https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=NikhilP&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:54:13ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_15&diff=867282008 AIME I Problems/Problem 152017-08-01T02:13:21Z<p>NikhilP: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
A square piece of paper has sides of length <math>100</math>. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance <math>\sqrt{17}</math> from the corner, and they meet on the diagonal at an angle of <math>60^{\circ}</math> (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upped edges, can be written in the form <math>\sqrt[n]{m}</math>, where <math>m</math> and <math>n</math> are positive integers, <math>m<1000</math>, and <math>m</math> is not divisible by the <math>n</math>th power of any prime. Find <math>m+n</math>.<br />
<center><asy>import cse5;<br />
size(200);<br />
pathpen=black;<br />
real s=sqrt(17);<br />
real r=(sqrt(51)+s)/sqrt(2);<br />
D((0,2*s)--(0,0)--(2*s,0));<br />
D((0,s)--r*dir(45)--(s,0));<br />
D((0,0)--r*dir(45));<br />
D((r*dir(45).x,2*s)--r*dir(45)--(2*s,r*dir(45).y));<br />
MP("30^\circ",r*dir(45)-(0.25,1),SW);<br />
MP("30^\circ",r*dir(45)-(1,0.5),SW);<br />
MP("\sqrt{17}",(0,s/2),W);<br />
MP("\sqrt{17}",(s/2,0),S);<br />
MP("\mathrm{cut}",((0,s)+r*dir(45))/2,N);<br />
MP("\mathrm{cut}",((s,0)+r*dir(45))/2,E);<br />
MP("\mathrm{fold}",(r*dir(45).x,s+r/2*dir(45).y),E);<br />
MP("\mathrm{fold}",(s+r/2*dir(45).x,r*dir(45).y));</asy></center><br />
<br />
__TOC__<br />
== Solution ==<br />
<center><asy><br />
import three; import math; import cse5;<br />
size(500);<br />
pathpen=blue;<br />
real r = (51^0.5-17^0.5)/200, h=867^0.25/100;<br />
triple A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0);<br />
triple F=B+(r,-r,h),G=(1,-r,h),H=(1+r,0,h),I=B+(0,0,h);<br />
draw(B--F--H--cycle); draw(B--F--G--cycle);<br />
draw(G--I--H); draw(B--I); draw(A--B--C--D--cycle);<br />
triple Fa=A+(-r,-r, h), Fc=C+(r,r, h), Fd=D+(-r,r, h);<br />
triple Ia = A+(0,0,h), Ic = C+(0,0,h), Id = D+(0,0,h);<br />
draw(Ia--I--Ic); draw(Fa--F--Fc--Fd--cycle);<br />
draw(A--Fa); draw(C--Fc); draw(D--Fd);<br />
</asy></center><br />
=== Solution 1 ===<br />
In the original picture, let <math>P</math> be the corner, and <math>M</math> and <math>N</math> be the two points whose distance is <math>\sqrt{17}</math> from <math>P</math>. Also, let <math>R</math> be the point where the two cuts intersect.<br />
<br />
Using <math>\triangle{MNP}</math> (a 45-45-90 triangle), <math>MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}</math>. <math>\triangle{MNR}</math> is [[equilateral triangle|equilateral]], so <math>MR = NR = \sqrt{34}</math>. (Alternatively, we could find this by the [[Law of Sines]].)<br />
<br />
The length of the perpendicular from <math>P</math> to <math>MN</math> in <math>\triangle{MNP}</math> is <math>\frac{\sqrt{17}}{\sqrt{2}}</math>, and the length of the perpendicular from <math>R</math> to <math>MN</math> in <math>\triangle{MNR}</math> is <math>\frac{\sqrt{51}}{\sqrt{2}}</math>. Adding those two lengths, <math>PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}</math>. (Alternatively, we could have used that <math>\tan 75^{\circ} = \tan (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}</math>.)<br />
<br />
Drop a [[perpendicular]] from <math>R</math> to the side of the square containing <math>M</math> and let the intersection be <math>G</math>.<br />
<br />
<cmath><br />
\begin{align*}PG&=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\<br />
MG=PG-PM&=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}</cmath><br />
<br />
<center><asy>import cse5;<br />
size(200);<br />
pathpen=black;<br />
real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2));<br />
pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0);<br />
D(2*N--P--2*M); D(N--R--M); D(P--R);<br />
D((R.x,2*N.y)--R--(2*M.x,R.y));<br />
MP("30^\circ",R-(0.25,1),SW);<br />
MP("30^\circ",R-(1,0.5),SW);<br />
MP("\sqrt{17}",N/2,W);<br />
MP("\sqrt{17}",M/2,S);<br />
D(N--M,dashed);<br />
D(G--R,dashed);<br />
MP("P",P,SW); MP("N",N,SW); MP("M",M,SW); MP("R",R,NE);<br />
MP("G",G,SW);<br />
</asy></center><br />
<br />
Let <math>ABCD</math> be the smaller square base of the tray and let <math>A'B'C'D'</math> be the larger square, such that <math>AA'</math>, etc, are edges. Let <math>F</math> be the foot of the perpendicular from <math>A</math> to plane <math>A'B'C'D'</math>.<br />
<br />
We know <math>AA'=MR=\sqrt{34}</math> and <math>A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}</math>. Now, use the Pythagorean Theorem on triangle <math>AFA'</math> to find <math>AF</math>:<br />
<br />
<cmath>\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&=34\\AF&=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&=\sqrt[4]{867}\end{align*}</cmath><br />
<br />
The answer is <math>867 + 4 = \boxed{871}</math>.<br />
<br />
=== Solution 2 ===<br />
In the final pyramid, let <math>ABCD</math> be the smaller square and let <math>A'B'C'D'</math> be the larger square such that <math>AA'</math>, etc. are edges.<br />
<br />
It is obvious from the diagram that <math>\angle A'AB = \angle A'AD = 105^\circ</math>.<br />
<br />
Let <math>AB</math> and <math>AD</math> be the positive <math>x</math> and <math>y</math> axes in a 3-d coordinate system such that <math>A'</math> has a positive <math>z</math> coordinate. Let <math>\alpha</math> be the angle made with the positive <math>x</math> axis. Define <math>\beta</math> and <math>\gamma</math> analogously.<br />
<br />
It is easy to see that if <math>P: = (x,y,z)</math>, then <math>x = AA'\cdot \cos\alpha</math>. Furthermore, this means that <math>\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1</math>.<br />
<br />
We have that <math>\alpha = \beta = 105^\circ</math>, so <math>\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt [4]{\frac {3}{4}}</math>.<br />
<br />
It is easy to see from the [[Law of Sines]] that <math>\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}</math>.<br />
<br />
