https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Junche&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T15:46:07ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_25&diff=2139282023 AMC 12B Problems/Problem 252024-01-31T16:14:46Z<p>Junche: /* Supplement (Calculating sin54/cos36 from Scratch) */</p>
<hr />
<div>{{duplicate|[[2023 AMC 10B Problems/Problem 25|2023 AMC 10B #25]] and [[2023 AMC 12B Problems/Problem 25|2023 AMC 12B #25]]}}<br />
<br />
==Problem==<br />
<br />
A regular pentagon with area <math>\sqrt{5}+1</math> is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?<br />
<br />
<math>\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}</math><br />
<br />
<br />
==Solution 1==<br />
<br />
[[File:Pentagon_2023_12B_Q25_dissmo.png|600px]]<br />
<br />
Let the original pentagon be <math>ABCDE</math> centered at <math>O</math>. The dashed lines represent the fold lines. WLOG, let's focus on vertex <math>A</math>.<br />
<br />
Since <math>A</math> is folded onto <math>O</math>, <math>AM = MO</math> where <math>M</math> is the intersection of <math>AO</math> and the creaseline between <math>A</math> and <math>O</math>. Note that the inner pentagon is regular, and therefore similar to the original pentagon, due to symmetry.<br />
<br />
Because of their similarity, the ratio of the inner pentagon's area to that of the outer pentagon can be represented by <br />
<br />
<math>(\frac{OM}{ON})^{2} = (\frac{\frac{OA}{2}}{OA\sin (\angle OAE)})^{2} = \frac{1}{4\sin^{2}54}</math> <br />
<br />
<br />
===Option 1: Knowledge===<br />
<br />
Remember that <math>\sin54 = \frac{1+\sqrt5}{4}</math>.<br />
<br />
<br />
===Option 2: Angle Identities===<br />
<br />
<math>\sin54 = \cos36</math><br />
<br />
<math>4\cos^{3}18-3\cos18 = 2\sin18\cos18</math><br />
<br />
<math>4(1-\sin^{2}18)-3-2\sin18=0</math><br />
<br />
<math>4\sin^{2}18+2\sin18-1=0</math><br />
<br />
<math>\sin18 = \frac{-1+\sqrt5}{4}</math><br />
<br />
<math>\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}</math><br />
<br />
<br />
<math>\sin^{2}54 =\frac{3+\sqrt5}{8}</math><br />
<br />
Let the inner pentagon be <math>Z</math>.<br />
<br />
<math>[Z] = \frac{1}{4\sin^{2}54}[ABCDE]</math><br />
<br />
<math> = \frac{2(1+\sqrt5)}{3+\sqrt5}</math><br />
<br />
<math> = \sqrt5-1</math><br />
<br />
<math>\boxed{B}</math><br />
<br />
-Dissmo<br />
<br />
==Solution 2==<br />
<br />
<br />
<asy><br />
unitsize(5cm);<br />
<br />
// Define the vertices of the pentagons<br />
pair A, B, C, D, E;<br />
pair F, G, H, I, J;<br />
<br />
// Calculate the vertices of the larger pentagon<br />
A = dir(90);<br />
B = dir(90 - 72);<br />
C = dir(90 - 2*72);<br />
D = dir(90 - 3*72);<br />
E = dir(90 - 4*72);<br />
<br />
<br />
// Draw the larger pentagon<br />
draw(A--B--C--D--E--cycle);<br />
<br />
pair O = (A+B+C+D+E)/5;<br />
pair AA,OO;<br />
real gap = 0.02;<br />
AA = A+(0,0);<br />
OO = O+(0,0);<br />
<br />
label("$O$", O, S);<br />
<br />
<br />
pair OOO, OAO;<br />
OOO = O+(gap,0);<br />
OAO = (O+A)/2 + (gap,0);<br />
<br />
dot(O);<br />
<br />
label("$A$", (0,1), E);<br />
label("$B$", B, S);<br />
label("$C$", C, S);<br />
label("$D$", D, S);<br />
label("$E$", E, W);<br />
<br />
<br />
real scaleFactor = 1/1.618; // Adjust this value as needed<br />
// Rotate the smaller pentagon by 180 degrees <br />
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);<br />
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);<br />
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);<br />
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);<br />
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);<br />
<br />
<br />
<br />
pair K, L, M, N, O, P, Q, R, S, T, U, V;<br />
<br />
real newScaleFactor = 0.8507;<br />
K = newScaleFactor*dir(270+18);<br />
L = newScaleFactor*dir(270+72+18);<br />
M = newScaleFactor*dir(270+72+72+18);<br />
N = newScaleFactor*dir(270+72+72+72+18);<br />
O = newScaleFactor*dir(270+72+72+72+72+18);<br />
P = newScaleFactor*dir(270-18);<br />
Q = newScaleFactor*dir(270+72-18);<br />
R = newScaleFactor*dir(270+72+72-18);<br />
S = newScaleFactor*dir(270+72+72+72-18);<br />
T = newScaleFactor*dir(270+72+72+72+72-18);<br />
label("$K$", K, S);<br />
label("$L$", L, S);<br />
label("$M$", M, S);<br />
label("$N$", N, S);<br />
label("$O$", O, W);<br />
label("$P$", P, S);<br />
label("$Q$", Q, E);<br />
label("$R$", R, S);<br />
label("$S$", S, S);<br />
label("$T$", T, W);<br />
draw(K--T, dashed);<br />
draw(S--O, dashed);<br />
draw(P--L, dashed);<br />
draw(Q--M, dashed);<br />
draw(R--N, dashed);<br />
<br />
label("$F$", F, S);<br />
label("$G$", G, S);<br />
label("$H$", H, S);<br />
label("$I$", I, S);<br />
label("$J$", J, S);<br />
<br />
<br />
// Draw the smaller pentagon<br />
<br />
draw(F--G--H--I--J--cycle,red);<br />
<br />
</asy><br />
<br />
We can find the area of the red pentagon by taking the area of the total pentagon and subtracting the area outside the red pentagon. <br />
<br />
The area outside the red pentagon is the sum of the larger isosceles triangles, but this double counts the overlapping regions of the small isosceles triangles, so we have to subtract those out.<br />
<br />
We have <math>[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK])</math><br />
<br />
Lets focus on finding the area of each individual triangle:<br />
<br />
<asy><br />
unitsize(5cm);<br />
<br />
// Define the vertices of the pentagons<br />
pair A, B, C, D, E;<br />
pair F, G, H, I, J;<br />
<br />
// Calculate the vertices of the larger pentagon<br />
A = dir(90);<br />
B = dir(90 - 72);<br />
C = dir(90 - 2*72);<br />
D = dir(90 - 3*72);<br />
E = dir(90 - 4*72);<br />
<br />
<br />
pair O = (A+B+C+D+E)/5;<br />
pair AA,OO;<br />
real gap = 0.02;<br />
AA = A+(0,0);<br />
OO = O+(0,0);<br />
<br />
label("$O$", O, S);<br />
<br />
pair OOO, OAO;<br />
OOO = O+(gap,0);<br />
OAO = (O+A)/2 + (gap,0);<br />
<br />
dot(O);<br />
<br />
label("$D$", D, S);<br />
<br />
<br />
real scaleFactor = 1/1.618; // Adjust this value as needed<br />
// Rotate the smaller pentagon by 180 degrees <br />
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);<br />
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);<br />
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);<br />
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);<br />
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);<br />
<br />
label("$F$", F, N);<br />
<br />
pair K, L, M, N, O, P, Q, R, S, T, U, V;<br />
<br />
real newScaleFactor = 0.8507;<br />
K = newScaleFactor*dir(270+18);<br />
T = newScaleFactor*dir(270+72+72+72+72-18);<br />
label("$K$", K, E);<br />
label("$T$", T, W);<br />
draw(K--T);<br />
draw(K--D);<br />
draw(D--T);<br />
<br />
<br />
</asy><br />
<br />
Notice that we have no information about the side length, so instead we let the side length be <math>s</math>. Now we can drop an altitude from <math>O</math> to the base of the triangle, and we know this altitude must split the base of the pentagon in half, so we can create a right triangle. Furthermore, draw a line from <math>O</math> to <math>D</math>. This must bisect angle <math>D</math> which is <math>108</math> degrees, so we create <math>36-54-90</math> triangles. Specifically, we know <math>\angle ODK = 54^{\circ}</math>, <math>\angle DOU = 36^{\circ}</math>, and <math>\angle DTK = 36^{\circ}</math> because <math>\triangle DTK</math> is isosceles and we know the vertex angle is <math>108^{\circ}</math>. We encode this information in the diagram below:<br />
<asy><br />
unitsize(5cm);<br />
<br />
// Define the vertices of the pentagons<br />
pair A, B, C, D, E;<br />
pair F, G, H, I, J;<br />
<br />
// Calculate the vertices of the larger pentagon<br />
A = dir(90);<br />
B = dir(90 - 72);<br />
C = dir(90 - 2*72);<br />
D = dir(90 - 3*72);<br />
E = dir(90 - 4*72);<br />
<br />
<br />
pair O = (A+B+C+D+E)/5;<br />
pair AA,OO;<br />
real gap = 0.02;<br />
AA = A+(0,0);<br />
OO = O+(0,0);<br />
<br />
label("$O$", O, E);<br />
<br />
pair OOO, OAO;<br />
OOO = O+(gap,0);<br />
OAO = (O+A)/2 + (gap,0);<br />
<br />
dot(O);<br />
<br />
label("$D$", D, S);<br />
<br />
<br />
real scaleFactor = 1/1.618; // Adjust this value as needed<br />
// Rotate the smaller pentagon by 180 degrees <br />
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);<br />
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);<br />
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);<br />
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);<br />
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);<br />
<br />
label("$F$", (0.1,-1/1.618), E);<br />
<br />
pair K, L, M, N, O, P, Q, R, S, T, U, V;<br />
<br />
real newScaleFactor = 0.8507;<br />
K = newScaleFactor*dir(270+18);<br />
T = newScaleFactor*dir(270+72+72+72+72-18);<br />
label("$K$", K, S);<br />
label("$T$", T, W);<br />
draw(K--T);<br />
draw(K--D);<br />
draw(D--T);<br />
pair U;<br />
U=(0,-0.809);<br />
label("$U$",(0,-0.9), S);<br />
draw(O--U);<br />
draw(O--D);<br />
<br />
pair V;<br />
V = midpoint(O--D);<br />
label("$V$", V+(0,0.05), N); <br />
markscalefactor = 0.005;<br />
draw(rightanglemark(D,U,O));<br />
draw(rightanglemark(F,V,O));<br />
draw(rightanglemark(K,U,F));<br />
<br />
draw(anglemark(U,D,O));<br />
<br />
label("$54^{\circ}$", D+(0.05,0),NE);<br />
<br />
draw(anglemark(D,O,U));<br />
<br />
label("$36^{\circ}$", O-(0,0.2),SW);<br />
<br />
draw(anglemark(D,T,F));<br />
<br />
label("$36^{\circ}$", T+(0.1,-0.17),SE);<br />
<br />
label("$\frac{s}{2}$", D+(0.3,-0.1), S);<br />
<br />
<br />
<br />
</asy><br />
<br />
