https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Rootthreeovertwo&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T08:09:28ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_14&diff=1047232008 AIME I Problems/Problem 142019-03-20T06:00:24Z<p>Rootthreeovertwo: </p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{AB}</math> be a diameter of circle <math>\omega</math>. Extend <math>\overline{AB}</math> through <math>A</math> to <math>C</math>. Point <math>T</math> lies on <math>\omega</math> so that line <math>CT</math> is tangent to <math>\omega</math>. Point <math>P</math> is the foot of the perpendicular from <math>A</math> to line <math>CT</math>. Suppose <math>AB = 18</math>, and let <math>m</math> denote the maximum possible length of segment <math>BP</math>. Find <math>m^{2}</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
size(250); defaultpen(0.70 + fontsize(10)); import olympiad;<br />
pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T);<br />
draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P);<br />
draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35));<br />
draw(B--P, blue);<br />
label("\(A\)",A,NW);<br />
label("\(B\)",B,NW);<br />
label("\(C\)",C,NW);<br />
label("\(O\)",O,NW);<br />
label("\(P\)",P,SE);<br />
label("\(T\)",T,SE);<br />
label("\(9\)",(O+A)/2,N);<br />
label("\(9\)",(O+B)/2,N);<br />
label("\(x-9\)",(C+A)/2,N);<br />
</asy></center><br />
<br />
Let <math>x = OC</math>. Since <math>OT, AP \perp TC</math>, it follows easily that <math>\triangle APC \sim \triangle OTC</math>. Thus <math>\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}</math>. By the [[Law of Cosines]] on <math>\triangle BAP</math>, <br />
<cmath>\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}</cmath><br />
where <math>\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}</math>, so: <br />
<cmath>\begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*}</cmath><br />
Let <math>m = \frac{2x-27}{x^2} \Longrightarrow mx^2 - 2x + 27 = 0</math>; this is a quadratic, and its [[discriminant]] must be nonnegative: <math>(-2)^2 - 4(m)(27) \ge 0 \Longleftrightarrow m \le \frac{1}{27}</math>. Thus,<br />
<cmath>BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}</cmath><br />
Equality holds when <math>x = 27</math>.<br />
<br />
===Solution 2===<br />
<center><asy><br />
unitsize(3mm);<br />
pair B=(0,13.5), C=(23.383,0);<br />
pair O=(7.794, 9), P=(2*7.794,0);<br />
pair T=(7.794,0), Q=(0,0);<br />
pair A=(2*7.794,4.5);<br />
<br />
draw(Q--B--C--Q);<br />
draw(O--T);<br />
draw(A--P);<br />
draw(Circle(O,9));<br />
<br />
dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q);<br />
label("\(B\)",B,NW);<br />
label("\(A\)",A,NE);<br />
label("\(\omega\)",O,N);<br />
label("\(P\)",P,S);<br />
label("\(T\)",T,S);<br />
label("\(Q\)",Q,S);<br />
label("\(C\)",C,E);<br />
label("\(\theta\)",C + (-1.7,-0.2), NW);<br />
label("\(9\)", (B+O)/2, N);<br />
label("\(9\)", (O+A)/2, N);<br />
label("\(9\)", (O+T)/2,W);<br />
</asy></center><br />
<br />
From the diagram, we see that <math>BQ = \omega T + B\omega\sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)</math>, and that <math>QP = BA\cos\theta = 18\cos\theta</math>.<br />
<br />
<cmath>\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \sin\theta)^2 + 18^2\cos^2\theta\\<br />
&= 9^2[1 + 2\sin\theta + \sin^2\theta + 4(1 - \sin^2\theta)]\\<br />
BP^2 &= 9^2[5 + 2\sin\theta - 3\sin^2\theta]\end{align*}</cmath><br />
<br />
This is a [[quadratic equation]], maximized when <math>\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}</math>. Thus, <math>m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}</math>.<br />
<br />
===Solution 3 (Calculus Bash)===<br />
<br />
<center><asy><br />
unitsize(3mm);<br />
pair B=(0,13.5), C=(23.383,0);<br />
pair O=(7.794, 9), P=(2*7.794,0);<br />
pair T=(7.794,0), Q=(0,0);<br />
pair A=(2*7.794,4.5);<br />
<br />
draw(Q--B--C--Q);<br />
draw(O--T);<br />
draw(A--P);<br />
draw(Circle(O,9));<br />
<br />
dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q);<br />
label("\(B\)",B,NW);<br />
label("\(A\)",A,NE);<br />
label("\(\omega\)",O,N);<br />
label("\(P\)",P,S);<br />
label("\(T\)",T,S);<br />
label("\(Q\)",Q,S);<br />
label("\(C\)",C,E);<br />
label("\(9\)", (B+O)/2, N);<br />
label("\(9\)", (O+A)/2, N);<br />
label("\(9\)", (O+T)/2,W);<br />
</asy></center><br />
<br />
(Diagram credit goes to Solution 2)<br />
<br />
We let <math>AC=x</math>. From similar triangles, we have that <math>PC=\frac{x\sqrt{x^2+18x}}{x+9}</math>. Similarly, <math>TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}</math>. Using the Pythagorean Theorem, <math>BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}</math>. Using the Pythagorean Theorem once again, <math>BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}</math>. After a large bashful simplification, <math>BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}</math>. The fraction is equivalent to <math>729\frac{2x-9}{(x+9)^2}</math>. Taking the derivative of the fraction and solving for x, we get that <math>x=18</math>. Plugging <math>x=18</math> back into the expression for <math>BP</math> yields <math>\sqrt{432}</math>, so the answer is <math>(\sqrt{432})^2=\boxed{432}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|num-b=13|num-a=15}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_10&diff=1046502009 AIME II Problems/Problem 102019-03-18T05:46:02Z<p>Rootthreeovertwo: /* Solution 4 */</p>
<hr />
<div><br />
==Problem==<br />
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. <br />
<br />
<br />
== Solution 1==<br />
<br />
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.<br />
<br />
==Solution 2==<br />
<br />
Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>.<br />
<br />
By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>:<br />
<br />
<cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\<br />
AD^2&=130-\dfrac{130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130(23^2-169)}{23^2}\\<br />
AD^2&=\dfrac{130(360)}{23^2}\\<br />
AD&=\dfrac{60\sqrt{13}}{23}\\<br />
\end{align*}</cmath><br />
<br />
and our final answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
==Solution 3==<br />
<br />
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.<br />
<br />
==Solution 4==<br />
<br />
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=(\frac{5\sqrt{13}}{3})(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
(Solution by RootThreeOverTwo)<br />
<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_10&diff=1046492009 AIME II Problems/Problem 102019-03-18T05:45:44Z<p>Rootthreeovertwo: </p>
<hr />
<div><br />
==Problem==<br />
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. <br />
<br />
<br />
== Solution 1==<br />
<br />
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.<br />
<br />
==Solution 2==<br />
<br />
Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>.<br />
<br />
By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>:<br />
<br />
<cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\<br />
AD^2&=130-\dfrac{130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130(23^2-169)}{23^2}\\<br />
AD^2&=\dfrac{130(360)}{23^2}\\<br />
AD&=\dfrac{60\sqrt{13}}{23}\\<br />
\end{align*}</cmath><br />
<br />
and our final answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
==Solution 3==<br />
<br />
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.<br />
<br />
==Solution 4==<br />
<br />
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=(\frac{5\sqrt{13}}{3})(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=1046152019 AIME I Problems/Problem 62019-03-17T06:21:59Z<p>Rootthreeovertwo: </p>
<hr />
<div>==Problem 6==<br />
In convex quadrilateral <math>KLMN</math> side <math>\overline{MN}</math> is perpendicular to diagonal <math>\overline{KM}</math>, side <math>\overline{KL}</math> is perpendicular to diagonal <math>\overline{LN}</math>, <math>MN = 65</math>, and <math>KL = 28</math>. The line through <math>L</math> perpendicular to side <math>\overline{KN}</math> intersects diagonal <math>\overline{KM}</math> at <math>O</math> with <math>KO = 8</math>. Find <math>MO</math>.<br />
<br />
<br />
==Solution 1 (Simple) ==<br />
Note that <math>KLMN</math> is cyclic with diameter <math>KN</math> since <math>\angle KLN = \angle KMN = \frac{\pi}{2}</math>. Also, note that we have <math>\triangle KML \sim \triangle KLO</math> by SS similarity. <br />
<br />
We see this by <math>\angle LKM = \angle OKL</math> and <math>\angle KLO = \angle KML</math>. <br />
The latter equality can be seen if we extend <math>LP</math> to point <math>L'</math> on <math>(KLMN)</math>. We know <math>LK = KL'</math> from which it follows <math>\angle KLO = \angle KML</math>.<br />
<br />
Let <math>MO = x</math>. By <math>\triangle KML \sim \triangle KLO</math> we have <br />
<br />
<cmath>\frac{KL}{KO} = \frac{KM}{KL} \Rightarrow \frac{28}{8} = \frac{x+8}{28}.</cmath><br />
<br />
<cmath>98 = x + 8 \Rightarrow x = \boxed{090}.</cmath><br />
<br />
Note: This solution does not use the condition <math>MN=65</math>.<br />
<br />
- gregwwl<br />
<br />
==Solution 2 (Trig)==<br />
Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>. <br />
<br />
Then, <math>KP=28\sin\beta=8\cos\alpha</math>.<br />
Furthermore, <math>KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha</math>.<br />
<br />
Dividing the equations gives<br />
<cmath>\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}</cmath><br />
<br />
Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>.<br />
<br />
==Solution 3 (Similar triangles)==<br />
<asy><br />
size(250);<br />
real h = sqrt(98^2+65^2);<br />
real l = sqrt(h^2-28^2);<br />
pair K = (0,0);<br />
pair N = (h, 0);<br />
pair M = ((98^2)/h, (98*65)/h);<br />
pair L = ((28^2)/h, (28*l)/h);<br />
pair P = ((28^2)/h, 0);<br />
pair O = ((28^2)/h, (8*65)/h);<br />
draw(K--L--N);<br />
draw(K--M--N--cycle);<br />
draw(L--M);<br />
label("K", K, SW);<br />
label("L", L, NW);<br />
label("M", M, NE);<br />
label("N", N, SE);<br />
draw(L--P);<br />
label("P", P, S);<br />
dot(O);<br />
label("O", shift((1,1))*O, NNE);<br />
label("28", scale(1/2)*L, W);<br />
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br />
</asy><br />
<br />
First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math> as shown above. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong \angle LKN</math>, <math>\triangle PKL \sim \triangle LKN</math> by AA similarity. Similarly, <math>\triangle KMN \sim \triangle KPO</math>. Using these similarities we see that<br />
<cmath>\frac{KP}{KL} = \frac{KL}{KN}</cmath><br />
<cmath>KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}</cmath><br />
and<br />
<cmath>\frac{KP}{KO} = \frac{KM}{KN}</cmath><br />
<cmath>KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}</cmath><br />
Combining the two equations, we get<br />
<cmath>\frac{8\cdot KM}{KN} = \frac{784}{KN}</cmath><br />
<cmath>8 \cdot KM = 28^2</cmath><br />
<cmath>KM = 98</cmath><br />
Since <math>KM = KO + MO</math>, we get <math>MO = 98 -8 = \boxed{090}</math>.<br />
<br />
Solution by vedadehhc<br />
<br />
==Solution 4 (Similar triangles, orthocenters)==<br />
Extend <math>KL</math> and <math>NM</math> past <math>L</math> and <math>M</math> respectively to meet at <math>P</math>. Let <math>H</math> be the intersection of diagonals <math>KM</math> and <math>LN</math> (this is the orthocenter of <math>\triangle KNP</math>).<br />
<br />
As <math>\triangle KOL \sim \triangle KHP</math> (as <math>LO \parallel PH</math>, using the fact that <math>H</math> is the orthocenter), we may let <math>OH = 8k</math> and <math>LP = 28k</math>.<br />
<br />
Then using similarity with triangles <math>\triangle KLH</math> and <math>\triangle KMP</math> we have<br />
<br />
<cmath>\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}</cmath><br />
