https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Alexog123&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T06:54:31ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=858732012 AIME I Problems/Problem 82017-05-28T00:29:52Z<p>Alexog123: /* Solution: think outside the box */</p>
<hr />
<div>==Problem 8==<br />
Cube <math>ABCDEFGH,</math> labeled as shown below, has edge length <math>1</math> and is cut by a plane passing through vertex <math>D</math> and the midpoints <math>M</math> and <math>N</math> of <math>\overline{AB}</math> and <math>\overline{CG}</math> respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form <math>\tfrac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math><br />
<br />
<center><asy>import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair A = (0,0), B = (3.8,0), C = (5.876,1.564), D = (2.076,1.564), E = (0,3.8), F = (3.8,3.8), G = (5.876,5.364), H = (2.076,5.364), M = (1.9,0), N = (5.876,3.465);<br />
pair[] dotted = {A,B,C,D,E,F,G,H,M,N};<br />
<br />
D(A--B--C--G--H--E--A);<br />
D(E--F--B);<br />
D(F--G);<br />
pathpen=dashed;<br />
D(A--D--H);<br />
D(D--C);<br />
<br />
dot(dotted);<br />
label("$A$",A,SW);<br />
label("$B$",B,S);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,W);<br />
label("$F$",F,SE);<br />
label("$G$",G,NE);<br />
label("$H$",H,NW);<br />
label("$M$",M,S);<br />
label("$N$",N,NE);<br />
<br />
</asy></center><br />
<br />
== Solution: think outside the box (pun intended) ==<br />
Define a coordinate system with <math>D</math> at the origin and <math>C,</math> <math>A,</math> and <math>H</math> on the <math>x</math>, <math>y</math>, and <math>z</math> axes respectively. The <math>D=(0,0,0),</math> <math>M=(.5,1,0),</math> and <math>N=(1,0,.5).</math> It follows that the plane going through <math>D,</math> <math>M,</math> and <math>N</math> has equation <math>2x-y-4z=0.</math> Let <math>Q = (1,1,.25)</math> be the intersection of this plane and edge <math>BF</math> and let <math>P = (1,2,0).</math> Now since <math>2(1) - 1(2) - 4(0) = 0,</math> <math>P</math> is on the plane. Also, <math>P</math> lies on the extensions of segments <math>DM,</math> <math>NQ,</math> and <math>CB</math> so the solid <math>DPCN = DMBCQN + MPBQ</math> is a right triangular pyramid. Note also that pyramid <math>MPBQ</math> is similar to <math>DPCN</math> with scale factor <math>.5</math> and thus the volume of solid <math>DMBCQN,</math> which is one of the solids bounded by the cube and the plane, is <math>[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].</math> But the volume of <math>DPCN</math> is simply the volume of a pyramid with base <math>1</math> and height <math>.5</math> which is <math>\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.</math> So <math>[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.</math> Note, however, that this volume is less than <math>.5</math> and thus this solid is the smaller of the two solids. The desired volume is then <math>[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}</math><br />
<br />
== Solution 2 ==<br />
Define a coordinate system with <math>D = (0,0,0)</math>, <math>M = (1, \frac{1}{2}, 0)</math>, <math>N = (0,1,\frac{1}{2})</math>. The plane formed by <math>D</math>, <math>M</math>, and <math>N</math> is <math>z = \frac{y}{2} - \frac{x}{4}</math>. It intersects the base of the unit cube at <math>y = \frac{x}{2}</math>. The z-coordinate of the plane never exceeds the height of the unit cube for <math>0 \leq x \leq 1, 0 \leq y \leq 1</math>. Therefore, the volume of one of the two regions formed by the plane is<br />
<cmath>\int_0^1 \int_{\frac{x}{2}}^1 \int_0^{\frac{y}{2}-\frac{x}{4}}dz\,dy\,dx = \frac{7}{48}</cmath><br />
Since <math>\frac{7}{48} < \frac{1}{2}</math>, our answer is <math>1-\frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}</math>.<br />
<br />
== Alternative Solution (using calculus) : think inside the box ==<br />
Let <math>Q</math> be the intersection of the plane with edge <math>FB,</math> then <math>\triangle MQB</math> is similar to <math>\triangle DNC</math> and the volume <math>[DNCMQB]</math> is a sum of areas of cross sections of similar triangles running parallel to face <math>ABFE.</math> Let <math>x</math> be the distance from face <math>ABFE,</math> let <math>h</math> be the height parallel to <math>AB</math> of the cross-sectional triangle at that distance, and <math>a</math> be the area of the cross-sectional triangle at that distance. <math>A(x)=\frac{h^2}{4},</math> and <math>h=\frac{x+1}{2},</math> then <math>A=\frac{(x+1)^2}{16}</math>, and the volume <math>[DNCMQB]</math> is <math>\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.</math> Thus the volume of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math><br />