Since <math>\triangle DKT</math> is isosceles, the area of <math>\triangle DVT</math> is half the area of <math>\triangle DKT</math>. Similarly, the area of <math>\triangle UFK</math> is half that of <math>\triangle PFK</math>. Thus:<br />
<br />
<br />
<cmath>[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK]) \implies [FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK])</cmath><br />
<br />
We also know that since we dropped an altitude from <math>O</math> to <math>U</math>, the area of <math>\triangle ODU</math> must be half of a fifth of the total area of the pentagon. Therefore we can rewrite the above equation as <cmath>[FGHIJ]=10 \cdot ([ODU]-[DVT]+[UFK])</cmath><br />
<br />
Now notice that <math>\triangle ODU ~ \triangle TDV ~ \triangle KFU</math> by AA similarity. Therefore, if we can write the areas of the latter two triangles as a ratio of the first triangle, we can express the whole equation in terms of <math>[ODU]</math>, and by extension <math>[ABCDE]</math>, which we know. To find these ratios, we can find the side length ratios and square them because the triangles are similar.<br />
<br />
We already know <math>DU = \frac{s}{2}</math>, so let's try to find it's analogous side for <math>\triangle TDV</math> and <math>\triangle KFU</math>. These sides are <math>DV</math> and <math>FU</math>, respectively. <br />
<br />
First, <math>\frac{s}{2} = OD \cdot cos(54^{\circ})</math>, so <math>OD = \frac{s}{2} \cdot sec(54^{\circ})</math>. Then notice that <math>DV = \frac{OD}{2}</math> because we have to fold <math>D</math> to hit <math>O</math>, so the folding crease has to be exactly halfway between <math>O</math> and <math>D</math>. Therefore, <cmath>DV = \frac{s}{4} \cdot sec(54^{\circ}) \implies \frac{DU}{DV} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot sec(54^{\circ})} = 2 \cdot cos(54^{\circ})</cmath><br />
<br />
Now the ratio between the area of two similar triangles is the square of the ratio of their analogous side lengths. Thus<br />
<cmath>\frac{[ODU]}{[TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{10 \cdot [ODU]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})</cmath><br />
<cmath> \implies \frac{[ABCDE]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{\sqrt{5}+1}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})</cmath><br />
<cmath>\implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4 \cdot cos^2(54^{\circ})} \implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})</cmath><br />
<br />
Now let's move on and calculate the ratio of the other side length. Calculating <math>FU</math> is slightly tricker. <br />
First, we find <math>TD</math>: <math>TD \cdot cos(54^{\circ}) = DV = \frac{s}{4} sec^54^{\circ}) \implies TD = \frac{s}{4} \cdot sec^2(54^{\circ})</math>. Now since <math>\triangle DTK</math> is isosceles, <math>TD = DK</math> and <math>UK = DK-DU = TD-DU = \frac{s}{4} \cdot sec^2(54^{\circ})-\frac{s}{2} = \frac{s}{4} \cdot (sec^2(54^{\circ})-2) = \frac{s}{4} \cdot (tan^2(54^{\circ})-1)</math>.<br />
<br />
Now <math>FU = UK \cdot tan(36^{\circ}) \implies FU \frac{s}{4} \cdot (tan^2(54^{\circ})-1) \cdot tan(36^{\circ})</math>. Now note that <math>tan(x) \cdot tan(90-x) = 1</math> because opposite over adjacent cancel each other out in a right triangle. Thus, <math>FU = \frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))</math> <br />
<br />
Now, <cmath>\frac{DU}{FU} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))} = \frac{2}{tan(54^{\circ})-tan(36^{\circ})}</cmath><br />
<br />
<cmath>\implies \frac{[DU]}{[FU]} = \frac{4}{(tan(54^{\circ})-tan(36^{\circ}))^2} \implies 10 \cdot [FU] = \frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2</cmath><br />
<br />
Now we go back to our first equation and plug in our values:<br />
<br />
<cmath> [FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK]) \implies [FGHIJ] = 1+\sqrt{5}-\frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})+\frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2</cmath><br />
<br />
<cmath>\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (4-sec^2(54^{\circ})+(tan(54^{\circ})-tan(36^{\circ}))^2)</cmath><br />
<br />
Note <math>(tan(x)-tan(90-x))^2 = tan^2(x)-2 \cdot tan(x) \cdot tan(90-x)+tan^2(90-x) = tan^2(x)-2+tan^2(90-x)</math>.<br />
<br />
Also note that <math>tan^2(x)+1 = sec^2(x)</math>.<br />
Thus <cmath>[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2-sec^2(54^{\circ})+tan^2(54^{\circ})+tan^2(36^{\circ}))</cmath><br />
<cmath>\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2+(-1)+tan^2(36^{\circ})) \implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ}))</cmath>.<br />
<br />
Now all that remains is to find <math>tan^2(36^{\circ})</math>. We can use the tan addition formula to find the general form of <math>tan(5x)</math> or remember question 25 from this year's AMC 12A. We have that <cmath>tan(5x) = \frac{5tan(x)-10tan^3(x)+tan^5(x)}{1-10tan^2(x)+5tan^4(x)}</cmath>. <br />
<br />
Plug in <math>x=36</math>. Then we have <cmath>tan(180^{\circ}) = \frac{5tan(36^{\circ})-10tan^3(36^{\circ})+tan^5(36^{\circ})}{1-10tan^2(36^{\circ})+5tan^4(36^{\circ})}</cmath><br />
Now let <math>y = tan(36^{\circ})</math>. We have the equation <cmath>\frac{5y-10y^3+y^5}{1-10y^2+5y^4} = 0 \implies 5y-10y^3+y^5 = 0</cmath><br />
<cmath>\implies 5-10y^2+y^4 = 0 \implies 5-10z+z^2 = 0</cmath><br />
Where we let <math>z = y^2</math>. Using the quadratic formula, we have <cmath>z = \frac{10 \pm \sqrt{80}}{2} = 5 \pm 2\sqrt{5}</cmath><br />
Now since <math>y = tan(36^{\circ})</math>, <math>z = tan^2(36^{\circ})</math>, which is what we were looking for. Notice that <math>tan(0^{\circ}) = 0</math> and <math>tan(45^{\circ}) = 1</math>, so <math>tan(36^{\circ})</math> is between <math>0</math> and <math>1</math>, and so is it's square. Thus <math>z = 5 - 2\sqrt{5}</math>, not the other root.<br />
<br />
Finally:<br />
<br />
<cmath>[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ})) = \frac{1+\sqrt{5}}{4} \cdot (1+5-2\sqrt{5})</cmath><br />
<cmath>\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (6-2\sqrt{5}) = \frac{4\sqrt{5}-4}{4} = \sqrt{5}-1</cmath><br />
<br />
Therefore, <cmath>[FGHIJ] = \sqrt{5}-1 = \boxed{B}</cmath><br />
<br />
<br />
~KingRavi<br />
<br />
==Solution 3==<br />
<br />
<asy><br />
unitsize(5cm);<br />
<br />
// Define the vertices of the pentagons<br />
pair A, B, C, D, E;<br />
pair F, G, H, I, J;<br />
<br />
// Calculate the vertices of the larger pentagon<br />
A = dir(90);<br />
B = dir(90 - 72);<br />
C = dir(90 - 2*72);<br />
D = dir(90 - 3*72);<br />
E = dir(90 - 4*72);<br />
<br />
// Draw the larger pentagon<br />
draw(A--B--C--D--E--cycle);<br />
<br />
pair O = (A+B+C+D+E)/5;<br />
pair AA,OO;<br />
real gap = 0.02;<br />
AA = A+(0,0);<br />
OO = O+(0,0);<br />
<br />
draw(AA--OO, blue);<br />
<br />
pair OOO, OAO;<br />
OOO = O+(gap,0);<br />
OAO = (O+A)/2 + (gap,0);<br />
<br />
draw(OOO--OAO,green);<br />
dot(O);<br />
dot((O+A)/2);<br />
<br />
label("$r_b$", (O+A)*.7, E,blue);<br />
label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green);<br />
label("$r_s$", O+(-0.175,0.2), E,pink);<br />
label("$A$", (0,0), E);<br />
<br />
<br />
real scaleFactor = 1/1.618; // Adjust this value as needed<br />
// Rotate the smaller pentagon by 180 degrees<br />
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);<br />
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);<br />
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);<br />
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);<br />
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);<br />
<br />
// Draw the smaller pentagon<br />
<br />
draw(F--G--H--I--J--cycle,red);<br />
<br />
draw(arc(O,(H+I)*.5*.6,H*.6));<br />
label("$36^\circ$",O+(+0.05,0.15),NW);<br />
draw(O--H,pink);<br />
</asy><br />
<br />
Let <math>r_b</math> and <math>r_s</math> be the circumradius of the big and small pentagon, respectively. Let <math>a_s</math> be the apothem of the smaller pentagon and <math>A_s</math> and <math>A_b</math> be the areas of the smaller and larger pentagon, respectively. <br />
<br />
From the diagram:<br />
<cmath>\begin{align*}<br />
\cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\<br />
a_s &= \dfrac{r_b}{2}\\<br />
A_s &= \left(\dfrac{r_s}{r_b}\right)^2A_b\\<br />
&=\left(\dfrac{a_s}{\cos{36^\circ} r_b}\right)^2 \left(1+\sqrt{5}\right)\\<br />
&=\left(\dfrac{a_s}{\dfrac{\phi}{2} r_b}\right)^2 \left(1+\sqrt{5}\right)\\<br />
&=\left(\dfrac{1}{2 \dfrac{\phi}{2}}\right)^2 \left(1+\sqrt{5}\right)\\<br />
&=\left(\dfrac{2}{\sqrt{5}+1}\right)^2 \left(1+\sqrt{5}\right)\\<br />
&=\dfrac{4}{\sqrt{5}+1} \\<br />
&=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)} \\<br />
&=\sqrt{5}-1<br />
\end{align*}</cmath><br />
<cmath>\boxed{\textbf{(B) }\sqrt{5}-1}</cmath><br />
~Technodoggo<br />
<br />
==Solution 4==<br />
Interestingly, we find that the pentagon we need is the one that is represented by the intersection of perpendicular bisectors of the connection from the center of the pentagon to one vertex. Through similar triangles and the golden ratio, we find that the side length ratio of the two pentagons is <math>\frac{\sqrt{5}-1}{2}</math> Thus, the answer is <math>\sqrt{5}+1 \cdot (\frac{\sqrt{5}-1}{2})^2 = \sqrt{5}-1</math>. <math>\boxed{\text{B}}</math><br />
~andliu766<br />
<br />
==Solution 5 (answer choices (not rigorous))==<br />
After drawing a decent diagram, we can see that the area of the inner pentagon is quite a bit smaller than half the area of the larger pentagon.<br />
<br />
Then, we can estimate the values of the answers and choose one that seems the closest to the smallest answer.<br />
<br />
We know that <math>\sqrt5 \approx 2.236</math>, so we'll use <math>\sqrt5=2.2</math> for our estimations.<br />