Now, <math>z = AA'\cdot \cos\gamma = \sqrt [4]{34^2\cdot \frac {3}{4}} = \sqrt [4]{867}</math>.<br />
<br />
It follows that the answer is <math>867 + 4 = \boxed{871}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_15&diff=867272008 AIME I Problems/Problem 152017-08-01T02:11:30Z<p>NikhilP: Changed the centering</p>
<hr />
<div>== Problem ==<br />
A square piece of paper has sides of length <math>100</math>. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance <math>\sqrt{17}</math> from the corner, and they meet on the diagonal at an angle of <math>60^{\circ}</math> (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upped edges, can be written in the form <math>\sqrt[n]{m}</math>, where <math>m</math> and <math>n</math> are positive integers, <math>m<1000</math>, and <math>m</math> is not divisible by the <math>n</math>th power of any prime. Find <math>m+n</math>.<br />
<center><asy>import cse5;<br />
size(200);<br />
pathpen=black;<br />
real s=sqrt(17);<br />
real r=(sqrt(51)+s)/sqrt(2);<br />
D((0,2*s)--(0,0)--(2*s,0));<br />
D((0,s)--r*dir(45)--(s,0));<br />
D((0,0)--r*dir(45));<br />
D((r*dir(45).x,2*s)--r*dir(45)--(2*s,r*dir(45).y));<br />
MP("30^\circ",r*dir(45)-(0.25,1),SW);<br />
MP("30^\circ",r*dir(45)-(1,0.5),SW);<br />
MP("\sqrt{17}",(0,s/2),W);<br />
MP("\sqrt{17}",(s/2,0),S);<br />
MP("\mathrm{cut}",((0,s)+r*dir(45))/2,N);<br />
MP("\mathrm{cut}",((s,0)+r*dir(45))/2,E);<br />
MP("\mathrm{fold}",(r*dir(45).x,s+r/2*dir(45).y),E);<br />
MP("\mathrm{fold}",(s+r/2*dir(45).x,r*dir(45).y));</asy></center><br />
<br />
__TOC__<br />
== Solution ==<br />
<center><asy><br />
import three; import math; import cse5;<br />
size(500);<br />
pathpen=blue;<br />
real r = (51^0.5-17^0.5)/200, h=867^0.25/100;<br />
triple A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0);<br />
triple F=B+(r,-r,h),G=(1,-r,h),H=(1+r,0,h),I=B+(0,0,h);<br />
draw(B--F--H--cycle); draw(B--F--G--cycle);<br />
draw(G--I--H); draw(B--I); draw(A--B--C--D--cycle);<br />
triple Fa=A+(-r,-r, h), Fc=C+(r,r, h), Fd=D+(-r,r, h);<br />
triple Ia = A+(0,0,h), Ic = C+(0,0,h), Id = D+(0,0,h);<br />
draw(Ia--I--Ic); draw(Fa--F--Fc--Fd--cycle);<br />
draw(A--Fa); draw(C--Fc); draw(D--Fd);<br />
</asy></center><br />
=== Solution 1 ===<br />
In the original picture, let <math>P</math> be the corner, and <math>M</math> and <math>N</math> be the two points whose distance is <math>\sqrt{17}</math> from <math>P</math>. Also, let <math>R</math> be the point where the two cuts intersect.<br />
<br />
Using <math>\triangle{MNP}</math> (a 45-45-90 triangle), <math>MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}</math>. <math>\triangle{MNR}</math> is [[equilateral triangle|equilateral]], so <math>MR = NR = \sqrt{34}</math>. (Alternatively, we could find this by the [[Law of Sines]].)<br />
<br />
The length of the perpendicular from <math>P</math> to <math>MN</math> in <math>\triangle{MNP}</math> is <math>\frac{\sqrt{17}}{\sqrt{2}}</math>, and the length of the perpendicular from <math>R</math> to <math>MN</math> in <math>\triangle{MNR}</math> is <math>\frac{\sqrt{51}}{\sqrt{2}}</math>. Adding those two lengths, <math>PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}</math>. (Alternatively, we could have used that <math>\tan 75^{\circ} = \tan (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}</math>.)<br />
<br />
Drop a [[perpendicular]] from <math>R</math> to the side of the square containing <math>M</math> and let the intersection be <math>G</math>.<br />
<br />
<cmath><br />
\begin{align*}PG&=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\<br />
MG=PG-PM&=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}</cmath><br />
<br />
<center><asy>import cse5;<br />
size(200);<br />
pathpen=black;<br />
real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2));<br />
pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0);<br />
D(2*N--P--2*M); D(N--R--M); D(P--R);<br />
D((R.x,2*N.y)--R--(2*M.x,R.y));<br />
MP("30^\circ",R-(0.25,1),SW);<br />
MP("30^\circ",R-(1,0.5),SW);<br />
MP("\sqrt{17}",N/2,W);<br />
MP("\sqrt{17}",M/2,S);<br />
D(N--M,dashed);<br />
D(G--R,dashed);<br />
MP("P",P,SW); MP("N",N,SW); MP("M",M,SW); MP("R",R,NE);<br />
MP("G",G,SW);<br />
</asy><center><br />
<br />
Let <math>ABCD</math> be the smaller square base of the tray and let <math>A'B'C'D'</math> be the larger square, such that <math>AA'</math>, etc, are edges. Let <math>F</math> be the foot of the perpendicular from <math>A</math> to plane <math>A'B'C'D'</math>.<br />
<br />
We know <math>AA'=MR=\sqrt{34}</math> and <math>A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}</math>. Now, use the Pythagorean Theorem on triangle <math>AFA'</math> to find <math>AF</math>:<br />
<br />
<cmath>\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&=34\\AF&=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&=\sqrt[4]{867}\end{align*}</cmath><br />
<br />
The answer is <math>867 + 4 = \boxed{871}</math>.<br />
<br />
=== Solution 2 ===<br />
In the final pyramid, let <math>ABCD</math> be the smaller square and let <math>A'B'C'D'</math> be the larger square such that <math>AA'</math>, etc. are edges.<br />
<br />
It is obvious from the diagram that <math>\angle A'AB = \angle A'AD = 105^\circ</math>.<br />
<br />
Let <math>AB</math> and <math>AD</math> be the positive <math>x</math> and <math>y</math> axes in a 3-d coordinate system such that <math>A'</math> has a positive <math>z</math> coordinate. Let <math>\alpha</math> be the angle made with the positive <math>x</math> axis. Define <math>\beta</math> and <math>\gamma</math> analogously.<br />
<br />
It is easy to see that if <math>P: = (x,y,z)</math>, then <math>x = AA'\cdot \cos\alpha</math>. Furthermore, this means that <math>\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1</math>.<br />
<br />
We have that <math>\alpha = \beta = 105^\circ</math>, so <math>\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt [4]{\frac {3}{4}}</math>.<br />
<br />
It is easy to see from the [[Law of Sines]] that <math>\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}</math>.<br />