<br />
== Alternative Solution : think inside the box with formula==<br />
<br />
If you memorized the formula for a frustum, then this problem is very trivial.<br />
<br />
The formula for a frustum is:<br />
<br />
<math>\frac{h_2b_2 -h_1b_1}3</math> where <math>b_i</math> is the area of the base and <math>h_i</math> is the height from the chopped of apex to the base.<br />
<br />
We can easily see that from symmetry, the area of the smaller front base is <math>\frac{1}{16}</math> and the area of the larger back base is <math>\frac{1}4</math><br />
<br />
Now to find the height of the apex.<br />
<br />
Extend the <math>DM</math> and (call the intersection of the plane with <math>FB</math> G) <math>NG</math> to meet at <math>x</math>. Now from similar triangles <math>XMG</math> and <math>XDN</math> we can easily find the total height of the triangle <math>XDN</math> to be <math>2</math><br />
<br />
Now straight from our formula, the volume is <math>\frac{7}{48}</math> Thus the answer is:<br />
<br />
<math>1-\text{Volume} \Longrightarrow \boxed{089}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}<br />
<br />
{{MAA Notice}}<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:3D Geometry Problems]]</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_8&diff=841452013 AIME II Problems/Problem 82017-02-21T23:55:01Z<p>Alexog123: /* Solution */</p>
<hr />
<div>==Problem 8==<br />
A hexagon that is inscribed in a circle has side lengths <math>22</math>, <math>22</math>, <math>20</math>, <math>22</math>, <math>22</math>, and <math>20</math> in that order. The radius of the circle can be written as <math>p+\sqrt{q}</math>, where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.<br />
<br />
==Solution==<br />
<asy><br />
import olympiad;<br />
import math;<br />
<br />
pair A,B,C,D,E,F;<br />
B=origin; C=(10,0); D=(12,-5); E=(10,-10); F=(0,-10); A=(-2, -5);<br />
draw(A--B);draw(B--C);draw(C--D);draw(D--E);draw(E--F);draw(F--A);<br />
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);<br />
draw(circumcircle(A, D, F));<br />
label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,W);label("$E$",E,SE);label("$F$",F,SW);<br />
<br />
</asy><br />
<br />
<br />
===Solution 1===<br />
Let us call the hexagon <math>ABCDEF</math>, where <math>AB=CD=DE=AF=22</math>, and <math>BC=EF=20</math>. <br />
We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler. <br />
Draw a line from the center of the circle, <math>O</math>, to the midpoint of <math>BC</math>, <math>E</math>. Now, draw a line from <math>O</math> to the midpoint of <math>AB</math>, <math>F</math>. Clearly, <math>\angle BEO=90^{\circ}</math>, because <math>BO=CO</math>, and <math>\angle BFO=90^{\circ}</math>, for similar reasons. Also notice that <math>\angle AOE=90^{\circ}</math>. <br />
Let us call <math>\angle BOF=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle BOE=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BF}{r}=\frac{11}{r}</cmath> <cmath>\sin{(90-2\theta)}=\frac{BE}{r}=\frac{10}{r}</cmath><br />
Now we can use simple trigonometry to solve for <math>r</math>.<br />
Recall that <math>\sin{(90-\alpha)}=\cos(\alpha)</math>: That means <math>\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}</math>.<br />
Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>.<br />
Let <math>\sin{\theta}=x</math>.<br />
Substitute to get <math>x=\frac{11}{r}</math> and <math>1-2x^2=\frac{10}{r}</math><br />
Now substitute the first equation into the second equation: <math>1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}</math><br />
Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath><br />
Using the quadratic equation to solve, we get that <math>r=5+\sqrt{267}</math> (because <math>5-\sqrt{267}</math> gives a negative value), so the answer is <math>5+267=\boxed{272}</math>.<br />
<br />
===Solution 2===<br />
Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242=0</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math>.<br />
<br />
===Solution 3===<br />
<br />