The area of the original pentagon is <math>\sqrt5+1\approx3.2</math>, so half of it is roughly <math>1.6</math>.<br />
<br />
A: <math>~4-\sqrt5\approx 1.8</math> clearly, this is wrong because it is greater than half the area of the pentagon.<br />
<br />
B: <math>\sqrt{5}-1 \approx 1.2</math> This answer could be right.<br />
<br />
C: <math>8-3\sqrt{5} \approx 1.4</math> This too.<br />
<br />
D: <math>\frac{\sqrt{5}+1}{2}</math> This answer is wrong, as it assumes that the area of the inner pentagon is exactly half the area of the larger one.<br />
<br />
E: <math>\frac{2+\sqrt{5}}{3}\approx1.4</math> This answer could be right.<br />
<br />
But, from our diagram, assume that the area of the pentagon is significantly less than the area half of the larger pentagon, so we choose the smallest answer choice, giving us <math>\boxed{\textbf{(B) }\sqrt{5}-1}</math>.<br />
~erics118<br />
<br />
==Supplement (Calculating sin54/cos36 from Scratch)==<br />
<br />
Method 1:<br />
<br />
[[File:2023AMC12BP25.png|center|250px]]<br />
<br />
Construct golden ratio triangle <math>\triangle ABC</math> with <math>\angle A = 36^{\circ}</math>, <math>\angle B = \angle C = 72^{\circ}</math> and <math>\triangle BCD</math> with <math>\angle C = 36^{\circ}</math>, <math>\angle DBC = \angle BDC = 72^{\circ}</math>. WLOG, let <math>AB = AC = 1</math>, <math>BC = CD = AD = a</math>, <math>BD = 1-a</math>. <math>\triangle ABC \sim \triangle BCD</math><br />
<br />
<cmath>\frac{AC}{BC} = \frac{BC}{BD}, \quad \frac{1}{a} = \frac{a}{1-a}, \quad 1-a=a^2, \quad a^2 + a - 1 = 0</cmath><br />
<br />
<cmath>a = \frac{ -1 + \sqrt{1^2 - 4(-1) } }{2} = \frac{ \sqrt{5} -1 }{2}</cmath><br />
<br />
<cmath>\cos 36^{\circ} = \cos \angle A = \frac{AE}{AC} = \frac{ 1-a }{2} + a = \frac{ a + 1 }{2} = \frac{ \frac{ \sqrt{5} -1 }{2} + 1 }{2} = \frac{ \sqrt{5} + 1 }{4}</cmath><br />
<br />
<cmath>\sin 54^{\circ} = \cos 36^{\circ} = \frac{ \sqrt{5} + 1 }{4}</cmath><br />
<br />
Method 2:<br />
<br />
As explained [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_of_why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 here], <math>\cos \frac{2 \pi}{5} + \cos \frac{4 \pi}{5} = - \frac12</math><br />
<br />
<cmath>\cos \frac{2 \pi}{5} - \cos \frac{\pi}{5} = - \frac12</cmath><br />
<br />
<cmath>2(\cos\frac{ \pi}{5})^2 - 1 - \cos \frac{\pi}{5} = -1/2</cmath><br />
<br />
<cmath>4(\cos \frac{\pi}{5})^2 - 2 \cos \frac{\pi}{5} - 1 = 0</cmath><br />
<br />
<cmath>\cos 36^{\circ} = \cos \frac{\pi}{5} = \frac{2 + \sqrt{2^2 + 4 \cdot 4} }{8} = \frac{1+\sqrt{5}}{4}</cmath><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Video Solution 1 by SpreadTheMathLove==<br />
https://www.youtube.com/watch?v=ROVjN3oYLbQ<br />
<br />
==Video Solution 2 by OmegaLearn==<br />
https://youtu.be/_WztOIk_2Q8<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/dGGPT9LYKxs<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==See also==<br />
{{AMC10 box|year=2023|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2023|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_24&diff=2042482021 Fall AMC 12B Problems/Problem 242023-11-17T04:54:40Z<p>Junche: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Triangle <math>ABC</math> has side lengths <math>AB = 11, BC=24</math>, and <math>CA = 20</math>. The bisector of <math>\angle{BAC}</math> intersects <math>\overline{BC}</math> in point <math>D</math>, and intersects the circumcircle of <math>\triangle{ABC}</math> in point <math>E \ne A</math>. The circumcircle of <math>\triangle{BED}</math> intersects the line <math>AB</math> in points <math>B</math> and <math>F \ne B</math>. What is <math>CF</math>?<br />
<br />
<math>\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}</math><br />
<br />
==Olympiad Solution using Spiral Similarity==<br />
Construct the <math>E</math>-antipode, <math>E^{\prime}\in(ABC)</math>. Notice <math>\triangle CE^{\prime}A\stackrel{+}{\sim}\triangle CBF</math> by spiral similarity at <math>C</math>, thus <math>CF=\dfrac{CB\cdot CA}{CE^{\prime}}=\frac{480}{CE^{\prime}}</math>. Let <math>CE^{\prime}=x</math>; by symmetry <math>BE^{\prime}=x</math> as well and <math>\cos\angle BE^{\prime}C=\cos\angle A=\tfrac{11^{2}+20^{2}-24^{2}}{2\cdot 11\cdot 20}=-\tfrac{1}{8}</math> from Law of Cosines in <math>\triangle ABC</math>, so by Law of Cosines in <math>\triangle BE^{\prime}C</math> we have <cmath>x^{2}+x^{2}+\left(2x^{2}\right)\left(-\dfrac{1}{8}\right)=24^{2}</cmath> from which <math>x=16</math>. Now, <math>CF=\dfrac{480}{16}=\boxed{\textbf{C}~\text{30}}</math>.<br />
<br />
[[File:AMC 12 2021B Fall-24 Geogebra Diagram.png|600px]]<br />
<br />
==Solution 1==<br />
<br />
<b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math><br />
<br />
<b>Proof:</b> Note that <math>\angle CAD = \angle CAE = \angle EAB</math> and <math>\angle DCA = \angle BCA = \angle BEA</math> meaning that our claim is true by AA similarity.<br />
<br />
Because of this similarity, we have that <cmath>\frac{AC}{AD} = \frac{AE}{AB} \Longrightarrow AB \cdot AC = AD \cdot AE = AB \cdot AF</cmath> by Power of a Point. Thus, <math>AC=AF=20.</math><br />
<br />
Two solution methods follow from here.<br />
<br />
===Solution 1.1 (Stewart's theorem)===<br />
<br />
Applying [[Stewart's theorem]] on <math>\triangle ABC</math> with cevian <math>\overline{CF}</math> using the [[Directed legnths|directed lengths]] <math>AF = AC = 20</math> and <math>FB = 11-20 = -9</math>, we obtain <cmath>\begin{align*} (20)(-9)(11) + (CF)(11)(CF) &= (24)(20)(24) + (20)(-9)(20) \\ 11CF^{2} - 1980 &= 11520 - 3600\end{align*}</cmath> so <math>CF=\sqrt{\frac{11520 - 3600 + 1980}{11}}=\sqrt{\frac{9900}{11}}=\sqrt{900}=\boxed{\textbf{(C) }30}</math>.<br />
<br />
===Solution 1.2 (Double Cosine Law)===<br />
<br />
Note that <math>\angle CAF = \angle CAB</math> so we may plug into Law of Cosines to find the angle's cosine: <cmath>AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB) = -\frac{1}{8}.</cmath><br />
<br />
So, we observe that we can use Law of Cosines again to find <math>CF</math>: <cmath>CF^2 = AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = 900 \to CF=\boxed{\textbf{(C) }30}</cmath> both ways.<br />
<br />
- Kevinmathz<br />
<br />
== Solution 2 ==<br />
This solution is based on this figure: [[file:2021_AMC_12B_(Nov)_Problem_24,_sol.png]]<br />
<br />
Denote by <math>O</math> the circumcenter of <math>\triangle BED</math>.<br />
Denote by <math>R</math> the circumradius of <math>\triangle BED</math>.<br />
<br />
In <math>\triangle BCF</math>, following from the law of cosines, we have<br />
<cmath><br />
\begin{align*}<br />
CF^2 & = BC^2 + BF^2 - 2 BC \cdot BF \cos \angle CBF \\<br />
& = BC^2 + BF^2 + 2 BC \cdot BF \cos \angle ABC . \hspace{1cm} (1)<br />
\end{align*}<br />
</cmath><br />
For <math>BF</math>, we have<br />
<cmath><br />
\begin{align*}<br />
BF & = 2 R \cos \angle FBO \\<br />
& = 2 R \cos \left( 180^\circ - \angle ABC - \angle CBO \right) \\<br />
& = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - \angle BOD}{2} \right) \\<br />
& = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BED}{2} \right) \\<br />
& = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BCA}{2} \right) \\<br />
& = 2 R \cos \left( 90^\circ - \angle ABC + \angle BCA \right) \\<br />
& = 2 R \sin \left( \angle ABC - \angle BCA \right) \\<br />
& = \frac{BD}{\sin \angle BED} \sin \left( \angle ABC - \angle BCA \right) \\<br />
& = \frac{BD}{\sin \angle BCA} \sin \left( \angle ABC - \angle BCA \right) \\<br />
& = BD \left( \sin \angle ABC \cot \angle BCA - \cos \angle ABC \right) . \hspace{1cm} (2)<br />
\end{align*}<br />
</cmath><br />
The fourth equality follows from the property that <math>B</math>, <math>D</math>, <math>E</math> are concyclic.<br />
The fifth and the ninth equalities follow from the property that <math>A</math>, <math>B</math>, <math>C</math>, <math>E</math> are concyclic.<br />
<br />
Because <math>AD</math> bisects <math>\angle BAC</math>, following from the angle bisector theorem, we have<br />
<cmath><br />
\[<br />
\frac{BD}{CD} = \frac{AB}{AC} .<br />
\]<br />
</cmath><br />
Hence, <math>BD = \frac{24 \cdot 11}{31}</math>.<br />
<br />
In <math>\triangle ABC</math>, following from the law of cosines, we have<br />
<cmath><br />
\begin{align*}<br />
\cos \angle ABC & = \frac{AB^2 + BC^2 - AC^2}{2 AB \cdot BC} \\<br />
& = \frac{9}{16}<br />
\end{align*}<br />
</cmath><br />
and<br />
<cmath><br />
\begin{align*}<br />
\cos \angle BCA & = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} \\<br />
& = \frac{57}{64} .<br />
\end{align*}<br />
</cmath><br />
Hence, <math>\sin \angle ABC = \frac{5 \sqrt{7}}{16}</math> and <math>\sin \angle BCA = \frac{11 \sqrt{7}}{64}</math>.<br />
Hence, <math>\cot \angle BCA = \frac{57}{11 \sqrt{7}}</math>.<br />
<br />
Now, we are ready to compute <math>BF</math> whose expression is given in Equation (2).<br />
We get <math>BF = 9</math>.<br />
<br />
Now, we can compute <math>CF</math> whose expression is given in Equation (1).<br />
We have <math>CF = 30</math>.<br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(C) }30}</math>.<br />
<br />
~Steven Chen (www.professorchenedu.com)<br />
<br />
==Solution 3==<br />
Denote <math>B=(0, 0)</math> and <math>C=(24, 0)</math>. Note that by Heron's formula the area of <math>\triangle ABC</math> is <math>\frac{165\sqrt{7}}{4}</math> so the <math>y</math>-coordinate of <math>A</math> (height of <math>A</math> above the <math>x</math>-axis) is easily computed by the base-height formula as <math>\frac{55\sqrt7}{16}</math>.<br />
<br />
Now, since <math>AB=11</math>, the <math>x</math>-coordinate of <math>A</math> satisfies <math>x^2+(\frac{55\sqrt7}{16})^2=11^2</math> and solving gives <math>x=\frac{99}{16}</math>.<br />
<br />