<br />
Now, <math>z = AA'\cdot \cos\gamma = \sqrt [4]{34^2\cdot \frac {3}{4}} = \sqrt [4]{867}</math>.<br />
<br />
It follows that the answer is <math>867 + 4 = \boxed{871}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=732652002 AMC 10A Problems/Problem 202015-11-27T20:18:24Z<p>NikhilP: </p>
<hr />
<div>==Problem==<br />
Points <math>A,B,C,D,E</math> and <math>F</math> lie, in that order, on <math>\overline{AF}</math>, dividing it into five segments, each of length 1. Point <math>G</math> is not on line <math>AF</math>. Point <math>H</math> lies on <math>\overline{GD}</math>, and point <math>J</math> lies on <math>\overline{GF}</math>. The line segments <math>\overline{HC}, \overline{JE},</math> and <math>\overline{AG}</math> are parallel. Find <math>HC/JE</math>.<br />
<br />
<math>\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2</math><br />
<br />
[asy]<br />
pair A,B,C,D,EE,F,G,H,J;<br />
A = (0,0);<br />
B = (0.2,0);<br />
C = 2*B;<br />
D = 3*B;<br />
EE = 4*B;<br />
F = 5*B;<br />
G = (-0.2,0.8);<br />
H = intersectionpoint(G--D,C -- (C + G));<br />
J = intersectionpoint(G--F,EE--(EE+G));<br />
draw(G--F--A--G--B);<br />
draw(H--C--G--D);<br />
draw(J--EE--G);<br />
label("<math>A</math>",A,SW);<br />
label("<math>B</math>",B,S);<br />
label("<math>C</math>",C,S);<br />
label("<math>D</math>",D,S);<br />
label("<math>E</math>",EE,S);<br />
label("<math>F</math>",F,SE);<br />
label("<math>J</math>",J,NE);<br />
label("<math>G</math>",G,N);<br />
label(scale(0.9)*"<math>H</math>",H,NE,UnFill(0.1mm));<br />
[/asy]<br />
<br />
==Solution==<br />
Solution #1:<br />
Since <math>AG</math> and <math>CH</math> are parallel, triangles <math>GAD</math> and <math>HCD</math> are similar. Hence, <math>CH/AG = CD/AD = 1/3</math>.<br />
<br />
Since <math>AG</math> and <math>JE</math> are parallel, triangles <math>GAF</math> and <math>JEF</math> are similar. Hence, <math>EJ/AG = EF/AF = 1/5</math>. Therefore, <math>CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}</math>. The answer is (D).<br />
<br />
Solution #2: <br />
As <math>\overline{JE}</math> is parallel to <math>\overline{AG}</math>, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, <math>\triangle AGF \sim \triangle EJF</math>; hence <math>\frac {AG}{JE} =5</math>. Similarly, <math>\frac {AG}{HC} = 3</math>. Thus, <math>\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_20&diff=732642002 AMC 10A Problems/Problem 202015-11-27T20:18:08Z<p>NikhilP: </p>
<hr />
<div>==Problem==<br />
Points <math>A,B,C,D,E</math> and <math>F</math> lie, in that order, on <math>\overline{AF}</math>, dividing it into five segments, each of length 1. Point <math>G</math> is not on line <math>AF</math>. Point <math>H</math> lies on <math>\overline{GD}</math>, and point <math>J</math> lies on <math>\overline{GF}</math>. The line segments <math>\overline{HC}, \overline{JE},</math> and <math>\overline{AG}</math> are parallel. Find <math>HC/JE</math>.<br />
<br />
<math>\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2</math><br />
<br />
<math>[asy]<br />
pair A,B,C,D,EE,F,G,H,J;<br />
A = (0,0);<br />
B = (0.2,0);<br />
C = 2*B;<br />
D = 3*B;<br />
EE = 4*B;<br />
F = 5*B;<br />
G = (-0.2,0.8);<br />
H = intersectionpoint(G--D,C -- (C + G));<br />
J = intersectionpoint(G--F,EE--(EE+G));<br />
draw(G--F--A--G--B);<br />
draw(H--C--G--D);<br />
draw(J--EE--G);<br />
label("</math>A<math>",A,SW);<br />
label("</math>B<math>",B,S);<br />
label("</math>C<math>",C,S);<br />
label("</math>D<math>",D,S);<br />
label("</math>E<math>",EE,S);<br />
label("</math>F<math>",F,SE);<br />
label("</math>J<math>",J,NE);<br />
label("</math>G<math>",G,N);<br />
label(scale(0.9)*"</math>H<math>",H,NE,UnFill(0.1mm));<br />
[/asy]</math><br />
<br />
==Solution==<br />
Solution #1:<br />
Since <math>AG</math> and <math>CH</math> are parallel, triangles <math>GAD</math> and <math>HCD</math> are similar. Hence, <math>CH/AG = CD/AD = 1/3</math>.<br />
<br />
Since <math>AG</math> and <math>JE</math> are parallel, triangles <math>GAF</math> and <math>JEF</math> are similar. Hence, <math>EJ/AG = EF/AF = 1/5</math>. Therefore, <math>CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}</math>. The answer is (D).<br />
<br />
Solution #2: <br />
As <math>\overline{JE}</math> is parallel to <math>\overline{AG}</math>, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, <math>\triangle AGF \sim \triangle EJF</math>; hence <math>\frac {AG}{JE} =5</math>. Similarly, <math>\frac {AG}{HC} = 3</math>. Thus, <math>\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=Joining_an_ARML_team&diff=69402Joining an ARML team2015-03-22T21:07:01Z<p>NikhilP: /* WW-P */</p>
<hr />
<div>Team selection for the [[American Regions Mathematics League]] varies from team to team.<br />
<br />
==Arizona==<br />
<br />
The Arizona ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==California==<br />
<br />
===Northern California===<br />
<br />
Practice meetings are held at UC Davis, Mathematical Sciences Building, room 2112, on Thursdays at 7pm. For more information, e-mail Eric Brattain-Morrin at [mailto:eric.brattain@gmail.com eric.brattain@gmail.com] and visit [http://www.math.ucdavis.edu/~exploration/arml/contact.php the team website].<br />
<br />
===San Diego===<br />
<br />
The SD team is entering just its third year of participation, so a permanent process of selecting team members has yet to be decided upon. The team is organized by and practices at the San Diego Math Circle (SDMC), and most of the students on last year's team were regular attendees at SDMC. Also, since the 2007 team contained no seniors, the organizers for the 2008 team are not planning on extending invites to new students unless scores on the AMC exams or the San Diego Math Olympiad (SDMO) are particularly high. <br />
<br />
A student who wishes to attend practice should try to take the SDMO. If a student wishes to speak with one of the coaches for the team, they may do so by contacting AoPS member 'Generating'.<br />
<br />
* [http://www.sdmathcircle.org/Welcome.php San Diego Math Circle website]<br />
