https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ramanconjecture&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T22:49:46ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_25&diff=1019752019 AMC 10A Problems/Problem 252019-02-13T03:08:14Z<p>Ramanconjecture: /* Solution */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #25]] and [[2019 AMC 12A Problems|2019 AMC 12A #24]]}}<br />
<br />
==Problem==<br />
<br />
For how many integers <math>n</math> between <math>1</math> and <math>50</math>, inclusive, is <cmath>\frac{(n^2-1)!}{(n!)^n}</cmath> an integer? (Recall that <math>0! = 1</math>.)<br />
<br />
<math>\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35</math><br />
<br />
==Solution==<br />
The main insight is that <br />
<br />
<cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath> <br />
<br />
is always an integer. This is true because it is precisely the number of ways to split up <math>n^2</math> objects into <math>n</math> unordered groups of size <math>n</math>, which has to be an integer. Thus, <br />
<br />
<cmath>\frac{(n^2)!}{(n!)^{n+1}}=\frac{n^2}{n!}\cdot\frac{(n^2-1)!}{(n!)^n}</cmath> <br />
<br />
is not an integer if <math>n^2 \nmid n!</math>. This happens only for <math>n=4</math> or <math>n</math> prime(by Wilson's Theorem). Thus, there are <math>16</math> integers that do not work, which means the answer is <math>\boxed{\mathbf{(D)}\ 34}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2019|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Ramanconjecturehttps://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_1&diff=930312010 USAMO Problems/Problem 12018-03-08T17:16:53Z<p>Ramanconjecture: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter<br />
<math>AB</math>. Denote by <math>P, Q, R, S</math> the feet of the perpendiculars from <math>Y</math> onto<br />
lines <math>AX, BX, AZ, BZ</math>, respectively. Prove that the acute angle<br />
formed by lines <math>PQ</math> and <math>RS</math> is half the size of <math>\angle XOZ</math>, where<br />
<math>O</math> is the midpoint of segment <math>AB</math>.<br />
<br />
==Solution==<br />
Let <math>\alpha = \angle BAZ</math>, <math>\beta = \angle ABX</math>.<br />
Since <math>XY</math> is a chord of the circle with diameter <math>AB</math>,<br />
<math>\angle XAY = \angle XBY = \gamma</math>. From the chord <math>YZ</math>,<br />
we conclude <math>\angle YAZ = \angle YBZ = \delta</math>.<br />
<center><br />
<asy><br />
import olympiad;<br />
<br />
// Scale<br />
unitsize(1inch);<br />
real r = 1.75;<br />
<br />
// Semi-circle: centre O, radius r, diameter A--B.<br />
pair O = (0,0); dot(O); label("$O$", O, plain.S);<br />
pair A = r * plain.W; dot(A); label("$A$", A, unit(A));<br />
pair B = r * plain.E; dot(B); label("$B$", B, unit(B));<br />
draw(arc(O, r, 0, 180)--cycle);<br />
<br />
// points X, Y, Z<br />
real alpha = 22.5;<br />
real beta = 15;<br />
real delta = 30;<br />
pair X = r * dir(180 - 2*beta); dot(X); label("$X$", X, unit(X));<br />
pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y));<br />
pair Z = r * dir(2*alpha); dot(Z); label("$Z$", Z, unit(Z));<br />
<br />
// Feet of perpendiculars from Y<br />
pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P);<br />
pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q);<br />
pair R = foot(Y, B, Z); dot(R); label("$S$", R, unit(R-Y)); dot(R);<br />
pair S = foot(Y, A, Z); dot(S); label("$R$", S, unit(B-S)); dot(S);<br />
pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T);<br />
<br />
// Segments<br />
draw(B--X); draw(B--Y); draw(B--R);<br />
draw(A--Z); draw(A--Y); draw(A--P);<br />
draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S);<br />
draw(R--T); draw(P--T);<br />
<br />
// Right angles<br />
draw(rightanglemark(A, X, B, 3));<br />
draw(rightanglemark(A, Y, B, 3));<br />
draw(rightanglemark(A, Z, B, 3));<br />
draw(rightanglemark(A, P, Y, 3));<br />
draw(rightanglemark(Y, R, B, 3));<br />
draw(rightanglemark(Y, S, A, 3));<br />
draw(rightanglemark(B, Q, Y, 3));<br />
<br />
// Acute angles<br />
import markers;<br />
void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0)<br />
{<br />
string sl = "$\scriptstyle{" + l + "}$";<br />
marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker;<br />
markangle(Label(sl), radius=r, n=n, A, B, C, m);<br />
}<br />
