https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jhggins&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T07:16:43ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=457382012 AIME I Problems/Problem 82012-03-24T19:42:09Z<p>Jhggins: /* Alternative Solution (using calculus): think inside 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 ==<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 />
== 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=\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}{dx}=\int^1_0{\frac{(x+1)^2}{16}}{dx}=\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 />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=457372012 AIME I Problems/Problem 82012-03-24T19:33:20Z<p>Jhggins: /* Alternative Solution (using calculus): think inside 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 ==<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 />
== 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 triangle <math>MQB</math> is similar to triangle <math>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=\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}{dx}=\int^1_0{\frac{(x+1)^2}{16}}{dx}=\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 />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=457102012 AIME I Problems/Problem 82012-03-23T20:29:04Z<p>Jhggins: /* Alternative Solution (using calculus) */</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 ==<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 />
== 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 triangle <math>MQB</math> is similar to triangle <math>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=\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}{dx}=\int^1_0{\frac{(x+1)^2}{16}}{dx}=\frac{7}{48}.</math> Thus the area of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=457092012 AIME I Problems/Problem 82012-03-23T20:28:51Z<p>Jhggins: /* Solution */</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 ==<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 />
== Alternative Solution (using calculus) ==<br />
Let <math>Q</math> be the intersection of the plane with edge <math>FB,</math> then triangle <math>MQB</math> is similar to triangle <math>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=\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}{dx}=\int^1_0{\frac{(x+1)^2}{16}}{dx}=\frac{7}{48}.</math> Thus the area of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=457082012 AIME I Problems/Problem 82012-03-23T20:15:55Z<p>Jhggins: /* Solution */</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 ==<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 />
== Alternative Solution (using calculus) ==<br />
Let <math>Q</math> be the intersection of the plane with edge <math>FB,</math> then triangle <math>MQB</math> is similar to triangle <math>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=\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}{dx}=\int^1_0{\frac{(x+1)^2}{16}}{dx}=\frac{7}{48}.</math> Thus the area of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=457072012 AIME I Problems/Problem 82012-03-23T19:43:49Z<p>Jhggins: /* Problem 8 */</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 ==<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 />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_24&diff=449222002 AMC 12A Problems/Problem 242012-02-21T17:04:28Z<p>Jhggins: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Find the number of ordered pairs of real numbers <math>(a,b)</math> such that <math>(a+bi)^{2002} = a-bi</math>.<br />
<br />
<math><br />
\text{(A) }1001<br />
\qquad<br />
\text{(B) }1002<br />
\qquad<br />
\text{(C) }2001<br />
\qquad<br />
\text{(D) }2002<br />
\qquad<br />
\text{(E) }2004<br />
</math><br />
<br />
== Solution ==<br />
<br />
Let <math>s=\sqrt{a^2+b^2}</math> be the magnitude of <math>a+bi</math>. Then the magnitude of <math>(a+bi)^{2002}</math> is <math>s^{2002}</math>, while the magnitude of <math>a-bi</math> is <math>s</math>. We get that <math>s^{2002}=s</math>, hence either <math>s=0</math> or <math>s=1</math>.<br />
<br />
For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>.<br />
<br />
Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math>rd complex roots of unity, and there are <math>2003</math> of those.<br />
<br />
The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>.<br />
<br />
<br />
== Solution 2 ==<br />
<br />
As in the other solution, split the problem into when <math>s=0</math> and when <math>s=1</math>. When <math>s=1</math> and <math>a+bi=\cos\theta+i\sin\theta</math>,<br />
<br />
<math>(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta) </math> <math> =a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)</math><br />
<br />
so we must have <math>2002\theta=-\theta+2\pi k</math> and hence <math>\theta=\frac{2\pi k}{2003}</math>. Since <math>\theta</math> is restricted to <math>[0,2\pi)</math>, <math>k</math> can range from <math>0</math> to <math>2002</math> inclusive, which is <math>2002-0+1=2003</math> values. Thus the total is <math>1+2003 = \boxed{\textbf{(E)}\ 2004}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2002|ab=A|num-b=23|num-a=25}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=Talk:2008_AMC_12B_Problems/Problem_16&diff=44623Talk:2008 AMC 12B Problems/Problem 162012-02-09T17:00:55Z<p>Jhggins: Created page with "The problem, as I found it, was stated incorrectly. The border around the inner rectangle should have width 1 foot, not 2 feet, and as the border lies on both sides of the rectan..."</p>
<hr />
<div>The problem, as I found it, was stated incorrectly. The border around the inner rectangle should have width 1 foot, not 2 feet, and as the border lies on both sides of the rectangle then the width of the inner rectangle is 2 feet less than the width of the outer rectangle, as the work in the solution indicates.<br />
<br />
The problem was reproduced correctly here [http://math.ncssm.edu/smc/tests/Test2011/2011_geo.pdf]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_16&diff=446222008 AMC 12B Problems/Problem 162012-02-09T16:56:46Z<p>Jhggins: Undo revision 44353 by JBL (talk)</p>
<hr />
<div>==Problem==<br />
A rectangular floor measures <math>a</math> by <math>b</math> feet, where <math>a</math> and <math>b</math> are positive integers with <math>b > a</math>. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width <math>1</math> foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair <math>(a,b)</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution==<br />
<math>A_{outer}=ab</math><br />
<br />
<math>A_{inner}=(a-2)(b-2)</math><br />
<br />
<math>A_{outer}=2A_{inner}</math><br />
<br />
<math>ab=2(a-2)(b-2)=2ab-4a-4b+8</math><br />
<br />
<math>0=ab-4a-4b+8</math><br />
<br />
By Simon's Favorite Factoring Trick:<br />
<br />
<math>8=ab-4a-4b+16=(a-4)(b-4)</math><br />
<br />
Since <math>8=1*8</math> and <math>8=2*4</math> are the only positive factorings of <math>8</math>.<br />
<br />
<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>2\Rightarrow \textbf{(B)}</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_23&diff=444882010 AMC 12B Problems/Problem 232012-02-07T02:36:23Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem 23 ==<br />
Monic quadratic polynomial <math>P(x)</math> and <math>Q(x)</math> have the property that <math>P(Q(x))</math> has zeros at <math>x=-23, -21, -17,</math> and <math>-15</math>, and <math>Q(P(x))</math> has zeros at <math>x=-59,-57,-51</math> and <math>-49</math>. What is the sum of the minimum values of <math>P(x)</math> and <math>Q(x)</math>? <br />
<br />
<math>\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0</math><br />
==Solution==<br />
<math> P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d</math>. Notice that <math> P(x)</math> has roots <math> a\pm \sqrt {b}</math>, so that the roots of <math> P(Q(x))</math> are the roots of <math> Q(x) = a + \sqrt {b}, a - \sqrt {b}</math>. For each individual equation, the sum of the roots will be <math> 2c</math> (symmetry or Vieta's). Thus, we have <math> 4c = - 23 - 21 - 17 - 15</math>, or <math> c = - 19</math>. Doing something similar for <math> Q(P(x))</math> gives us <math> a = - 54</math>.<br />
We now have <math> P(x) = (x + 54)^2 - b, Q(x) = (x + 19)^2 - d</math>. Since <math> Q</math> is monic, the roots of <math> Q(x) = a + \sqrt {b}</math> are "farther" from the axis of symmetry than the roots of <math> Q(x) = a - \sqrt {b}</math>. Thus, we have <math> Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}</math>, or <math> 16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}</math>. Adding these gives us <math> 20 - 2d = - 108</math>, or <math> d = 64</math>. Plugging this into <math> 16 - d = - 54 + \sqrt {b}</math>, we get <math> b = 36</math>.<br />
