<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&diff=865592017 AIME I Problems/Problem 22017-07-24T20:38:43Z<p>Stormstar--: </p>
<hr />
<div>==Problem 2==<br />
When each of <math>702</math>, <math>787</math>, and <math>855</math> is divided by the positive integer <math>m</math>, the remainder is always the positive integer <math>r</math>. When each of <math>412</math>, <math>722</math>, and <math>815</math> is divided by the positive integer <math>n</math>, the remainder is always the positive integer <math>s \neq r</math>. Find <math>m+n+r+s</math>.<br />
<br />
==Solution==<br />
Let's tackle the first part of the problem first. We can safely assume: <cmath>702 = xm + r</cmath> <cmath>787 = ym + r</cmath> <cmath>855 = zm + r</cmath><br />
Now, if we subtract two values: <cmath>787-702 = 85 = 17\cdot5</cmath><br />
which also equals <cmath>(ym+r)-(xm+r) = m\cdot(y-x)</cmath><br />
Similarly, <cmath>855-787 = 68 = 17\cdot4; (zm+r)-(ym+r) = m\cdot(z-y)</cmath><br />
Since <math>17</math> is the only common factor, we can assume that <math>m=17</math>, and through simple division, that <math>r=5</math>.<br />
<br />
Using the same method on the second half: <cmath>412 = an + s</cmath> <cmath>722 = bn + s</cmath> <cmath>815 = cn + s</cmath><br />
Then. <cmath>722-412 = 310 = 31\cdot10; (bn+s)-(an+s) = n\cdot(b-a)</cmath> <cmath>815-722 = 93 = 31\cdot3; (cn+s)-(bn+s) = n\cdot(c-b)</cmath><br />
The common factor is <math>31</math>, so <math>n=31</math> and through division, <math>s=9</math>.<br />
<br />
The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{62}</math><br />
<br />
~IYN~<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_2&diff=864952002 AMC 12B Problems/Problem 22017-07-21T22:07:50Z<p>Stormstar--: </p>
<hr />
<div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #2]] and [[2002 AMC 10B Problems|2002 AMC 10B #4]]}}<br />
== Problem ==<br />
<br />
What is the value of <math>(3x - 2)(4x + 1) - (3x - 2)4x + 1</math> when <math>x=4</math>?<br />
<br />
<math>\mathrm{(A)}\ 0<br />
\qquad\mathrm{(B)}\ 1<br />
\qquad\mathrm{(C)}\ 10<br />
\qquad\mathrm{(D)}\ 11<br />
\qquad\mathrm{(E)}\ 12</math><br />
== Solution ==<br />
By the distributive property, <br />
<br />
<cmath>(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}</cmath><br />
<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|ab=B|num-b=3|num-a=5}}<br />
{{AMC12 box|year=2002|ab=B|num-b=1|num-a=3}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_11&diff=864702010 AMC 8 Problems/Problem 112017-07-20T00:07:52Z<p>Stormstar--: </p>
<hr />
<div>== Problem ==<br />
The top of one tree is <math>16</math> feet higher than the top of another tree. The heights of the two trees are in the ratio <math>3:4</math>. In feet, how tall is the taller tree? <br />
<br />
<math> \textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112 </math><br />
== Solution ==<br />
Let the height of the taller tree be <math>h</math> and let the height of the smaller tree be <math>h-16</math>. Since the ratio of the smaller tree to the larger tree is <math>\frac{3}{4}</math>, we have <math>\frac{h-16}{h}=\frac{3}{4}</math>. Solving for <math>h</math> gives us <math>h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}</math><br />
<br />
==Solution 2 ==<br />
To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers are consecutive, if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get <math>h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}</math><br />
<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_8&diff=864692010 AMC 8 Problems/Problem 82017-07-20T00:06:00Z<p>Stormstar--: </p>
<hr />
<div>==Problem==<br />
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction <math>1/2</math> mile in front of her. After she passes him, she can see him in her rear mirror until he is <math>1/2</math> mile behind her. Emily rides at a constant rate of <math>12</math> miles per hour, and Emerson skates at a constant rate of <math>8</math> miles per hour. For how many minutes can Emily see Emerson? <br />
<br />