Join the diameter of the circle <math>AD</math> and let the length be <math>d</math>. By [[Ptolemy's Theorem]] on trapezoid <math>ADEF</math>, <math>(AD)(EF) + (AF)(DE) = (AE)(DF)</math>. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to <math>x</math> each. Then<br />
<br />
<cmath>20d + 22^2 = x^2</cmath><br />
<br />
Since <math>\angle AED</math> is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right <math>\triangle AED</math>:<br />
<br />
<cmath>(AE)^2 + (ED)^2 = (AD)^2</cmath><br />
<cmath>x^2 + 22^2 = d^2</cmath><br />
<br />
From the above equations, we have:<br />
<cmath>x^2 = d^2 - 22^2 = 20d + 22^2</cmath><br />
<cmath>d^2 - 20d = 2\times22^2</cmath><br />
<cmath>d^2 - 20d + 100 = 968+100 = 1068</cmath><br />
<cmath>(d-10) = \sqrt{1068}</cmath><br />
<cmath>d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)</cmath><br />
<br />
Since the radius is half the diameter, it is <math>\sqrt{267}+5</math>, so the answer is <math>5+267 \Rightarrow \boxed{272}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2013|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_8&diff=841442013 AIME II Problems/Problem 82017-02-21T23:53:11Z<p>Alexog123: /* Solution 1 */</p>
<hr />
<div>==Problem 8==<br />
A hexagon that is inscribed in a circle has side lengths <math>22</math>, <math>22</math>, <math>20</math>, <math>22</math>, <math>22</math>, and <math>20</math> in that order. The radius of the circle can be written as <math>p+\sqrt{q}</math>, where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.<br />
<br />
==Solution==<br />
<asy><br />
import olympiad;<br />
import math;<br />
<br />
pair A,B,C,D,E,F;<br />
B=origin; C=(10,0); D=(12,-5); E=(10,-10); F=(0,-10); A=(-2, -5);<br />
draw(A--B);draw(B--C);draw(C--D);draw(D--E);draw(E--F);draw(F--A);<br />
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);<br />
draw(circumcircle(A, D, F));<br />
label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,W);label("$E$",E,SE);label("$F$",F,SW);<br />
<br />
</asy><br />
<br />
<br />
===Solution 1===<br />
Let us call the hexagon <math>ABCDEF</math>, where <math>AB=CD=DE=AF=22</math>, and <math>BC=EF=20</math>. <br />
We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler. <br />
Draw a line from the center of the circle, <math>O</math>, to the midpoint of <math>BC</math>, <math>G</math>. Now, draw a line from <math>O</math> to the midpoint of <math>AB</math>, <math>H</math>. Clearly, <math>\angle BEO=90^{\circ}</math>, because <math>BO=CO</math>, and <math>\angle BFO=90^{\circ}</math>, for similar reasons. Also notice that <math>\angle AOE=90^{\circ}</math>. <br />
Let us call <math>\angle BOF=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle BOE=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BF}{r}=\frac{11}{r}</cmath> <cmath>\sin{(90-2\theta)}=\frac{BE}{r}=\frac{10}{r}</cmath><br />
Now we can use simple trigonometry to solve for <math>r</math>.<br />
Recall that <math>\sin{(90-\alpha)}=\cos(\alpha)</math>: That means <math>\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}</math>.<br />
Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>.<br />
Let <math>\sin{\theta}=x</math>.<br />
Substitute to get <math>x=\frac{11}{r}</math> and <math>1-2x^2=\frac{10}{r}</math><br />
Now substitute the first equation into the second equation: <math>1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}</math><br />
Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath><br />
Using the quadratic equation to solve, we get that <math>r=5+\sqrt{267}</math> (because <math>5-\sqrt{267}</math> gives a negative value), so the answer is <math>5+267=\boxed{272}</math>.<br />
<br />
===Solution 2===<br />
Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242=0</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math>.<br />
<br />
===Solution 3===<br />
<br />
Join the diameter of the circle <math>AD</math> and let the length be <math>d</math>. By [[Ptolemy's Theorem]] on trapezoid <math>ADEF</math>, <math>(AD)(EF) + (AF)(DE) = (AE)(DF)</math>. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to <math>x</math> each. Then<br />
<br />
<cmath>20d + 22^2 = x^2</cmath><br />
<br />
Since <math>\angle AED</math> is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right <math>\triangle AED</math>:<br />
<br />
<cmath>(AE)^2 + (ED)^2 = (AD)^2</cmath><br />
<cmath>x^2 + 22^2 = d^2</cmath><br />
<br />
From the above equations, we have:<br />