The minimum value of <math> P(x)</math> is <math> - b</math>, and the minimum value of <math> Q(x)</math> is <math> - d</math>. Thus, our answer is <math> - (b + d) = - 100</math>, or answer <math> \boxed{\textbf{(A)}}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_22&diff=444682011 AMC 12A Problems/Problem 222012-02-05T19:01:21Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Let <math>R</math> be a square region and <math>n \geq 4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called ''n-ray partitional'' if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are <math>100</math>-ray partitional but not <math>60</math>-ray partitional?<br />
<br />
<math><br />
\textbf{(A)}\ 1500 \qquad<br />
\textbf{(B)}\ 1560 \qquad<br />
\textbf{(C)}\ 2320 \qquad<br />
\textbf{(D)}\ 2480 \qquad<br />
\textbf{(E)}\ 2500 </math><br />
<br />
== Solution ==<br />
<br />
First, notice that there must be four rays emanating from <math>X</math> that intersect the four corners of the square region. Depending on the location of <math>X</math>, the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is <math>100</math>-ray partitional (let this point be the bottom-left-most point). We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining <math>96</math> rays are divided among the other two triangular sectors, each sector with <math>48</math> rays, thus dividing these two sectors into <math>49</math> triangles of equal areas. Let the distance from this corner point to the closest side be <math>a</math> and the side of the square be <math>s</math>. From this, we get the equation <math>\frac{a\times s}{2}=\frac{(s-a)\times s}{2}\times\frac1{49}</math>. Solve for <math>a</math> to get <math>a=\frac s{50}</math>. Therefore, point <math>X</math> is <math>\frac1{50}</math> of the side length away from the two sides it is closest to. By moving <math>X</math> <math>\frac s{50}</math> to the right, we also move one ray from the right sector to the left sector, which determines another <math>100</math>-ray partitional point. We can continue moving <math>X</math> right and up to derive the set of points that are <math>100</math>-ray partitional. In the end, we get a square grid of points each <math>\frac s{50}</math> apart from one another. Since this grid ranges from a distance of <math>\frac s{50}</math> from one side to <math>\frac{49s}{50}</math> from the same side, we have a <math>49\times49</math> grid, a total of <math>2401</math> <math>100</math>-ray partitional points. To find the overlap from the <math>60</math>-ray partitional, we must find the distance from the corner-most <math>60</math>-ray partitional point to the sides closest to it. Since the <math>100</math>-ray partitional points form a <math>49\times49</math> grid, each point <math>\frac s{50}</math> apart from each other, we can deduce that the <math>60</math>-ray partitional points form a <math>29\times29</math> grid, each point <math>\frac s{30}</math> apart from each other. To find the overlap points, we must find the common divisors of <math>30</math> and <math>50</math> which are <math>1, 2, 5,</math> and <math>10</math>. Therefore, the overlapping points will form grids with points <math>s</math>, <math>\frac s{2}</math>, <math>\frac s{5}</math>, and <math>\frac s{10}</math> away from each other respectively. Since the grid with points <math>\frac s{10}</math> away from each other includes the other points, we can disregard the other grids. The total overlapping set of points is a <math>9\times9</math> grid, which has <math>81</math> points. Subtract <math>81</math> from <math>2401</math> to get <math>2401-81=\boxed{\textbf{(C)}\ 2320}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=21|num-a=23|ab=A}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_22&diff=444672009 AMC 12A Problems/Problem 222012-02-05T18:57:21Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A regular [[octahedron]] has side length <math>1</math>. A [[plane]] [[parallel]] to two of its opposite faces cuts the octahedron into the two congruent solids. The [[polygon]] formed by the intersection of the plane and the octahedron has area <math>\frac {a\sqrt {b}}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, <math>a</math> and <math>c</math> are relatively prime, and <math>b</math> is not divisible by the square of any prime. What is <math>a + b + c</math>?<br />
<br />
<math>\textbf{(A)}\ 10\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 14</math><br />
<br />
== Solution ==<br />
<center><asy><br />
import three; currentprojection = orthographic(0.5,-3,1.4); pen g = rgb(0.8,1,0.8);<br />
triple[] P = {(1,0,0),(0,1,0),(-1,0,0),(0,-1,0),(0,0,1),(0,0,-1)};<br />
<br />
void drawFrontFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); } <br />
void drawBackFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); } <br />
void fillFace(int x, int y, int z, pen c) {fill(P[x] -- P[y] -- P[z] -- cycle, c);}<br />
pair midpt(int x,int y){ return (P[x] + P[y])/2;}<br />
<br />
path planecut = midpt(1,0)--midpt(1,5)--midpt(2,5)--midpt(2,3)--midpt(4,3)--midpt(4,0)--cycle; <br />
fillFace(0,3,5,g);fillFace(1,2,4,g);fill(planecut,rgb(0.8,0.8,1));<br />
drawFrontFace(0,1,4);drawFrontFace(1,2,4);drawFrontFace(0,1,5);drawFrontFace(1,2,5);drawBackFace(2,3,4);drawBackFace(3,0,4);drawBackFace(2,3,5);drawBackFace(3,0,5);<br />
draw(planecut,linetype("4 4")+linewidth(0.7)); dot((0,0,0));<br />
</asy></center><br />
<br />
If the plane divides the octahedron into two congruent solids, it goes through the center of the octahedron. As it is parallel to two opposite faces (colored above in green), it passes through the midpoints of the edges connecting the corresponding vertices of the faces. The distance between the center and any of the midpoints, as well as the distance between any consecutive midpoints, is found to be <math>1/2</math> (by midline and so forth). Thus, the intersection of the plane and the octahedron is a regular hexagon, and the answer is <math>6 \times \left(\frac {\left(\frac {1}{2}\right)^2 \sqrt {3}}{4}\right) = \frac {3\sqrt {3}}{8}</math>, and <math>a + b + c = 14\ \mathbf{(E)}</math>.<br />
<br />
Since the code above was unparseable, this diagram should be more useful: [http://a.images.memegenerator.net/instances/500x/14091624.jpg]<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=Rational_Root_Theorem&diff=44430Rational Root Theorem2012-02-02T01:04:38Z<p>Jhggins: /* Problems */</p>
<hr />
<div>{{stub}}<br />
<br />
<br />
Given a [[polynomial]] <math>P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \ldots + a_1 x + a_0</math> with [[integer | integral]] [[coefficient]]s, <math>a_n \neq 0</math>. The '''Rational Root Theorem''' states that if <math>P(x)</math> has a [[rational number| rational]] [[root]] <math>r = \pm\frac pq</math> with <math>p, q</math> [[relatively prime]] [[positive integer]]s, <math>p</math> is a [[divisor]] of <math>a_0</math> and <math>q</math> is a divisor of <math>a_n</math>.<br />
<br />
As a consequence, every rational root of a [[monic polynomial]] with integral coefficients must be integral.<br />
<br />
This gives us a relatively quick process to find all "nice" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check.<br />
<br />
== Proof ==<br />
<br />
Given <math>\frac{p}{q}</math> is a rational root of a polynomial <math>f(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_0</math>, we wish to show that <math>p|a_0</math> and <math>q|a_n</math>. Since <math>\frac{p}{q}</math> is a root, <cmath>0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0</cmath> Multiplying by <math>q^n</math>, we have: <cmath>0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n</cmath> Examining this in modulo <math>p</math>, we have <math>a_0q^n\equiv 0\pmod p</math>. As <math>q</math> and <math>p</math> are relatively prime, <math>p|a_0</math>. With the same logic, but with modulo <math>q</math>, we have <math>q|a_n</math>, and we are done.<br />
<br />
==Problems==<br />
<br />
===Easy===<br />
<br />
Factor the polynomial <math>x^3-5x^2+2x+8</math>.<br />
<br />
===Intermediate===<br />
<br />
Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>.<br />
<br />
Prove that <math>\sqrt{2}</math> is irrational, using the Rational Root Theorem.</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_19&diff=443952004 AMC 12A Problems/Problem 192012-01-29T19:15:08Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem 19 ==<br />
[[Circle]]s <math>A, B</math> and <math>C</math> are externally tangent to each other, and internally tangent to circle <math>D</math>. Circles <math>B</math> and <math>C</math> are congruent. Circle <math>A</math> has radius <math>1</math> and passes through the center of <math>D</math>. What is the [[radius]] of circle <math>B</math>?<br />
<br />
<center><asy><br />
unitsize(15mm);<br />
pair A=(-1,0),B=(2/3,8/9),C=(2/3,-8/9),D=(0,0);<br />
<br />
draw(Circle(D,2));<br />
draw(Circle(A,1));<br />
draw(Circle(B,8/9));<br />
draw(Circle(C,8/9));<br />
<br />
label("\(A\)", A);<br />
label("\(B\)", B);<br />
label("\(C\)", C);<br />
label("D", (-1.2,1.8));<br />
</asy></center><br />
<br />
<math>\text {(A)} \frac23 \qquad \text {(B)} \frac {\sqrt3}{2} \qquad \text {(C)}\frac78 \qquad \text {(D)}\frac89 \qquad \text {(E)}\frac {1 + \sqrt3}{3}</math><br />