The circumcircle of <math>\triangle ABC</math> has radius <math>\frac{abc}{4A}=\frac{11\cdot 24\cdot 20}{165\sqrt7}=\frac{32}{\sqrt7}</math>. We know by the perpendicular bisector rule that the circumcenter <math>O</math> is located directly below the midpoint of <math>\overline{BC}</math> (<math>x</math>-coordinate <math>12</math>).<br />
<br />
So, the negative <math>y</math> coordinate of <math>O</math> satisfies <math>12^2+y^2=(\frac{32}{\sqrt7})^2</math> and solving gives <math>y=-\frac{4}{\sqrt7}</math>.<br />
<br />
It's also clear that point <math>E</math> is going to be located directly below <math>O</math> on the circle, because the angle bisector intersects the circumcircle at the midpoint of the arc (Fact 5). Since the radius of the circle is <math>\frac{32}{\sqrt7}</math>, we have the coordinates of <math>E=(12, -\frac{36}{\sqrt7})</math><br />
<br />
Solving for point <math>D</math> (the point on the <math>x</math>-axis between <math>A</math> and <math>E</math>), we get that <math>D=(\frac{264}{31}, 0)</math>.<br />
<br />
So now we know six of the critical points: <math>A=(\frac{99}{16}, \frac{55\sqrt7}{16})</math>; <math>B=(0, 0)</math>; <math>C=(24, 0)</math>; <math>D=(\frac{264}{31}, 0)</math>; <math>E=(12, -\frac{36}{\sqrt7})</math>; <math>O=(12, -\frac{4}{\sqrt7})</math>.<br />
<br />
We are now ready to add in the circumcircle of <math>\triangle BDE</math>, which has radius <math>\frac{BD\cdot DE\cdot BE}{4[BDE]}</math>. From the above information, <math>BD=\frac{264}{31}</math>, <math>DE=\sqrt{(\frac{108}{31})^2+(\frac{36}{\sqrt7})^2}</math>, and <math>BE=\sqrt{12^2+(\frac{36}{\sqrt7})^2}</math>.<br />
<br />
After a bit of simplification we end up with <math>DE=\frac{1152}{31\sqrt7}</math> and <math>BE=\frac{48}{\sqrt7}</math>.<br />
<br />
For the area of <math>\triangle BDE</math>, the altitude dropped from vertex <math>E</math> has height <math>\frac{36}{\sqrt7}</math>, and the base <math>\overline{BD}</math> has length <math>\frac{264}{31}</math>, so its area is <math>\frac12\cdot\frac{36}{\sqrt7}\cdot\frac{264}{31}=\frac{4752}{31\sqrt7}</math>.<br />
<br />
Thus, <math>\frac{BD\cdot DE\cdot BE}{4[BDE]}=\frac{\tfrac{264}{31}\cdot\tfrac{1152}{31\sqrt7}\cdot\tfrac{48}{\sqrt{7}}}{4\cdot\tfrac{4752}{31\sqrt7}}</math> which after tons of cancellations becomes <math>\frac{768}{31\sqrt7}</math>.<br />
<br />
We know from the perpendicular bisector rule that the circumcenter <math>P</math> of <math>\triangle BDE</math> is located directly below the midpoint of <math>\overline{BD}</math> (<math>x</math>-coordinate <math>\frac{132}{31}</math>).<br />
<br />
So, the negative <math>y</math>-coordinate of <math>P</math> satisfies <math>(\frac{132}{31})^2+y^2=(\frac{768}{31\sqrt7})^2</math>, and solving gives <math>y=-\frac{684}{31\sqrt7}</math>. Thus, the equation of the circumcircle of <math>\triangle BDE</math> is <math>(x-\frac{132}{31})^2+(y+\frac{684}{31\sqrt7})^2=(\frac{768}{31\sqrt7})^2</math>.<br />
<br />
Point <math>F</math> is the intersection of this circle and the line <math>\overline{AB}</math>, which has equation <math>y=\frac{5\sqrt7}{9}x</math>. So, we substitute <math>y=\frac{5\sqrt7}{9}x</math> into the equation of the circle to get <math>(x-\frac{132}{31})^2+(\frac{5\sqrt7}{9}x+\frac{684}{31\sqrt7})^2=(\frac{768}{31\sqrt7})^2</math>.<br />
<br />
After simplifying, we have <math>\frac{256}{81}x^2+16x=0</math> (the <math>\frac{768}{31\sqrt7}</math>'s cancel out), whose solutions are <math>x=0</math> and <math>x=-\frac{81}{16}</math>. The first corresponds to the origin, and the second corresponds to point <math>F</math>. Thus the coordinates of <math>F</math> are <math>(-\frac{81}{16}, \frac{5\sqrt7}{9}\cdot\frac{-81}{16})=(-\frac{81}{16}, -\frac{45\sqrt7}{16})</math>.<br />
<br />
The coordinates of <math>C</math> are <math>(24, 0)</math>, so <cmath>CF=\sqrt{(24+\frac{81}{16})^2+(\frac{45\sqrt7}{16})^2}=\sqrt{(\frac{465}{16})^2+(\frac{45\sqrt7}{16})^2}=\frac{\sqrt{465^2+(45\sqrt7)^2}}{16}=\frac{\sqrt{(15\cdot 31)^2+(15\cdot 3\sqrt7)^2}}{16}=\frac{15\sqrt{31^2+(3\sqrt7)^2}}{16}=\frac{15\sqrt{961+63}}{16}=\frac{15\sqrt{1024}}{16}=\frac{15}{16}\cdot 32=30.</cmath><br />
<br />
==Video Solution by Power of Logic(Trig and Power of a point)==<br />
https://youtu.be/tEVbTtJlZjA<br />
<br />
~math2718281828459<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021 Fall|ab=B|num-a=25|num-b=23}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=User:Isabelchen&diff=176046User:Isabelchen2022-07-19T16:19:25Z<p>Junche: /* AIME */</p>
<hr />
<div>==AOPS Contributions==<br />
<br />
=== AHSME ===<br />
<br />
=== AMC 8 ===<br />
<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_25#Solution_4_.28Dynamic_Programming.29 2022 AMC8 Problem 25 Solution 4] (Probability)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_25#Remark 2022 AMC8 Problem 25 Remark] (Probability)<br />
<br />
===AMC 10===<br />
<br />
# [https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_24#Graph_Theory_Insight 2013 AMC10A Problem 24 Graph Theory Insight] (Graph Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25#Solution_5_.28Case_Work_with_Drawing.29 2013 AMC10A Problem 25 Solution 5] (Discrete Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_22#Remark 2013 AMC10B Problem 22 Remark] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18#Solution_2 2014 AMC10A Problem 18 Solution 2] (Analytic Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18#Solution_3 2014 AMC10A Problem 18 Solution 3] (Analytic Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22#Solution_6_.28Pure_Euclidian_Geometry.29 2014 AMC10A Problem 22 Solution 6] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22#Solution_7_.28Pure_Euclidian_Geometry.29 2014 AMC10A Problem 22 Solution 7] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21#Solution_3 2015 AMC10A Problem 21 Solution 3] (3D Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_22#Solution_4_.28Recursion.29 2015 AMC10A Problem 22 Solution 4] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_24#Solution_3 2015 AMC10A Problem 24 Solution 3] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_19#Solution_3 2015 AMC10B Problem 19 Solution 3] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_20#Solution_4_.28Casework.29 2016 AMC10A Problem 20 Solution 4] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_19#Solution_4_.28Area.29 2016 AMC10B Problem 19 Solution 4] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_19#Solution_5_.28Area.29 2016 AMC10B Problem 19 Solution 5] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22#Solution_4_.28Aggregate_Counting.29 2016 AMC10B Problem 22 Solution 4] (Graph Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_25#Supplement 2016 AMC10B Problem 25 Solution 1 Supplement] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_25#Solution_3_.28Casework.29 2016 AMC10B Problem 25 Solution 3] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_25#Solution_4 2016 AMC10B Problem 25 Solution 4] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_25#Remark 2016 AMC10B Problem 25 Remark] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17#Solution_4 2017 AMC10B Problem 17 Solution 4] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_25#Solution_4_.28Working_Backwards.29 2017 AMC10B Problem 25 Solution 4] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_10#Solution_8_.28Analytic_Geometry.29 2018 AMC10A Problem 10 Solution 8] (Analytic Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22#Solution_6 2018 AMC10A Problem 22 Solution 6] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18#Solution_6_.28Derangement.29 2018 AMC10B Problem 18 Solution 6] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_24#Solution_4_.28Area_Subtraction.29 2018 AMC10B Problem 24 Solution 4] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25#Solution_6_.28General_Equation.29 2018 AMC10B Problem 25 Solution 6] (Algebra)<br />
# [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_22#Solution_3_.28with_Table.29 2019 AMC10A Problem 22 Solution 3] (Probability)<br />
# [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_24#Solution_2_.28Pure_Elementary_Algebra.29 2019 AMC10A Problem 24 Solution 2] (Algebra)<br />
# [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_22#Solution_4_.28Markov_Chain.29 2019 AMC10B Problem 22 Solution 4] (Probability, Markov Chain)<br />
# [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25#Solution_6_.28Recursion.29 2019 AMC10B Problem 25 Solution 6] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_20#Solution_8_.28Solving_Equations.29 2020 AMC10A Problem 20 Solution 8] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24#Solution_9_.28Euclidean_Algorithm.29 2020 AMC10A Problem 24 Solution 9] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24#Solution_10_.28Diophantine_Equations.29 2020 AMC10A Problem 24 Solution 10] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24#Solution_11_.28Chinese_Remainder_Theorem.29 2020 AMC10A Problem 24 Solution 11] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_13#Solution_3_.28Find_the_Pattern.29 2020 AMC10B Problem 13 Solution 3] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_21#Solution_6 2020 AMC10B Problem 21 Solution 6] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_23#Solution_7_.28Group_Theory.29 2020 AMC10B Problem 23 Solution 7] (Group Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24#Solution_8_.28General_Equation.29 2020 AMC10B Problem 24 Solution 8] (Algebra)<br />
# [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25#Solution_7_.28Integer_Partition.29 2020 AMC10B Problem 25 Solution 7] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_23#Solution_8_.28Markov_Chain.29 2021 Spring AMC10A Problem 23 Solution 8] (Probability, Markov Chain)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_24#Solution_3_.28Slopes_and_Intercepts.29 2021 Spring AMC10A Problem 24 Solution 3] (Analytic Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25#Solution_4_.28Casework_and_Symmetry.29 2021 Spring AMC10A Problem 25 Solution 4] (Combinatorics)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_25#Solution_5 2021 Spring AMC10B Problem 25 Solution 5] (Analytic Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_25#Solution_6 2021 Spring AMC10B Problem 25 Solution 6] (Analytic Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_25#Remark 2021 Spring AMC10B Problem 25 Remark] (Analytic Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24#Solution_4 2021 Fall AMC10A Problem 24 Solution 4] (Graph Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_7#Solution_1_.28Completing_the_Square.29 2021 Fall AMC10B Problem 12 Solution 1] (Algebra)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_20#Remark_.28Bayes.27_Theorem.29 2021 Fall AMC10B Problem 20 Remark] (Probability)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_23#Solution_2_.28Ramsey.27s_Theorem.29 2021 Fall AMC10B Problem 23 Solution 2] (Graph Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_20#Solution_1_.28Graph_Theory.29 2021 Fall AMC10B Problem 24 Solution 1] (Graph Theory)<br />
<br />