<br />
Cross-multiplying and dividing by <math>4+4k</math> gives <math>2(8+8k+HM) = 28 \cdot 7 = 196</math> so <math>MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}</math>. (Solution by scrabbler94)<br />
<br />
==Solution 5 (Algebraic Bashing)==<br />
First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. We can use the right triangles in the problem to create equations. Let <math>a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,</math> and <math>g=NC.</math> We are trying to find <math>d.</math> We can find <math>7</math> equations. They are<br />
<cmath>4225+d^2=c^2,</cmath><br />
<cmath>4225+d^2+16d+64=a^2+2ab+b^2,</cmath><br />
<cmath>a^2+e^2=c^2,</cmath><br />
<cmath>b^2+e^2=64,</cmath><br />
<cmath>b^2+e^2+2ef+f^2=784,</cmath><br />
<cmath>a^2+e^2+2ef+f^2=g^2,</cmath><br />
and <cmath>g^2+784=a^2+2ab+b^2.</cmath><br />
We can subtract the fifth equation from the sixth equation to get <math>a^2-b^2=g^2-784.</math> We can subtract the fourth equation from the third equation to get <math>a^2-b^2=c^2-64.</math> Combining these equations gives <math>c^2-64=g^2-784</math> so <math>g^2=c^2+720.</math> Substituting this into the seventh equation gives <math>c^2+1504=a^2+2ab+b^2.</math> Substituting this into the second equation gives <math>4225+d^2+16d+64=c^2+1504</math>. Subtracting the first equation from this gives <math>16d+64=1504.</math> Solving this equation, we find that <math>d=\boxed{090}.</math><br />
(Solution by DottedCaculator)<br />
<br />
==Solution 6 (5-second PoP)==<br />
<br />
<asy><br />
size(8cm);<br />
pair K, L, M, NN, X, O;<br />
K=(-sqrt(98^2+65^2)/2, 0);<br />
NN=(sqrt(98^2+65^2)/2, 0);<br />
L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));<br />
M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));<br />
X=foot(L, K, NN);<br />
O=extension(L, X, K, M);<br />
draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K));<br />
draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed);<br />
<br />
draw(rightanglemark(K, L, NN, 100));<br />
draw(rightanglemark(K, M, NN, 100));<br />
draw(rightanglemark(L, X, NN, 100));<br />
dot("$K$", K, SW);<br />
dot("$L$", L, unit(L));<br />
dot("$M$", M, unit(M));<br />
dot("$N$", NN, SE);<br />
dot("$X$", X, S);<br />
</asy><br />
Notice that <math>KLMN</math> is inscribed in the circle with diameter <math>\overline{KN}</math> and <math>XOMN</math> is inscribed in the circle with diameter <math>\overline{ON}</math>. Furthermore, <math>(XLN)</math> is tangent to <math>\overline{KL}</math>. Then, <cmath>KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,</cmath>and <math>MO=KM-KO=\boxed{090}</math>.<br />
<br />
(Solution by TheUltimate123)<br />
<br />
==Solution 7 (Alternative PoP)==<br />
<br />
<asy><br />
size(250);<br />
real h = sqrt(98^2+65^2);<br />
real l = sqrt(h^2-28^2);<br />
pair K = (0,0);<br />
pair N = (h, 0);<br />
pair M = ((98^2)/h, (98*65)/h);<br />
pair L = ((28^2)/h, (28*l)/h);<br />
pair P = ((28^2)/h, 0);<br />
pair O = ((28^2)/h, (8*65)/h);<br />
draw(K--L--N);<br />
draw(K--M--N--cycle);<br />
draw(L--M);<br />
label("K", K, SW);<br />
label("L", L, NW);<br />
label("M", M, NE);<br />
label("N", N, SE);<br />
draw(L--P);<br />
label("P", P, S);<br />
dot(O);<br />
label("O", shift((1,1))*O, NNE);<br />
label("28", scale(1/2)*L, W);<br />
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br />
</asy><br />
<br />
(Diagram by vedadehhc)<br />
<br />
Call the base of the altitude from <math>L</math> to <math>NK</math> point <math>P</math>. Let <math>PO=x</math>. Now, we have that <math>KO=\sqrt{64-x^2}</math> by the Pythagorean Theorem. Once again by Pythagorean, <math>LO=\sqrt{720+x^2}-x</math>. Using Power of a Point, we have <br />
<br />
<cmath>(KO)(OM)=(LO)(OQ)</cmath> (<math>Q</math> is the intersection of <math>OL</math> with the circle <math>\neq L</math>)<br />
<br />
<cmath>8(MO)=(\sqrt{720+x^2}+x)(\sqrt{720+x^2}-x)</cmath><br />
<br />
<cmath>8(MO)=720</cmath><br />
<br />
<cmath>MO=\boxed{090}</cmath>.<br />
<br />
(Solution by RootThreeOverTwo)<br />
==Video Solution==<br />
Video Solution:<br />
https://www.youtube.com/watch?v=0AXF-5SsLc8<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_15&diff=1008482014 AIME II Problems/Problem 152019-01-26T05:41:27Z<p>Rootthreeovertwo: </p>
<hr />
<div>==Problem==<br />
For any integer <math>k\geq 1</math>, let <math>p(k)</math> be the smallest prime which does not divide <math>k.</math> Define the integer function <math>X(k)</math> to be the product of all primes less than <math>p(k)</math> if <math>p(k)>2</math>, and <math>X(k)=1</math> if <math>p(k)=2.</math> Let <math>\{x_n\}</math> be the sequence defined by <math>x_0=1</math>, and <math>x_{n+1}X(x_n)=x_np(x_n)</math> for <math>n\geq 0.</math> Find the smallest positive integer <math>t</math> such that <math>x_t=2090.</math><br />
<br />
==Solution==<br />
Note that for any <math>x_i</math>, for any prime <math>p</math>, <math>p^2 \nmid x_i</math>. This provides motivation to translate <math>x_i</math> into a binary sequence <math>y_i</math>. <br />
<br />
Let the prime factorization of <math>x_i</math> be written as <math>p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots</math>, where <math>p_i</math> is the <math>i</math>th prime number. Then, for every <math>p_{a_k}</math> in the prime factorization of <math>x_i</math>, place a <math>1</math> in the <math>a_k</math>th digit of <math>y_i</math>. This will result in the conversion <math>x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots</math>.<br />
<br />
Multiplication for the sequence <math>x_i</math> will translate to addition for the sequence <math>y_i</math>. Thus, we see that <math>x_{n+1}X(x_n) = x_np(x_n)</math> translates into <math>y_{n+1} = y_n+1</math>. Since <math>x_0=1</math>, and <math>y_0=0</math>, <math>x_i</math> corresponds to <math>y_i</math>, which is <math>i</math> in binary. Since <math>x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090</math>, <math>t = 10010101_2</math> = <math>\boxed{149}</math>.<br />
<br />
==Solution 2 (Painful Bash)==<br />
We go through the terms and look for a pattern. We find that<br />
<br />
<math>x_0 = 1</math> <math>x_8 = 7</math><br />
<br />
<math>x_1 = 2</math> <math>x_9 = 14</math><br />
<br />
<math>x_2 = 3</math> <math>x_{10} = 21</math><br />
<br />
<math>x_3 = 6</math> <math>x_{11} = 42</math><br />
<br />
<math>x_4 = 5</math> <math>x_{12} = 35</math><br />
<br />
<math>x_5 = 10</math> <math>x_{13} = 70</math><br />
<br />
<math>x_6 = 15</math> <math>x_{14} = 105</math><br />
<br />
<math>x_7 = 30</math> <math>x_{15} = 210</math><br />
<br />
Commit to the bash. Eventually, you will recieve that <math>x_{149} = 2090</math>, so <math>\boxed{149}</math> is the answer. Trust me, this is worth the 10 index points.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
== See also ==<br />
{{AIME box|year=2014|n=II|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_14&diff=997662018 AIME II Problems/Problem 142018-12-26T06:36:31Z<p>Rootthreeovertwo: /* Solution 3 (Combination of Law of Sine and Law of Cosine) */</p>
<hr />
<div>==Problem==<br />
<br />
The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other intersection of <math>\overline{AX}</math> with <math>\omega</math>. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>\overline{PQ}</math> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
<br />
==Solution 1==<br />
Let sides <math>\overline{AB}</math> and <math>\overline{AC}</math> be tangent to <math>\omega</math> at <math>Z</math> and <math>W</math>, respectively. Let <math>\alpha = \angle BAX</math> and <math>\beta = \angle AXC</math>. Because <math>\overline{PQ}</math> and <math>\overline{BC}</math> are both tangent to <math>\omega</math> and <math>\angle YXC</math> and <math>\angle QYX</math> subtend the same arc of <math>\omega</math>, it follows that <math>\angle AYP = \angle QYX = \angle YXC = \beta</math>. By equal tangents, <math>PZ = PY</math>. Applying the Law of Sines to <math>\triangle APY</math> yields <cmath>\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.</cmath>Similarly, applying the Law of Sines to <math>\triangle ABX</math> gives <cmath>\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.</cmath>It follows that <cmath>2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,</cmath>implying <math>AZ = \tfrac{21}5</math>. Applying the same argument to <math>\triangle AQY</math> yields <cmath>2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),</cmath>from which <math>AQ = \tfrac{168}{59}</math>. The requested sum is <math>168 + 59 = \boxed{227}</math>.<br />
<br />
==Solution 2 (Projective)==<br />
Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math>. By Brianchon's theorem on tangential hexagons <math>QNCBMP</math> and <math>PYQCXB</math>, we know that <math>MN,CP,BQ</math> and <math>XY</math> are concurrent at a point <math>O</math>. Let <math>PQ \cap BC = Z</math>. Then by La Hire's <math>A</math> lies on the polar of <math>Z</math> so <math>Z</math> lies on the polar of <math>A</math>. Therefore, <math>MN</math> also passes through <math>Z</math>. Then projecting through <math>Z</math>, we have<br />
<cmath> -1 = (A,O;Y,X) \stackrel{Z}{=} (A,M;P,B) \stackrel{Z}{=} (A,N;Q,C).</cmath>Therefore, <math>\frac{AP \cdot MB}{MP \cdot AB} = 1 \implies \frac{3 \cdot MB}{MP \cdot 7} = 1</math>. Since <math>MB+MP=4</math> we know that <math>MB = \frac{6}{5}</math> and <math>MB = \frac{14}{5}</math>. Therefore, <math>AN = AM = \frac{21}{5}</math> and <math>NC = 8 - \frac{21}{5} = \frac{19}{5}</math>. Since <math>(A,N;Q,C) = -1</math>, we also have <math>\frac{AQ \cdot NC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1</math>. Solving for <math>AQ</math>, we obtain <math>AQ = \frac{168}{59} \implies m+n = \boxed{227}</math>.<br />
-Vfire<br />
<br />
==Solution 3 (Combination of Law of Sine and Law of Cosine)==<br />
Let the center of the incircle of <math>\triangle ABC</math> be <math>O</math>. Link <math>OY</math> and <math>OX</math>. Then we have <math>\angle OYP=\angle OXB=90^{\circ}</math><br />
<br />
<math>\because</math> <math>OY=OX</math><br />
<br />
<math>\therefore</math> <math>\angle OYX=\angle OXY</math><br />
<br />
<math>\therefore</math> <math>\angle PYX=\angle YXB</math><br />
<br />
<math>\therefore</math> <math>\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA</math><br />
<br />
Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math>, let <math>MP=YP=x</math> and <math>NQ=YQ=y</math>.<br />
<br />
Use Law of Sine in <math>\triangle APY</math> and <math>\triangle AXB</math>, we have<br />
<br />
<math>\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}</math><br />
<br />
<math>\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}</math><br />
<br />
therefore we have<br />
<br />
<math>\frac{3}{x}=\frac{7}{4-x}</math><br />
<br />
Solve this equation, we have <math>x=\frac{6}{5}</math><br />
<br />
As a result, <math>MB=4-x=\frac{14}{5}=BX</math>, <math>AM=x+3=\frac{21}{5}=AN</math>, <math>NC=8-AN=\frac{19}{5}=XC</math>, <math>AQ=\frac{21}{5}-y</math>, <math>PQ=\frac{6}{5}+y</math><br />
<br />
So, <math>BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}</math><br />
<br />
Use Law of Cosine in <math>\triangle BAC</math> and <math>\triangle PAQ</math>, we have<br />
<br />
<math>\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}</math><br />
<br />
<math>\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}</math><br />
<br />
And we have<br />
<br />
<math>\cos \angle BAC=\cos \angle PAQ</math><br />
<br />
So<br />
<br />
<math>\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}</math><br />
<br />
Solve this equation, we have <math>y=\frac{399}{295}=QN</math><br />
<br />
As a result, <math>AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}</math><br />
<br />