<br />
==Solution==<br />
<br />
=== Solution 1 ===<br />
<br />
<center><asy><br />
unitsize(15mm);<br />
pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3);<br />
<br />
draw(Circle(D,2));<br />
draw(Circle(A,1));<br />
draw(Circle(B,8/9));<br />
draw(Circle(C,8/9));<br />
draw(A--B--C--A);<br />
draw(B--D--C);<br />
draw(A--E);<br />
<br />
dot(A);dot(B);dot(C);dot(D);dot(E);<br />
<br />
label("\(D\)", D,NW);<br />
label("\(A\)", A,N);<br />
label("\(B\)", B,W);<br />
label("\(C\)", C,E);<br />
label("\(E\)", E,S);<br />
label("\(1\)",(-.4,.7));<br />
label("\(1\)",(0,0.5),W);<br />
label("\(r\)", (-.8,-.1));<br />
label("\(r\)", (-4/9,-2/3),S);<br />
label("\(h\)", (0,-1/3), W);<br />
</asy></center><br />
<br />
Note that <math>BD= 2-r</math> since <math>D</math> is the center of the larger circle of radius <math>2</math>. Using the Pythagorean Theorem on <math>\triangle BDE</math>,<br />
<br />
<cmath><br />
\begin{align*}<br />
r^2 + h^2 &= (2-r)^2 \\<br />
r^2 + h^2 &= 4 - 4r + r^2 \\<br />
h^2 &= 4 - 4r \\<br />
h &= 2\sqrt{1-r} \end{align*}</cmath><br />
<br />
Now using the [[Pythagorean Theorem]] on <math>\triangle BAE</math>,<br />
<br />
<cmath><br />
\begin{align*}<br />
r^2 + (h+1)^2 &= (r+1)^2 \\<br />
r^2 + h^2 + 2h + 1 &= r^2 + 2r + 1 \\<br />
h^2 + 2h &= 2r \end{align*} </cmath><br />
<br />
Substituting <math>h</math>,<br />
<br />
<cmath><br />
\begin{align*}<br />
(4-4r) + 4\sqrt{1-r} &= 2r \\<br />
4\sqrt{1-r} &= 6r - 4 \\<br />
16-16r &= 36r^2 - 48r + 16 \\<br />
0 &= 36r^2 - 32r \\<br />
r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}</cmath><br />
<br />
=== Solution 2 ===<br />
<br />
We can apply [[Descartes' Circle Formula]].<br />
<br />
The four circles have curvatures <math>-\frac{1}{2}, 1, \frac{1}{r}</math>, and <math>\frac{1}{r}</math>.<br />
<br />
We have <math>2((-\frac{1}{2})^2+1^2+\frac {1}{r^2}+\frac{1}{r^2})=(-\frac{1}{2}+1+\frac{1}{r}+\frac{1}{r})^2</math><br />
<br />
Simplifying, we get <math>\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}</math><br />
<br />
<math>\frac{2}{r}=\frac{9}{4}</math><br />
<br />
<math>r=\frac{8}{9} \Longrightarrow \qquad \textbf{(D)}</math><br />
<br />
==See Also== <br />
{{AMC12 box|year=2004|ab=A|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_18&diff=443752001 AMC 12 Problems/Problem 182012-01-26T07:17:38Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is<br />
<br />
<asy><br />
unitsize(0.75cm);<br />
pair A=(0,1), B=(4,4);<br />
dot(A); dot(B);<br />
draw( circle(A,1) );<br />
draw( circle(B,4) );<br />
draw( (-1.5,0)--(8.5,0) );<br />
draw( A -- (A+(-1,0)) );<br />
label("$1$", A -- (A+(-1,0)), N );<br />
draw( B -- (B+(4,0)) );<br />
label("$4$", B -- (B+(4,0)), N );<br />
label("$A$",A,E);<br />
label("$B$",B,W);<br />
<br />
filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black );<br />
dot( (12/9,4/9) );<br />
</asy><br />
<br />
<math><br />
\text{(A) }\frac {1}{3}<br />
\qquad<br />
\text{(B) }\frac {2}{5}<br />
\qquad<br />
\text{(C) }\frac {5}{12}<br />
\qquad<br />
\text{(D) }\frac {4}{9}<br />
\qquad<br />
\text{(E) }\frac {1}{2}<br />
</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
<asy><br />
unitsize(1cm);<br />
pair A=(0,1), B=(4,4), C=(4,1);<br />
dot(A); dot(B);<br />
draw( circle(A,1) );<br />
draw( circle(B,4) );<br />
draw( (-1.5,0)--(8.5,0) );<br />
draw( (A+(4,0)) -- A -- (A+(0,-1)) );<br />
draw( A -- B -- (B+(0,-4)) );<br />
label("$A$",A,N);<br />
label("$B$",B,N);<br />
label("$C$",C,E);<br />
<br />
filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black );<br />
dot( (12/9,4/9) );<br />
draw( rightanglemark(A,C,B) );<br />
</asy><br />
<br />
In the triangle <math>ABC</math> we have <math>AB = 1+4 = 5</math> and <math>BC=4-1 = 3</math>, thus by the [[Pythagorean theorem]] we have <math>AC=4</math>.<br />
<br />
We can now pick a coordinate system where the common tangent is the <math>x</math> axis and <math>A</math> lies on the <math>y</math> axis. <br />
In this coordinate system we have <math>A=(0,1)</math> and <math>B=(4,4)</math>. <br />
<br />
Let <math>r</math> be the radius of the small circle, and let <math>s</math> be the <math>x</math>-coordinate of its center <math>S</math>. We then know that <math>S=(s,r)</math>, as the circle is tangent to the <math>x</math> axis. Moreover, the small circle is tangent to both other circles, hence we have <math>SA=1+r</math> and <math>SB=4+r</math>.<br />
<br />
We have <math>SA = \sqrt{s^2 + (1-r)^2}</math> and <math>SB=\sqrt{(4-s)^2 + (4-r)^2}</math>. Hence we get the following two equations:<br />
<br />
<cmath><br />
\begin{align*}<br />
s^2 + (1-r)^2 & = (1+r)^2<br />
\\<br />
(4-s)^2 + (4-r)^2 & = (4+r)^2<br />
\end{align*}<br />
</cmath><br />
<br />
Simplifying both, we get<br />
<br />
<cmath><br />
\begin{align*}<br />
s^2 & = 4r<br />
\\<br />
(4-s)^2 & = 16r<br />
\end{align*}<br />
</cmath><br />
<br />
As in our case both <math>r</math> and <math>s</math> are positive, we can divide the second one by the first one to get <math>\left( \frac{4-s}s \right)^2 = 4</math>.<br />
<br />
Now there are two possibilities: either <math>\frac{4-s}s=-2</math>, or <math>\frac{4-s}s=2</math>. In the first case clearly <math>s<0</math>, hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the <math>x</math> axis - a large circle whose center is somewhere to the left of <math>A</math>.) The second case solves to <math>s=\frac 43</math>. We then have <math>4r = s^2 = \frac {16}9</math>, hence <math>r = \boxed{\frac 49}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
The horizontal line is the equivalent of a circle of curvature <math>0</math>, thus we can apply [[Descartes' Circle Formula]].<br />
<br />
The four circles have curvatures <math>0, 1, \frac 14</math>, and <math>\frac 1r</math>.<br />
<br />
We have <math>2(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2})=(0+1+\frac 14+\frac 1r)^2</math><br />
<br />
Simplifying, we get <math>\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}</math><br />
<br />
<math>\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0</math><br />
<br />
<math>\frac{16}{r^2}-\frac{40}{r}+9=0</math><br />
<br />
<math>(\frac{4}{r}-9)(\frac{4}{r}-1)=0</math><br />
<br />
Obviously <math>r</math> cannot equal <math>4</math>, therefore <math>r = \boxed{\frac 49}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2001|num-b=17|num-a=19}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=443742010 AMC 12B Problems/Problem 182012-01-26T04:12:15Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
<br />
The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in, the remaining places that the frog can jump to is a circle of radius 2 from where he landed. Every point in the circle of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want to land in is enclosed in this larger circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=443732010 AMC 12B Problems/Problem 182012-01-26T03:53:24Z<p>Jhggins: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br />
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath><br />
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
===Solution 2===<br />
<br />
The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in, the remaining places that the frog can jump to is a circle of radius 2 from where he landed. Every point in the circle of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want to land in is enclosed in this larger circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=443722010 AMC 12B Problems/Problem 182012-01-26T03:41:37Z<p>Jhggins: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br />
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath><br />
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
===Solution 2===<br />
<br />
The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in, the remaining places that the frog can jump to is a circle of radius 2 from where he landed. Every point in the circle of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want to land in is enclosed in this circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=443712010 AMC 12B Problems/Problem 182012-01-26T03:41:10Z<p>Jhggins: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br />
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath><br />
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
===Solution 2===<br />
<br />
The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in, the remaining places that the frog can jump to is a circle of radius two from where he landed. Every point in the circle of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want to land in is enclosed in this circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_16&diff=443522008 AMC 12B Problems/Problem 162012-01-24T01:18:58Z<p>Jhggins: /* Problem */</p>
<hr />
<div>==Problem==<br />
A rectangular floor measures <math>a</math> by <math>b</math> feet, where <math>a</math> and <math>b</math> are positive integers with <math>b > a</math>. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width <math>1</math> foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair <math>(a,b)</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution==<br />