langle(B, A, Z, "\alpha" );<br />
langle(X, B, A, "\beta", n=2);<br />
langle(Y, A, X, "\gamma", nm=1);<br />
langle(Y, B, X, "\gamma", nm=1);<br />
langle(Z, A, Y, "\delta", nm=2);<br />
langle(Z, B, Y, "\delta", nm=2);<br />
langle(R, S, Y, "\alpha+\delta", r=23);<br />
langle(Y, Q, P, "\beta+\gamma", r=23);<br />
langle(R, T, P, "\chi", r=15);<br />
</asy><br />
</center><br />
<br />
Triangles <math>BQY</math> and <math>APY</math> are both right-triangles, and share the<br />
angle <math>\gamma</math>, therefore they are similar, and so the ratio <math>PY :<br />
YQ = AY : YB</math>. Now by [[Thales' theorem]] the angles <math>\angle AXB =<br />
\angle AYB = \angle AZB</math> are all right-angles. Also, <math>\angle PYQ</math>,<br />
being the fourth angle in a quadrilateral with 3 right-angles is<br />
again a right-angle. Therefore <math>\triangle PYQ \sim \triangle AYB</math> and<br />
<math>\angle YQP = \angle YBA = \gamma + \beta</math>.<br />
Similarly, <math>RY : YS = AY : YB</math>, and so <math>\angle YRS = \angle YAB = \alpha + \delta</math>.<br />
<br />
Now <math>RY</math> is perpendicular to <math>AZ</math> so the direction <math>RY</math> is <math>\alpha</math> counterclockwise from the vertical, and since <math>\angle YRS = \alpha + \delta</math> we see that <math>SR</math> is <math>\delta</math> clockwise from the vertical. (Draw an actual vertical line segment if necessary.)<br />
<br />
Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> counterclockwise from the vertical.<br />
<br />
Therefore the lines <math>PQ</math> and <math>RS</math> intersect at an angle <math>\chi = \gamma<br />
+ \delta</math>. Now by the central angle theorem <math>2\gamma = \angle XOY</math><br />
and <math>2\delta = \angle YOZ</math>, and so <math>2(\gamma + \delta) = \angle XOZ</math>,<br />
and we are done.<br />
<br />
''Note that <math>RTQY</math> is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?''<br />
<br />
===Footnote===<br />
We can prove a bit more. Namely, the extensions of the segments<br />
<math>RS</math> and <math>PQ</math> meet at a point on the diameter <math>AB</math> that is vertically<br />
below the point <math>Y</math>.<br />
<br />
Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise<br />
from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math><br />
horizontally to the right of <math>Y</math>.<br />
<br />
Now <math>AS = AY \cos(\delta)</math>, so <math>S</math> is <math>AS \sin(\alpha) = AY<br />
\cos(\delta)\sin(\alpha)</math> vertically above the diameter <math>AB</math>. Also,<br />
the segment <math>SR</math> is inclined <math>\delta</math> clockwise from the vertical,<br />
so if we extend it down from <math>S</math> towards the diameter <math>AB</math> it will<br />
meet the diameter at a point which is<br />
<math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math><br />
horizontally to the left of <math>S</math>. This places the intersection point<br />
of <math>RS</math> and <math>AB</math> vertically below <math>Y</math>.<br />
<br />
Similarly, and by symmetry the intersection point of <math>PQ</math> and <math>AB</math><br />
is directly below <math>Y</math> on <math>AB</math>, so the lines through <math>PQ</math> and <math>RS</math><br />
meet at a point <math>T</math> on the diameter that is vertically below <math>Y</math>.<br />
<br />
===Footnote to the Footnote===<br />
The Footnote's claim is more easily proved as follows.<br />
<br />
Note that because <math>\angle{QPY}</math> and <math>\angle{YAB}</math> are both complementary to <math>\beta + \gamma</math>, they must be equal. Now, let <math>PQ</math> intersect diameter <math>AB</math> at <math>T'</math>. Then <math>PYT'A</math> is cyclic and so <math>\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ</math>. Hence <math>T'YSB</math> is cyclic as well, and so we deduce that <math>\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.</math> Hence <math>S, R, T'</math> are collinear and so <math>T = T'</math>. This proves the Footnote.<br />
<br />
===Footnote to the Footnote to the Footnote===<br />
The Footnote's claim can be proved even more easily as follows.<br />
<br />