So, the final answer of this question is <math>168+59=\boxed {227}</math><br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
{{AIME box|year=2018|n=II|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_11&diff=997642018 AIME II Problems/Problem 112018-12-26T05:38:30Z<p>Rootthreeovertwo: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
Find the number of permutations of <math>1, 2, 3, 4, 5, 6</math> such that for each <math>k</math> with <math>1</math> <math>\leq</math> <math>k</math> <math>\leq</math> <math>5</math>, at least one of the first <math>k</math> terms of the permutation is greater than <math>k</math>.<br />
<br />
==Solution 1==<br />
<br />
If the first number is <math>6</math>, then there are no restrictions. There are <math>5!</math>, or <math>120</math> ways to place the other <math>5</math> numbers.<br />
<br />
<br />
If the first number is <math>5</math>, <math>6</math> can go in four places, and there are <math>4!</math> ways to place the other <math>4</math> numbers. <math>4 \cdot 4! = 96</math> ways.<br />
<br />
<br />
If the first number is <math>4</math>, ....<br />
<br />
4 6 _ _ _ _ <math>\implies</math> 24 ways<br />
<br />
4 _ 6 _ _ _ <math>\implies</math> 24 ways<br />
<br />
4 _ _ 6 _ _ <math>\implies</math> 24 ways<br />
<br />
4 _ _ _ 6 _ <math>\implies</math> 5 must go between <math>4</math> and <math>6</math>, so there are <math>3 \cdot 3! = 18</math> ways.<br />
<br />
<math>24 + 24 + 24 + 18 = 90</math> ways if 4 is first.<br />
<br />
<br />
If the first number is <math>3</math>, ....<br />
<br />
3 6 _ _ _ _ <math>\implies</math> 24 ways<br />
<br />
3 _ 6 _ _ _ <math>\implies</math> 24 ways<br />
<br />
3 1 _ 6 _ _ <math>\implies</math> 4 ways <br />
<br />
3 2 _ 6 _ _ <math>\implies</math> 4 ways<br />
<br />
3 4 _ 6 _ _ <math>\implies</math> 6 ways<br />
<br />
3 5 _ 6 _ _ <math>\implies</math> 6 ways<br />
<br />
3 5 _ _ 6 _ <math>\implies</math> 6 ways<br />
<br />
3 _ 5 _ 6 _ <math>\implies</math> 6 ways<br />
<br />
3 _ _ 5 6 _ <math>\implies</math> 4 ways<br />
<br />
<math>24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84</math> ways<br />
<br />
<br />
If the first number is <math>2</math>, ....<br />
<br />
2 6 _ _ _ _ <math>\implies</math> 24 ways<br />
<br />
2 _ 6 _ _ _ <math>\implies</math> 18 ways <br />
<br />
2 3 _ 6 _ _ <math>\implies</math> 4 ways<br />
<br />
2 4 _ 6 _ _ <math>\implies</math> 6 ways<br />
<br />
2 5 _ 6 _ _ <math>\implies</math> 6 ways<br />
<br />
2 5 _ _ 6 _ <math>\implies</math> 6 ways<br />
<br />
2 _ 5 _ 6 _ <math>\implies</math> 4 ways<br />
<br />
2 4 _ 5 6 _ <math>\implies</math> 2 ways<br />
<br />
2 3 4 5 6 1 <math>\implies</math> 1 way<br />
<br />
<br />
<math>24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71</math> ways<br />
<br />
<br />
Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math><br />
<br />
==Solution 2==<br />
If <math>6</math> is the first number, then there are no restrictions. There are <math>5!</math>, or <math>120</math> ways to place the other <math>5</math> numbers.<br />
<br />
<br />
If <math>6</math> is the second number, then the first number can be <math>2, 3, 4,</math> or <math>5</math>, and there are <math>4!</math> ways to place the other <math>4</math> numbers. <math>4 \cdot 4! = 96</math> ways.<br />
<br />
<br />
If <math>6</math> is the third number, then we cannot have the following:<br />
<br />
1 _ 6 _ _ _ <math>\implies</math> 24 ways<br />
<br />
2 1 6 _ _ _ <math>\implies</math> 6 ways<br />
<br />
<math>120 - 24 - 6 = 90</math> ways<br />
<br />
If <math>6</math> is the fourth number, then we cannot have the following:<br />
1 _ _ 6 _ _ <math>\implies</math> 24 ways<br />
<br />
2 1 _ 6 _ _ <math>\implies</math> 6 ways<br />
<br />
2 3 1 6 _ _ <math>\implies</math> 2 ways<br />
<br />
3 1 2 6 _ _ <math>\implies</math> 2 ways<br />
<br />
3 2 1 6 _ _ <math>\implies</math> 2 ways<br />
<br />
<math>120 - 24 - 6 - 2 - 2 - 2 = 84</math> ways<br />
<br />
If <math>6</math> is the fifth number, then we cannot have the following:<br />
<br />
_ _ _ _ 6 5 <math>\implies</math> 24 ways<br />
<br />
1 5 _ _ 6 _ <math>\implies</math> 6 ways<br />
<br />
1 _ 5 _ 6 _ <math>\implies</math> 6 ways<br />
<br />
2 1 5 _ 6 _ <math>\implies</math> 2 ways<br />
<br />
1 _ _ 5 6 _ <math>\implies</math> 6 ways<br />
<br />
2 1 _ 5 6 _ <math>\implies</math> 2 ways<br />
<br />
2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 <math>\implies</math> 3 ways<br />
<br />
<math>120 - 24 - 6 - 6 - 2 - 6 - 2 - 3 = 71</math> ways<br />
<br />
Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math><br />
<br />
==Solution 3 (needs explanation)==<br />
The answer is <math>\frac{6!}{2} + 5! - 4! + 3! - 2! + 1! = \boxed{\boxed{461}}</math>.<br />
<br />
==Solution 4 (General Case, and you won't get 458, 459, 460, 462, 465, 467, etc. with this method!!!)==<br />
<br />
First let us look at the General Case of this kind of Permutation: Consider this kind of Permutation of set <cmath>S=\{1,2,...,n\}</cmath> for arbitrary <math>n \in N</math><br />
<br />
It is easy to count the total number of the permutation (<math>N</math>) of <math>S</math>: <cmath>N=n!</cmath> For every <math>i \in S</math>, we can divide <math>S</math> into two subsets: <cmath>S_{1\to i}=\{1,2,...i\}; S_{i+1\to n}=\{i+1,i+2,...,n\}</cmath> Define permutation <math>P</math> as the permutation satisfy the condition of this problem. Then according to the condition of this problem, for each <math>i\in \{1,2,...,n-1\}</math>, <math>P</math> is not a permutation of set <math>S_{1\to i}</math>. For each <math>i\in \{1,2,...,n\}</math>, mark the number of permutation <math>P</math> of set <math>S</math> as <math>P_{k}</math>, where <math>k=i</math>, mark the number of permutation <math>P</math> for set <math>S_{i+1\to n}</math> as <math>P_{i}</math>; then, according to the condition of this problem, the permutation for <math>S_{i+1\to n}</math> is unrestricted, so the number of the unrestricted permutation of <math>S_{i+1\to n}</math> is <math>(n-i)!</math>. As a result, for each <math>i\in \{1,2,...,n\}</math>, the total number of permutation <math>P</math> is <cmath>P_{k}=P_{i}(n-i)!</cmath> Notice that according to the condition of this problem, if you sum all <math>P_{k}</math> up, you will get the total number of permutation of <math>S</math>, that is, <cmath>N=\sum^{n}_{k=1}{P_{k}}=\sum^{n}_{i=1}{P_{i}(n-i)!}=n!</cmath> Put <math>n=1,2,3,...,6</math>, we will have <cmath>P_{1}=1</cmath> <cmath>P_{2}=1</cmath> <cmath>P_{3}=3</cmath> <cmath>P_{4}=13</cmath> <cmath>P_{5}=71</cmath> <cmath>P_{6}=461</cmath> So the total number of permutations satisify this problem is <math>P_{6}=\boxed{461}</math>.<br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
{{AIME box|year=2018|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_5&diff=997272018 AIME II Problems/Problem 52018-12-24T22:06:46Z<p>Rootthreeovertwo: </p>
<hr />
<div>==Problem==<br />
<br />
Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>.<br />
<br />
==Solution 1==<br />
<br />
First we evaluate the magnitudes. <math>|xy|=80\sqrt{17}</math>, <math>|yz|=60</math>, and <math>|zx|=24\sqrt{17}</math>. Therefore, <math>|x^2y^2z^2|=17\cdot80\cdot60\cdot24</math>, or <math>|xyz|=240\sqrt{34}</math>. Divide to find that <math>|z|=3\sqrt{2}</math>, <math>|x|=40\sqrt{34}</math>, and <math>|y|=10\sqrt{2}</math>.<br />
<asy><br />
draw((0,0)--(4,0));<br />
dot((4,0),red);<br />
draw((0,0)--(-4,0));<br />
draw((0,0)--(0,-4));<br />
draw((0,0)--(-4,1));<br />
dot((-4,1),red);<br />
draw((0,0)--(-1,-4));<br />
dot((-1,-4),red);<br />
draw((0,0)--(4,4),red);<br />
draw((0,0)--(4,-4),red);<br />
</asy><br />
This allows us to see that the argument of <math>y</math> is <math>\frac{\pi}{4}</math>, and the argument of <math>z</math> is <math>-\frac{\pi}{4}</math>. We need to convert the polar form to a standard form. Simple trig identities show <math>y=10+10i</math> and <math>z=3-3i</math>. More division is needed to find what <math>x</math> is. <cmath>x=-20-12i</cmath> <cmath>x+y+z=-7-5i</cmath> <cmath>(-7)^2+(-5)^2=\boxed{074}</cmath><br />
<cmath>QED\blacksquare</cmath><br />
Written by [[User:A1b2|a1b2]]<br />
==Solution 2==<br />
Dividing the first equation by the second equation given, we find that <math>\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)</math>. Substituting this into the third equation, we get <math>z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i</math>. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of <math>y</math> is the negative of that of <math>z</math>, and their magnitudes multiply to <math>60</math>. Thus we have <math>z=\sqrt{-18i}=3-3i</math> and <math>3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i</math>. To find <math>x</math>, we can use the previous substitution we made to find that <math>x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i</math><br />
Therefore, <math>x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{074}</math><br />
Solution by ktong<br />
<br />
==Solution 3 ==<br />
<br />
We are given that <math>xy=-80-320i</math>. Thus <math>y=\frac{-80-320i}{x}</math>. We are also given that <math>xz= -96+24i</math>. Thus <math>z=\frac{-96+24i}{x}</math>. We are also given that <math>yz</math> = <math>60</math>. Substitute <math>y=\frac{-80-320i}{x}</math> and <math>z=\frac{-96+24i}{x}</math> into <math>yz</math> = <math>60</math>. We have <math> \frac{(-80-320i)(-96+24i)}{x^2}=60</math>. Multiplying out <math>(-80-320i)(-96+24i)</math> we get <math>(1920)(8+15i)</math>. Thus <math>\frac{1920(8+15i)}{x^2} =60</math>. Simplifying this fraction we get <math>\frac{32(8+15i)}{x^2}=1</math>. Cross-multiplying the fractions we get <math>x^2=32(8+15i)</math> or <math>x^2= 256+480i</math>. Now we can rewrite this as <math>x^2-256=480i</math>. Let <math>x= (a+bi)</math>.Thus <math>x^2=(a+bi)^2</math> or <math>a^2+2abi-b^2</math>. We can see that <math>a^2+2abi-b^2-256=480i</math> and thus <math>2abi=480i</math> or <math>ab=240</math>.We also can see that <math>a^2-b^2-256=0</math> because there is no real term in <math>480i</math>. Thus <math>a^2-b^2=256</math> or <math>(a+b)(a-b)=256</math>. Using the two equations <math>ab=240</math> and <math>(a+b)(a-b)=256</math> we solve by doing system of equations that <math>a=-20</math> and <math>b=-12</math>. And <math>x=a+bi</math> so <math>x=-20-12i</math>. Because <math>y=\frac{-80-320i}{x}</math>, then <math>y=\frac{-80-320i}{-20-12i}</math>. Simplifying this fraction we get <math>y=\frac{-80(1+4i)}{-4(5+3i)}</math> or <math>y=\frac{20(1+4i)}{(5+3i)}</math>. Multiplying by the conjugate of the denominator (<math>5-3i</math>) in the numerator and the denominator and we get <math>y=\frac{20(17-17i)}{34}</math>. Simplifying this fraction we get <math>y=10-10i</math>. Given that <math>yz</math> = <math>60</math> we can substitute <math>(10-10i)(z)=60</math> We can solve for z and get <math>z=3+3i</math>. Now we know what <math>x</math>, <math>y</math>, and <math>z</math> are, so all we have to do is plug and chug. <math>x+y+z= (-20-12i)+(10+10i)+(3-3i)</math> or <math>x+y+z= -7-5i</math> Now <math>a^2 +b^2=(-7)^2+(-5)^2</math> or <math>a^2 +b^2 = 74</math>. Thus <math>074</math> is our final answer.(David Camacho)<br />
<br />
==Solution 4 ==<br />
<br />
We observe that by multiplying <math>xy,</math> <math>yz,</math> and <math>zx,</math> we get <math>(xyz)^2=(-80-320i)(60)(-96+24i).</math> Next, we divide <math>(xyz)^2</math> by <math>(yz)^2</math> to <br />
<br />
get <math>x^2.</math> We have <math>x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.</math> We can write <math>x</math> in the form of <math>a+bi,</math> so we get <br />
<br />
<math>(a+bi)^2=256+480i.</math> Then, <math>a^2-b^2+2abi=256+480i,</math> <math>a^2-b^2=256,</math> and <math>2ab=480.</math> Solving this system of equations is relatively <br />
<br />
simple. We have two cases, <math>a=20, b=12,</math> and <math>a=-20, b=-12.</math> <br />