<math>A_{outer}=ab</math><br />
<br />
<math>A_{inner}=(a-2)(b-2)</math><br />
<br />
<math>A_{outer}=2A_{inner}</math><br />
<br />
<math>ab=2(a-2)(b-2)=2ab-4a-4b+8</math><br />
<br />
<math>0=ab-4a-4b+8</math><br />
<br />
By Simon's Favorite Factoring Trick:<br />
<br />
<math>8=ab-4a-4b+16=(a-4)(b-4)</math><br />
<br />
Since <math>8=1*8</math> and <math>8=2*4</math> are the only positive factorings of <math>8</math>.<br />
<br />
<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>2\Rightarrow \textbf{(B)}</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_16&diff=443512008 AMC 12B Problems/Problem 162012-01-24T01:18:46Z<p>Jhggins: /* Problem */</p>
<hr />
<div>==Problem==<br />
A rectangular floor measures <math>a</math> by <math>b</math> feet, where <math>a</math> and <math>b</math> are positive integers with <math>b > a</math>. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width <math></math> foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair <math>(a,b)</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
==Solution==<br />
<math>A_{outer}=ab</math><br />
<br />
<math>A_{inner}=(a-2)(b-2)</math><br />
<br />
<math>A_{outer}=2A_{inner}</math><br />
<br />
<math>ab=2(a-2)(b-2)=2ab-4a-4b+8</math><br />
<br />
<math>0=ab-4a-4b+8</math><br />
<br />
By Simon's Favorite Factoring Trick:<br />
<br />
<math>8=ab-4a-4b+16=(a-4)(b-4)</math><br />
<br />
Since <math>8=1*8</math> and <math>8=2*4</math> are the only positive factorings of <math>8</math>.<br />
<br />
<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>2\Rightarrow \textbf{(B)}</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_15&diff=443422010 AMC 12B Problems/Problem 152012-01-22T21:37:55Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem 15 ==<br />
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?<br />
<br />
<math>\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235</math><br />
<br />
== Solution ==<br />
We have either <math>i^{x}=(1+i)^{y}\neq z</math>, <math>i^{x}=z\neq(1+i)^{y}</math>, or <math>(1+i)^{y}=z\neq i^x</math>.<br />
<br />
For <math>i^{x}=(1+i)^{y}</math>, this only occurs at <math>1</math>. <math>(1+i)^{y}=1</math> has only one solution, namely, <math>y=0</math>. <math>i^{x}=1</math> has five solutions between zero and nineteen, <math>x=0, x=4, x=8, x=12</math>, and <math>x=16</math>. <math>z\neq 1</math> has nineteen integer solutions between zero and nineteen. So for <math>i^{x}=(1+i)^{y}\neq z</math>, we have <math>5\times 1\times 19=95</math> ordered triples.<br />
<br />
For <math>i^{x}=z\neq(1+i)^{y}</math>, again this only occurs at <math>1</math>. <math>(1+i)^{y}\neq 1</math> has nineteen solutions, <math>i^{x}=1</math> has five solutions, and <math>z=1</math> has one solution, so again we have <math>5\times 1\times 19=95</math> ordered triples.<br />
<br />
For <math>(1+i)^{y}=z\neq i^x</math>, this occurs at <math>1</math> and <math>16</math>. <math>(1+i)^{y}=1</math> and <math>z=1</math> both have one solution while <math>i^{x}\neq 1</math> has fifteen solutions. <math>(1+i)^{y}=16</math> and <math>z=16</math> both have one solution, namely, <math>y=8</math> and <math>z=16</math>, while <math>i^{x}\neq 16</math> has twenty solutions. So we have <math>15\times 1\times 1+20\times 1\times 1=35</math> ordered triples.<br />
<br />
In total we have <math>{95+95+35=225}</math> ordered triples <math>\Rightarrow \boxed{D}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=14|num-a=16|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_14&diff=443372008 AMC 12A Problems/Problem 142012-01-22T18:53:29Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem ==<br />
What is the area of the region defined by the [[inequality]] <math>|3x-18|+|2y+7|\le3</math>?<br />
<br />
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ \frac {7}{2}\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 5</math><br />
<br />
== Solution ==<br />
Area is invariant under translation, so after translating left <math>6</math> and up <math>7/2</math> units, we have the inequality<br />
<br />
<cmath>|3x| + |2y|\leq 3</cmath><br />
<br />
which forms a [[rhombus|diamond]] centered at the [[origin]] and vertices at <math>(\pm 1, 0), (0, \pm 1.5)</math>. Thus the diagonals are of length <math>2</math> and <math>3</math>. Using the formula <math>A = \frac 12 d_1 d_2</math>, the answer is <math>\frac{1}{2}(2)(3) = 3 \Rightarrow \mathrm{(A)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2008|num-b=13|num-a=15|ab=A}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
[[Category:Introductory Geometry Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_13&diff=443082008 AMC 12B Problems/Problem 132012-01-19T19:16:28Z<p>Jhggins: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Vertex <math>E</math> of equilateral <math>\triangle{ABE}</math> is in the interior of unit square <math>ABCD</math>. Let <math>R</math> be the region consisting of all points inside <math>ABCD</math> and outside <math>\triangle{ABE}</math> whose distance from <math>AD</math> is between <math>\frac{1}{3}</math> and <math>\frac{2}{3}</math>. What is the area of <math>R</math>? <br />
<br />
<math>\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad<br />
\textbf{(B)}\ \frac{12-5\sqrt3}{36} \qquad<br />
\textbf{(C)}\ \frac{\sqrt3}{18} \qquad<br />
\textbf{(D)}\ \frac{3-\sqrt3}{9} \qquad<br />
\textbf{(E)}\ \frac{\sqrt3}{12}</math><br />
<br />
==Solution==<br />
The region is the shaded area:<br />
<br />
<center><asy><br />
pair A,B,C,D,E;<br />
A=(0,1);<br />
B=(1,1);<br />
C=(1,0);<br />
D=(0,0);<br />
E=(1/2,1-sqrt(3)/2);<br />
draw(A--B--C--D--cycle);<br />
label("A",A,NW);<br />
dot(A);<br />
label("B",B,NE);<br />
dot(B);<br />
label("C",C,SE);<br />
dot(C);<br />
label("D",D,SW);<br />
dot(D);<br />
draw(A--E--B--cycle);<br />
label("E",E,S);<br />
dot(E);<br />
draw((1/3,0)--(1/3,1));<br />
draw((2/3,0)--(2/3,1));<br />
fill((1/3,0)--(1/3,1-sqrt(3)/3)--E--(2/3,1-sqrt(3)/3)--(2/3,0)--cycle,Black);</asy><br />
</center><br />
We can find the area of the shaded region by subtracting the pentagon from the middle third of the square. The area of the middle third of the square is <math>\left(\frac13\right)(1)=\frac13</math>. The pentagon can be split into a rectangle and an equilateral triangle.<br />
<br />
The base of the equilateral triangle is <math>\frac13</math> and the height is <math>\left(\frac13\right)\left(\frac12\right)(\sqrt{3})=\frac{\sqrt{3}}{6}</math>. Thus, the area is <math>\left(\frac{\sqrt3}{6}\right)\left(\frac13\right)\left(\frac12\right)=\frac{\sqrt3}{36}</math>.<br />
<br />
The base of the rectangle is <math>\frac13</math> and the height is the height of the equilateral triangle minus the height of the smaller equilateral triangle. This is:<br />
<math>\frac{\sqrt3}{2}-\frac{\sqrt3}{6}=\frac{\sqrt3}{3}</math><br />
Therefore, the area of the shaded region is<br />
<center><math>\frac13-\frac{\sqrt3}{9}-\frac{\sqrt3}{36}=\boxed{\text{(B) }\frac{12-5\sqrt3}{36}}.</math></center><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2008|ab=B|num-b=12|num-a=14}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=441562011 AMC 12A Problems/Problem 122012-01-07T04:51:56Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
A power boat and a raft both left dock <math>A</math> on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock <math>B</math> downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock <math>A.</math> How many hours did it take the power boat to go from <math>A</math> to <math>B</math>?<br />
<br />
<math><br />
\textbf{(A)}\ 3 \qquad<br />
\textbf{(B)}\ 3.5 \qquad<br />
\textbf{(C)}\ 4 \qquad<br />
\textbf{(D)}\ 4.5 \qquad<br />
\textbf{(E)}\ 5 </math><br />
<br />
== Solution ==<br />
Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of <math>0</math>. In this case, when the powerboat travels from <math>A</math> to <math>B</math>, the raft remains at <math>A</math>. Thus the trip from <math>A</math> to <math>B</math> takes the same time as the trip from <math>B</math> to the raft. Since these times are equal and sum to <math>9</math> hours, the trip from <math>A</math> to <math>B</math> must take half this time, or <math>4.5</math> hours. The answer is thus <math>\boxed{\textbf{D}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_3&diff=441412007 AMC 12A Problems/Problem 32012-01-04T18:59:35Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The larger of two consecutive odd integers is three times the smaller. What is their sum?<br />
<br />
<math>\displaystyle \mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20</math><br />
<br />
== Solution ==<br />
'''Solution 1'''<br />
Let <math>n</math> be the smaller term. Then <math>n+2=3n \Longrightarrow 2n = 2 \Longrightarrow n=1</math><br />
*Thus, the answer is <math>1+(1+2)=4 \mathrm{(A)}</math><br />
<br />
----<br />
'''Solution 2'''<br />