===AMC 12===<br />
<br />
#[https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_23#Remark_.28Chinese_Remainder_Theorem.29 2010 AMC12A Problem 23 Remark] (Number Theory)<br />
#[https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_22#Remark_.28Markov_Chain.29 2014 AMC12B Problem 22 Remark] (Probability, Markov Chain)<br />
# [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24#Remark_.28Morley.27s_Trisector_Theorem.29 2016 AMC12A Problem 21 Remark] (Geometry)<br />
<br />
===AIME===<br />
<br />
# [https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4#Solution_6_.28Trigonometry.29 1983 AIME Problem 4 Solution 6] (Trigonometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12#Solution_5_.28Dynamic_Programming.29 1985 AIME Problem 12 Solution 5] (Probability)<br />
# [https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9#Solution_4 1997 AIME Problem 9 Solution 4] (Algebra)<br />
#[https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_12#Solution_2_.28Markov_Chain.29 2021 AIME I Problem 12 Solution 2] (Probability, Markov Chain)<br />
#[https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_12#Remark_.28Markov_Chain.29 2021 AIME I Problem 12 Remark] (Probability, Markov Chain)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_8#Solution_2_.28Markov_Chain_and_Dynamic_Programming.29 2021 AIME II Problem 8 Solution 2] (Probability, Markov Chain)<br />
# [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_8#Remark_.28Markov_Chain.29 2021 AIME II Problem 8 Remark] (Probability, Markov Chain)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8#Solution_2_.28Euclidean_Geometry.29 2022 AIME I Problem 8 Solution 2] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_2#Solution_1 2022 AIME II Problem 2 Solution 1] (Probability)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_5#Solution_1 2022 AIME II Problem 5 Solution 1] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6#Solution_1 2022 AIME II Problem 6 Solution 1] (Algebra)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7#Solution_1 2022 AIME II Problem 7 Solution 1] (Geometry)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8#Solution_2_Supplement 2022 AIME II Problem 8 Solution 2 Supplement] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8#Solution_3_Supplement 2022 AIME II Problem 8 Solution 3 Supplement] (Number Theory)<br />
# [https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_14#Solution_1 2022 AIME II Problem 14 Solution 1] (Algebra)</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_10&diff=1728101985 AIME Problems/Problem 102022-03-20T15:36:30Z<p>Junche: </p>
<hr />
<div>== Problem ==<br />
How many of the first 1000 [[positive integer]]s can be expressed in the form<br />
<br />
<math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor</math>,<br />
<br />
where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer]] less than or equal to <math>z</math>?<br />
__TOC__<br />
<br />
== Solution 1 ==<br />
Noting that all of the numbers are even, we can reduce this to any real number <math>x</math> between <math>0</math> to <math>\frac 12</math>, as this will be equivalent to <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10. <br />
<br />
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):<br />
<br />
We can match up the greatest integer functions with one of the partitions of the integer. If we let <math>x = \frac 12</math> then we get the solution <math>10</math>; now consider when <math>x < \frac 12</math>: <math>\lfloor 2x \rfloor = 0</math>, <math>\lfloor 4x \rfloor \le 1</math>, <math>\lfloor 6x \rfloor \le 2</math>, <math>\lfloor 8x \rfloor \le 3</math>. But according to this the maximum we can get is <math>1+2+3 = 6</math>, so we only need to try the first 6 numbers.<br />
<br />
*<math>1</math>: Easily possible, for example try plugging in <math>x =\frac 18</math>. <br />
*<math>2</math>: Also simple, for example using <math>\frac 16</math>.<br />
*<math>3</math>: The partition must either be <math>1+1+1</math> or <math>1+2</math>. If <math>\lfloor 4x \rfloor = 1</math>, then <math>x \ge \frac 14</math>, but then <math>\lfloor 8x \rfloor \ge 2</math>; not possible; and vice versa to show that the latter partition doesn't work. So we cannot obtain <math>3</math>.<br />
*<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case we see that <math>\frac 14</math> works.<br />
*<math>5</math>: We can partition as <math>1+2+2</math>, from which we find that <math>\frac 13</math> works.<br />
*<math>6</math>: We can partition as <math>1+2+3</math>, from which we find that <math>\frac 38</math> works.<br />
<br />
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers <math>1,2,4,5,6,10</math>; hence our solution is <math>6 \cdot 100 = \boxed{600}</math>.<br />
<br />
== Solution 2 ==<br />
As we change the value of <math>x</math>, the value of our [[expression]] changes only when <math>x</math> crosses [[rational number]] of the form <math>\frac{m}{n}</math>, where <math>n</math> is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form <math>\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}</math>. This gives us 24 calculations to make; we summarize the results here:<br />
<br />
<math>\frac{1}{24}, \frac{2}{24} \to 0</math><br />
<br />
<math>\frac{3}{24} \to 1</math><br />
<br />
<math>\frac{4}{24}, \frac{5}{24} \to 2</math><br />
<br />
<math>\frac{6}{24}, \frac{7}{24} \to 4</math><br />
<br />
<math>\frac{8}{24} \to 5</math><br />
<br />
<math>\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6</math><br />
<br />
<math>\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10</math><br />
<br />
<math>\frac{15}{24} \to 11</math><br />
<br />
<math>\frac{16}{24},\frac{17}{24} \to 12</math><br />
<br />
<math>\frac{18}{24}, \frac{19}{24} \to 14</math><br />
<br />
<math>\frac{20}{24}\to 15</math><br />
<br />
<math>\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16</math><br />
<br />
<math>\frac{24}{24} \to 20</math><br />
<br />
Thus, we hit 12 of the first 20 integers and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.<br />
<br />
===Solution 2 Shortcut===<br />
<br />
Because <math>2,4,6,8</math> are all multiples of <math>2</math>, we can speed things up. We only need to check up to <math>\frac{12}{24}</math>, and the rest should repeat. As shown before, we hit 6 integers (<math>1,2,4,5,6,10</math>) from <math>\frac{1}{24}</math> to <math>\frac{12}{24}</math>. Similarly, this should repeat 100 times, for <math>\boxed{600}</math><br />
<br />
~[[User:N828335|N828335]]<br />
<br />
==Solution 3==<br />
<br />
Recall from Hermite's Identity that <math>\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor</math>. Then we can rewrite <math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor</math><br />
<math>+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor</math>. There are <math>12</math> terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from <math>20</math>). Starting from every integer <math>x</math>, we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by <math>20</math> while only achieving <math>12</math> of those <math>20</math> values. We can conveniently shift the <math>1000</math> (since it can be achieved) to the position of the <math>0</math> so that there are only complete cycles of <math>20</math>, and the answer is <math>\frac {12}{20}\cdot1000 = \boxed{600}</math>.<br />
<br />
== Solution 4 ==<br />
<br />
Let <math>x=\lfloor x\rfloor+\{x\}</math> then<br />
<cmath>\begin{align*}<br />
\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor&=\lfloor 2(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 4(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 6(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 8(\lfloor x\rfloor+\{x\})\rfloor\\<br />
&=2\lfloor x\rfloor+4\lfloor x\rfloor+6\lfloor x\rfloor+8\lfloor x\rfloor+\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor\\<br />
&=20\lfloor x\rfloor+(\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor)<br />
\end{align*}</cmath><br />
Similar to the previous solutions, the value of <math>\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor</math> changes when <math>\{x\}=\frac{m}{n}</math>, where <math>m\in\{1,2,3,...,n-1\}</math>, <math>n\in\{2,4,6,8\}</math>. Using [[Euler's Totient Function]]<br />