<br />
Case 1: <math>a=20, b=12,</math> so <math>x=20+12i.</math> We solve for <math>y</math> and <math>z</math> by plugging in <math>x</math> to the two equations. We see<br />
<br />
<math>y=\frac{-80-320i}{20+12i}=-10-10i</math> and <math>z=\frac{-96+24i}{20+12i}=-3+3i.</math> <math>x+y+z=7+5i,</math> so <math>a=7</math> and <math>b=5.</math> Solving, we end up with <br />
<br />
<math>7^2+5^2=\boxed{074}</math> as our answer. <br />
<br />
Case 2: <math>a=-20, b=-12,</math> so <math>x=-20-12i.</math> Again, we solve for <math>y</math> and <math>z.</math> We find <math>y=\frac{-80-320i}{-20-12i}=10+10i,</math> <br />
<br />
<math>z=\frac{-96+24i}{-20-12i}=3-3i,</math> so <math>x+y+z=-7-5i.</math> We again have <math>(-7)^2+(-5)^2=\boxed{074}.</math><br />
<br />
Solution by <math>Airplane50</math><br />
<br />
==Solution 5 (Based on advanced mathematical knowledge)==<br />
According to the Euler's Theory, we can rewrite <math>x</math>, <math>y</math> and <math>z</math> as <cmath>x=r_{1}e^{i{\theta}_1}</cmath> <cmath>y=r_{2}e^{i{\theta}_2}</cmath> <cmath>x=r_{3}e^{i{\theta}_3}</cmath> As a result, <cmath>|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}</cmath> <cmath>|yz|=r_{2}r_{3}=60</cmath> <cmath>|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}</cmath> Also, it is clear that <cmath>yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60</cmath> So <math>{\theta}_2+{\theta}_3=0</math>, or <cmath>{\theta}_2=-{\theta}_3</cmath> Also, we have <cmath>xy=-80\sqrt{17}e^{i\arctan{4}}</cmath> <cmath>yz=60</cmath> <cmath>xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}</cmath> So now we have <math>r_{1}r_{2}=80\sqrt{17}</math>, <math>r_{2}r_{3}=60</math>, <math>r_{1}r_{3}=24\sqrt{17}</math>, <math>{\theta}_1+{\theta}_2=\arctan{4}</math> and <math>{\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}</math>. Solve these above, we get <cmath>r_{1}=4\sqrt{34}</cmath> <cmath>r_{2}=10\sqrt{2}</cmath> <cmath>r_{3}=3\sqrt{2}</cmath> <cmath>{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}</cmath> So we can get <cmath>y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i</cmath> <cmath>z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i</cmath> Use <math>xy=-80-320i</math> we can find that <cmath>x=-20-12i</cmath> So <cmath>x+y+z=-20-12i+10+10i+3-3i=-7-5i</cmath> So we have <math>a=-7</math> and <math>b=-5</math>.<br />
<br />
As a result, we finally get <cmath>a^2+b^2=(-7)^2+(-5)^2=\boxed{074}</cmath><br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
==Solution 6==<br />
<br />
We can turn the expression <math>x+y+z</math> into <math>\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}</math>, and this would allow us to plug in the values after some computations.<br />
<br />
Based off of the given products, we have<br />
<br />
<math>xy^2z=60(-80-320i)</math><br />
<br />
<math>xyz^2=60(-96+24i)</math><br />
<br />
<math>x^2yz=(-96+24i)(-80-320i)</math>.<br />
<br />
Dividing by the given products, we have<br />
<br />
<math>y^2=\frac{60(-80-320i)}{-96+24i}</math><br />
<br />
<math>z^2=\frac{60(-96+24i)}{-80-320i}</math><br />
<br />
<math>x^2=\frac{(-96+24i)(-80-320i)}{60}</math>.<br />
<br />
Simplifying, we get that this expression becomes <math>\sqrt{24+70i}</math>. This equals <math>\pm{(7+5i)}</math>, so the answer is <math>7^2+5^2=\boxed{074}</math>.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
<br />
{{AIME box|year=2018|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_16&diff=994242017 AMC 12B Problems/Problem 162018-12-13T06:35:06Z<p>Rootthreeovertwo: </p>
<hr />
<div>==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
==Solution==<br />
If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math>.After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>.<br />
<br />
Solution by: vedadehhc<br />
<br />
==Solution 2==<br />
We can write <math>21!</math> as its prime factorization:<br />
<cmath>21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19</cmath><br />
<br />
Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; <math>2^{18}</math> is going to have <math>19</math> factors: <math>2^0, 2^1, 2^2,...\text{ }2^{18}</math>, and the other exponents will behave identically. <br />
<br />
In other words, <math>21!</math> has <math>(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)</math> factors. <br />
<br />
We are looking for the probability that a randomly chosen factor of <math>21!</math> will be odd--numbers that do not contain multiples of <math>2</math> as factors.<br />
<br />
From our earlier observation, the only factors of <math>21!</math> that are even are ones with at least one multiplier of <math>2</math>, so our probability of finding an odd factor becomes the following:<br />
<cmath>P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}</cmath><br />
<br />
Solution submitted by [[User:TrueshotBarrage|David Kim]]<br />
<br />
==Solution 3 (Time Shortage)==<br />
<br />
<math>\textbf{WARNING}</math>: Only use this solution if you are running out of time.<br />
<br />
Prime factor <math>21!</math>. It is <math>2^{18}\cdot3^9\cdot5^4\cdot7^3\cdot11\cdot13\cdot17\cdot19</math>. <br />
<br />
The number of factors is <math>19\cdot10\cdot5\cdot4\cdot2\cdot2\cdot2\cdot2</math>. <br />
Looking at the choices and using some quick computation, one can figure that the only sensical answer is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
[[Category:Introductory Probability Problems]]</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=990642017 AIME II Problems/Problem 62018-11-25T21:44:56Z<p>Rootthreeovertwo: /* Solution 6 */</p>
<hr />
<div>==Problem==<br />
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.<br />
<br />
==Solution 1==<br />
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 2==<br />
Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so<br />
<br />
<cmath>n=\frac{x^2-2017}{85-2x}.</cmath><br />
<br />
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 3 (Abuse the discriminant)==<br />
<br />
Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br />
<br />
Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.<br />
<br />
The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.<br />
<br />
==Solution 4==<br />
<br />
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.<br />
<br />
==Solution 5 (Using factors)==<br />
<br />
Let the expression be equal to <math>a</math>. This expression can be factored into <math>\sqrt{(n+44)^2-3n+81}</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.<br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 6==<br />
<br />
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>71^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_13&diff=990631984 AIME Problems/Problem 132018-11-25T21:42:42Z<p>Rootthreeovertwo: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
Find the value of <math>10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).</math><br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
We know that <math>\tan(\arctan(x)) = x</math> so we can repeatedly apply the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, <math>c=\cot^{-1}(13)</math>, and <math>d=\cot^{-1}(21)</math>. We have<br />
<br />
<center><p><math>\tan(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}</math>,</p></center><br />
<br />
So<br />
<br />
<center><p><math>\tan(a+b) = \frac{\frac{1}{3}+\frac{1}{7}}{1-\frac{1}{21}} = \frac{1}{2}</math></p></center><br />
<br />
and<br />
<br />
<center><p><math>\tan(c+d) = \frac{\frac{1}{13}+\frac{1}{21}}{1-\frac{1}{273}} = \frac{1}{8}</math>,</p></center><br />
<br />
so<br />
<br />
<center><p><math>\tan((a+b)+(c+d)) = \frac{\frac{1}{2}+\frac{1}{8}}{1-\frac{1}{16}} = \frac{2}{3}</math>.</p></center><br />
<br />
Thus our answer is <math>10\cdot\frac{3}{2}=15</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. Using it twice on the inside, the desired sum becomes <math>\cot (\cot^{-1}2+\cot^{-1}8)</math>. This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.<br />
<br />
=== Solution 3 ===<br />
<br />
On the coordinate plane, let <math>O=(0,0)</math>, <math>A_1=(3,0)</math>, <math>A_2=(3,1)</math>, <math>B_1=(21,7)</math>, <math>B_2=(20,10)</math>, <math>C_1=(260,130)</math>, <math>C_2=(250,150)</math>, <math>D_1=(5250,3150)</math>, <math>D_2=(5100,3400)</math>, and <math>H=(5100,0)</math>. We see that <math>\cot^{-1}(\angle A_2OA_1)=3</math>, <math>\cot^{-1}(\angle B_2OB_1)=7</math>, <math>\cot^{-1}(\angle C_2OC_1)=13</math>, and <math>\cot^{-1}(\angle D_2OD_1)=21</math>. The sum of these four angles forms the angle of triangle <math>OD_2H</math>, which has a cotangent of <math>\frac{5100}{3400}=\frac{3}{2}</math>, which must mean that <math> \cot( \cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)=\frac{3}{2}</math>. So the answer is <math>10*\left(\frac{3}{2}\right)=\boxed{015}</math>.<br />
<br />
<br />
===Solution 4===<br />
Recall that <math>\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta</math> and that <math>\arg(a + bi) = \tan^{-1}\frac{b}{a}</math>. Then letting <math>w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,</math> and <math>z = 1 + 21i</math>, we are left with<br />
<br />
<cmath>10\cot(\frac{\pi}{2} - \arg w + \frac{\pi}{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)</cmath><br />
<cmath>= -10\cot(\arg wxyz).</cmath><br />
<br />
Expanding <math>wxyz</math>, we are left with<br />
<cmath>(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)</cmath><br />
<cmath>= (2+i)(13+i)(21+i)</cmath><br />
<cmath>= (25+15i)(21+i)</cmath><br />
<cmath>= (5+3i)(21+i)</cmath><br />
<cmath>= (102+68i)</cmath><br />
<cmath>= (3+2i)</cmath><br />
<cmath>= 10\cot \tan^{-1}\frac{2}{3}</cmath><br />
<cmath> = 10 \cdot \frac{3}{2} = \boxed{015}</cmath><br />
<br />
== See also ==<br />
{{AIME box|year=1984|num-b=12|num-a=14}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_13&diff=990621984 AIME Problems/Problem 132018-11-25T21:42:04Z<p>Rootthreeovertwo: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
Find the value of <math>10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).</math><br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
We know that <math>\tan(\arctan(x)) = x</math> so we can repeatedly apply the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, <math>c=\cot^{-1}(13)</math>, and <math>d=\cot^{-1}(21)</math>. We have<br />
<br />
<center><p><math>\tan(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}</math>,</p></center><br />
<br />
So<br />
<br />
<center><p><math>\tan(a+b) = \frac{\frac{1}{3}+\frac{1}{7}}{1-\frac{1}{21}} = \frac{1}{2}</math></p></center><br />
<br />
and<br />
<br />
<center><p><math>\tan(c+d) = \frac{\frac{1}{13}+\frac{1}{21}}{1-\frac{1}{273}} = \frac{1}{8}</math>,</p></center><br />
<br />
so<br />
<br />
<center><p><math>\tan((a+b)+(c+d)) = \frac{\frac{1}{2}+\frac{1}{8}}{1-\frac{1}{16}} = \frac{2}{3}</math>.</p></center><br />
<br />
Thus our answer is <math>10\cdot\frac{3}{2}=15</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. Using it twice on the inside, the desired sum becomes <math>\cot (\cot^{-1}2+\cot^{-1}8)</math>. This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.<br />
<br />
=== Solution 3 ===<br />
<br />
On the coordinate plane, let <math>O=(0,0)</math>, <math>A_1=(3,0)</math>, <math>A_2=(3,1)</math>, <math>B_1=(21,7)</math>, <math>B_2=(20,10)</math>, <math>C_1=(260,130)</math>, <math>C_2=(250,150)</math>, <math>D_1=(5250,3150)</math>, <math>D_2=(5100,3400)</math>, and <math>H=(5100,0)</math>. We see that <math>\cot^{-1}(\angle A_2OA_1)=3</math>, <math>\cot^{-1}(\angle B_2OB_1)=7</math>, <math>\cot^{-1}(\angle C_2OC_1)=13</math>, and <math>\cot^{-1}(\angle D_2OD_1)=21</math>. The sum of these four angles forms the angle of triangle <math>OD_2H</math>, which has a cotangent of <math>\frac{5100}{3400}=\frac{3}{2}</math>, which must mean that <math> \cot( \cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)=\frac{3}{2}</math>. So the answer is <math>10*\left(\frac{3}{2}\right)=\boxed{015}</math>.<br />