* By trial and error, 1 and 3 work. 1+3=4.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2007|ab=A|num-b=2|num-a=4}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_1&diff=441402003 AMC 12B Problems/Problem 12012-01-04T18:40:36Z<p>Jhggins: /* Solution */</p>
<hr />
<div>==Problem==<br />
Which of the following is the same as<br />
<br />
<cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}</cmath>?<br />
<br />
<math><br />
\text {(A) } -1 \qquad \text {(B) } -\frac{2}{3} \qquad \text {(C) } \frac{2}{3} \qquad \text {(D) } 1 \qquad \text {(E) } \frac{14}{3}<br />
</math><br />
<br />
==Solution==<br />
<cmath><br />
\begin{align*}<br />
2-4+6-8+10-12+14=-2-2-2+14&=8\\<br />
3-6+9-12+15-18+21=-3-3-3+21&=12\\<br />
\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}&=\frac{8}{12}=\frac{2}{3} \Rightarrow \text {(C)}<br />
\end{align*}</cmath><br />
<br />
Alternatively, notice that each term in the numerator is <math>\frac{2}{3}</math> of a term in the denominator, so the quotient has to be <math>\frac{2}{3}</math>.<br />
<br />
==See also==<br />
{{AMC12 box|year=2003|ab=B|before=First Question|num-a=2}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_6&diff=441182010 AMC 12B Problems/Problem 62012-01-02T20:52:44Z<p>Jhggins: /* Problem 6 */</p>
<hr />
<div>== Problem 6 ==<br />
At the beginning of the school year, <math>50\%</math> of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and <math>50\%</math> answered "No." At the end of the school year, <math>70\%</math> answered "Yes" and <math>30\%</math> answered "No." Altogether, <math>x\%</math> of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80</math><br />
<br />
== Solution ==<br />
Clearly, the minimum possible value would be <math>70 - 50 = 20\%</math>. The maximum possible value would be <math>30 + 50 = 80\%</math>. The difference is <math>80 - 20 = 60</math> <math>(D)</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=5|num-a=7|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_25&diff=441012011 AMC 12A Problems/Problem 252012-01-02T02:43:30Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>\angle BAC = 60^{\circ}</math>, <math>\angle CBA \leq 90^{\circ}</math>, <math>BC=1</math>, and <math>AC \geq AB</math>. Let <math>H</math>, <math>I</math>, and <math>O</math> be the orthocenter, incenter, and circumcenter of <math>\triangle ABC</math>, respectively. Assume that the area of pentagon <math>BCOIH</math> is the maximum possible. What is <math>\angle CBA</math>?<br />
<br />
<math><br />
\textbf{(A)}\ 60^{\circ} \qquad<br />
\textbf{(B)}\ 72^{\circ} \qquad<br />
\textbf{(C)}\ 75^{\circ} \qquad<br />
\textbf{(D)}\ 80^{\circ} \qquad<br />
\textbf{(E)}\ 90^{\circ} </math><br />
<br />
== Solution ==<br />
25) Answer: (D) 80 degree<br />
<br />
Given: <math>BC = 1</math>, <math>\angle BAC = 60^{\circ}</math>, <math>\angle CBA \le 90^{\circ}</math>, <math>AC \ge BC</math><br />
<br />
<math>H</math>, <math>I</math>, <math>O</math> are orthocenter, incenter, and circumcenter. and <math>BOIHC</math> has maximum area.<br />
<br />
Find <math>\angle CBA</math>.<br />
<br />
<br /><br />
'''Solution:'''<br />
<br />
1) Let's draw a circle with center <math>O</math> (which will be the circumcircle of <math>\triangle ABC</math>. Since <math>\angle BAC = 60^{\circ}</math>, <math>\overline{BC}</math> is a chord that intercept an arc of <math>120 ^{\circ}</math><br />
<br />
2) Draw any chord that can be <math>BC</math>, and lets define that as unit length.<br />
<br />
3) Draw the diameter <math>\perp</math> to <math>BC</math>. Let's call the interception of the diameter with <math>BC</math> <math>M</math> (because it is the midpoint) and interception with the circle <math>X</math>.<br />
<br />
4) Note that OMB and XMC is fixed, hence the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area.<br />
<br />
<br /><br />
'''Lemma:''' <br />
<br />
<math>m\angle BOC = m \angle BIC = m \angle BHC = 120^{\circ}</math><br />
<br />
For <math>m\angle BOC</math>, we fixed it to <math>120^{\circ}</math> when we drew the diagram.<br />
<br />
Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</math><br />
<br />
<br /><br />
Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>.<br />
<br />
<math>m\angle IBC = \frac{\beta}{2}</math>, <math>m\angle ICB = \frac{\gamma}{2}</math><br />
<br />
<math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - \frac{180^{\circ} - 120^{\circ}}{2} = 120 ^{\circ}</math><br />
<br />
<br /><br />
Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>.<br />
<br />
<math>m\angle HBC = \beta - 30^{\circ}</math>, <math>m\angle HCB = \gamma - 30^{\circ}</math><br />
<br />
<math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math><br />
<br />
<br /><br />
Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.<br />
<br />
Since we got that XOIHC also achieved maximum area,<br />
<br />
Let <math>m\angle XOI = x_1</math>, <math>m\angle OIH = x_2</math>, <math>m\angle IHC = x_3</math>, and the radius is <math>R</math> (which will drop out.)<br />
<br />
then area = <math>\frac{r^2}{2}(\sin x_1 + \sin x_2 + \sin x_3)</math>, where <math>x_1 + x_2 + x_3 = 60^\circ</math><br />
<br />
So we want to maximize <math>f(x_1, x_2) = \sin x_1 + \sin x_2 + \sin x_3</math>, Note that <math>x_3 = 60 ^\circ - x_1 - x_2</math>.<br />
<br />
Let's do some multi-variable calculus.<br />
<br />
<math>f_{x_1} = \cos x_1 - \cos (x_3)</math>, <math>f_{x_2} = \cos x_2 - \cos (x_3)</math><br />
<br />
If both partial is zero, then <math>x_1 = x_2 = x_3 = 20^\circ</math>, and it is very easy to show that <math>f(x_1, x_2)</math> is maximum here with second derivative test (left for the reader).<br />
<br />
<br /><br />
Now, we need to verify that such situation exist and find the angle for this situation.<br />
<br />
Let's extend <math>AI</math> to the direction of <math>X</math>, since <math>AI</math> is the angle bisector, <math>AI</math> should intersection the midpoint of the arc, which is <math>X</math>. Hence, if such case exist, <math>m\angle AXB = m \angle ACB = 40 ^\circ</math>, which yield that <math>m\angle CBA = 80 ^\circ</math>.<br />
<br />
If the angle is <math>80 ^\circ</math>, it is clear that since <math>I</math> and <math>H</math> are on the second circle (follow from lemma). <math>I</math> will be at the right place. <math>H</math> can be easily verified too.<br />
<br />
<br /><br />
Hence, the answer is <math>(D)</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_23&diff=441002003 AMC 12A Problems/Problem 232012-01-02T02:17:47Z<p>Jhggins: /* Problem */</p>
<hr />
<div><br />
==Problem==<br />
<br />
How many perfect squares are divisors of the product <math>1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!</math>?<br />
<br />
<math> \textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008 </math><br />
<br />
==Solution==<br />
<br />
We want to find the number of perfect square factors in the product of all the factorials of numbers from 1 - 9. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. Anyway, this comes out to be equal to 2^30 * 3^13 * 5^5 * 7^3. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even. This gives us 16*7*3*2 = 672 (B).<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_21&diff=440972010 AMC 12A Problems/Problem 212012-01-02T02:01:24Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
The graph of <math>y=x^6-10x^5+29x^4-4x^3+ax^2</math> lies above the line <math>y=bx+c</math> except at three values of <math>x</math>, where the graph and the line intersect. What is the largest of these values?<br />
<br />
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math><br />
<br />
== Solution ==<br />
The <math>x</math> values in which <math>y=x^6-10x^5+29x^4-4x^3+ax^2</math> intersect at <math>y=bx+c</math> are the same as the zeros of <math>y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c</math>.<br />
<br />
Since there are <math>3</math> zeros and the function is never negative, all <math>3</math> zeros must be double roots because the function's degree is <math>6</math>.<br />
<br />
Suppose we let <math>p</math>, <math>q</math>, and <math>r</math> be the roots of this function, and let <math>x^3-ux^2+vx-w</math> be the cubic polynomial with roots <math>p</math>, <math>q</math>, and <math>r</math>.<br />
<br />
<cmath>\begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\<br />
(x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\<br />
\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*}</cmath><br />
<br />
In order to find <math>\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}</math> we must first expand out the terms of <math>(x^3-ux^2+vx-w)^2</math>.<br />
<br />
<cmath>(x^3-ux^2+vx-w)^2</cmath><br />
<math>= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2</math><br />
<br />
[Quick note: Since we don't know <math>a</math>, <math>b</math>, and <math>c</math>, we really don't even need the last 3 terms of the expansion.]<br />
<br />
<cmath>\begin{align*}&2u = 10\\<br />
u^2+2v &= 29\\<br />
2uv+2w &= 4\\<br />
u &= 5\\<br />
v &= 2\\<br />