<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math><br />
<br />
==Solution==<br />
Because they are skating in the same direction, Emily is skating relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is not moving at all[on his skateboard] and Emily is riding at <math>4</math> mph. It takes her<br />
<br />
<cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath><br />
<br />
to skate the <math>1/2</math> mile to reach him, and then the same amount of time to be <math>1/2</math> mile ahead of him. This totals to<br />
<br />
<cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath><br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_6&diff=862052010 AMC 10B Problems/Problem 62017-06-28T22:01:33Z<p>Stormstar--: removed vandalism</p>
<hr />
<div>==Problem==<br />
<br />
A circle is centered at <math>O</math>, <math>\overline{AB}</math> is a diameter and <math>C</math> is a point on the circle with <math>\angle COB = 50^\circ</math>. What is the degree measure of <math>\angle CAB</math>?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math><br />
<br />
==Solution==<br />
Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since <math>O</math> is the center, <math>OC</math> and <math>OA</math> are radii and they are congruent. Thus, <math>\triangle COA</math> is an isosceles triangle. Also, note that <math>\angle COB</math> and <math>\angle COA</math> are supplementary, then <math>\angle COA = 180 - 50 = 130^{\circ}</math>. Since <math>\triangle COA</math> is isosceles, then <math>\angle OCA \cong \angle OAC</math>. They also sum to <math>50^{\circ}</math>, so each angle is <math>\boxed{\textbf{(B)}\ 25}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_1&diff=860152016 USAMO Problems/Problem 12017-06-11T22:11:45Z<p>Stormstar--: Removed the second box</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>X_1, X_2, \ldots, X_{100}</math> be a sequence of mutually distinct nonempty subsets of a set <math>S</math>. Any two sets <math>X_i</math> and <math>X_{i+1}</math> are disjoint and their union is not the whole set <math>S</math>, that is, <math>X_i\cap X_{i+1}=\emptyset</math> and <math>X_i\cup X_{i+1}\neq S</math>, for all <math>i\in\{1, \ldots, 99\}</math>. Find the smallest possible number of elements in <math>S</math>.<br />
<br />
== Solution 1==<br />
<br />
The answer is that <math>|S| \ge 8</math>.<br />
<br />
First, we provide a inductive construction for <math>S = \left\{ 1, \dots, 8 \right\}</math>. Actually, for <math>n \ge 4</math> we will provide a construction for <math>S = \left\{ 1, \dots, n \right\}</math> which has <math>2^{n-1} + 1</math> elements in a line. (This is sufficient, since we then get <math>129</math> for <math>n = 8</math>.) The idea is to start with the following construction for <math>|S| = 4</math>: <cmath> \begin{array}{ccccccccc} 34 & 1 & 23 & 4 & 12 & 3 & 14 & 2 & 13 \end{array}. </cmath>Then inductively, we do the following procedure to move from <math>n</math> to <math>n+1</math>: take the chain for <math>n</math> elements, delete an element, and make two copies of the chain (which now has even length). Glue the two copies together, joined by <math>\varnothing</math> in between. Then place the element <math>n+1</math> in alternating positions starting with the first (in particular, this hits <math>n+1</math>). For example, the first iteration of this construction gives: <cmath> \begin{array}{ccccccccc} 345 & 1 & 235 & 4 & 125 & 3 & 145 & 2 & 5 \\ 34 & 15 & 23 & 45 & 12 & 35 & 14 & 25 & \end{array} </cmath><br />
Now let's check <math>|S| \ge 8</math> is sufficient. Consider a chain on a set of size <math>|S| = 7</math>. (We need <math>|S| \ge 7</math> else <math>2^{|S|} < 100</math>.) Observe that there are sets of size <math>\ge 4</math> can only be neighbored by sets of size <math>\le 2</math>, of which there are <math>\binom 71 + \binom 72 = 28</math>. So there are <math>\le 30</math> sets of size <math>\ge 4</math>. Also, there are <math>\binom 73 = 35</math> sets of size <math>3</math>. So the total number of sets in a chain can be at most <math>30 + 28 + 35 = 93 < 100</math>.<br />