<br />
<br />
===Solution 4===<br />
Recall that <math>\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta</math> and that <math>\arg(a + bi) = \tan^{-1}\frac{b}{a}</math>. Then letting <math>w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,</math> and <math>z = 1 + 21i</math>, we are left with<br />
<br />
<cmath>10\cot(\frac{\pi}{2} - \arg w + \frac{\pi}{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)</cmath><br />
<cmath>= -10\cot(\arg wxyz).</cmath><br />
<br />
Expanding <math>wxyz</math>, we are left with<br />
<cmath>(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)</cmath><br />
<cmath>= (2+i)(13+i)(21+i)</cmath><br />
<cmath>= (25+15i)(21+i)</cmath><br />
<cmath>= (5+3i)(21+i)</cmath><br />
<cmath>= (102+68i)</cmath><br />
<cmath>= (3+2i)</cmath><br />
<cmath>= 10\cot \tan^{-1}\frac{2}{3}</cmath><br />
<cmath> = 10\cdot\frac{3}{2} = \boxed{015}</cmath><br />
<br />
== See also ==<br />
{{AIME box|year=1984|num-b=12|num-a=14}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_13&diff=986471992 AIME Problems/Problem 132018-11-21T05:54:48Z<p>Rootthreeovertwo: </p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>AB=9</math> and <math>BC: AC=40: 41</math>. What's the largest area that this triangle can have?<br />
<br />
== Solution ==<br />
===Solution 1===<br />
First, consider the [[triangle]] in a [[coordinate system]] with [[vertex|vertices]] at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>. Applying the [[distance formula]], we see that <math>\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}</math>.<br />
<br />
We want to maximize <math>b</math>, the height, with <math>9</math> being the base.<br />
<br />
Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>.<br />
<br />
To maximize <math>b</math>, we want to maximize <math>b^2</math>. So if we can write: <math>b^2=-(a+n)^2+m</math>, then <math>m</math> is the maximum value of <math>b^2</math> (this follows directly from the [[trivial inequality]], because if <math>{x^2 \ge 0}</math> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>).<br />
<br />
<math>b^2=-a^2 -\frac{3200}{9}a +1600=-\left(a +\frac{1600}{9}\right)^2 +1600+\left(\frac{1600}{9}\right)^2</math>.<br />
<br />
<math>\Rightarrow b\le\sqrt{1600+\left(\frac{1600}{9}\right)^2}=40\sqrt{1+\frac{1600}{81}}=\frac{40}{9}\sqrt{1681}=\frac{40\cdot 41}{9}</math>.<br />
<br />
Then the area is <math>9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}</math>.<br />
<br />
===Solution 2===<br />
Let the three sides be <math>9,40x,41x</math>, so the area is <math>\frac14\sqrt {(81^2 - 81x^2)(81x^2 - 1)}</math> by Heron's formula. By AM-GM, <math>\sqrt {(81^2 - 81x^2)(81x^2 - 1)}\le\frac {81^2 - 1}2</math>, and the maximum possible area is <math>\frac14\cdot\frac {81^2 - 1}2 = \frac18(81 - 1)(81 + 1) = 10\cdot82 = \boxed{820}</math>. This occurs when <math>81^2 - 81x^2 = 81x^2 - 1\implies x = \frac {4\sqrt {205}}9</math>.<br />
<br />
===Solution 3===<br />
Let <math>A, B</math> be the endpoints of the side with length <math>9</math>. Let <math>\Gamma</math> be the Apollonian Circle of <math>AB</math> with ratio <math>40:41</math>; let this intersect <math>AB</math> at <math>P</math> and <math>Q</math>, where <math>P</math> is inside <math>AB</math> and <math>Q</math> is outside. Then because <math>(A, B; P, Q)</math> describes a harmonic set, <math>AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{\frac{40}{9}}{BQ}\implies BQ=360</math>. Finally, this means that the radius of <math>\Gamma</math> is <math>\dfrac{360+\frac{40}{9}}{2}=180+\dfrac{20}{9}</math>.<br />
<br />
Since the area is maximized when the altitude to <math>AB</math> is maximized, clearly we want the last vertex to be the highest point of <math>\Gamma</math>, which just makes the altitude have length <math>180+\dfrac{20}{9}</math>. Thus, the area of the triangle is <math>\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}</math><br />
<br />
===Solution 4 (Involves Basic Calculus)===<br />
We can apply Heron's on this triangle after letting the two sides equal <math>40x</math> and <math>41x</math>. Heron's gives<br />
<br />
<math>\sqrt{(\frac{81x+9}{2})(\frac{81x-9}{2})(\frac{x+9}{2})(\frac{-x+9}{2})}</math>.<br />
<br />
This can be simplified to<br />
<br />
<math>\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}</math>.<br />
<br />
We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.<br />
<br />
We have that <math>-324x^3+13124x=0</math>, so <math>x=\frac{\sqrt{3281}}{9}</math>.<br />
<br />
Plugging this into the expression, we have that the area is <math>\boxed{820}</math>.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
== See also ==<br />
{{AIME box|year=1992|num-b=12|num-a=14}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_9&diff=986461990 AIME Problems/Problem 92018-11-21T05:40:35Z<p>Rootthreeovertwo: </p>
<hr />
<div>== Problem ==<br />
A [[fair]] coin is to be tossed <math>10_{}^{}</math> times. Let <math>i/j^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>. <br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Clearly, at least <math>5</math> tails must be flipped; any less, then by the [[Pigeonhole Principle]] there will be heads that appear on consecutive tosses. <br />
<br />
Consider the case when <math>5</math> tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled <math>(H)</math>:<br />
<br />
:<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</math><br />
<br />
There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, using [[stars-and-bars]] there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.<br />
<br />
=== Solution 2 ===<br />
Call the number of ways of flipping <math>n</math> coins and not receiving any consecutive heads <math>S_n</math>. Notice that tails must be received in at least one of the first two flips.<br />
<br />
If the first coin flipped is a T, then the remaining <math>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.<br />
<br />
If the first coin flipped is a H, then the second coin must be a T. There are then <math>S_{n-2}</math> configurations.<br />
<br />
Thus, <math>S_n = S_{n-1} + S_{n-2}</math>. By counting, we can establish that <math>S_1 = 2</math> and <math>S_2 = 3</math>. Therefore, <math>S_3 = 5,\ S_4 = 8</math>, forming the [[Fibonacci sequence]]. Listing them out, we get <math>2,3,5,8,13,21,34,55,89,144</math>, and the 10th number is <math>144</math>. Putting this over <math>2^{10}</math> to find the probability, we get <math>\frac{9}{64}</math>. Our solution is <math>9+64=\boxed{073}</math>.<br />
<br />
=== Solution 3 ===<br />
We can also split the problem into casework.<br />
<br />
Case 1: 0 Heads<br />
<br />
There is only one possibility.<br />
<br />
Case 2: 1 Head<br />
<br />
There are 10 possibilities.<br />
<br />
Case 3: 2 Heads<br />
<br />
There are 36 possibilities.<br />
<br />
Case 4: 3 Heads<br />
<br />
There are 56 possibilities.<br />
<br />
Case 5: 4 Heads<br />
<br />
There are 35 possibilities.<br />
<br />
Case 6: 5 Heads<br />
<br />
There are 6 possibilities.<br />
<br />
We have <math>1+10+36+56+35+6=144</math>, and there are <math>1024</math> possible outcomes, so the probability is <math>\frac{144}{1024}=\frac{9}{64}</math>, and the answer is <math>\boxed{073}</math>.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
== See also ==<br />
{{AIME box|year=1990|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_9&diff=986451990 AIME Problems/Problem 92018-11-21T05:40:08Z<p>Rootthreeovertwo: </p>
<hr />
<div>== Problem ==<br />
A [[fair]] coin is to be tossed <math>10_{}^{}</math> times. Let <math>i/j^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>. <br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Clearly, at least <math>5</math> tails must be flipped; any less, then by the [[Pigeonhole Principle]] there will be heads that appear on consecutive tosses. <br />
<br />
Consider the case when <math>5</math> tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled <math>(H)</math>:<br />
<br />
:<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</math><br />
<br />
There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, using [[stars-and-bars]] there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.<br />
<br />
=== Solution 2 ===<br />
Call the number of ways of flipping <math>n</math> coins and not receiving any consecutive heads <math>S_n</math>. Notice that tails must be received in at least one of the first two flips.<br />
<br />
If the first coin flipped is a T, then the remaining <math>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.<br />
<br />
If the first coin flipped is a H, then the second coin must be a T. There are then <math>S_{n-2}</math> configurations.<br />
<br />
Thus, <math>S_n = S_{n-1} + S_{n-2}</math>. By counting, we can establish that <math>S_1 = 2</math> and <math>S_2 = 3</math>. Therefore, <math>S_3 = 5,\ S_4 = 8</math>, forming the [[Fibonacci sequence]]. Listing them out, we get <math>2,3,5,8,13,21,34,55,89,144</math>, and the 10th number is <math>144</math>. Putting this over <math>2^{10}</math> to find the probability, we get <math>\frac{9}{64}</math>. Our solution is <math>9+64=\boxed{073}</math>.<br />
<br />
=== Solution 3 ===<br />
We can also split the problem into casework.<br />
<br />
Case 1: 0 Heads<br />
<br />
There is only one possibility.<br />
<br />
Case 2: 1 Head<br />
<br />
There are 10 possibilities.<br />
<br />
Case 3: 2 Heads<br />
<br />
There are 36 possibilities.<br />
<br />
Case 4: 3 Heads<br />
<br />
There are 56 possibilities.<br />
<br />
Case 5: 4 Heads<br />
<br />
There are 35 possibilities.<br />
<br />
Case 6: 5 Heads<br />
<br />
There are 6 possibilities.<br />
<br />
We have <math>1+10+36+56+35+6=144</math>, and there are <math>1024</math> possible outcomes, so the probability is <math>\frac{144}{1024}=\frac{9}{64}</math>, so the answer is <math>\boxed{073}</math>.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
== See also ==<br />
{{AIME box|year=1990|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_9&diff=986441990 AIME Problems/Problem 92018-11-21T05:38:38Z<p>Rootthreeovertwo: </p>
<hr />
<div>== Problem ==<br />
A [[fair]] coin is to be tossed <math>10_{}^{}</math> times. Let <math>i/j^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>. <br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Clearly, at least <math>5</math> tails must be flipped; any less, then by the [[Pigeonhole Principle]] there will be heads that appear on consecutive tosses. <br />
<br />
Consider the case when <math>5</math> tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled <math>(H)</math>:<br />
<br />
:<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</math><br />
<br />
There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, using [[stars-and-bars]] there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.<br />
<br />
=== Solution 2 ===<br />
Call the number of ways of flipping <math>n</math> coins and not receiving any consecutive heads <math>S_n</math>. Notice that tails must be received in at least one of the first two flips.<br />
<br />
If the first coin flipped is a T, then the remaining <math>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.<br />
<br />
If the first coin flipped is a H, then the second coin must be a T. There are then <math>S_{n-2}</math> configurations.<br />
<br />
Thus, <math>S_n = S_{n-1} + S_{n-2}</math>. By counting, we can establish that <math>S_1 = 2</math> and <math>S_2 = 3</math>. Therefore, <math>S_3 = 5,\ S_4 = 8</math>, forming the [[Fibonacci sequence]]. Listing them out, we get <math>2,3,5,8,13,21,34,55,89,144</math>, and the 10th number is <math>144</math>. Putting this over <math>2^{10}</math> to find the probability, we get <math>\frac{9}{64}</math>. Our solution is <math>9+64=\boxed{073}</math>.<br />