<br />
===San Francisco Bay Area===<br />
<br />
The San Francisco Bay Area ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml]. Teams are coordinated by Tim Sanders, AoPS user farmertim. Coaching and practices are run by Moor Xu, AoPS user annoyingpi. One can contact them for further information, although nearly all questions are answered by the link provided above. <br />
In SFBA, anyone is welcome to sign up for an ARML team. The number of teams that will be going is determined by the amount of interest, and it very rare that someone who signs up will be excluded. Only the top three teams are selected by merit, and the other teams are chosen by geographical placement. The three top teams usually all place in the top 20, often even in the top 15 or 10.<br />
<br />
===Southern California===<br />
<br />
The Southern California team is open to residents of the following Southern California counties: Santa Barbara, Ventura, Los Angeles, Orange, Kern, San Bernardino, and Riverside. The organization fields three or four teams (45 or 60 students) and competes at the western ARML site in Las Vegas.<br />
<br />
Practices are held throughout the school year, approximately once a month, on the campus of California State University, Long Beach. Practices are normally held on Saturday afternoons. In addition, there is a Santa Barbara area group that meets and practices in Goleta and becomes part of the Southern California team.<br />
<br />
Team selection is based on all of the following criteria:<br />
<br />
* Attendence at practice sessions.<br />
* Performance on problems at practice sessions.<br />
* Performance at ARML itself in previous years.<br />
* Performance on AMC and AIME, including current and previous years.<br />
* Performance at CSULB Math Day at the Beach, a contest held in March.<br />
<br />
The coach is Dr. Kent Merryfield, a professor at CSULB. His AoPS user name is Kent Merryfield. Please contact him for further information.<br />
<br />
* [http://www.csulb.edu/depts/math/?q=node/20 Math Day at the Beach website.]<br />
* [http://www.csulb.edu/depts/math/?q=node/32 SoCal ARML website.]<br />
* [[Southern California ARML | SoCal ARML wiki page.]]<br />
<br />
== Connecticut ==<br />
Connecticut team selection is based on performance in math leagues across the state, from which a few upperclassmen can automatically qualify, and a runoff which is held in early March. To participate in the runoff, either have the math team coach at your school contact the league director in February to get the date or, if you don't participate in a league, contact Daniel Bochicchio, the current coach, at [mailto:dbochicchio@eosmith.org dbochicchio@eosmith.org]. Connecticut sends three teams.<br />
<br />
== Florida ==<br />
Florida ARML sends three teams to ARML each year. The selection criteria for the Florida ARML team takes into consideration several factors:<br />
<br />
*[[AMC]] and [[AIME]] performance<br />
*Past [[AMC]], [[AIME]], and [[USAMO]] scores<br />
*Past [[ARML]] performance<br />
*FAMAT-designated competitions<br />
*An annual statewide tryout test <br />
<br />
Florida ARML is organized by the Florida Student Association of Mathematics. Email [mailto:eliross2@aol.com eliross2@aol.com] for more information.<br />
<br />
== Georgia ==<br />
<br />
[[Georgia ARML]] has sent at least two teams to ARML each year since 1991. Beginning in 2008, we took advantage of the opening of the new Southern Site to add a third team (the "C" team). In 2011, we added a fourth team (the "D" team). In the past, the A team competed in the A division and the B team in the B division, but recently, the A team, the B team and the C team compete in the A division. We also field a few alternates as well. Students interested in participation should do well in [[Georgia mathematics competitions | local tournaments]] and the [[AMC]] and [[AIME]]. In addition to schools invited to the annual [[GCTM State Math Tournament | varsity state tournament]], the Georgia ARML coaches invite other individuals that are under serious consideration for the ARML team. The coaches select the team members during the state tournament based on [[AMC | USAMO index]], performance in local tournaments, and score at the state tournament. It is vital that any student under consideration has what we call "verified ability". This can be described as having good results from more than one contest or tournament. "Good results" can be described as placing higher than someone who was on the ARML team last year -- that is, doing better than someone who we know is good. <br />
<br />
The coaches attempt to select the best 30 students in the state (regardless of age or grade) to comprise the A and B teams -- those selected usually consist of USAMO qualifiers, the top 12 at the state tournament, and team veterans. (Of course, these groups are represented by overlapping Venn diagrams.) It is rare that a first-time senior is selected, although it does happen. Then the coaches select the best 15 young students (10th grade and below) to comprise a C team -- a team specifically comprised of up-and-comers who the coaches hope will be A and B team material in the future. Finally, the coaches allow anyone not selected, but who is interested in competing at ARML, to form a team of "walk-ons" which becomes the D team. It is rare for a student from the walk-ons to be so good that they move up to the A or B teams, but it has happened.<br />
<br />
The team usually practices on Sundays from the state tournament until the trip to ARML. The specific compositions of the A, B, C, and D teams are not usually determined until immediately before ARML. A member's team placement (on A, B, C, or D) depends on the person's performance against other team members in practice and the coaches's discretion.<br />
<br />
== Maine ==<br />
<br />
The two Maine ARML Teams consist of approximately the top 30 scorers on 5 [[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42."<br />
<br />
* [http://www.maml.net MAML Website]<br />