Drop an altitude from <math>Y</math> to <math>AB</math> at point <math>T</math>. Notice that <math>P, Q, T</math> are collinear because they form the Simson line of <math>\triangle AXB</math> from <math>Y</math>. Also notice that <math>P, Q, T</math> are collinear because they form the Simson line of <math>\triangle AZB</math> from <math>Y</math>. Since <math>T</math> is at the diameter <math>AB</math>, lines <math>PQ</math> and <math>SR</math> must intersect at the diameter.<br />
<br />
== See Also ==<br />
{{USAMO newbox|year=2010|before=First problem|num-a=2}}<br />
{{USAJMO newbox|year=2010|num-b=2|num-a=4}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Ramanconjecturehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_14&diff=857151983 AIME Problems/Problem 142017-05-19T02:18:47Z<p>Ramanconjecture: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In the adjoining figure, two circles with radii <math>8</math> and <math>6</math> are drawn with their centers <math>12</math> units apart. At <math>P</math>, one of the points of intersection, a line is drawn in such a way that the chords <math>QP</math> and <math>PR</math> have equal length. Find the square of the length of <math>QP</math>. <br />
<br />
<asy>size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);</asy><br />
<br />
__TOC__<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, notice that if we reflect <math>R</math> over <math>P</math> we get <math>Q</math>. Since we know that <math>R</math> is on [[circle]] <math>B</math> and <math>Q</math> is on circle <math>A</math>, we can reflect circle <math>B</math> over <math>P</math> to get another circle (centered at a new point <math>C</math> with radius <math>6</math>) that intersects circle <math>A</math> at <math>Q</math>. The rest is just finding lengths:<br />
<br />
Since <math>P</math> is the midpoint of segment <math>BC</math>, <math>AP</math> is a median of triangle <math>ABC</math>. Because we know that <math>AB=12</math>, <math>BP=PC=6</math>, and <math>AP=8</math>, we can find the third side of the triangle using [[Stewart's Theorem]] or similar approaches. We get <math>AC = \sqrt{56}</math>. So now we have a kite <math>AQCP</math> with <math>AQ=AP=8</math>, <math>CQ=CP=6</math>, and <math>AC=\sqrt{56}</math>, and all we need is the length of the other diagonal <math>PQ</math>. The easiest way it can be found is with the [[Pythagorean Theorem]]. Let <math>2x</math> be the length of <math>PQ</math>. Then<br />
<br />
<center><math>\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.</math></center><br />
<br />
Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math><br />
=== Solution 2 (easiest and quickest)===<br />
<asy><br />
size(0,5cm);<br />
pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0);<br />
draw(b--r--n--b--a--m--n);<br />
draw(a--q--m);<br />
draw(circumcircle(origin,q,p));<br />
draw(circumcircle((14,0),p,r));<br />
draw(rightanglemark(a,m,n,24));<br />
draw(rightanglemark(b,n,r,24));<br />
label("$A$",a,S);<br />
label("$B$",b,S);<br />
label("$M$",m,NE);<br />
label("$N$",n,NE);<br />
label("$P$",p,N);<br />
label("$Q$",q,NW);<br />
label("$R$",r,E);<br />
label("$12$",(14,0),SW);<br />
label("$6$",(23,0),S);<br />
</asy><br />
''' <br />
<br />
Draw additional lines as indicated. Note that since triangles <math>AQP</math> and <math>BPR</math> are isosceles, the altitudes are also bisectors, so let <math>QM=MP=PN=NR=x</math>.<br />
<br />
Since <math>\frac{AR}{MR}=\frac{BR}{NR},</math> triangles <math>BNR</math> and <math>AMR</math> are similar. If we let <math>y=BN</math>, we have <math>AM=3BN=3y</math>. <br />
<br />
Applying the Pythagorean Theorem on triangle <math>BNR</math>, we have <math>x^2+y^2=36</math>. Similarly, for triangle <math>QMA</math>, we have <math>x^2+9y^2=64</math>.<br />
<br />
Subtracting, <math>8y^2=28\Rightarrow y^2=\frac72\Rightarrow x^2=\frac{65}2\Rightarrow QP^2=4x^2=\boxed{130}</math>.<br />
<br />
=== Solution 3 ===<br />
Let <math>QP=PR=x</math>. Angles <math>QPA</math>, <math>APB</math>, and <math>BPR</math> must add up to <math>180^{\circ}</math>. By the [[Law of Cosines]], <math>\angle APB=\cos^{-1}(-11/24)</math>. Also, angles <math>QPA</math> and <math>BPR</math> equal <math>\cos^{-1}(x/16)</math> and <math>\cos^{-1}(x/12)</math>. So we have <center><math>\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).</math></center> Taking the <math>\cos</math> of both sides and simplifying using the cosine addition identity gives <math>x^2=130</math>.<br />
<br />
===Solution 4(The actual quickest solution) ===<br />