<br />
=== Solution 3 ===<br />
We can also split the problem into casework.<br />
\newline Case 1: 0 Heads<br />
\newline There is only one possibility.<br />
\newline Case 2: 1 Head<br />
\newline There are 10 possibilities.<br />
\newline Case 3: 2 Heads<br />
\newline There are 36 possibilities.<br />
\newline Case 4: 3 Heads<br />
\newline There are 56 possibilities.<br />
\newline Case 5: 4 Heads<br />
\newline There are 35 possibilities.<br />
\newline Case 6: 5 Heads<br />
\newline There are 6 possibilities.<br />
\newline We have <math>1+10+36+56+35+6=144</math>, and there are <math>1024</math> possible outcomes, so the probability is <math>\frac{144}{1024}=\frac{9}{64}</math>, so the answer is <math>\boxed{073}</math>.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
== See also ==<br />
{{AIME box|year=1990|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_9&diff=986431990 AIME Problems/Problem 92018-11-21T05:37:27Z<p>Rootthreeovertwo: </p>
<hr />
<div>== Problem ==<br />
A [[fair]] coin is to be tossed <math>10_{}^{}</math> times. Let <math>i/j^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>. <br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Clearly, at least <math>5</math> tails must be flipped; any less, then by the [[Pigeonhole Principle]] there will be heads that appear on consecutive tosses. <br />
<br />
Consider the case when <math>5</math> tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled <math>(H)</math>:<br />
<br />
:<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</math><br />
<br />
There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, using [[stars-and-bars]] there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.<br />
<br />
=== Solution 2 ===<br />
Call the number of ways of flipping <math>n</math> coins and not receiving any consecutive heads <math>S_n</math>. Notice that tails must be received in at least one of the first two flips.<br />
<br />
If the first coin flipped is a T, then the remaining <math>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.<br />
<br />
If the first coin flipped is a H, then the second coin must be a T. There are then <math>S_{n-2}</math> configurations.<br />
<br />
Thus, <math>S_n = S_{n-1} + S_{n-2}</math>. By counting, we can establish that <math>S_1 = 2</math> and <math>S_2 = 3</math>. Therefore, <math>S_3 = 5,\ S_4 = 8</math>, forming the [[Fibonacci sequence]]. Listing them out, we get <math>2,3,5,8,13,21,34,55,89,144</math>, and the 10th number is <math>144</math>. Putting this over <math>2^{10}</math> to find the probability, we get <math>\frac{9}{64}</math>. Our solution is <math>9+64=\boxed{073}</math>.<br />
<br />
=== Solution 3 ===<br />
We can also split the problem into casework.<br />
Case 1: 0 Heads<br />
There is only one possibility.<br />
Case 2: 1 Head<br />
There are 10 possibilities.<br />
Case 3: 2 Heads<br />
There are 36 possibilities.<br />
Case 4: 3 Heads<br />
There are 56 possibilities.<br />
Case 5: 4 Heads<br />
There are 35 possibilities.<br />
Case 6: 5 Heads<br />
There are 6 possibilities.<br />
We have <math>1+10+36+56+35+6=144</math>, and there are <math>1024</math> possible outcomes, so the probability is <math>\frac{144}{1024}=\frac{9}{64}</math>, so the answer is <math>\boxed{073}</math>.<br />
<br />
\textbf{-RootThreeOverTwo}<br />
== See also ==<br />
{{AIME box|year=1990|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_13&diff=986421990 AIME Problems/Problem 132018-11-21T04:38:36Z<p>Rootthreeovertwo: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Let <math>T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}</math>. Given that <math>9^{4000}_{}</math> has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of <math>T_{}^{}</math> have 9 as their leftmost digit?<br />
<br />
. <br />
<br />
Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, <math>4000 - 3816 = \boxed{184}</math> numbers have 9 as their leftmost digits.<br />
<br />
== See also ==<br />
{{AIME box|year=1990|num-b=12|num-a=14}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=961552018 AIME I Problems/Problem 42018-07-13T02:04:19Z<p>Rootthreeovertwo: /* Solution 5 (Fastest using Law of Cosines) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 3 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 4==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 5 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot cos A</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot cos A</math><br />
<br />
Solving for <math>cos A</math> in both equations, we get<br />
<br />
<math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=944882018 AIME I Problems/Problem 52018-05-15T02:36:43Z<p>Rootthreeovertwo: /* Note */</p>
<hr />
<div>For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>.<br />
<br />
==Solution==<br />
<br />
Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. <br />
That gives <math>x^2+xy-2y^2=0</math> upon simplification and division by <math>3</math>. Then, <math>x=y</math> or <math>x=-2y</math>.<br />
From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>.<br />
<br />
-expiLnCalc<br />
<br />
==Note==<br />
<br />
The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=944872017 AIME I Problems/Problem 42018-05-15T02:34:41Z<p>Rootthreeovertwo: /* Solution 2 (Coordinates) */</p>
<hr />
<div>==Problem 4==<br />
A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. We find that the altitude to side <math>\overline {AB}</math> is <math>16</math>, so the area of <math>\triangle ABC</math> is <math>(24*16)/2 = 192</math>.<br />
<br />
Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C</math>. Then,<br />
<br />
<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath><br />
<br />
Let <math>\overline {OM} = d</math>.<br />
Equation <math>(1)</math>:<br />
<cmath>d + \sqrt {d^2 + 144} = 16</cmath><br />
<br />
Squaring both sides, we have<br />
<br />
<cmath>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</cmath><br />
<br />
<cmath>2d^2 + 2d\sqrt {d^2+144} = 112</cmath><br />
<br />
<cmath>2d(d + \sqrt {d^2+144}) = 112</cmath><br />
<br />
Substituting with equation <math>(1)</math>:<br />
<br />
<cmath>2d(16) = 112</cmath><br />
<br />
<cmath>d = 7/2</cmath><br />
<br />
We now find that <math>\sqrt{d^2 + 144} = 25/2</math>.<br />
<br />
Let the distance <math>\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by SSS):<br />
<br />
<cmath>25^2 = h^2 + (25/2)^2</cmath><br />
<br />
<cmath>625 = h^2 + 625/4</cmath><br />
<br />
<cmath>1875/4 = h^2</cmath><br />
<br />
<cmath>25\sqrt {3} / 2 = h</cmath><br />
<br />
<br />
Finally, by the formula for volume of a pyramid,<br />
<br />
<cmath>V = Bh/3</cmath><br />
<br />
<cmath>V = (192)(25\sqrt{3}/2)/3</cmath><br />
This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.<br />
<br />
==Shortcut==<br />
Here is a shortcut for finding the radius <math>R</math> of the circumcenter of <math>\triangle ABC</math>.<br />
<br />
As before, we find that the foot of the altitude from <math>P</math> lands on the circumcenter of <math>\triangle ABC</math>. Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>.<br />
Then we write the area of <math>\triangle ABC</math> in two ways:<br />
<cmath>[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}</cmath><br />
<br />
Plugging in <math>20</math>, <math>20</math>, and <math>24</math> for <math>a</math>, <math>b</math>, and <math>c</math> respectively, and solving for <math>R</math>, we obtain <math>R= \frac{25}{2}=OA=OB=OC</math>.<br />
<br />
Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid.<br />
<br />
==Solution 2 (Coordinates)==<br />
<br />
We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length <math>24</math> is at the origin, or <math>(0, 0, 0)</math>. Then, the two other vertices can be <math>(-12, -16, 0)</math> and <math>(12, -16, 0)</math>. Let the fourth vertex have coordinates of <math>(x, y, z)</math>. We have the following <math>3</math> equations from the distance formula.<br />
<br />
<cmath>x^2+y^2+z^2=625</cmath><br />
<br />
<cmath>(x+12)^2+(y+16)^2+z^2=625</cmath><br />
<br />
<cmath>(x-12)^2+(y+16)^2+z^2=625</cmath><br />
<br />
Adding the last two equations and substituting in the first equation, we get that <math>y=-\frac{25}{2}</math>. If you drew a good diagram, it should be obvious that <math>x=0</math>. Now, solving for <math>z</math>, we get that <math>z=\frac{25\sqrt{3}}{2}</math>. So, the height of the pyramid is <math>\frac{25\sqrt{3}}{2}</math>. The base is equal to the area of the triangle, which is <math>\frac{1}{2} \cdot 24 \cdot 16 = 192</math>. The volume is <math>\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}</math>. Thus, the answer is <math>800+3 = \boxed{803}</math>.<br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=944652018 AIME I Problems/Problem 42018-05-13T21:24:06Z<p>Rootthreeovertwo: /* Solution 5 (Fastest [Law of Cosines]) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 3 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 4==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 5 (Fastest using Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot cos A</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot cos A</math><br />
<br />
Solving for <math>cos A</math> in both equations, we get<br />
<br />
<math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=944632018 AIME I Problems/Problem 52018-05-13T21:23:05Z<p>Rootthreeovertwo: /* Note */</p>
<hr />
<div>For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>.<br />
<br />
==Solution==<br />
<br />
Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. <br />
That gives <math>x^2+xy-2y^2=0</math> upon simplification and division by <math>3</math>. Then, <math>x=y</math> or <math>x=-2y</math>.<br />
From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>.<br />
<br />
-expiLnCalc<br />
<br />
==Note==<br />
<br />
The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>.<br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=944622017 AIME II Problems/Problem 62018-05-13T21:22:43Z<p>Rootthreeovertwo: /* Solution 5 (Using factors) */</p>
<hr />
<div>==Problem==<br />
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.<br />
<br />
==Solution 1==<br />
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 2==<br />
Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so<br />
<br />
<cmath>n=\frac{x^2-2017}{85-2x}.</cmath><br />
<br />
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 3 (Abuse the discriminant)==<br />
<br />
Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br />
<br />
Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.<br />
<br />
The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.<br />
<br />
==Solution 4==<br />
<br />
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.<br />
<br />
==Solution 5 (Using factors)==<br />
<br />
Let the expression be equal to <math>a</math>. This expression can be factored into <math>\sqrt{(n+44)^2-3n+81}</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.<br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 6==<br />
<br />
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>61^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=944612017 AIME I Problems/Problem 42018-05-13T21:22:01Z<p>Rootthreeovertwo: /* Solution 2 (Coordinates) */</p>
<hr />
<div>==Problem 4==<br />
A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. We find that the altitude to side <math>\overline {AB}</math> is <math>16</math>, so the area of <math>\triangle ABC</math> is <math>(24*16)/2 = 192</math>.<br />
<br />
Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C</math>. Then,<br />
<br />
<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath><br />
<br />
Let <math>\overline {OM} = d</math>.<br />