<br />
== Montgomery (Montgomery County, MD) ==<br />
<br />
Montgomery County typically sends four teams of high-schoolers and one team of middle-schoolers to ARML, with Montgomery A, B and C competing in division A and the D and Junior teams competing in division B. <br />
<br />
Top scorers from the Montgomery County high school math league are invited to ARML practices after the regular season, which typically ends around February. Top scorers from the middle school math league are also invited. However, interested students who did not participate in the league or weren't invited are still welcome to join and should contact Eric Walstein at Montgomery Blair High School.<br />
<br />
Practices are usually held at Montgomery Blair High School on Thursdays from 6:00pm-9:30pm. Team selection is done by individual scores at practice and at the discretion of the coaches.<br />
<br />
==Minnesota==<br />
<br />
Minnesota sends two teams to ARML each year, with the Gold and Maroon teams usually competing in divisions A and B, respectively.<br />
<br />
Roughly 35 students are invited to ARML practices, which take place on three consecutive Saturdays in May. There is no practice Memorial Day weekend. Invitations to the ARML team are extended to the top 10 state scorers on the AMC12, the top 10 regular-season scorers on the Minnesota High School Math League, and the top 10 scorers on the Invitational Event at the statewide math league tournament (held in March).<br />
<br />
Since these lists tend to overlap quite a bit, invitations are usually given to students "further down" these lists until enough invites have been given to fill two 15-person teams. <br />
<br />
In addition, an extra 5 or so younger students (typically in grades 8 through 10) are invited to be ARML "students in training". They may or may not go to ARML, but often serve as alternates (in the event that other students cannot attend). The expectation is that a student in training will learn from the practices, and the following year will be on one of the two teams. Since the creation of the AMC8 and AMC10 exams, the top scorers from these exams have typically been invited to ARML practices, either as team members or students in training.<br />
<br />
The selection of the Gold and Maroon teams is determined by students' performance at practices, and is not announced until the night before the competition.<br />
<br />
Usually at least one student in training is invited to go to ARML. This is to prevent a last-minute no-show (due to illness or emergency, for example) from crippling one of the teams.<br />
<br />
Invitations to participate on the MN team are usually sent out shortly after the MN State High School Math League statewide tournament in March. <br />
<br />
* [http://www.macalester.edu/mathleague/index.htm Minnesota State High School Math League site]<br />
<br />
==Missouri==<br />
<br />
Missouri sends two teams, Red and Blue, and a few alternates for a total of 36-37 students to the Iowa site ARML competition.<br />
<br />
Open practices are held usually once a month during the school year in two places, Springfield and St. Louis. Attendance is encouraged but not required. There are usually one or two "all day practices" a year at Missouri University of Science and Technology.<br />
<br />
Applications are based on AMC/AIME scores, GPML scores, other math competitions, and past ARML experience. Most if not all applicants are selected for the team.<br />
<br />
Since ARML is NOT a strictly a “state” team competition and because neither Kansas nor southern Illinois have ARML teams, ARML has given the “Missouri” team permission to include students from Kansas and from western Illinois (eg Edwardsville, etc.) on the MO ARML team. (eg. the 2008 team fielded 6 from KS, and 1 from IL)<br />
<br />
[http://math.missouristate.edu/MissouriARML.htm MO ARML homepage]<br />
<br />
==Nevada==<br />
<br />
The Nevada ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New Jersey==<br />
New Jersey sends three different ARML teams; the four being Central Jersey (which typically sends an A team and a B team), BCA/AAST, and WW-P. <br />
<br />
Also, many NJ students are on the Lehigh Valley teams.<br />
<br />
=== BCA/AAST ===<br />
AAST (a school in Bergen County Academies) sent 6 teams in 2012, all named after Greek Gods. Although there is a testing process to determine who's on which team, anyone in the AAST Math Team is permitted to participate in ARML.<br />
<br />
=== Central Jersey ===<br />
Qualification for the Central Jersey ARML team requires that<br />
*the student has participated (and chosen as a representative for the school) in at least 4 of the Central Jersey Math League meets,<br />
*and the average score meets a certain minimum which varies per year (this year, it is 3.0).<br />
<br />
Afterwards, interested students have to attend training sessions which take place at Highland Park High School or Hillsborough.<br />
<br />
More information can be found at [http://uzza.us/cjml/arml.html CJML website].<br />
<br />
=== WW-P ===<br />
WW-P, short for West Windsor-Plainsboro, is a team founded in 2012. Currently, it consists of students only from the WW-P school district, but students in close areas may be welcome as well. The selection process is a series of individual tests, and other experience is taken into consideration if needed. Two teams were sent in 2013; the A1 team ranked 3rd nationally. Also they got 1st nationally in the B-Division two times in a row. More teams may be formed if more students join the program.<br />
<br />
==New Mexico==<br />
<br />
The New Mexico ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New York City==<br />
For information about the New York City Math Team, please visit [http://www.nycmathteam.com the NYC Math Team homepage].<br />
<br />
==North Carolina==<br />
<br />