<cmath>\sum\limits_{k=0}^4 \phi(2k)</cmath><br />
to obtain <math>12</math> different values for <math>\{x\}=\frac{m}{n}</math>. (note that here Euler's Totient Function counts the number of <math>\{x\}=\frac{m}{n}</math> where <math>m</math>, <math>n</math> are relatively prime so that the values of <math>\{x\}</math> won't overlap.). <br />
<br />
Thus if <math>k</math> can be expressed as <math>\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor</math>, then <math>k=20a+b</math> for some non-negative integers <math>a</math>, <math>b</math>, where there are <math>12</math> values for <math>b</math>. <br />
<br />
Exclusively, there are <math>49</math> values for <math>a</math> in the range <math>0<k<1000</math>, or <math>49\cdot12=588</math> ordered pairs <math>(a,b)</math>.<br />
<br />
If <math>a=0</math>, <math>b\neq0</math>, which includes <math>11</math> ordered pairs.<br />
<br />
If <math>a=50</math>, <math>b=0</math>, which includes <math>1</math> ordered pair.<br />
<br />
In total, there are <math>588+11+1=\boxed{600}</math> values for <math>k</math>.<br />
<br />
~ Nafer<br />
<br />
== Solution 5 ==<br />
<br />
To simplify the question, let <math>y = 2x</math>. Then, the expression in the question becomes <math>\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor</math>. <br />
<br />
Let <math>\{x\}</math> represent the non-integer part of <math>x</math> (For example, <math>\{2.8\} = 0.8</math>). Then, <br />
<br />
<cmath>\begin{align*}<br />
\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor &= y - \{y\} + 2y - \{2y\} + 3y - \{3y\} + 4y - \{4y\} \\<br />
&= 10y - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\<br />
&= 10(\lfloor y \rfloor + \{y\}) - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\<br />
&= 10\lfloor y \rfloor + 10\{y\} - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\<br />
&= 10\lfloor y \rfloor + 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) \\<br />
\end{align*}</cmath><br />
<br />
Since <math>\lfloor y \rfloor</math> is always an integer, <math>10\lfloor y \rfloor</math> will be a multiple of 10. Thus, we look for the range of the other part of the expression. We will be able to reach the same numbers when <math>y</math> ranges from <math>0</math> to <math>1</math>, because the curly brackets (<math>\{\}</math>) gets rid of any integer part. Let the combined integer part of <math>2y</math>, <math>3y</math>, and <math>4y</math> be <math>k</math> (In other words, <math>k = \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor</math>). Then, <br />
<br />
<cmath>\begin{align*}<br />
9\{y\} - (\{2y\} + \{3y\} + \{4y\}) &= 9\{y\} - (2\{y\} + 3\{y\} + 4\{y\} - k) \\<br />
&= 9\{y\} - (9\{y\} - k) \\<br />
&= k<br />
\end{align*}</cmath><br />
<br />
The maximum value of <math>k</math> will be when <math>y</math> is slightly less than <math>1</math>, which means <math>k = 1 + 2 + 3 = 6</math>. As <math>y</math> increases from <math>0</math> to <math>1</math>, <math>k</math> will increase whenever <math>2y</math>, <math>3y</math>, or <math>4y</math> is an integer, which happens when <math>y</math> hits one of the numbers in the set <math>\left\{\dfrac14, \dfrac13, \dfrac12, \dfrac23, \dfrac34 \right\}</math>. When <math>y</math> reaches <math>\dfrac12</math>, both <math>2y</math> and <math>4y</math> will be an integer, so <math>k</math> will increase by <math>2</math>. For all the other numbers in the set, <math>k</math> increases by <math>1</math> since only <math>1</math> number in the set will be an integer. Thus, the possible values of <math>k</math> are <math>\{0, 1, 2, 4, 5, 6\}</math>. <br />
<br />
Finally, looking back at the original expression, we plug in <math>k</math> to get a multiple of <math>10</math> plus any number in <math>\{0, 1, 2, 4, 5, 6\}</math>. Thus, we hit all numbers ending in <math>\{0, 1, 2, 4, 5, 6\}</math>, of which there are <math>\boxed{600}</math>. <br />
<br />
~Owen1204<br />
<br />
==Solution 6==<br />
<br />
Imagine that we gradually increase <math>x</math> from <math>0</math> to <math>1</math>. At the beginning, the value of our expression is <math>0</math>, at the end it is <math>2+4+6+8=20</math>. Note that every time <math>x=\frac{a}{b}</math> for some positive integer <math>a</math> and a positive multiple <math>b</math> of either <math>2, 4, 6,</math> or <math>8</math>. Thus, we have been able to express 12 of the integers from 1 through 20 when <math>0<x<1</math>, namely when <cmath>x = \frac{1}{2}, \frac{2}{2}=1, \frac{1}{4}, \frac{3}{4}, \frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \frac{5}{6}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}</cmath>.<br />
<br />
<br />
Notice that <cmath>\lfloor 2(x+n) \rfloor + \lfloor 4(x+n)\rfloor + \lfloor 6(x+n) \rfloor+ \lfloor8(x+n) \rfloor= \lfloor 2x \rfloor+ \lfloor 4x \rfloor+ \lfloor 6x \rfloor+ \lfloor 8x \rfloor+ 20n</cmath>. This conceptually means that the number of integers that can be expressed in the range <math>(i, j)</math> is the same as the number of integers that can be expressed in the range <math>(i+x, j+x)</math>. Thus, because we can express <math>12</math> integers in the range <math>(1, 20)</math>, we can cover <math>12*50 = \boxed{600}</math> from 1 to 1000.<br />
<math>-\text{thinker123}</math><br />
<br />
==Solution 7==<br />
<br />
After observing, we can see that there are <math>6</math> values of can be evaluated through the expression every <math>10</math> numbers, so our answer is <math>6*100=600</math> ~bluesoul<br />
<br />
== See also ==<br />
* [[Floor function]]<br />
{{AIME box|year=1985|num-b=9|num-a=11}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_12&diff=1722492021 AIME I Problems/Problem 122022-03-02T04:18:48Z<p>Junche: /* Remark (Markov Chain) */</p>
<hr />
<div>==Problem==<br />
Let <math>A_1A_2A_3\ldots A_{12}</math> be a dodecagon (<math>12</math>-gon). Three frogs initially sit at <math>A_4,A_8,</math> and <math>A_{12}</math>. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
Define the <b>distance</b> between two frogs as the number of sides between them that do not contain the third frog.<br />
<br />
Let <math>E(a,b,c)</math> denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are <math>a,b,</math> and <math>c</math> (in either clockwise or counterclockwise order). Without the loss of generality, assume that <math>a\leq b\leq c.</math><br />
<br />
We wish to find <math>E(4,4,4).</math> Note that:<br />
<ol style="margin-left: 1.5em;"><br />
<li>At any moment before the frogs stop jumping, the only possibilities for <math>(a,b,c)</math> are <math>(4,4,4),(2,4,6),</math> and <math>(2,2,8).</math></li><p><br />
<li><math>E(a,b,c)</math> does not depend on the actual positions of the frogs, but depends on the distances between the frogs.</li><p><br />
<li>At the end of each minute, each frog has <math>2</math> outcomes. So, there are <math>2^3=8</math> outcomes in total.</li><p><br />
</ol><br />
We have the following system of equations:<br />
<cmath>\begin{align*}<br />
E(4,4,4)&=1+\frac{2}{8}E(4,4,4)+\frac{6}{8}E(2,4,6), \\<br />
E(2,4,6)&=1+\frac{4}{8}E(2,4,6)+\frac{1}{8}E(4,4,4)+\frac{1}{8}E(2,2,8), \\<br />
E(2,2,8)&=1+\frac{2}{8}E(2,2,8)+\frac{2}{8}E(2,4,6).<br />
\end{align*}</cmath><br />
Rearranging and simplifying each equation, we get<br />
<cmath>\begin{align*}<br />
E(4,4,4)&=\frac{4}{3}+E(2,4,6), &(1) \\<br />
E(2,4,6)&=2+\frac{1}{4}E(4,4,4)+\frac{1}{4}E(2,2,8), &\hspace{12.75mm}(2) \\<br />
E(2,2,8)&=\frac{4}{3}+\frac{1}{3}E(2,4,6). &(3)<br />
\end{align*}</cmath><br />
Substituting <math>(1)</math> and <math>(3)</math> into <math>(2),</math> we obtain <cmath>E(2,4,6)=2+\frac{1}{4}\left[\frac{4}{3}+E(2,4,6)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(2,4,6)\right],</cmath> from which <math>E(2,4,6)=4.</math> Substituting this into <math>(1)</math> gives <math>E(4,4,4)=\frac{16}{3}.</math><br />
<br />
Therefore, the answer is <math>16+3=\boxed{019}.</math><br />
<br />
~Ross Gao (Fundamental Logic)<br />
<br />
~MRENTHUSIASM (Reconstruction)<br />
<br />
===Solution 1 Supplement (Markov Chain)===<br />
<br />
The above solution can be represented by the following [https://en.wikipedia.org/wiki/Markov_chain Markov Chain]:<br />
<br />
[[File:Markov Chain Frog AIME.png| 800px |center]]<br />
<br />
From state <math>(4, 4, 4)</math> to state <math>(4, 4, 4)</math>: the <math>3</math> frogs must jump in the same direction, <math>2 \cdot \frac18 = \frac14</math>.<br />
<br />
From state <math>(4, 4, 4)</math> to state <math>(2, 4, 6)</math>: <math>2</math> frogs must jump in the same direction, and the other must jump in the opposite direction, <math>\binom32 \cdot 2 \cdot \frac18 = \frac34</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(4, 4, 4)</math>: the <math>2</math> frogs with a distance of <math>4</math> in between must jump in the same direction so that they will be further away from the other frog, and the other frog must jump in the opposite direction as those <math>2</math> frogs, <math>\frac18</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(2, 4, 6)</math>: the <math>3</math> frogs can all jump in the same direction; or the <math>2</math> frogs with a distance of <math>2</math> in between jumps away from each other and the other frog jumps closer to the closest frog; or the <math>2</math> frogs with a distance of <math>6</math> in between jump closer to each other and away from the third frog, the third frog jumps closer to the closest frog; <math>(2 + 1 + 1) \cdot \frac18 = \frac12</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(2, 2, 8)</math>: the <math>2</math> frogs with a distance of <math>2</math> in between must jump closer to the other frog and the other frog must jump close to those <math>2</math> frogs, <math>\frac18</math>.<br />