<cmath>x^2 = d^2 - 22^2 = 20d + 22^2</cmath><br />
<cmath>d^2 - 20d = 2\times22^2</cmath><br />
<cmath>d^2 - 20d + 100 = 968+100 = 1068</cmath><br />
<cmath>(d-10) = \sqrt{1068}</cmath><br />
<cmath>d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)</cmath><br />
<br />
Since the radius is half the diameter, it is <math>\sqrt{267}+5</math>, so the answer is <math>5+267 \Rightarrow \boxed{272}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2013|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_7&diff=841432013 AIME II Problems/Problem 72017-02-21T23:37:00Z<p>Alexog123: /* Solution */</p>
<hr />
<div>==Problem 7==<br />
A group of clerks is assigned the task of sorting <math>1775</math> files. Each clerk sorts at a constant rate of <math>30</math> files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in <math>3</math> hours and <math>10</math> minutes. Find the number of files sorted during the first one and a half hours of sorting.<br />
<br />
==Solution==<br />
There are <math>x</math> clerks at the beginning, and <math>t</math> clerks are reassigned to another task at the end of each hour. So, <math>30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775</math>, and simplify that we get <math>19x-21t=355</math>.<br />
Now the problem is to find a reasonable integer solution. Now we know <math>x= \frac{355+21t}{19}</math>, so <math>19</math> divides <math>355+21t</math>, AND as long as <math>t</math> is a integer, <math>19</math> must divide <math>2t+355</math>. Now, we suppose that <math>19m=2t+355</math>, similarly we get <math>t=\frac{19m-355}{2}</math>, and so in order to get a minimum integer solution for <math>t</math>, it is obvious that <math>m=19</math> works. So we get <math>t=3</math> and <math>x=22</math>. One and a half hour's work should be <math>30x+15(x-t)</math>, so the answer is <math>\boxed{945}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2013|n=II|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_22&diff=826152016 AMC 10B Problems/Problem 222017-01-28T23:43:34Z<p>Alexog123: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won <math>10</math> games and lost <math>10</math> games; there were no ties. How many sets of three teams <math>\{A, B, C\}</math> were there in which <math>A</math> beat <math>B</math>, <math>B</math> beat <math>C</math>, and <math>C</math> beat <math>A?</math><br />
<br />
<math>\textbf{(A)}\ 385 \qquad<br />
\textbf{(B)}\ 665 \qquad<br />
\textbf{(C)}\ 945 \qquad<br />
\textbf{(D)}\ 1140 \qquad<br />
\textbf{(E)}\ 1330</math><br />
<br />
==Solution==<br />
There are <math>21</math> teams. Any of the <math>\tbinom{21}3=1330</math> sets of three teams must either be a fork (in which one team beat both the others) or a cycle:<br />
<br />
<asy>size(7cm);label("X",(5,5));label("Z",(10,0));label("Y",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow);<br />
label("X",(20,5));label("Z",(25,0));label("Y",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow);<br />
</asy><br />
But we know that every team beat exactly <math>10</math> other teams, so for each possible <math>X</math> at the head of a fork, there are always exactly <math>\tbinom{10}2</math> choices for <math>Y</math> and <math>Z</math>. Therefore there are <math>21\cdot\tbinom{10}2=945</math> forks, and all the rest must be cycles.<br />
<br />
Thus the answer is <math>1330-945=385</math> which is <math>\textbf{(A)}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_18&diff=825482008 AMC 10B Problems/Problem 182017-01-24T00:14:20Z<p>Alexog123: </p>
<hr />
<div>==Problem==<br />
Bricklayer Brenda would take nine hours to build a chimney alone, and bricklayer Brandon would take <math>10</math> hours to build it alone. When they work together, they talk a lot, and their combined output decreases by <math>10</math> bricks per hour. Working together, they build the chimney in <math>5</math> hours. How many bricks are in the chimney?<br />
<br />
<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math><br />
<br />
==Solution 1==<br />
Let <math>x</math> be the number of bricks in the chimney. Using <math>distance = rate \cdot time</math>, we get<br />