North Carolina sends two 15-person ARML teams to compete at Penn State. The 2006 NC "A" team placed 1st in Division A. Invitations to spring team practice sessions are extended based on performances at the [http://courses.ncssm.edu/goebel/statecon/state.htm NC State Math Contest], AMC, AIME, Duke Math Meet and other math competitions. The top 15 or so finishers at the NC State Math Contest (Comprehensive) and all USAMO qualifiers are among those typically invited to practice sessions. Performance at these practice sessions and math contests determine team assignments. <br />
<br />
[http://courses.ncssm.edu/goebel/statecon/Inform/sites.htm NC State Math Contest qualifying events] are held throughout the state in February and March. Archie Benton, John Noland and several others coach and prepare the team with e-mail problems and the spring practice sessions held at the NC School of Science and Mathematics in Durham.<br />
<br />
==Ohio==<br />
<br />
Invitation is based on OCTM (a state-wide competition) and AMC scores. Also, at the OHMIO (second level of OCTM) the offer to join is extended to anyone who is interested. Team placement is based on a combination of OCTM scores, AMC scores, and how well the person does in practices. Ohio normally sends two teams, but is sending three this year because enough students were interested. Also, starting this year, the Ohio A team is competing in division A. The other two teams are competing in division B.<br />
<br />
The first practice is Sunday, April 22, and the second is Saturday, May 19.<br />
<br />
==Pennsylvania==<br />
===Pittsburgh===<br />
Visit the [http://www.westernpaarml.org/ Western PA ARML website] and feel free to contact a coach or member to join the team.<br />
===Lehigh Valley===<br />
For information about the Lehigh Valley ARML teams, please visit [http://www.lehigh.edu/~dmd1/arml.html the Lehigh Valley ARML homepage] and [http://www.lehigh.edu/~dmd1/logistics.html the Lehigh Valley ARML logistics page.].<br />
<br />
==South Carolina==<br />
<br />
To join the SC All-State Team, one must take a preliminary exam administered through their school. For more information, please contact [mailto:coach@scall-statemathteam.com coach@scall-statemathteam.com].<br />
<br />
The preliminary exam is composed of 25 questions (non multiple choice), and is usually composed of easy to mid range AMC-12 level questions. From this exam, approximately 50-60 (in 2006 it was 49) of the top scorers from the state are selected into the South Carolina All State Mathematics Team. The qualifying floor this year was 11 out of the 25 questions. After an individual is accepted into the SC All State Team, he or she is invited to one or two ARML practices which are usually composed of individual tests, team tests, and a power round test.<br />
<br />
* [[South Carolina ARML | SC ARML wiki page]]<br />
<br />
==Texas==<br />
<br />
The coach, Sam Baethge chooses mathletes around Texas (or schools in Texas) who have done well in various math contests. The keys are the AMCs, and Mathcounts. TXML also helps, as he runs the competition. The TML, or Texas Math League takes the team to ARML in Iowa. Three teams of 15 students each go. There is a gold, silver, and unofficial team.<br />
<br />
==West Virginia==<br />
<br />
The West Virginia team is selected using the top 15 winners in West Virginia State Math Field Day. Winners 16-30 are used as potential alternates for the team. West Virginia State Math Field Day uses a similar format as the ARML, having an Individual Exam, Individual Short Answer Section, Team Questions, a Team Power Question, and 2 sets of relays of 5 each (there are 10 members in each team).<br />
<br />
==See Other==<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=820893#820893 Newer discussion on AoPS message boards.]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40434 Older discussion on AoPS message boards.]<br />
* [[ARML]]<br />
[[Category:ARML]]</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_2&diff=689802011 AIME I Problems/Problem 22015-03-15T21:30:06Z<p>NikhilP: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=12</math> and <math>BC=10</math>. Points <math>E</math> and <math>F</math> lie inside rectangle <math>ABCD</math> so that <math>BE=9</math>,<math>DF=8</math>,<math>\overline{BE}||\overline{DF}</math>,<math>\overline{EF}||\overline{AB}</math>, and line <math>BE</math> intersects segment <math>\overline{AD}</math>. The length <math>EF</math> can be expressed in the form <math>m\sqrt{n}-p</math>, where <math>m</math>,<math>n</math>, and <math>p</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.<br />
<br />
<br />
== Solution 1 ==<br />
<!--<br />
<asy><br />
unitsize(0.5cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B=(12,0), C=(12,10), D=(0,10);<br />
draw(A--B--C--D--cycle);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
<br />
pair E=(3,5), F=(9,5);<br />
dot("$E$",E,N);<br />
dot("$F$",F,N);<br />
pair G = extension(A, D, E, F);<br />
pair H = extension(B, C, E, F);<br />
draw(G--H);<br />
dot("$G$",G,W);<br />
dot("$H$", H,E);<br />
</asy><br />
--><br />
<br />
Let us call the point where <math>\overline{EF}</math> intersects <math>\overline{AD}</math> point <math>G</math>, and the point where <math>\overline{EF}</math> intersects <math>\overline{BC}</math> point <math>H</math>. Since angles <math>FHB</math> and <math>EGA</math> are both right angles, and angles <math>BEF</math> and <math>DFE</math> are congruent due to parallelism, right triangles <math>BHE</math> and <math>DGF</math> are similar. This implies that <math>\frac{BH}{GD} = \frac{9}{8}</math>. Since <math>BC=10</math>, <math>BH+GD=BH+HC=BC=10</math>. (<math>HC</math> is the same as <math>GD</math> because they are opposite sides of a rectangle.) Now, we have a system:<br />
<br />
<math>\frac{BH}{GD}=\frac{9}8</math><br />
<br />
<math>BH+GD=10</math><br />
<br />
Solving this system (easiest by substitution), we get that:<br />
<br />
<math>BH=\frac{90}{17}</math><br />
<br />
<math>GD=\frac{80}{17}</math><br />
<br />
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:<br />
<br />
<math>\sqrt{9^2-(\frac{90}{17})^2}</math> and <math>\sqrt{8^2-(\frac{80}{17})^2} </math><br />
<br />