Equation <math>(1)</math>:<br />
<cmath>d + \sqrt {d^2 + 144} = 16</cmath><br />
<br />
Squaring both sides, we have<br />
<br />
<cmath>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</cmath><br />
<br />
<cmath>2d^2 + 2d\sqrt {d^2+144} = 112</cmath><br />
<br />
<cmath>2d(d + \sqrt {d^2+144}) = 112</cmath><br />
<br />
Substituting with equation <math>(1)</math>:<br />
<br />
<cmath>2d(16) = 112</cmath><br />
<br />
<cmath>d = 7/2</cmath><br />
<br />
We now find that <math>\sqrt{d^2 + 144} = 25/2</math>.<br />
<br />
Let the distance <math>\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by SSS):<br />
<br />
<cmath>25^2 = h^2 + (25/2)^2</cmath><br />
<br />
<cmath>625 = h^2 + 625/4</cmath><br />
<br />
<cmath>1875/4 = h^2</cmath><br />
<br />
<cmath>25\sqrt {3} / 2 = h</cmath><br />
<br />
<br />
Finally, by the formula for volume of a pyramid,<br />
<br />
<cmath>V = Bh/3</cmath><br />
<br />
<cmath>V = (192)(25\sqrt{3}/2)/3</cmath><br />
This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.<br />
<br />
==Shortcut==<br />
Here is a shortcut for finding the radius <math>R</math> of the circumcenter of <math>\triangle ABC</math>.<br />
<br />
As before, we find that the foot of the altitude from <math>P</math> lands on the circumcenter of <math>\triangle ABC</math>. Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>.<br />
Then we write the area of <math>\triangle ABC</math> in two ways:<br />
<cmath>[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}</cmath><br />
<br />
Plugging in <math>20</math>, <math>20</math>, and <math>24</math> for <math>a</math>, <math>b</math>, and <math>c</math> respectively, and solving for <math>R</math>, we obtain <math>R= \frac{25}{2}=OA=OB=OC</math>.<br />
<br />
Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid.<br />
<br />
==Solution 2 (Coordinates)==<br />
<br />
We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length <math>24</math> is at the origin, or <math>(0, 0, 0)</math>. Then, the two other vertices can be <math>(-12, -16, 0)</math> and <math>(12, -16, 0)</math>. Let the fourth vertex have coordinates of <math>(x, y, z)</math>. We have the following <math>3</math> equations from the distance formula.<br />
<br />
<cmath>x^2+y^2+z^2=625</cmath><br />
<br />
<cmath>(x+12)^2+(y+16)^2+z^2=625</cmath><br />
<br />
<cmath>(x-12)^2+(y+16)^2+z^2=625</cmath><br />
<br />
Adding the last two equations and substituting in the first equation, we get that <math>y=-\frac{25}{2}</math>. If you drew a good diagram, it should be obvious that <math>x=0</math>. Now, solving for <math>z</math>, we get that <math>z=\frac{25\sqrt{3}}{2}</math>. So, the height of the pyramid is <math>\frac{25\sqrt{3}}{2}</math>. The base is equal to the area of the triangle, which is <math>\frac{1}{2} \cdot 24 \cdot 16 = 192</math>. The volume is therefore <math>\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}</math>. Thus, the answer is <math>800+3 = \boxed{803}</math>.<br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_4&diff=944592017 AIME I Problems/Problem 42018-05-13T19:29:18Z<p>Rootthreeovertwo: </p>
<hr />
<div>==Problem 4==<br />
A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. We find that the altitude to side <math>\overline {AB}</math> is <math>16</math>, so the area of <math>\triangle ABC</math> is <math>(24*16)/2 = 192</math>.<br />
<br />
Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C</math>. Then,<br />
<br />
<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath><br />
<br />
Let <math>\overline {OM} = d</math>.<br />
Equation <math>(1)</math>:<br />
<cmath>d + \sqrt {d^2 + 144} = 16</cmath><br />
<br />
Squaring both sides, we have<br />
<br />
<cmath>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</cmath><br />
<br />
<cmath>2d^2 + 2d\sqrt {d^2+144} = 112</cmath><br />
<br />
<cmath>2d(d + \sqrt {d^2+144}) = 112</cmath><br />
<br />
Substituting with equation <math>(1)</math>:<br />
<br />
<cmath>2d(16) = 112</cmath><br />
<br />
<cmath>d = 7/2</cmath><br />
<br />
We now find that <math>\sqrt{d^2 + 144} = 25/2</math>.<br />
<br />
Let the distance <math>\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by SSS):<br />
<br />
<cmath>25^2 = h^2 + (25/2)^2</cmath><br />
<br />
<cmath>625 = h^2 + 625/4</cmath><br />
<br />
<cmath>1875/4 = h^2</cmath><br />
<br />
<cmath>25\sqrt {3} / 2 = h</cmath><br />
<br />
<br />
Finally, by the formula for volume of a pyramid,<br />
<br />
<cmath>V = Bh/3</cmath><br />
<br />
<cmath>V = (192)(25\sqrt{3}/2)/3</cmath><br />
This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.<br />
<br />
==Shortcut==<br />
Here is a shortcut for finding the radius <math>R</math> of the circumcenter of <math>\triangle ABC</math>.<br />
<br />
As before, we find that the foot of the altitude from <math>P</math> lands on the circumcenter of <math>\triangle ABC</math>. Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>.<br />
Then we write the area of <math>\triangle ABC</math> in two ways:<br />
<cmath>[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}</cmath><br />
<br />
Plugging in <math>20</math>, <math>20</math>, and <math>24</math> for <math>a</math>, <math>b</math>, and <math>c</math> respectively, and solving for <math>R</math>, we obtain <math>R= \frac{25}{2}=OA=OB=OC</math>.<br />
<br />
Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid.<br />
<br />
==Solution 2 (Coordinates)==<br />
<br />
We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length <math>24</math> is at the origin, or <math>(0, 0, 0)</math>. Then, the two other vertices can be <math>(-12, -16, 0)</math> and <math>(12, -16, 0)</math>. Let the fourth vertex have coordinates of <math>(x, y, z)</math>. We have the following <math>3</math> equations from the distance formula.<br />
<br />
<cmath>x^2+y^2+z^2=625</cmath><br />
<br />
<cmath>(x+12)^2+(y+16)^2+z^2=625</cmath><br />
<br />
<cmath>(x-12)^2+(y+16)^2+z^2=625</cmath><br />
<br />
Adding the last two equations and substituting in the first equation, we get that <math>y=-\frac{25}{2}</math>. If you drew a good diagram, it should be obvious that <math>x=0</math>. Now, solving for <math>z</math>, we get that <math>z=\frac{25\sqrt{3}}{2}</math>. So, the height of the pyramid is <math>\frac{25\sqrt{3}}{2}</math>. The base is equal to the area of the triangle, which is <math>\frac{1}{2} \cdot 24 \cdot 16 = 192</math>. The volume is therefore <math>\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}</math>. Thus, the answer is <math>800+3 = \boxed{803}</math>.<br />
<br />
-RootThreeOverTwo<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=944572017 AIME II Problems/Problem 62018-05-13T07:12:28Z<p>Rootthreeovertwo: /* Solution 5 (Introducing a variable) */</p>
<hr />
<div>==Problem==<br />
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.<br />
<br />
==Solution 1==<br />
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 2==<br />
Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so<br />
<br />
<cmath>n=\frac{x^2-2017}{85-2x}.</cmath><br />
<br />
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 3 (Abuse the discriminant)==<br />
<br />
Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br />
<br />
Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.<br />
<br />
The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.<br />
<br />
==Solution 4==<br />
<br />
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.<br />
<br />
==Solution 5 (Using factors)==<br />
<br />
Let the expression be equal to <math>a</math>. This expression can be factored into <math>\sqrt{(n+44)^2-3n+81}</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.<br />
<br />
-RootThreeOverTwo<br />
<br />
==Solution 6==<br />
<br />
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>61^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=944562017 AIME II Problems/Problem 62018-05-13T07:12:00Z<p>Rootthreeovertwo: /* Solution 5 (Introducing a variable) */</p>
<hr />
<div>==Problem==<br />
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.<br />
<br />
==Solution 1==<br />
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 2==<br />
Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so<br />
<br />
<cmath>n=\frac{x^2-2017}{85-2x}.</cmath><br />
<br />
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 3 (Abuse the discriminant)==<br />
<br />
Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br />
<br />
Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.<br />
<br />
The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.<br />
<br />
==Solution 4==<br />
<br />
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.<br />
<br />
==Solution 5 (Introducing a variable)==<br />
<br />
Let the expression be equal to <math>a</math>. This expression can be factored into <math>\sqrt{(n+44)^2-3n+81}</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.<br />
<br />
-RootThreeOverTwo<br />
<br />
==Solution 6==<br />
<br />
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>61^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=944552017 AIME II Problems/Problem 62018-05-13T07:09:00Z<p>Rootthreeovertwo: /* Solution 5 (Introducing a Variable) */</p>
<hr />
<div>==Problem==<br />
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.<br />
<br />
==Solution 1==<br />
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 2==<br />
Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so<br />
<br />
<cmath>n=\frac{x^2-2017}{85-2x}.</cmath><br />
<br />
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 3 (Abuse the discriminant)==<br />
<br />
Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br />
<br />
Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.<br />
<br />
The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.<br />
<br />
==Solution 4==<br />
<br />
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.<br />
<br />
==Solution 5 (Introducing a variable)==<br />
<br />
Let the expression be equal to <math>a</math>. This expression can be factored into <math>(n+44)^2-3n+81</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.<br />
<br />
-RootThreeOverTwo<br />
<br />
==Solution 6==<br />
<br />
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>61^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=944542017 AIME II Problems/Problem 62018-05-13T07:08:41Z<p>Rootthreeovertwo: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.<br />
<br />
==Solution 1==<br />
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 2==<br />
Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so<br />
<br />
<cmath>n=\frac{x^2-2017}{85-2x}.</cmath><br />
<br />
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 3 (Abuse the discriminant)==<br />
<br />
Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br />
<br />
Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.<br />
<br />
The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.<br />
<br />
==Solution 4==<br />
<br />
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.<br />
<br />
==Solution 5 (Introducing a Variable)==<br />
<br />
Let the expression be equal to <math>a</math>. This expression can be factored into <math>(n+44)^2-3n+81</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.<br />
<br />
-RootThreeOverTwo<br />
<br />
==Solution 6==<br />
<br />
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>61^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_6&diff=944532017 AIME II Problems/Problem 62018-05-13T07:06:07Z<p>Rootthreeovertwo: </p>
<hr />
<div>==Problem==<br />
Find the sum of all positive integers <math>n</math> such that <math>\sqrt{n^2+85n+2017}</math> is an integer.<br />
<br />
==Solution 1==<br />
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 2==<br />