Let <math>QP = PR = x.</math> Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. <br />
<br />
The line segment consisting of R and the first intersection of the larger circle has length 10. <br />
The length of the diameter of the larger circle be16.<br />
<br />
Through power of a point,<br />
<cmath>x \cdot 2x = 10 \cdot 26.</cmath><br />
<cmath>x^2 = 130.</cmath><br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=13|num-a=15}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Ramanconjecturehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_19&diff=835772012 AMC 10A Problems/Problem 192017-02-15T03:08:34Z<p>Ramanconjecture: /* Solution */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #13]] and [[2012 AMC 10A Problems|2012 AMC 10A #19]]}}<br />
<br />
==Problem 19==<br />
<br />
Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?<br />
<br />
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60 </math><br />
<br />
==Solution==<br />
Let Paula work at a rate of <math>p</math>, the two helpers work at a combined rate of <math>h</math>, and the time it takes to eat lunch be <math>L</math>, where <math>p</math> and <math>h</math> are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:<br />
<br />
<cmath>(8-L)(p+h)=50</cmath><br />
<br />
<cmath>(6.2-L)h=24</cmath><br />
<br />
<cmath>(11.2-L)p=26</cmath><br />
<br />
With three equations and three variables, we need to find the value of <math>L</math>.<br />
Adding the second and third equations together gives us <math>6.2h+11.2p-L(p+h)=.50</math>. Subtracting the first equation from this new one gives us <math>-1.8h+3.2p=0</math>, so we get <math>h=\frac{16}{9}p</math>. <br />
Plugging into the second equation:<br />
<br />
<cmath>(6.2-L)\frac{16}{9}p=24</cmath><br />
<cmath>(6.2-L)p=\frac{27}{2}</cmath><br />
<br />
We can then subtract this from the third equation:<br />
<br />
<cmath>5p=26-\frac{27}{2}</cmath><br />
<cmath>p=\frac{5}{2}</cmath><br />
Plugging <math>p</math> into our third equation gives: <cmath>L=\frac{4}{5}</cmath><br />
<br />
Converting <math>L</math> from hours to minutes gives us <math>L=48</math> minutes, which is <math>\boxed{\textbf{(D)}\ 48}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2012|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Ramanconjecturehttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=834792013 AMC 10A Problems/Problem 202017-02-12T18:37:07Z<p>Ramanconjecture: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
A unit square is rotated <math>45^\circ</math> about its center. What is the area of the region swept out by the interior of the square? <br />
<br />
<br />
<math> \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1==<br />
<br />
First, we need to see what this looks like. Below is a diagram.<br />
<br />
<asy><br />
size(200);<br />
defaultpen(linewidth(0.8));<br />
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br />
fill(square^^square2,grey);<br />
for(int i=0;i<=3;i=i+1)<br />
{<br />
path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br />
draw(arcrot);<br />
fill(arcrot--(0,0)--cycle,grey);<br />
draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br />
}<br />
draw(square^^square2);</asy><br />
<br />
For this square with side length 1, the distance from center to vertex is <math>r = \frac{1}{\sqrt{2}}</math>, hence the area is composed of a semicircle of radius <math>r</math>, plus <math>4</math> times a parallelogram with height <math>\frac{1}{2}</math> and base <math>\frac{\sqrt{2}}{2(1+\sqrt{2})}</math>. That is to say, the total area is <math>\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}</math>.<br />
<br />
<asy><br />
size(150);defaultpen(linewidth(0.8));<br />
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;//fill(square^^square2,grey);<br />
for(int i=0;i<=3;i=i+1){path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));draw(arcrot);<br />
fill(arcrot--(0,0)--cycle,grey);}<br />
//draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);}<br />
draw(square^^square2);<br />
//draw((-.5,.5)--(.5,-.5)^^(0,sqrt(.5))--(0,-sqrt(.5)),dotted);draw((.5,.5)--(-.5,-.5),dotted);<br />
</asy><br />
(To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.)<br />
<asy><br />
size(150,Aspect);real r=sqrt(2);real b=2-2/r;<br />