Notice that adding these two sides would give us twelve plus the overlap <math>EF</math>. This means that:<br />
<br />
<math>EF= \sqrt{9^2-(\frac{90}{17})^2}+\sqrt{8^2-(\frac{80}{17})^2}-12=3\sqrt{21}-12</math><br />
<br />
Since <math>21</math> isn't divisible by any perfect square, our answer is:<br />
<br />
<math>3+21+12=\boxed{36}</math><br />
<br />
<br />
== Solution 2 ==<br />
<br />
Extend lines <math>BE</math> and <math>CD</math> to meet at point <math>G</math>.<br />
Draw the altitude <math>GH</math> from point <math>G</math> to line <math>BA</math> extended.<br />
<br />
<math>GE=CF=8</math><br />
<math>GB=17</math><br />
<br />
In right <math>\bigtriangleup GHB</math>, <math>GH=10</math>, <math>GB=17</math>, thus by Pythagoras Theorem we have:<br />
<math>HB=\sqrt{17^2-10^2}=3\sqrt{21}</math><br />
<br />
<math>HA=EF=3\sqrt{21}-12</math><br />
<br />
Thus our answer is:<br />
<math>3+21+12=\boxed{36}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_2&diff=689792011 AIME I Problems/Problem 22015-03-15T21:28:19Z<p>NikhilP: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=12</math> and <math>BC=10</math>. Points <math>E</math> and <math>F</math> lie inside rectangle <math>ABCD</math> so that <math>BE=9</math>,<math>DF=8</math>,<math>\overline{BE}||\overline{DF}</math>,<math>\overline{EF}||\overline{AB}</math>, and line <math>BE</math> intersects segment <math>\overline{AD}</math>. The length <math>EF</math> can be expressed in the form <math>m\sqrt{n}-p</math>, where <math>m</math>,<math>n</math>, and <math>p</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.<br />
<br />
<br />
== Solution 1 ==<br />
<!--<br />
<asy><br />
unitsize(0.5cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B=(12,0), C=(12,10), D=(0,10);<br />
draw(A--B--C--D--cycle);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
<br />
pair E=(3,5), F=(9,5);<br />
dot("$E$",E,N);<br />
dot("$F$",F,N);<br />
pair G = extension(A, D, E, F);<br />
pair H = extension(B, C, E, F);<br />
draw(G--H);<br />
dot("$G$",G,W);<br />
dot("$H$", H,E);<br />
</asy><br />
--><br />
<br />
Let us call the point where <math>\overline{EF}</math> intersects <math>\overline{AD}</math> point <math>G</math>, and the point where <math>\overline{EF}</math> intersects <math>\overline{BC}</math> point <math>H</math>. Since angles <math>FHB</math> and <math>EGA</math> are both right angles, and angles <math>BEF</math> and <math>DFE</math> are congruent due to parallelism, right triangles <math>BHE</math> and <math>DGF</math> are similar. This implies that <math>\frac{BH}{GD} = \frac{9}{8}</math>. Since <math>BC=10</math>, <math>BH+GD=BH+HC=BC=10</math>. (<math>HC</math> is the same as <math>GD</math> because they are opposite sides of a rectangle.) Now, we have a system:<br />
<br />
<math>\frac{BH}{GD}=\frac{9}8</math><br />
<br />
<math>BH+GD=10</math><br />
<br />
Solving this system (easiest by substitution), we get that:<br />
<br />
<math>BH=\frac{90}{17}</math><br />
<br />
<math>GD=\frac{80}{17}</math><br />
<br />
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:<br />
<br />
<math>\sqrt{9^2-(\frac{90}{17})^2}</math> and <math>\sqrt{8^2-(\frac{80}{17})^2} </math><br />
<br />
Notice that adding these two sides would give us twelve plus the overlap <math>EF</math>. This means that:<br />
<br />
<math>EF= \sqrt{9^2-(\frac{90}{17})^2}+\sqrt{8^2-(\frac{80}{17})^2}-12=3\sqrt{21}-12</math><br />
<br />
Since <math>21</math> isn't divisible by any perfect square, our answer is:<br />
<br />
<math>3+21+12=\boxed{36}</math><br />
<br />
<br />
== Solution 2 ==<br />
<br />
Extend lines <math>BE</math> and <math>CD</math> to meet at point <math>G</math>.<br />
Draw the altitude <math>GH</math> from point <math>G</math> to line <math>BA</math> extended.<br />
<br />
<math>GE=CF=8</math><br />
<math>GB=17</math><br />
<br />
In right <math>\bigtriangleup GHB</math>, <math>GH=10</math>, <math>GB=17</math>, thus by Pythagoras Theorem we have:<br />
<math>HB=\sqrt{17^2-10^2}=3\sqrt{21}</math><br />
<br />
<math>EF=GD=3\sqrt{21}-12</math><br />
<br />
Thus our answer is:<br />
<math>3+21+12=\boxed{36}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_16&diff=673932015 AMC 10A Problems/Problem 162015-02-04T23:11:58Z<p>NikhilP: </p>
<hr />
<div>==Problem==<br />
<br />
If <math>y+4 = (x-2)^2, x+4 = (y-2)^2</math>, and <math>x \neq y</math>, what is the value of <math>x^2+y^2</math>?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30} </math><br />
<br />
==Solution==<br />
Our equations simplify to (after subtracting 4 from both sides):<br />
<cmath>y = x^2 - 4x,</cmath><br />
<cmath>x = y^2 - 4y.</cmath><br />
Subtract the equations to obtain <math>y - x = x^2 - y^2 - 4x + 4y</math>, so <math>x^2 - y^2 = 3x - 3y</math>. This factors as <math>(x - y)(x + y) = 3(x - y)</math>, and so because <math>x \neq y</math>, we have <math>x + y = 3</math>.<br />
<br />
Add the equations to yield <math>x + y = x^2 + y^2 - 4(x + y)</math>. Hence, <math>x^2 + y^2 = 5(x + y) = 15</math>, so our answer is <math>\boxed{\textbf{(B)}}</math>.</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_16&diff=673862015 AMC 10A Problems/Problem 162015-02-04T23:01:33Z<p>NikhilP: </p>
<hr />
<div>==Problem==<br />
<br />
If <math>y+4 = (x-2)^2, x+4 = (y-2)^2</math>, and <math>x \neq y</math>, what is the value of <math>x^2+y^2</math>?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30} </math><br />
<br />
==Solution==<br />
First simplify the equations <br />
<br />
<math>y+4=(x-2)^2</math><br />
<br />
<math>y+4=x^2-4x+4</math><br />
<br />
<math>y=x^2-4x</math> and the the other equation will become <math>x=y^2-4y</math><br />
<br />
Substitute <math>y</math> into <math>x=y^2-4y</math> to get<br />
<br />
<math>x=(x^2-4x)^2-4(x^2-4x)</math><br />
<br />
<math>x=(x^2-4x)(x^2-4x-4)</math><br />
<br />
<math>x=x(x-4)(x^2-4x-4)</math><br />
<br />
<math>1=x^3-8x^2+12x+16</math><br />
<br />
<math>0=x^3-8x^2+12x+15</math></div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_15&diff=673792015 AMC 10A Problems/Problem 152015-02-04T22:46:20Z<p>NikhilP: </p>