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math><br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_18&diff=825442008 AMC 10B Problems/Problem 182017-01-24T00:12:43Z<p>Alexog123: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Bricklayer Brenda would take nine hours to build a chimney alone, and bricklayer Brandon would take <math>10</math> hours to build it alone. When they work together, they talk a lot, and their combined output decreases by <math>10</math> bricks per hour. Working together, they build the chimney in <math>5</math> hours. How many bricks are in the chimney?<br />
<br />
<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math><br />
<br />
==Solution==<br />
Let <math>x</math> be the number of bricks in the chimney. Using <math>distance = rate \cdot time</math>, we get<br />
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_18&diff=825422008 AMC 10B Problems/Problem 182017-01-24T00:12:17Z<p>Alexog123: /* Solution */</p>
<hr />
<div>==Problem==<br />
Bricklayer Brenda would take nine hours to build a chimney alone, and bricklayer Brandon would take <math>10</math> hours to build it alone. When they work together, they talk a lot, and their combined output decreases by <math>10</math> bricks per hour. Working together, they build the chimney in <math>5</math> hours. How many bricks are in the chimney?<br />
<br />
<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math><br />
<br />
==Solution 1==<br />
Let <math>x</math> be the number of bricks in the chimney. Using <math>distance = rate \cdot time</math>, we get<br />
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=Principle_of_Inclusion-Exclusion&diff=80578Principle of Inclusion-Exclusion2016-10-09T22:37:34Z<p>Alexog123: /* Examples */</p>
<hr />
<div>The '''Principle of Inclusion-Exclusion''' (abbreviated PIE) provides an organized method/formula to find the number of [[element]]s in the [[union]] of a given group of [[set]]s, the size of each set, and the size of all possible [[intersection]]s among the sets.<br />
<br />
== Remarks ==<br />
Sometimes it is also useful to know that, if you take into account only the first <math>m\le n</math> sums on the right, then you will get an overestimate if <math>m</math> is [[odd integer | odd]] and an underestimate if <math>m</math> is [[even integer | even]].<br />
So, <br />
<br />
and so on.<br />
<br />
== Examples ==<br />
We don't have any. ;D<br />
<br />
== See also ==<br />
* [[Combinatorics]]<br />
* [[Overcounting]]<br />
<br />
[[Category:Combinatorics]]</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=Principle_of_Inclusion-Exclusion&diff=80577Principle of Inclusion-Exclusion2016-10-09T22:37:10Z<p>Alexog123: /* Examples */</p>
<hr />
<div>The '''Principle of Inclusion-Exclusion''' (abbreviated PIE) provides an organized method/formula to find the number of [[element]]s in the [[union]] of a given group of [[set]]s, the size of each set, and the size of all possible [[intersection]]s among the sets.<br />
<br />
== Remarks ==<br />
Sometimes it is also useful to know that, if you take into account only the first <math>m\le n</math> sums on the right, then you will get an overestimate if <math>m</math> is [[odd integer | odd]] and an underestimate if <math>m</math> is [[even integer | even]].<br />
So, <br />
<br />
and so on.<br />
<br />
== Examples ==<br />
We don't have any. <math>;D</math><br />
<br />
== See also ==<br />
* [[Combinatorics]]<br />
* [[Overcounting]]<br />
<br />
[[Category:Combinatorics]]</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=Vieta%27s_Formulas&diff=79661Vieta's Formulas2016-07-21T02:13:58Z<p>Alexog123: /* Introduction */</p>
<hr />
<div>'''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.<br />
<br />
== Introduction ==<br />
<br />
Vieta's Formulas were discovered by the French mathematician [[François Viète]].<br />
<br />
Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as<br />
<br />
<center><math>x^2+ax+b=(x-p)(x-q)</math></center><br />
<br />
(Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we get<br />
<br />
<center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center> <br />
<br />
We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.<br />
<br />
A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.<br />
<br />
We can state Vieta's formulas more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,<br />
where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that<br />
<br />
<center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center><br />
<br />
Expanding out the right hand side gives us<br />
<br />