<br />
From state <math>(2, 2, 8)</math> to state <math>(2, 4, 6)</math>: <math>2</math> frogs that have no frogs in between must both jump in the same direction away from the other frog, the other frog must also jump away from those <math>2</math> frogs, <math>2 \cdot \frac18 = \frac14</math>.<br />
<br />
From state <math>(2, 2, 8)</math> to state <math>(2, 2, 8)</math>: the <math>3</math> frogs must all jump in the same direction, <math>2 \cdot \frac18 = \frac14</math>.<br />
<br />
From state <math>(2, 2, 8)</math> to state <math>(0, x, y)</math>: frogs with a distance of <math>2</math> must jump closer to each other, the other frog can jump in any direction, <math>\frac12 \cdot \frac12 \cdot 2 = \frac12</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(0, x, y)</math>: the frogs with a distance of <math>2</math> must jump closer to each other, the other frog can jump in any direction, <math>\frac12 \cdot \frac12 = \frac14</math>.<br />
<br />
<br />
Because <math>a + b + c = 12</math>, we can use <math>(a, b)</math> to represent the states. The following system of equations is the same as above.<br />
<br />
<cmath>\begin{align*}<br />
E(4, 4) &= \frac14(E(4, 4)+1) + \frac34(E(2, 4)+1)\\<br />
E(2, 4) &= \frac14 + \frac12(E(2, 4)+1) + \frac18(E(4, 4)+1) + \frac18(E(2, 2)+1) \\<br />
E(2, 2) &= \frac12 + \frac14(E(2, 2)+1) + \frac14(E(2, 4)+1) <br />
\end{align*}</cmath><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Solution 2 (Markov Chain)==<br />
<br />
We can solve the problem by removing <math>1</math> frog, and calculate the expected time for the remaining <math>2</math> frogs. In the original problem, when the movement stops, <math>2</math> of the <math>3</math> frogs meet. Because the <math>3</math> frogs cannot meet at one vertex, the probability that those two specific frogs meet is <math>\frac13</math>. If the expected time for the two frog problem is <math>E'</math>, then the expected time for the original problem is <math>\frac{E'}{3}</math>.<br />
<br />
The distance between the two frogs can only be <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>. We use the distances as the states to draw the following [https://en.wikipedia.org/wiki/Markov_chain Markov Chain]. This Markov Chain is much simpler than that of Solution <math>1</math>.<br />
<br />
[[File:Markov Chain Frog AIME copy.png| 900px |center]]<br />
<br />
<cmath>\begin{align*}<br />
E(2) &= 1 + \frac12 \cdot E(2) + \frac14 \cdot E(4)\\<br />
E(4) &= 1 + \frac14 \cdot E(2) + \frac12 \cdot E(4) + \frac14 \cdot E(6)\\<br />
E(6) &= 1 + \frac12 \cdot E(4) + \frac12 \cdot E(6)<br />
\end{align*}</cmath><br />
<br />
By solving the above system of equations, <math>E(4) = 16</math>. The answer for the original problem is <math>\frac{16}{3}</math>, <math>16 + 3 = \boxed{\textbf{019}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Remark (Markov Chain)==<br />
<br />
The above <math>2</math> Markov Chains are [https://en.wikipedia.org/wiki/Absorbing_Markov_chain Absorbing Markov Chain]. The state of <math>2</math> frogs meeting is the absorbing state. This problem asks for the [https://en.wikipedia.org/wiki/Absorbing_Markov_chain#Expected_number_of_steps Expected Number of Steps] before being absorbed in the absorbing state.<br />
<br />
Let <math>p_{ij} = P(X_{n+1} = j | X_n = i)</math>, the probability that state <math>i</math> transits to state <math>j</math> on the next step. <br />
<br />
Let <math>e_i</math> be the expected number of steps before being absorbed in the absorbing state when starting from transient state <math>i</math>.<br />
<br />
<math>e_i = 1 + \sum_{j} (p_{ij} \cdot e_{j})</math><br />
<br />
<math>e_i</math> is <math>1</math> more than the sum of the products of <math>p_{ij}</math> and <math>s_j</math> of all the next state <math>j</math>.<br />
<br />
[https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_22 2014 AMC12B Problem 22] is a similar problem with simpler states, both problem can be solved by Absorbing Markov Chain.<br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==See Also==<br />
{{AIME box|year=2021|n=I|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_12&diff=1722482021 AIME I Problems/Problem 122022-03-02T04:17:47Z<p>Junche: /* Remark (Markov Chain) */</p>
<hr />
<div>==Problem==<br />
Let <math>A_1A_2A_3\ldots A_{12}</math> be a dodecagon (<math>12</math>-gon). Three frogs initially sit at <math>A_4,A_8,</math> and <math>A_{12}</math>. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
Define the <b>distance</b> between two frogs as the number of sides between them that do not contain the third frog.<br />
<br />
Let <math>E(a,b,c)</math> denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are <math>a,b,</math> and <math>c</math> (in either clockwise or counterclockwise order). Without the loss of generality, assume that <math>a\leq b\leq c.</math><br />
<br />
We wish to find <math>E(4,4,4).</math> Note that:<br />
<ol style="margin-left: 1.5em;"><br />
<li>At any moment before the frogs stop jumping, the only possibilities for <math>(a,b,c)</math> are <math>(4,4,4),(2,4,6),</math> and <math>(2,2,8).</math></li><p><br />
<li><math>E(a,b,c)</math> does not depend on the actual positions of the frogs, but depends on the distances between the frogs.</li><p><br />
<li>At the end of each minute, each frog has <math>2</math> outcomes. So, there are <math>2^3=8</math> outcomes in total.</li><p><br />
</ol><br />
We have the following system of equations:<br />
<cmath>\begin{align*}<br />
E(4,4,4)&=1+\frac{2}{8}E(4,4,4)+\frac{6}{8}E(2,4,6), \\<br />
E(2,4,6)&=1+\frac{4}{8}E(2,4,6)+\frac{1}{8}E(4,4,4)+\frac{1}{8}E(2,2,8), \\<br />
E(2,2,8)&=1+\frac{2}{8}E(2,2,8)+\frac{2}{8}E(2,4,6).<br />
\end{align*}</cmath><br />
Rearranging and simplifying each equation, we get<br />
<cmath>\begin{align*}<br />
E(4,4,4)&=\frac{4}{3}+E(2,4,6), &(1) \\<br />
E(2,4,6)&=2+\frac{1}{4}E(4,4,4)+\frac{1}{4}E(2,2,8), &\hspace{12.75mm}(2) \\<br />
E(2,2,8)&=\frac{4}{3}+\frac{1}{3}E(2,4,6). &(3)<br />
\end{align*}</cmath><br />
Substituting <math>(1)</math> and <math>(3)</math> into <math>(2),</math> we obtain <cmath>E(2,4,6)=2+\frac{1}{4}\left[\frac{4}{3}+E(2,4,6)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(2,4,6)\right],</cmath> from which <math>E(2,4,6)=4.</math> Substituting this into <math>(1)</math> gives <math>E(4,4,4)=\frac{16}{3}.</math><br />
<br />
Therefore, the answer is <math>16+3=\boxed{019}.</math><br />
<br />
~Ross Gao (Fundamental Logic)<br />
<br />
~MRENTHUSIASM (Reconstruction)<br />
<br />
===Solution 1 Supplement (Markov Chain)===<br />
<br />
The above solution can be represented by the following [https://en.wikipedia.org/wiki/Markov_chain Markov Chain]:<br />
<br />
[[File:Markov Chain Frog AIME.png| 800px |center]]<br />
<br />
From state <math>(4, 4, 4)</math> to state <math>(4, 4, 4)</math>: the <math>3</math> frogs must jump in the same direction, <math>2 \cdot \frac18 = \frac14</math>.<br />
<br />
From state <math>(4, 4, 4)</math> to state <math>(2, 4, 6)</math>: <math>2</math> frogs must jump in the same direction, and the other must jump in the opposite direction, <math>\binom32 \cdot 2 \cdot \frac18 = \frac34</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(4, 4, 4)</math>: the <math>2</math> frogs with a distance of <math>4</math> in between must jump in the same direction so that they will be further away from the other frog, and the other frog must jump in the opposite direction as those <math>2</math> frogs, <math>\frac18</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(2, 4, 6)</math>: the <math>3</math> frogs can all jump in the same direction; or the <math>2</math> frogs with a distance of <math>2</math> in between jumps away from each other and the other frog jumps closer to the closest frog; or the <math>2</math> frogs with a distance of <math>6</math> in between jump closer to each other and away from the third frog, the third frog jumps closer to the closest frog; <math>(2 + 1 + 1) \cdot \frac18 = \frac12</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(2, 2, 8)</math>: the <math>2</math> frogs with a distance of <math>2</math> in between must jump closer to the other frog and the other frog must jump close to those <math>2</math> frogs, <math>\frac18</math>.<br />
<br />
From state <math>(2, 2, 8)</math> to state <math>(2, 4, 6)</math>: <math>2</math> frogs that have no frogs in between must both jump in the same direction away from the other frog, the other frog must also jump away from those <math>2</math> frogs, <math>2 \cdot \frac18 = \frac14</math>.<br />
<br />
From state <math>(2, 2, 8)</math> to state <math>(2, 2, 8)</math>: the <math>3</math> frogs must all jump in the same direction, <math>2 \cdot \frac18 = \frac14</math>.<br />
<br />
From state <math>(2, 2, 8)</math> to state <math>(0, x, y)</math>: frogs with a distance of <math>2</math> must jump closer to each other, the other frog can jump in any direction, <math>\frac12 \cdot \frac12 \cdot 2 = \frac12</math>.<br />
<br />
From state <math>(2, 4, 6)</math> to state <math>(0, x, y)</math>: the frogs with a distance of <math>2</math> must jump closer to each other, the other frog can jump in any direction, <math>\frac12 \cdot \frac12 = \frac14</math>.<br />
<br />
<br />
Because <math>a + b + c = 12</math>, we can use <math>(a, b)</math> to represent the states. The following system of equations is the same as above.<br />
<br />
<cmath>\begin{align*}<br />
E(4, 4) &= \frac14(E(4, 4)+1) + \frac34(E(2, 4)+1)\\<br />
E(2, 4) &= \frac14 + \frac12(E(2, 4)+1) + \frac18(E(4, 4)+1) + \frac18(E(2, 2)+1) \\<br />
E(2, 2) &= \frac12 + \frac14(E(2, 2)+1) + \frac14(E(2, 4)+1) <br />