<br />
== Solution 2 ==<br />
My proof that <math>|S|\ge 8</math> is basically the same as the one above. Here is another construction for <math>|S| = 8</math> that I like because it works with remainders and it's pretty intuitive. The basic idea is to assign different subsets to different remainders when divided by particular numbers, and then to use the Chinese Remainder Theorem to show that all of the subsets are distinct. The motivation for this comes from the fact that we want <math>X_i</math> and <math>X_{i+1}</math> to always be disjoint, so remainders are a great way to systematically make that happen, since <math>i</math> and <math>i+1</math> do not have the same remainder modulo any positive integer greater than <math>1.</math> Anyway, here is the construction:<br />
<br />
Let <math>S = \left\{ {1, 2, ..., 8} \right\}.</math> For <math>i = 1, 2, ..., 98,</math> we will choose which elements of the set <math>\left\{ {1, 2, 3, 4} \right\}</math> belong to <math>X_i</math> based on the remainder of <math>i</math> modulo <math>9,</math> and we will choose which elements of the set <math>\left\{ {5, 6, 7, 8} \right\}</math> belong to <math>X_i</math> based on the remainder of <math>i</math> modulo <math>11.</math> We do this as follows:<br />
<cmath>X_i\cap\left\{ {1, 2, 3, 4} \right\} =<br />
\begin{array}{ll}<br />
\emptyset & i\equiv 0 \text{ (mod } 9\text{)} \\<br />
\left\{ {1, 2} \right\} & i\equiv 1 \text{ (mod } 9\text{)} \\<br />
\left\{ {3} \right\} & i\equiv 2 \text{ (mod } 9\text{)} \\<br />
\left\{ {1, 4} \right\} & i\equiv 3 \text{ (mod } 9\text{)} \\<br />
\left\{ {2} \right\} & i\equiv 4 \text{ (mod } 9\text{)} \\<br />
\left\{ {3, 4} \right\} & i\equiv 5 \text{ (mod } 9\text{)} \\<br />
\left\{ {1} \right\} & i\equiv 6 \text{ (mod } 9\text{)} \\<br />
\left\{ {2, 3} \right\} & i\equiv 7 \text{ (mod } 9\text{)} \\<br />
\left\{ {4} \right\} & i\equiv 8 \text{ (mod } 9\text{)} \\<br />
\end{array}</cmath><br />
<cmath>X_i\cap\left\{ {5, 6, 7, 8} \right\} =<br />
\begin{array}{ll}<br />
\emptyset & i\equiv 0 \text{ (mod } 11\text{)} \\<br />
\left\{ {5, 6} \right\} & i\equiv 1 \text{ (mod } 11\text{)} \\<br />
\left\{ {7, 8} \right\} & i\equiv 2 \text{ (mod } 11\text{)} \\<br />
\left\{ {5} \right\} & i\equiv 3 \text{ (mod } 11\text{)} \\<br />
\left\{ {6, 8} \right\} & i\equiv 4 \text{ (mod } 11\text{)} \\<br />
\left\{ {5, 7} \right\} & i\equiv 5 \text{ (mod } 11\text{)} \\<br />
\left\{ {6} \right\} & i\equiv 6 \text{ (mod } 11\text{)} \\<br />
\left\{ {5, 8} \right\} & i\equiv 7 \text{ (mod } 11\text{)} \\<br />
\left\{ {6, 7} \right\} & i\equiv 8 \text{ (mod } 11\text{)} \\<br />
\left\{ {8} \right\} & i\equiv 9 \text{ (mod } 11\text{)} \\<br />
\left\{ {7} \right\} & i\equiv 10 \text{ (mod } 11\text{)} \\<br />
\end{array}</cmath><br />
Finally, we specially define <math>X_{99} = \left\{ {1, 2, 3} \right\}</math> and <math>X_{100} = \left\{ {5, 6, 7} \right\}.</math><br />
<br />
It is relatively easy to see that this configuration satisfies all of the desired conditions. We see that <math>X_{98} = \left\{ {4, 7} \right\},</math> so <math>X_{98}</math> and <math>X_{99}</math> are disjoint, as are <math>X_{99}</math> and <math>X_{100}.</math> The remainder configuration above takes care of the rest, so any two consecutive sets are disjoint. Then, by the Chinese Remainder Theorem, no two integers from <math>1</math> to <math>98</math> have the same combination of residues modulo <math>9</math> and modulo <math>11,</math> so all of the sets <math>X_i</math> are distinct for <math>i = 1, 2, ..., 98.</math> It is also easy to verify that none of these match <math>X_{99}</math> or <math>X_{100},</math> since they all have at most two elements of <math>\left\{ {1, 2, 3, 4} \right\}</math> and at most two elements of <math>\left\{ {5, 6, 7, 8} \right\},</math> whereas <math>X_{99}</math> and <math>X_{100}</math> do not satisfy this; hence all of the sets are distinct. Finally, notice that, for any pair of consecutive sets, at least one of them has at most <math>3</math> elements, while the other has at most <math>4.</math> Thus, their union always has at most <math>7</math> elements, so <math>X_i\cup X_{i+1}\neq S</math> for all <math>i = 1, 2, ..., 99.</math><br />