<math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math><br />
<br />
The coefficient of <math> x^k </math> in this expression will be the <math>k</math>-th [[elementary symmetric sum]] of the <math>r_i</math>. <br />
<br />
We now have two different expressions for <math>P(x)</math>. These must be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that<br />
<br />
<center><math>a_n = a_n</math></center><br />
<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center><br />
<center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center><br />
<center><math>\vdots</math></center><br />
<center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center><br />
<br />
More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).<br />
<br />
If we denote <math>\sigma_k</math> as the <math>k</math>-th elementary symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.<br />
<br />
==Problems==<br />
===Introduction===<br />
* Let <math>r_1,r_2,</math> and <math>r_3</math> be the three roots of the cubic <math>x^3 + 3x^2 + 4x - 4</math>. Find the value of <math>r_1r_2+r_1r_3+r_2r_3</math>.<br />
* Suppose the polynomial <math>5x^3 + 4x^2 - 8x + 6</math> has three real roots <math>a,b</math>, and <math>c</math>. Find the value of <math>a(1+b+c)+b(1+a+c)+c(1+a+b)</math>.<br />
* Let <math>m</math> and <math>n</math> be the roots of the quadratic equation <math>4x^2 + 5x + 3 = 0</math>. Find <math>(m + 7)(n + 7)</math>.<br />
===Intermediate===<br />
<br />
* Let <math>a</math>, <math>b</math>, and <math>c</math> be positive real numbers with <math>a<b<c</math> such that <math>a+b+c=12</math>, <math>a^2+b^2+c^2=50</math>, and <math>a^3+b^3+c^3=216</math>. Find <math>a+2b+3c</math>.<br />
* (USAMTS 2010) Find <math>c>0</math> such that if <math>r</math>, <math>s</math>, and <math>t</math> are the roots of the cubic <cmath>f(x)=x^3-4x^2+6x-c,</cmath> then <cmath>1=\dfrac1{r^2+s^2}+\dfrac1{s^2+t^2}+\dfrac1{t^2+r^2}.</cmath><br />
* (HMMT 2007) The complex numbers <math>\alpha_1</math>, <math>\alpha_2</math>, <math>\alpha_3</math>, and <math>\alpha_4</math> are the four distinct roots of the equation <math>x^4+2x^3+2=0</math>. Determine the unordered set <cmath>\{\alpha_1\alpha_2+\alpha_3\alpha_4,\,\alpha_1\alpha_3+\alpha_2\alpha_4,\,\alpha_1\alpha_4+\alpha_2\alpha_3\}.</cmath><br />
<br />
== See Also ==<br />
<br />
* [[Algebra]]<br />
* [[Polynomials]]<br />
* [[Newton's Sums]]<br />
<br />
== External Links ==<br />
*[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article]<br />
<br />
[[Category:Elementary algebra]]</div>Alexog123https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_19&diff=792772009 AMC 10A Problems/Problem 192016-07-12T23:15:27Z<p>Alexog123: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Circle <math>A</math> has radius <math>100</math>. Circle <math>B</math> has an integer radius <math>r<100</math> and remains internally tangent to circle <math>A</math> as it rolls once around the circumference of circle <math>A</math>. The two circles have the same points of tangency at the beginning and end of cirle <math>B</math>'s trip. How many possible values can <math>r</math> have?<br />
<br />
<math><br />
\mathrm{(A)}\ 4\<br />
\qquad<br />
\mathrm{(B)}\ 8\<br />
\qquad<br />
\mathrm{(C)}\ 9\<br />
\qquad<br />
\mathrm{(D)}\ 50\<br />
\qquad<br />
\mathrm{(E)}\ 90\<br />
\qquad<br />
</math><br />
<br />
[[Category: Introductory Geometry Problems]]<br />
<br />
== Solution ==<br />
<br />
The circumference of circle A is <math>200\pi</math>, and the circumference of circle B with radius <math>r</math> is <math>2r\pi</math>. Since circle B makes a complete revolution and ''ends up on the same point'', the circumference of A must be a multiple of the circumference of B, therefore the quotient must be an integer.<br />
<br />
Thus, <math>\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math><br />
<br />
Therefore <math>r</math> must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). <math>100\: =\: 2^2\; \cdot \; 5^2</math>. Therefore 100 has <math>(2+1)\; \cdot \; (2+1)\;</math> factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is <math>\boxed{8}</math>.<br />
<br />
*The number of factors of <math>a^x\: \cdot \: b^y\: \cdot \: c^z\;...</math> and so on, where <math>a, b,</math> and <math>c</math> are prime numbers, is <math>(x+1)(y+1)(z+1)...</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Alexog123