w &= -8\\<br />
&\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*}</cmath><br />
<br />
All that's left is to find the largest root of <math>x^3-5x^2+2x+8</math>.<br />
<br />
<cmath>\begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\<br />
&\boxed{\textbf{(A)}\ 4}\end{align*}</cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=20|num-a=22|ab=A}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_21&diff=440962006 AMC 12B Problems/Problem 212012-01-02T01:59:56Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Rectangle <math>ABCD</math> has area <math>2006</math>. An ellipse with area <math>2006\pi</math> passes through <math>A</math> and <math>C</math> and has foci at <math>B</math> and <math>D</math>. What is the perimeter of the rectangle? (The area of an ellipse is <math>ab\pi</math> where <math>2a</math> and <math>2b</math> are the lengths of the axes.)<br />
<br />
<math><br />
\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi}<br />
\qquad<br />
\mathrm{(B)}\ \frac {1003}4<br />
\qquad<br />
\mathrm{(C)}\ 8\sqrt {1003}<br />
\qquad<br />
\mathrm{(D)}\ 6\sqrt {2006}<br />
\qquad<br />
\mathrm{(E)}\ \frac {32\sqrt {1003}}\pi<br />
</math><br />
<br />
== Solution ==<br />
<br />
''This solution needs a picture. Please help add it.''<br />
<br />
Let the rectangle have side lengths <math>l</math> and <math>w</math>. Let the axis of the ellipse on which the foci lie have length <math>2a</math>, and let the other axis have length <math>2b</math>. We have<br />
<cmath>lw=ab=2006</cmath><br />
From the definition of an ellipse, <math>l+w=2a\Longrightarrow \frac{l+w}{2}=a</math>. Also, the diagonal of the rectangle has length <math>\sqrt{l^2+w^2}</math>. Comparing the lengths of the axes and the distance from the foci to the center, we have<br />
<cmath>a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b=\sqrt{1003}</cmath><br />
Since <math>ab=2006</math>, we now know <math>a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}</math> and because <math>a=\frac{l+w}{2}</math>, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of <math>\boxed{8\sqrt{1003}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=20|num-a=22}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_21&diff=440952006 AMC 12B Problems/Problem 212012-01-02T01:58:50Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Rectange <math>ABCD</math> has area <math>2006</math>. An ellipse with area <math>2006\pi</math> passes through <math>A</math> and <math>C</math> and has foci at <math>B</math> and <math>D</math>. What is the perimeter of the rectangle? (The area of an ellipse is <math>ab\pi</math> where <math>2a</math> and <math>2b</math> are the lengths of the axes.)<br />
<br />
<math><br />
\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi}<br />
\qquad<br />
\mathrm{(B)}\ \frac {1003}4<br />
\qquad<br />
\mathrm{(C)}\ 8\sqrt {1003}<br />
\qquad<br />
\mathrm{(D)}\ 6\sqrt {2006}<br />
\qquad<br />
\mathrm{(E)}\ \frac {32\sqrt {1003}}\pi<br />
</math><br />
<br />
== Solution ==<br />
<br />
''This solution needs a picture. Please help add it.''<br />
<br />
Let the rectangle have side lengths <math>l</math> and <math>w</math>. Let the axis of the ellipse on which the foci lie have length <math>2a</math>, and let the other axis have length <math>2b</math>. We have<br />
<cmath>lw=ab=2006</cmath><br />
From the definition of an ellipse, <math>l+w=2a\Longrightarrow \frac{l+w}{2}=a</math>. Also, the diagonal of the rectangle has length <math>\sqrt{l^2+w^2}</math>. Comparing the lengths of the axes and the distance from the foci to the center, we have<br />
<cmath>a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b=\sqrt{1003}</cmath><br />
Since <math>ab=2006</math>, we now know <math>a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}</math> and because <math>a=\frac{l+w}{2}</math>, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of <math>\boxed{8\sqrt{1003}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=20|num-a=22}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_19&diff=440852003 AMC 12A Problems/Problem 192012-01-02T01:27:34Z<p>Jhggins: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A parabola with equation <math>y=ax^2+bx+c</math> is reflected about the <math>x</math>-axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of <math>y=f(x)</math> and <math>y=g(x)</math>, respectively. Which of the following describes the graph of <math>y=(f+g)x</math>?<br />
<br />
<math> \textbf{(A)}\ \text{a parabola tangent to the }x\text{-axis} </math><br />
<math> \textbf{(B)}\ \text{a parabola not tangent to the }x\text{-axis}\qquad\textbf{(C)}\ \text{a horizontal line} </math><br />
<math> \textbf{(D)}\ \text{a non-horizontal line}\qquad\textbf{(E)}\ \text{the graph of a cubic function} </math><br />
<br />
==Solution==<br />
<br />
If we take the parabola <math>ax^2 + bx + c</math> and reflect it over the x - axis, we have the parabola <math>-ax^2 - bx - c</math>. Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then:<br />
<br />
<cmath> \begin{align*} f(x) &= a(x+5)^2 + b(x+5) + c = ax^2 + (10a+b)x + 25a + 5b + c \\ g(x) &= -a(x-5)^2 - b(x-5) - c = -ax^2 + 10ax -bx - 25a + 5b - c \end{align*} </cmath> <br />
<br />
Adding them up produces: <cmath> (f + g)(x) &= ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c &= 20ax + 10b </cmath> <br />
<br />
This is a line with slope <math>20a</math>. Since <math>a</math> cannot be <math>0</math> (because <math>ax^2 + bx + c</math> would be a line) we end up with <math>\boxed{\textbf{(D)} \text{ a non-horizontal line }}</math></div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=440842010 AMC 12B Problems/Problem 182012-01-02T01:22:01Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br />
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath><br />
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
===Solution 2===<br />
<br />
The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in, the remaining places that the frog can jump to is a circle of radius two from where he landed. The entirety of the circle you want to land in is enclosed in this circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_16&diff=440772003 AMC 12A Problems/Problem 162012-01-01T23:59:28Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<br />
A point P is chosen at random in the interior of equilateral triangle <math>ABC</math>. What is the probability that <math>\triangle ABP</math> has a greater area than each of <math>\triangle ACP</math> and <math>\triangle BCP</math>?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
== Solution==<br />
===Solution 1===<br />
<br />
After we pick point <math>P</math>, we realize that <math>ABC</math> is symmetric for this purpose, and so the probability that <math>ACP</math> is the greatest area, or <math>ABP</math> or <math>BCP</math>, are all the same. Since they add to <math>1</math>, the probability that <math>ACP</math> has the greatest area is <math>\boxed{\mathrm{(C)}\ \dfrac{1}{3}}</math><br />
<br />
===Solution 2===<br />
<br />
We will use an approach of geometric probability to solve this problem. Let us take point P, and draw the perpendiculars to AB, BC, and AC, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply 1/2 * AC * PF. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ACP greater than ABP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the orthocenter of ABC. Let us call the orthocenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QFCE. So our probability (using geometric probability) is the area of QFCE divided by the area of ABC. We will now show that the three quadrilaterals, QFCE, QEBD, and QDAF are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. Using the Pythagorean theorem, BD = DA = AF = FC = CE = EB. Also, angles BDQ, BEQ, CEQ, CFQ, AFQ, and ADQ are all equal to 90 degrees by the definition of an altitude. Also, angles DBE, FCE, DAF are all equal to 60 degrees as equilateral triangles are also equiangular. It is now clear that QFCE, QFAD, QEBD are all congruent. Summing up these areas gives us the area of ABC. QFCE contributes to a third of that area, as they are all congruent, so the ratio of the areas of QFCE to ABC is 1/3 (C).<br />
<br />
==See Also==<br />
{{AMC12 box|year=2003|ab=A|num-b=15|num-a=17}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_15&diff=440762011 AMC 12B Problems/Problem 152012-01-01T23:52:14Z<p>Jhggins: /* Solution */</p>
<hr />
<div>==Problem 15==<br />
<br />
How many positive two-digits integers are factors of <math>2^{24}-1</math>?<br />
<br />
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math><br />
<br />
<br />
== Solution ==<br />
<br />
From repeated application of difference of squares:<br />
<br />
<math>2^{24}-1 = (2^{12} + 1)(2^{6} + 1)(2^{3} + 1)(2^{3} - 1)</math><br />
<br />
<math>2^{24}-1 = (2^{12} + 1) * 65 * 9 * 7</math><br />
<br />
<math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math><br />
<br />
Aplying sum of cubes: <br />
<br />
<math>2^{12}+1 = (2^4 + 1)(2^8 - 2^4 + 1) </math><br />