<br />
All of the conditions are satisfied, so this configuration works. We thus conclude that <math>\text{min}\left(\left|S\right|\right) = 8.</math><br />
<br />
{{MAA Notice}}<br />
<br />
==See also==<br />
{{USAMO newbox|year=2016|beforetext=|before=First Problem|num-a=2}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_7&diff=859792014 AMC 10B Problems/Problem 72017-06-08T19:12:42Z<p>Stormstar--: </p>
<hr />
<div><br />
==Problem==<br />
<br />
Suppose <math>A>B>0</math> and A is <math>x</math>% greater than <math>B</math>. What is <math>x</math>?<br />
<br />
<math> \textbf {(A) } 100(\frac{A-B}{B}) \qquad \textbf {(B) } 100(\frac{A+B}{B}) \qquad \textbf {(C) } 100(\frac{A+B}{A})\qquad \textbf {(D) } 100(\frac{A-B}{A}) \qquad \textbf {(E) } 100(\frac{A}{B})</math><br />
<br />
==Solution==<br />
We have that A is <math>x\%</math> greater than B, so <math>A=\frac{100+x}{100}(B)</math>. We solve for <math>x</math>. We get <br />
<br />
<math>\frac{A}{B}=\frac{100+x}{100}</math><br />
<br />
<math>100\frac{A}{B}=100+x</math><br />
<br />
<math>100(\frac{A}{B}-1)=x</math><br />
<br />
<math>100(\frac{A-B}{B})=x</math>. <math>\boxed{(\textbf{A})}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_7&diff=859772014 AMC 10B Problems/Problem 72017-06-08T19:09:10Z<p>Stormstar--: </p>
<hr />
<div><br />
==Problem==<br />
<br />
Suppose <math>A>B>0</math> and A is <math>x</math>% greater than <math>B</math>. What is <math>x</math>?<br />
<br />
<math> \textbf {(A) } 100(\frac{A-B}{B}) \qquad \textbf {(B) } 100(\frac{A+B}{B}) \qquad \textbf {(C) } 100(\frac{A+B}{A})\qquad \textbf {(D) } 100(\frac{A-B}{A}) \qquad \textbf {(E) } 100(\frac{A}{B})</math><br />
<br />
==Solution==<br />
We have that A is <math>x\%</math> greater than B, so <math>A=\frac{100+x}{100}(B)</math>. We solve for <math>x</math>. We get <br />
<br />
<math>\frac{A}{B}=\frac{100+x}{100}</math><br />
<br />
<math>100\frac{A}{B}=100+x</math><br />
<br />
<math>100(\frac{A}{B}-1)=x</math><br />
<br />
<math>(\textbf{A}) \boxed{100(\frac{A-B}{B}) }=x</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Stormstar--https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&diff=859562017 AIME I Problems/Problem 22017-06-06T00:42:50Z<p>Stormstar--: </p>
<hr />
<div>==Problem 2==<br />
When each of <math>702</math>, <math>787</math>, and <math>855</math> is divided by the positive integer <math>m</math>, the remainder is always the positive integer <math>r</math>. When each of <math>412</math>, <math>722</math>, and <math>815</math> is divided by the positive integer <math>n</math>, the remainder is always the positive integer <math>s \neq r</math>. Find <math>m+n+r+s</math>.<br />
<br />
==Solution==<br />
Let's tackle the first part of the problem first. We can safely assume: <cmath>702 = xm + r</cmath> <cmath>787 = ym + r</cmath> <cmath>855 = zm + r</cmath><br />
Now, if we subtract two values: <cmath>787-702 = 85 = 17\cdot5</cmath><br />
which also equals <cmath>(ym+r)-(xm+r) = m\cdot(y-x)</cmath><br />
Similarly, <cmath>855-787 = 68 = 17\cdot4; (zm+r)-(ym+r) = m\cdot(z-y)</cmath><br />
Since <math>17</math> is the only common factor, we can assume that <math>m=17</math>, and through simple division, that <math>r=5</math>.<br />
<br />
Using the same method on the second half: <cmath>412 = an + s</cmath> <cmath>722 = bn + s</cmath> <cmath>815 = cn + s</cmath><br />
Then. <cmath>722-412 = 310 = 31\cdot10; (bn+s)-(an+s) = n\cdot(b-a)</cmath> <cmath>815-722 = 93 = 31\cdot3; (cn+s)-(bn+s) = n\cdot(c-b)</cmath><br />
The common factor is <math>31</math>, so <math>n=31</math> and through division, <math>s=9</math>.<br />
<br />
The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{62}</math><br />
<br />
~IYN~<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Stormstar--