Clearly, the result when <math>n</math> is plugged into the given expression is larger than <math>n</math> itself. Let <math>x</math> be the positive difference between that result and <math>n</math>, so that <math>\sqrt{n^2+85n+2017}=n+x</math>. Squaring both sides and canceling the <math>n^2</math> terms gives <math>85n+2017=2xn+x^2</math>. Combining like terms, <math>(85-2x)n=x^2-2017</math>, so<br />
<br />
<cmath>n=\frac{x^2-2017}{85-2x}.</cmath><br />
<br />
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.<br />
<br />
==Solution 3 (Abuse the discriminant)==<br />
<br />
Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)<br />
<br />
Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.<br />
<br />
The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.<br />
<br />
==Solution 4==<br />
<br />
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.<br />
<br />
==Solution 5==<br />
<br />
Let the expression be equal to <math>a</math>. This expression can be factored into <math>(n+44)^2-3n+81</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.<br />
<br />
-RootThreeOverTwo <br />
<br />
==Solution 6==<br />
<br />
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>61^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=944522018 AIME I Problems/Problem 52018-05-13T06:31:32Z<p>Rootthreeovertwo: /* Note */</p>
<hr />
<div>For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>.<br />
<br />
==Solution==<br />
<br />
Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. <br />
That gives <math>x^2+xy-2y^2=0</math> upon simplification and division by <math>3</math>. Then, <math>x=y</math> or <math>x=-2y</math>.<br />
From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>.<br />
<br />
-expiLnCalc<br />
<br />
==Note==<br />
<br />
The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>.<br />
<br />
-RootThreeOverTwo<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=944512018 AIME I Problems/Problem 52018-05-13T06:29:52Z<p>Rootthreeovertwo: </p>
<hr />
<div>For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>.<br />
<br />
==Solution==<br />
<br />
Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. <br />
That gives <math>x^2+xy-2y^2=0</math> upon simplification and division by <math>3</math>. Then, <math>x=y</math> or <math>x=-2y</math>.<br />
From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>.<br />
<br />
-expiLnCalc<br />
<br />
==Note==<br />
<br />
The cases <math>x=y</math> or <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0</math>.<br />
<br />
-RootThreeOverTwo<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=944502018 AIME I Problems/Problem 42018-05-13T05:59:12Z<p>Rootthreeovertwo: /* Solution 5 (Fastest [Law of Cosines]) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 3 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 4==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 5 (Fastest [Law of Cosines])==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot cos A</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot cos A</math><br />
<br />
Solving for <math>cos A</math> in both equations, we get<br />
<br />
<math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
-RootThreeOverTwo<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=944492018 AIME I Problems/Problem 42018-05-13T05:54:03Z<p>Rootthreeovertwo: /* Solution 5 (Fastest [Law of Cosines]) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 3 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 4==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 5 (Fastest [Law of Cosines])==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot cos A</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot cos A</math><br />
<br />
Solving for <math>cos A</math> in both equations, we get<br />
<br />
<math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
~RootThreeOverTwo<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=944482018 AIME I Problems/Problem 42018-05-13T05:35:58Z<p>Rootthreeovertwo: /* Solution 5 (Fastest (Law of Cosines)) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 3 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 4==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 5 (Fastest [Law of Cosines])==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot cos A</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot cos A</math><br />
<br />
Solving for <math>cos A</math> in both equations, we get<br />
<br />
<math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x^2)}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
~RootThreeOverTwo<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=944472018 AIME I Problems/Problem 42018-05-13T05:35:39Z<p>Rootthreeovertwo: /* Solution 5 (Faster Law of Cosines) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 3 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 4==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 5 (Fastest (Law of Cosines))==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot cos A</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot cos A</math><br />
<br />
Solving for <math>cos A</math> in both equations, we get<br />
<br />
<math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x^2)}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
~RootThreeOverTwo<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=944462018 AIME I Problems/Problem 42018-05-13T05:34:38Z<p>Rootthreeovertwo: </p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 3 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 4==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 5 (Faster Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot cos A</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot cos A</math><br />
<br />
Solving for <math>cos A</math> in both equations, we get<br />
<br />
<math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x^2)}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
~RootThreeOverTwo<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Rootthreeovertwohttps://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_14&diff=943232016 AIME I Problems/Problem 142018-05-06T20:04:19Z<p>Rootthreeovertwo: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
<br />
Centered at each lattice point in the coordinate plane are a circle radius <math>\frac{1}{10}</math> and a square with sides of length <math>\frac{1}{5}</math> whose sides are parallel to the coordinate axes. The line segment from <math>(0,0)</math> to <math>(1001, 429)</math> intersects <math>m</math> of the squares and <math>n</math> of the circles. Find <math>m + n</math>.<br />
<br />
== Solution 1 ==<br />
<br />
First note that <math>1001 = 143 \cdot 7</math> and <math>429 = 143 \cdot 3</math> so every point of the form <math>(7k, 3k)</math> is on the line. Then consider the line <math>l</math> from <math>(7k, 3k)</math> to <math>(7(k + 1), 3(k + 1))</math>. Translate the line <math>l</math> so that <math>(7k, 3k)</math> is now the origin. There is one square and one circle that intersect the line around <math>(0,0)</math>. Then the points on <math>l</math> with an integral <math>x</math>-coordinate are, since <math>l</math> has the equation <math>y = \frac{3x}{7}</math>:<br />
<br />
<cmath> (0,0), (1, \frac{3}{7}), (2, \frac{6}{7}), (3, 1 + \frac{2}{7}), (4, 1 + \frac{5}{7}), (5, 2 + \frac{1}{7}), (6, 2 + \frac{4}{7}), (7,3). </cmath><br />
<br />
We claim that the lower right vertex of the square centered at <math>(2,1)</math> lies on <math>l</math>. Since the square has side length <math>\frac{1}{5}</math>, the lower right vertex of this square has coordinates <math>(2 + \frac{1}{10}, 1 - \frac{1}{10}) = (\frac{21}{10}, \frac{9}{10})</math>. Because <math>\frac{9}{10} = \frac{3}{7} \cdot \frac{21}{10}</math>, <math>(\frac{21}{10}, \frac{9}{10})</math> lies on <math>l</math>. Since the circle centered at <math>(2,1)</math> is contained inside the square, this circle does not intersect <math>l</math>. Similarly the upper left vertex of the square centered at <math>(5,2)</math> is on <math>l</math>. Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between <math>(0,0)</math> and <math>(7,3)</math> that intersect <math>l</math>. Since there are <math>\frac{1001}{7} = \frac{429}{3} = 143</math> segments from <math>(7k, 3k)</math> to <math>(7(k + 1), 3(k + 1))</math>, the above count is yields <math>143 \cdot 2 = 286</math> squares. Since every lattice point on <math>l</math> is of the form <math>(3k, 7k)</math> where <math>0 \le k \le 143</math>, there are <math>144</math> lattice points on <math>l</math>. Centered at each lattice point, there is one square and one circle, hence this counts <math>288</math> squares and circles. Thus <math>m + n = 286 + 288 = \boxed{574}</math>.<br />
<br />
(Solution by gundraja)<br />
<asy>size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted);<br />
for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));}<br />
</asy><br />
<br />
==Solution 2==<br />
<br />
This is mostly a clarification to Solution 1, but let's take the diagram for the origin to <math>(7,3)</math>. We have the origin circle and square intersected, then two squares, then the circle and square at <math>(7,3)</math>. If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segment (with its circles and squares) end to end from <math>(0,0)</math> to <math>(1001,429)</math>, which forms the line we need without overlapping. Since <math>143</math> of these segments are needed to do this, and <math>3</math> squares and <math>1</math> circle are intersected with each, there are <math>143 \cdot (3+1) = 572</math> squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are <math>572+2=\boxed{574}</math> squares and circles intersected in total.<br />
<br />
==Solution 3==<br />
This solution is a more systematic approach for finding when the line intersects the squares and circles. Because <math>1001 = 7*11*13</math> and <math>429=3*11*13</math>, the slope of our line is <math>\frac{3}{7}</math>, and we only need to consider the line in the rectangle from the origin to <math>(7,3)</math>, and we can iterate the line <math>11*13=143</math> times. First, we consider how to figure out if the line intersects a square. Given a lattice point <math>(x_1, y_1)</math>, we can think of representing a square centered at that lattice point as all points equal to <math>(x_1 \pm a, y_1 \pm b)</math> s.t. <math>0 \leq a,b \leq \frac{1}{10}</math>. If the line <math>y = \frac{3}{7}x</math> intersects the square, then we must have <math>\frac{y_1 + b}{x_1 + a} = \frac{3}{7}</math>. The line with the least slope that intersects the square intersects at the bottom right corner and the line with the greatest slope that intersects the square intersects at the top left corner; thus we must have that <math>\frac{3}{7}</math> lies in between these slopes, or that <math>\frac{y_1-\frac{1}{10}}{x_1+\frac{1}{10}} \leq \frac{3}{7} \leq \frac{y_1+\frac{1}{10}}{x_1-\frac{1}{10}}</math>. Simplifying, <math>3x_1 - 1 \leq 7y_1 \leq 3x_1 + 1</math>. Because <math>y</math> can only equal <math>0, 1, 2, 3</math>, we just do casework based on the values of <math>y</math> and find that the points <math>(2, 1)</math> and <math>(5, 2)</math> are intersected just at the corner of the square and <math>(0, 0), (7, 3)</math> are intersected through the center of the square. However, we disregard one of <math>(0, 0)</math> and <math>(7, 3)</math>, WLOG <math>(0, 0)</math>, since we just use it in our count for the next of the 143 segments. Therefore, in one of our "segments", 3 squares are intersected and 1 circle is intersected giving 4 total. Thus our answer is <math>143*4 = 572</math>. HOWEVER, we cannot forget that we ignored <math>(0, 0)</math>, which contributes another square and circle to our count, making the final answer <math>572 + 2 = \boxed{574}</math>.<br />
<br />
-Patrick4President<br />
<br />
== See also ==<br />
{{AIME box|year=2016|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Rootthreeovertwo