draw((0,0)--(-1,1)--(b-1,1)--(0,r)--cycle);draw((0,1)--(b-1,1)--(b/2-1,1-b/2));draw((0,0)--(b-1,1),dashed);<br />
fill((2,0)--(b+1,1)--(b+2,0)--cycle,lightgray);draw((.5,.5)--(1,.5),EndArrow);<br />
draw((2,0)--(1,1)--(b+1,1)--(b+2,0)--(2,0)^^(b+1,1)--(b/2+1,1-b/2)^^(2,0)--(2+b/2,b/2));<br />
draw((2,0)--(b+1,1),dashed);<br />
</asy><br />
<br />
==Solution 2==<br />
[[Image:AMC 10A 2013 20.jpg]]<br />
<br />
Let <math>O</math> be the center of the square and <math>C</math> be the intersection of <math>OB</math> and <math>AD</math>. The desired area consists of the unit square, plus <math>4</math> regions congruent to the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math>, plus <math>4</math> triangular regions congruent to right triangle <math>BCD</math>. The area of the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math> is <math>\frac{\text{Area of Circle}-\text{Area of Square}}{8}</math>. Since the circle has radius <math>\dfrac{1}{\sqrt {2}}</math>, the area of the region is <math>\dfrac{\dfrac{\pi}{2}-1}{8}</math>, so 4 times the area of that region is <math>\dfrac{\pi}{4}-\dfrac{1}{2}</math>. Now we find the area of <math>\triangle BCD</math>. <math>BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}</math>. Since <math>\triangle BCD</math> is a <math>45-45-90</math> right triangle, the area of <math>\triangle BCD</math> is <math>\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}</math>, so 4 times the area of <math>\triangle BCD</math> is <math>\dfrac{3}{2}-\sqrt {2}</math>. Finally, the area of the whole region is <math>1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}</math>, which we can rewrite as <math>\boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}</math>.<br />
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==See Also==<br />
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{{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Ramanconjecturehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_3&diff=826371983 AIME Problems/Problem 32017-01-31T00:17:58Z<p>Ramanconjecture: /* Solution 2 */</p>
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<div>== Problem ==<br />
What is the product of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>?<br />
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== Solutions ==<br />
=== Solution 1 ===<br />
If we expand by squaring, we get a quartic [[polynomial]], which isn't always the easiest thing to deal with.<br />
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Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> and our equation becomes <math>y=2\sqrt{y+15}</math>.<br />
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Now we can square; solving for <math>y</math>, we get <math>y=10</math> or <math>y=-6</math> (The second solution is extraneous since <math>2\sqrt{y+15}</math> is positive (plugging in <math>6</math> as <math>y</math>, we get <math>-</math><math>6</math> <math>=</math> <math>6</math>, which is obviously not true)).So, we have <math>y=10</math> as the only solution for <math>y</math>. Substituting <math>x^2+18x+30</math> back in for <math>y</math>,<br />
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<center><math>x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.</math></center> By [[Vieta's formulas]], the product of the roots is <math>\boxed{020}</math>.<br />
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=== Solution 2 ===<br />
We begin by noticing that the polynomial on the left is <math>15</math> less than the polynomial under the radical sign. Thus: <cmath>(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.</cmath> Letting <math>n = \sqrt{x^2+18x+45}</math>, we have <math>n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0</math>. Because the square root of a real number can't be negative, the only possible <math>n</math> is <math>5</math>.<br />
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Substituting that in, we have <cmath>\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.</cmath><br />
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And by [[Vieta's formulas]], the product of the roots is <math>\boxed{020}</math>.<br />
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== See Also ==<br />
{{AIME box|year=1983|num-b=2|num-a=4}}<br />
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[[Category:Intermediate Algebra Problems]]</div>Ramanconjecture