<hr />
<div>==Problem==<br />
<br />
Consider the set of all fractions <math>\frac{x}{y}</math>, where <math>x</math> and <math>y</math> are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by <math>1</math>, the value of the fraction is increased by <math>10\%</math>?<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}</math><br />
<br />
==Solution==<br />
<br />
You can create the equation <br />
<math>\frac{x+1}{y+1}=(1.1)(\frac{x}{y})</math><br />
<br />
<math>\frac{x+1}{y+1}=\frac{1.1x}{y}</math><br />
<br />
<math>(x+1)(y)=(1.1x)(y+1)</math><br />
<br />
<math>xy+y=1.1xy+1.1x</math><br />
<br />
<math>y=.1xy+1.1x</math><br />
<br />
<math>10y=xy+11x</math></div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_15&diff=673772015 AMC 10A Problems/Problem 152015-02-04T22:42:42Z<p>NikhilP: </p>
<hr />
<div>==Problem==<br />
<br />
Consider the set of all fractions <math>\frac{x}{y}</math>, where <math>x</math> and <math>y</math> are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by <math>1</math>, the value of the fraction is increased by <math>10\%</math>?<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}</math><br />
<br />
==Solution==<br />
<br />
You can create the equation <math>\frac{x+1}{y+1}=(1.1)(\frac{x}{y})</math></div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_15&diff=673762015 AMC 10A Problems/Problem 152015-02-04T22:38:43Z<p>NikhilP: Created page with "You can create the equation <math>\frac{x+1}{y+1}=(1.1)(\frac{x}{y})</math>"</p>
<hr />
<div>You can create the equation <math>\frac{x+1}{y+1}=(1.1)(\frac{x}{y})</math></div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_24&diff=659342009 AMC 12A Problems/Problem 242014-11-17T01:51:37Z<p>NikhilP: i deleted a unessecary parenthesis</p>
<hr />
<div>== Problem ==<br />
The ''tower function of twos'' is defined recursively as follows: <math>T(1) = 2</math> and <math>T(n + 1) = 2^{T(n)}</math> for <math>n\ge1</math>. Let <math>A = (T(2009))^{T(2009)}</math> and <math>B = (T(2009))^A</math>. What is the largest integer <math>k</math> such that<br />
<br />
<center><cmath>\underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}</cmath></center><br />
<br />
is defined?<br />
<br />
<math>\textbf{(A)}\ 2009\qquad \textbf{(B)}\ 2010\qquad \textbf{(C)}\ 2011\qquad \textbf{(D)}\ 2012\qquad \textbf{(E)}\ 2013</math> <br />
<br />
== Solution ==<br />
We just look at the last three logarithms for the moment, and use the fact that <math>\log_2 T(k) = T(k - 1)</math>. We wish to find:<br />
<center><cmath><br />
\begin{align*}<br />
& \log_2\left(\log_2\left(\log_2 \left({T(2009)^{\left({T(2009)}}^{T(2009)}\right)}\right)\right)\right) \\<br />
&= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009))) \\<br />
&= \log_2(T(2009)\log_2(T(2009)T(2008))) \\<br />
&= \log_2(T(2009)(T(2008) + T(2007)))</cmath></center><br />
<br />
Now we realize that <math>T(n - 1)</math> is much smaller than <math>T(n)</math>. So we approximate this, remembering we have rounded down, as:<br />
<br />
<center><cmath>\log_2(T(2009)) = T(2008)</cmath></center><br />
We have used <math>3</math> logarithms so far. Applying <math>2007</math> more to the left of our expression, we get <math>T(1) = 2</math>. Then we can apply the logarithm <math>2</math> more times, until we get to <math>0</math>. So our answer is approximately <math>3 + 2007 + 2 = 2012</math>. But we rounded down, so that means that after <math>2012</math> logarithms we get a number slightly greater than <math>0</math>, so we can apply logarithms one more time. We can be sure it is small enough so that the logarithm can only be applied <math>1</math> more time since <math>2012 + 1 = 2013</math> is the largest answer choice. So the answer is <math>\mathbf{(E)}</math>.<br />
<br />
== Alternative Solution ==<br />
Let <math>L_k(x)=\log_2\log_2...\log_2(x)</math> where there are <math>k</math> <math>\log_2</math>'s. <math>L_k(B)</math> is defined iff <math>L_{k-1}(B) > 0</math> iff <math>L_{k-2}(B) > 1</math>. Note <math>\log_2 T(k) = T(k - 1)</math>, so <math>L_{k-2}(T(k-2))=1</math>. Thus, we seek the largest <math>k</math> such that <math>B > T(k-2)</math>. Now note that<br />
<br />
<math>T(2009)^{T(2009)^{T(2009)}} > 2^{2^{T(2009)}} = T(2011)</math><br />
<br />
so <math>k=2013</math> satisfies the inequality. Since it is the largest choice, it is the answer.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>NikhilPhttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_8&diff=622742012 AIME I Problems/Problem 82014-06-14T19:16:47Z<p>NikhilP: </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(x)=\frac{h^2}{4},</math> and <math>h=\frac{x+1}{2},</math> then <math>A=\frac{(x+1)^2}{16}</math>, and the volume <math>[DNCMQB]</math> is <math>\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.</math> Thus the volume of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math><br />
<br />
== Alternative Solution : think inside the box like a total nerd==<br />
<br />
If you memorized the formula for a frustum, then this problem is very trivial.<br />
<br />
The formula for a frustum is:<br />
<br />
<math>\frac{h_2b_2 -h_1b_1}3</math> where <math>b_i</math> is the area of the base and <math>h_i</math> is the height from the chopped of apex to the base.<br />
<br />
We can easily see that from symmetry, the area of the smaller front base is <math>\frac{1}{16}</math> and the area of the larger back base is <math>\frac{1}4</math><br />
<br />
Now to find the height of the apex.<br />
<br />
Extend the <math>DM</math> and (call the intersection of the plane with <math>FB</math> G) <math>NG</math> to meet at <math>x</math>. Now from similar triangles <math>XMG</math> and <math>XDN</math> we can easily find the total height of the triangle <math>XDN</math> to be <math>2</math><br />
<br />
Now straight from our formula, the area is <math>\frac{7}{48}</math> Thus the answer is:<br />
<br />
<math>1-Area \Longrightarrow \boxed{089}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}<br />
<br />
{{MAA Notice}}</div>NikhilP