<br />
<math>2^{12}+1 = 17 * 241</math><br />
<br />
<br />
A quick check shows <math>241</math> is prime. Thus, the only factors to be concerned about are <math>3^2 * 5 * 7 * 13 * 17</math>, since multiplying by <math>241</math> will make any factor too large.<br />
<br />
Multiply <math>17</math> by <math>3</math> or <math>5</math> will give a two digit factor; <math>17</math> itself will also work. The next smallest factor, <math>7</math>, gives a three digit number. Thus, there are <math>3</math> factors which are multiples of <math>17</math>.<br />
<br />
Multiply <math>13</math> by <math>3, 5</math> or <math>7</math> will also give a two digit factor, as well as <math>13</math> itself. Higher numbers will not work, giving an additional <math>4</math> factors.<br />
<br />
Multiply <math>7</math> by <math>3, 5, </math> or <math> 3^2</math> for a two digit factor. There are no mare factors to check, as all factors which include <math>13</math> are already counted. Thus, there are an additional <math>3</math> factors.<br />
<br />
Multiply <math>5</math> by <math>3</math> or <math>3^2</math> for a two digit factor. All higher factors have been counted already, so there are <math>2</math> more factors.<br />
<br />
Thus, the total number of factors is <math>3 + 4 + 3 + 2 = \boxed{12 \textbf{ (D)}}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|ab=B|num-b=14|num-a=16}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_14&diff=440672010 AMC 12B Problems/Problem 142012-01-01T23:32:25Z<p>Jhggins: /* Problem 14 */</p>
<hr />
<div>== Problem 14 ==<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sum <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>?<br />
<br />
<math>\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804</math><br />
<br />
== Solution ==<br />
We want to try make <math>a+b</math>, <math>b+c</math>, <math>c+d</math>, and <math>d+e</math> as close as possible so that <math>M</math>, the maximum of these, if smallest.<br />
<br />
Notice that <math>2010=670+670+670</math>. In order to express <math>2010</math> as a sum of <math>5</math> numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): <math>2010=670+1+670+1+668</math> or <math>2010=670+1+669+1+669</math>. We see that in both cases, the value of <math>M</math> is <math>671</math>, so the answer is <math>671 \Rightarrow \boxed{B}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_14&diff=440662004 AMC 12B Problems/Problem 142012-01-01T23:27:16Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<br />
In <math>\triangle ABC</math>, <math>AB=13</math>, <math>AC=5</math>, and <math>BC=12</math>. Points <math>M</math> and <math>N</math> lie on <math>AC</math> and <math>BC</math>, respectively, with <math>CM=CN=4</math>. Points <math>J</math> and <math>K</math> are on <math>AB</math> so that <math>MJ</math> and <math>NK</math> are perpendicular to <math>AB</math>. What is the area of pentagon <math>CMJKN</math>? <br />
<br />
<asy><br />
unitsize(0.5cm);<br />
defaultpen(0.8);<br />
pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0);<br />
pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) );<br />
pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) );<br />
draw( A--B--C--cycle );<br />
draw( M--J );<br />
draw( N--K );<br />
label("$A$",A,NW);<br />
label("$B$",B,SE);<br />
label("$C$",C,SW);<br />
label("$M$",M,SW);<br />
label("$N$",N,S);<br />
label("$J$",J,NE);<br />
label("$K$",K,NE);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ 15<br />
\qquad<br />
\mathrm{(B)}\ \frac{81}{5}<br />
\qquad<br />
\mathrm{(C)}\ \frac{205}{12}<br />
\qquad<br />
\mathrm{(D)}\ \frac{240}{13}<br />
\qquad<br />
\mathrm{(E)}\ 20<br />
</math><br />
<br />
== Solution ==<br />
<br />
The triangle <math>ABC</math> is clearly a right triangle, its area is <math>\frac{5\cdot 12}2 = 30</math>. If we knew the areas of triangles <math>AMJ</math> and <math>BNK</math>, we could subtract them to get the area of the pentagon.<br />
<br />
Draw the height <math>CL</math> from <math>C</math> onto <math>AB</math>. As <math>AB=13</math> and the area is <math>30</math>, we get <math>CL=\frac{60}{13}</math>. The situation is shown in the picture below:<br />
<br />
<asy><br />
unitsize(0.5cm);<br />
defaultpen(0.8);<br />
pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0);<br />
pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) );<br />
pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) );<br />
pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) );<br />
draw( A--B--C--cycle );<br />
draw( M--J );<br />
draw( N--K );<br />
draw( C--L, dashed );<br />
label("$A$",A,NW);<br />
label("$B$",B,SE);<br />
label("$C$",C,SW);<br />
label("$M$",M,SW);<br />
label("$N$",N,S);<br />
label("$J$",J,NE);<br />
label("$K$",K,NE);<br />
label("$L$",L,NE);<br />
</asy><br />
<br />
Now note that the triangles <math>ABC</math>, <math>AMJ</math>, <math>ACL</math>, <math>CBL</math> and <math>NBK</math> all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio <math>k</math>, their areas have ratio <math>k^2</math>. We will use this fact repeatedly.<br />
Below we will use <math>[XYZ]</math> to denote the area of the triangle <math>XYZ</math>.<br />
<br />
We have <math>\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}</math>, hence <math>[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}</math>.<br />
<br />
Also, <math>\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}</math>, hence <math>[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}</math>.<br />
<br />
Now for the smaller triangles:<br />
<br />
We know that <math>\frac{AM}{AC}=\frac 15</math>, hence <math>[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}</math>. <br />
<br />
Similarly, <math>\frac{BN}{BC}=\frac 8{12} = \frac 23</math>, hence <math>[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}</math>.<br />
<br />
Finally, the area of the pentagon is <math>30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}</math>.<br />
<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2004|ab=B|num-b=13|num-a=15}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_13&diff=440632006 AMC 12B Problems/Problem 132012-01-01T23:08:25Z<p>Jhggins: /* Problem */</p>
<hr />
<div>{{empty}}<br />
<br />
== Problem ==<br />
<br />
Rhombus <math> ABCD</math> is similar to rhombus <math> BFDE</math>. The area of rhombus <math> ABCD</math> is 24, and <math> \angle BAD \equal{} 60^\circ</math>. What is the area of rhombus <math> BFDE</math>?<br />
<br />
<asy> defaultpen(linewidth(0.7)+fontsize(11));<br />
pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3));<br />
pair point=(3/2, sqrt(3)/2);<br />
draw(B--C--D--A--B--F--D--E--B);<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$D$", D, dir(point--D));<br />
label("$E$", E, dir(point--E));<br />
label("$F$", F, dir(point--F));<br />
</asy><br />
<br />
<math> \textrm{(A) } 6 \qquad \textrm{(B) } 4\sqrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } 6\sqrt {3}</math><br />
<br />
[[2006 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
== Solution ==<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=12|num-a=14}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_5&diff=438902008 AMC 12A Problems/Problem 52011-12-25T22:44:17Z<p>Jhggins: /* Problem */</p>
<hr />
<div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #5]] and [[2008 AMC 10A Problems/Problem 9|2008 AMC 10A #9]]}}<br />
==Problem==<br />
Suppose that<br />
<cmath>\frac{2x}{3}-\frac{x}{6}</cmath><br />
is an integer. Which of the following statements must be true about <math>x</math>?<br />
<br />
<math>\mathrm{(A)}\ \text{It is negative.}\\\qquad\mathrm{(B)}\ \text{It is even, but not necessarily a multiple of 3.}\\\qquad\mathrm{(C)}\ \text{It is a multiple of 3, but not necessarily even.}\\\qquad\mathrm{(D)}\ \text{It is a multiple of 6, but not necessarily a multiple of 12.}\\\qquad\mathrm{(E)}\ \text{It is a multiple of 12.}</math><br />
<br />
==Solution==<br />
<cmath>\frac{2x}{3}-\frac{x}{6}\quad\Longrightarrow\quad\frac{4x}{6}-\frac{x}{6}\quad\Longrightarrow\quad\frac{3x}{6}\quad\Longrightarrow\quad\frac{x}{2}</cmath><br />
For <math>\frac{x}{2}</math> to be an integer, <math>x</math> must be even, but not necessarily divisible by <math>3</math>. Thus, the answer is <math>\mathrm{(B)}</math>.<br />
<br />
==See also==<br />
{{AMC12 box|year=2008|ab=A|num-b=4|num-a=6}}<br />
{{AMC10 box|year=2008|ab=A|num-b=8|num-a=10}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_5&diff=438892005 AMC 12A Problems/Problem 52011-12-25T22:39:59Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
The average ([[mean]]) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?<br />
<br />
<math><br />
(\mathrm {A}) \ 23 \qquad (\mathrm {B}) \ 24 \qquad (\mathrm {C})\ 25 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 27<br />
</math><br />
<br />
== Solution ==<br />
<math>\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|num-b=4|num-a=6|ab=A}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_5&diff=438882004 AMC 12A Problems/Problem 52011-12-25T22:37:04Z<p>Jhggins: /* Problem */</p>
<hr />
<div>==Problem==<br />
The graph of the line <math>y=mx+b</math> is shown. Which of the following is true?<br />
<br />
[[Image:2004 AMC 12A Problem 5.png]]<br />
<br />
<math>\mathrm {(A)} mb<-1 \qquad \mathrm {(B)} -1<mb<0 \qquad \mathrm {(C)} mb=0 \qquad \mathrm {(D)}</math> <math> 0<mb<1 \qquad \mathrm {(E)} mb>1</math><br />
<br />
==Solution==<br />
It looks like it has a slope of <math>-\dfrac{1}{2}</math> and is shifted <math>\dfrac{4}{5}</math> up.<br />
<br />
<math>\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}</math><br />
<br />
==See also==<br />
{{AMC12 box|year=2004|ab=A|num-b=4|num-a=6}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_4&diff=438872002 AMC 12B Problems/Problem 42011-12-25T22:05:44Z<p>Jhggins: /* Problem */</p>
<hr />
<div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #4]] and [[2002 AMC 10B Problems|2002 AMC 10B #7]]}}<br />
== Problem ==<br />
Let <math>n</math> be a positive [[integer]] such that <math>\frac 12 + \frac 13 + \frac 17 + \frac 1n</math> is an integer. Which of the following statements is '''not ''' true:<br />
<br />
<math>\mathrm{(A)}\ 2\ \text{divides\ }n<br />
\qquad\mathrm{(B)}\ 3\ \text{divides\ }n<br />
\qquad\mathrm{(C)}</math> <math>\ 6\ \text{divides\ }n <br />
\qquad\mathrm{(D)}\ 7\ \text{divides\ }n<br />
\qquad\mathrm{(E)}\ n > 84</math><br />
<br />
== Solution ==<br />
Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>,<br />
<br />
<cmath>0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2</cmath><br />
<br />
From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. Thus the answer is <math>\boxed{\mathrm{(E)}\ n>84}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2002|ab=B|num-b=3|num-a=5}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_4&diff=438862002 AMC 12B Problems/Problem 42011-12-25T22:01:25Z<p>Jhggins: /* Problem */</p>
<hr />
<div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #4]] and [[2002 AMC 10B Problems|2002 AMC 10B #7]]}}<br />
== Problem ==<br />
Let <math>n</math> be a positive [[integer]] such that <math>\frac 12 + \frac 13 + \frac 17 + \frac 1n</math> is an integer. Which of the following statements is '''not ''' true:<br />
<br />
<math>\mathrm{(A)}\ 2\ \text{divides\ }n<br />
\qquad\mathrm{(B)}\ 3\ \text{divides\ }n<br />
\qquad\mathrm{(C)}\ 6\ \text{divides\ }n <br />
\qquad\mathrm{(D)}\ 7\ \text{divides\ }n<br />
\qquad\mathrm{(E)}\ n > 84</math><br />
<br />
== Solution ==<br />
Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>,<br />
<br />
<cmath>0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2</cmath><br />
<br />
From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. Thus the answer is <math>\boxed{\mathrm{(E)}\ n>84}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2002|ab=B|num-b=3|num-a=5}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_15&diff=372182010 AMC 12B Problems/Problem 152011-03-02T23:54:41Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem 15 ==<br />
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?<br />
<br />
<math>\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235</math><br />
<br />
== Solution ==<br />
We have either <math>i^{x}=(1+i)^{y}\neq z</math>, <math>i^{x}=z\neq(1+i)^{y}</math>, or <math>(1+i)^{y}=z\neq i^x</math>.<br />
<br />
For <math>i^{x}=(1+i)^{y}</math>, this only occurs at <math>1</math>. <math>(1+i)^{y}=1</math> has only one solution, <math>i^{x}=1</math> has five solutions between zero and nineteen, and <math>z\neq 1</math> has nineteen integer solutions between zero and nineteen. So for <math>i^{x}=(1+i)^{y}\neq z</math>, we have <math>5\times 1\times 19=95</math> ordered triples.<br />
<br />
For <math>i^{x}=z\neq(1+i)^{y}</math>, again this only occurs at <math>1</math>. <math>(1+i)^{y}\neq 1</math> has nineteen solutions, <math>i^{x}=1</math> has five solutions, and <math>z=1</math> has one solution, so again we have <math>5\times 1\times 19=95</math> ordered triples.<br />
<br />
For <math>(1+i)^{y}=z\neq i^x</math>, this occurs at <math>1</math> and <math>16</math>. <math>(1+i)^{y}=1</math> and <math>z=1</math> both have one solution while <math>i^{x}\neq 1</math> has fifteen solutions. <math>(1+i)^{y}=16</math> and <math>z=16</math> both have one solution while <math>i^{x}\neq 16</math> has twenty solutions. So we have <math>15\times 1\times 1+20\times 1\times 1=35</math> ordered triples.<br />
<br />
In total we have <math>{95+95+35=225}</math> ordered triples <math>\Rightarrow \boxed{D}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_20&diff=365752005 AMC 12B Problems/Problem 202011-02-04T19:20:27Z<p>Jhggins: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>a,b,c,d,e,f,g</math> and <math>h</math> be distinct elements in the set <math>\{-7,-5,-3,-2,2,4,6,13\}.</math><br />
<br />
What is the minimum possible value of <math>(a+b+c+d)^{2}+(e+f+g+h)^{2}?</math><br />
<br />
<math><br />
\mathrm{(A)}\ 30 \qquad<br />
\mathrm{(B)}\ 32 \qquad<br />
\mathrm{(C)}\ 34 \qquad<br />
\mathrm{(D)}\ 40 \qquad<br />
\mathrm{(E)}\ 50<br />
</math><br />
<br />
== Solution ==<br />
The sum of the set is <math>-7-5-3-2+2+4+6+13=8</math>, so if we could have the sum in each set of parenthesis be <math>4</math> then the minimum value would be <math>2(4^2)=32</math>. Considering the set of four terms containing <math>13</math>, this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be <math>13-7-5-3=-2</math>, and with two odd terms then its minimum value is <math>13-7+2-2=6</math>, so we cannot achieve two sums of <math>4</math>. The closest we could have to <math>4</math> and <math>4</math> is <math>3</math> and <math>5</math>, which can be achieved through <math>13-7-5+2</math> and <math>6-3-2+4</math>. So the minimum possible value is <math>3^2+5^2=34\Rightarrow\boxed{C}</math>.<br />
<br />
== See also ==<br />
* [[2005 AMC 12B Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_20&diff=365742005 AMC 12B Problems/Problem 202011-02-04T19:10:46Z<p>Jhggins: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>a,b,c,d,e,f,g</math> and <math>h</math> be distinct elements in the set <math>\{-7,-5,-3,-2,2,4,6,13\}.</math><br />
<br />
What is the minimum possible value of <math>(a+b+c+d)^{2}+(e+f+g+h)^{2}?</math><br />
<br />
<math><br />
\mathrm{(A)}\ 30 \qquad<br />
\mathrm{(B)}\ 32 \qquad<br />
\mathrm{(C)}\ 34 \qquad<br />
\mathrm{(D)}\ 40 \qquad<br />
\mathrm{(E)}\ 50<br />
</math><br />
<br />
== Solution ==<br />
<br />
== See also ==<br />
* [[2005 AMC 12B Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_21&diff=365732005 AMC 12B Problems/Problem 212011-02-04T19:06:37Z<p>Jhggins: </p>
<hr />
<div>== Problem ==<br />
A positive integer <math>n</math> has <math>60</math> divisors and <math>7n</math> has <math>80</math> divisors. What is the greatest integer <math>k</math> such that <math>7^k</math> divides <math>n</math>?<br />
<br />
<math>\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}</math><br />
<br />
== Solution ==<br />
If <math>n</math> has <math>60</math> factors, then <math>n</math> is a product of <math>2\times2\times3\times5</math> powers of (not necessarily distinct) primes. When multiplied by <math>7</math>, the amount of factors of <math>n</math> increased by <math>\frac{80}{60}=\frac{4}{3}</math>, so there are <math>4</math> possible powers of <math>7</math> in the factorization of <math>7n</math>, and <math>3</math> possible powers of <math>7</math> in the factorization of <math>n</math>, which would be <math>7^0</math>, <math>7^1</math>, and <math>7^2</math>. Therefore the highest power of <math>7</math> that could divide <math>n</math> is <math>2\Rightarrow\boxed{C}</math>.<br />
<br />
== See also ==<br />
* [[2005 AMC 12B Problems]]</div>Jhgginshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_21&diff=365722005 AMC 12B Problems/Problem 212011-02-04T19:05:57Z<p>Jhggins: /* Solution */</p>
<hr />
<div>{{empty}}<br />
== Problem ==<br />
A positive integer <math>n</math> has <math>60</math> divisors and <math>7n</math> has <math>80</math> divisors. What is the greatest integer <math>k</math> such that <math>7^k</math> divides <math>n</math>?<br />
<br />
<math>\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}</math><br />
<br />
== Solution ==<br />
If <math>n</math> has <math>60</math> factors, then <math>n</math> is a product of <math>2\times2\times3\times5</math> powers of (not necessarily distinct) primes. When multiplied by <math>7</math>, the amount of factors of <math>n</math> increased by <math>\frac{80}{60}=\frac{4}{3}</math>, so there are <math>4</math> possible powers of <math>7</math> in the factorization of <math>7n</math>, and <math>3</math> possible powers of <math>7</math> in the factorization of <math>n</math>, which would be <math>7^0</math>, <math>7^1</math>, and <math>7^2</math>. Therefore the highest power of <math>7</math> that could divide <math>n</math> is <math>2\Rightarrow\boxed{C}</math>.<br />
<br />
== See also ==<br />
* [[2005 AMC 12B Problems]]</div>Jhggins