\end{align*}</cmath><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Solution 2 (Markov Chain)==<br />
<br />
We can solve the problem by removing <math>1</math> frog, and calculate the expected time for the remaining <math>2</math> frogs. In the original problem, when the movement stops, <math>2</math> of the <math>3</math> frogs meet. Because the <math>3</math> frogs cannot meet at one vertex, the probability that those two specific frogs meet is <math>\frac13</math>. If the expected time for the two frog problem is <math>E'</math>, then the expected time for the original problem is <math>\frac{E'}{3}</math>.<br />
<br />
The distance between the two frogs can only be <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>. We use the distances as the states to draw the following [https://en.wikipedia.org/wiki/Markov_chain Markov Chain]. This Markov Chain is much simpler than that of Solution <math>1</math>.<br />
<br />
[[File:Markov Chain Frog AIME copy.png| 900px |center]]<br />
<br />
<cmath>\begin{align*}<br />
E(2) &= 1 + \frac12 \cdot E(2) + \frac14 \cdot E(4)\\<br />
E(4) &= 1 + \frac14 \cdot E(2) + \frac12 \cdot E(4) + \frac14 \cdot E(6)\\<br />
E(6) &= 1 + \frac12 \cdot E(4) + \frac12 \cdot E(6)<br />
\end{align*}</cmath><br />
<br />
By solving the above system of equations, <math>E(4) = 16</math>. The answer for the original problem is <math>\frac{16}{3}</math>, <math>16 + 3 = \boxed{\textbf{019}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Remark (Markov Chain)==<br />
<br />
The above <math>2</math> Markov Chains are [https://en.wikipedia.org/wiki/Absorbing_Markov_chain Absorbing Markov Chain]. The state of <math>2</math> frogs meeting is the absorbing state. This problem asks for the [https://en.wikipedia.org/wiki/Absorbing_Markov_chain#Expected_number_of_steps Expected Number of Steps] before being absorbed in the absorbing state.<br />
<br />
Let <math>p_{ij} = P(X_{n+1} = j | X_n = i)</math>, the probability that state <math>i</math> transits to state <math>j</math> on the next step. <br />
<br />
Let <math>e_i</math> be the expected number of steps before being absorbed in the absorbing state when starting from transient state <math>i</math>.<br />
<br />
<math>e_i = 1 + \sum_{j} (p_{ij} \cdot e_{j})</math><br />
<br />
<math>e_i</math> is the sum of the products of <math>p_{ij}</math> and <math>s_j</math> of all the next state <math>j</math>.<br />
<br />
[https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_22 2014 AMC12B Problem 22] is a similar problem with simpler states, both problem can be solved by Absorbing Markov Chain.<br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==See Also==<br />
{{AIME box|year=2021|n=I|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_15&diff=1713922022 AIME II Problems/Problem 152022-02-18T16:51:29Z<p>Junche: /* See Also */</p>
<hr />
<div>==Problem==<br />
<br />
Two externally tangent circles <math>\omega_1</math> and <math>\omega_2</math> have centers <math>O_1</math> and <math>O_2</math>, respectively. A third circle <math>\Omega</math> passing through <math>O_1</math> and <math>O_2</math> intersects <math>\omega_1</math> at <math>B</math> and <math>C</math> and <math>\omega_2</math> at <math>A</math> and <math>D</math>, as shown. Suppose that <math>AB = 2</math>, <math>O_1O_2 = 15</math>, <math>CD = 16</math>, and <math>ABO_1CDO_2</math> is a convex hexagon. Find the area of this hexagon.<br />
<asy><br />
import geometry;<br />
size(10cm);<br />
point O1=(0,0),O2=(15,0),B=9*dir(30);<br />
circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B);<br />
point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0];<br />
filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black);<br />
draw(w1);<br />
draw(w2);<br />
draw(O1--O2,dashed);<br />
draw(o);<br />
dot(O1);<br />
dot(O2);<br />
dot(A);<br />
dot(D);<br />
dot(C);<br />
dot(B);<br />
label("$\omega_1$",8*dir(110),SW);<br />
label("$\omega_2$",5*dir(70)+(15,0),SE);<br />
label("$O_1$",O1,W);<br />
label("$O_2$",O2,E);<br />
label("$B$",B,N+1/2*E);<br />
label("$A$",A,N+1/2*W);<br />
label("$C$",C,S+1/4*W);<br />
label("$D$",D,S+1/4*E);<br />
label("$15$",midpoint(O1--O2),N);<br />
label("$16$",midpoint(C--D),N);<br />
label("$2$",midpoint(A--B),S);<br />
label("$\Omega$",o.C+(o.r-1)*dir(270));<br />
</asy><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_12&diff=1710941985 AIME Problems/Problem 122022-02-07T16:38:19Z<p>Junche: Guess does not fit mathematical rigor.</p>
<hr />
<div>== Problem ==<br />
Let <math>A</math>, <math>B</math>, <math>C</math> and <math>D</math> be the vertices of a regular tetrahedron, each of whose edges measures <math>1</math> meter. A bug, starting from vertex <math>A</math>, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let <math>p = \frac{n}{729}</math> be the probability that the bug is at vertex <math>A</math> when it has crawled exactly <math>7</math> meters. Find the value of <math>n</math>.<br />
<br />
== Solution 1 (Single Variable Recursion) ==<br />
For all nonnegative integers <math>k,</math> let <math>P(k)</math> be the probability that the bug is at vertex <math>A</math> when it has crawled exactly <math>k</math> meters. We wish to find <math>p=P(7).</math><br />
<br />
Clearly, we have <math>P(0)=1.</math> For all <math>k\geq1,</math> note that after <math>k-1</math> crawls:<br />
<ol style="margin-left: 1.5em;"><br />
<li>The probability that the bug is at vertex <math>A</math> is <math>P(k-1),</math> and the probability that it crawls to vertex <math>A</math> on the next move is <math>0.</math></li><p><br />
<li>The probability that the bug is not at vertex <math>A</math> is <math>1-P(k-1),</math> and the probability that it crawls to vertex <math>A</math> on the next move is <math>\frac13.</math></li><p><br />
</ol><br />
Together, the recursive formula for <math>P(k)</math> is<br />
<cmath>P(k) = \begin{cases}<br />
1 & \mathrm{if} \ k=0 \\<br />
\frac13(1-P(k-1)) & \mathrm{if} \ k\geq1<br />
\end{cases}.</cmath><br />
Two solutions follow from here:<br />
<br />
=== Solution 1.1 (Recursive Formula) ===<br />
We evaluate <math>P(7)</math> recursively:<br />
<cmath>\begin{alignat*}{6}<br />
P(0)&=1, \\<br />
P(1)&=\frac13(1-P(0))&&=0, \\<br />
P(2)&=\frac13(1-P(1))&&=\frac13, \\<br />
P(3)&=\frac13(1-P(2))&&=\frac29, \\<br />
P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\<br />
P(5)&=\frac13(1-P(4))&&=\frac{20}{81}, \\<br />
P(6)&=\frac13(1-P(5))&&=\frac{61}{243},\\<br />
P(7)&=\frac13(1-P(6))&&=\frac{182}{729}.<br />
\end{alignat*}</cmath><br />
Therefore, the answer is <math>n=\boxed{182}.</math><br />
<br />
~Azjps (Fundamental Logic)<br />
<br />
~MRENTHUSIASM (Reconstruction)<br />
<br />
=== Solution 1.2 (Explicit Formula) ===<br />
Let <math>P(k)=Q(k)+c</math> for some function <math>Q(k)</math> and constant <math>c.</math> For all <math>k\geq1,</math> the recursive formula for <math>P(k)</math> becomes <cmath>Q(k)+c=\frac13(1-(Q(k-1)+c))=\frac13-\frac13Q(k-1)-\frac13c.</cmath> Solving for <math>Q(k),</math> we get <cmath>Q(k)=\frac13-\frac13Q(k-1)-\frac43c.</cmath><br />
For simplicity purposes, we set <math>c=\frac14,</math> which gives <cmath>Q(k)=-\frac13Q(k-1).</cmath><br />
Clearly, <math>Q(0),Q(1),Q(2),\ldots</math> is a geometric sequence with the common ratio <math>\frac{Q(k)}{Q(k-1)}=-\frac13.</math> Substituting <math>P(0)=1</math> and <math>c=\frac14</math> into <math>P(0)=Q(0)+c</math> produces <math>Q(0)=\frac34,</math> the first term of the geometric sequence. <br />
<br />
For all nonnegative integers <math>k,</math> the explicit formula for <math>Q(k)</math> is <cmath>Q(k)=\frac34\left(-\frac13\right)^k,</cmath> and the explicit formula for <math>P(k)</math> is <cmath>P(k)=\frac34\left(-\frac13\right)^k+\frac14.</cmath> Finally, the requested probability is <math>p=P(7)=\frac{182}{729},</math> from which <math>n=\boxed{182}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
== Solution 2 (Multivariable Recursion by Algebra) ==<br />
<u><b>Denominator</b></u><br />
<br />
There are <math>3^7</math> ways for the bug to make <math>7</math> independent crawls without restrictions.<br />
<br />
<u><b>Numerator</b></u><br />
<br />
Let <math>V_k</math> denote the number of ways for the bug to crawl exactly <math>k</math> meters starting from vertex <math>V</math> and ending at vertex <math>A,</math> where <math>V\in\{A,B,C,D\}</math> and <math>k</math> is a positive integer. We wish to find <math>A_7.</math><br />
<br />
Since the bug must crawl to vertex <math>B,C,</math> or <math>D</math> on the first move, we have<br />
<cmath>\begin{align*}<br />
A_7&=B_6+C_6+D_6 \\<br />
&=(A_5+C_5+D_5)+(A_5+B_5+D_5)+(A_5+B_5+C_5) \\<br />
&=A_5+2(A_5+B_5+C_5+D_5) \\<br />
&=A_5+2S_5,<br />
\end{align*}</cmath><br />
where <math>S_k=A_k+B_k+C_k+D_k.</math><br />
<br />
More generally, we get <cmath>A_{k+2}=A_k+2S_k. \qquad\qquad (\spadesuit)</cmath><br />
For <math>S_k,</math> note that<br />
<ol style="margin-left: 1.5em;"><br />
<li>Base Case:</li><br />
<cmath>\begin{align*}<br />
S_1&=A_1+B_1+C_1+D_1 \\<br />
&=0+1+1+1 \\<br />
&=3.<br />
\end{align*}</cmath><br />
<li>Recursive Case:</li><br />
<cmath>\begin{align*}<br />
S_{k+1}&=A_{k+1}+B_{k+1}+C_{k+1}+D_{k+1} \\<br />
&=(B_k+C_k+D_k)+(A_k+C_k+D_k)+(A_k+B_k+D_k)+(A_k+B_k+C_k) \\<br />
&=3(A_k+B_k+C_k+D_k) \\<br />
&=3S_k.<br />
\end{align*}</cmath><br />
</ol><br />
Clearly, <math>S_1,S_2,S_3,\ldots</math> is a geometric sequence with the first term <math>S_1=3</math> and the common ratio <math>\frac{S_{k+1}}{S_k}=3.</math> Therefore, its explicit formula is <cmath>S_k=3^k. \qquad\qquad (\clubsuit)</cmath><br />
Recall that <math>A_1=0.</math> By <math>(\spadesuit)</math> and <math>(\clubsuit),</math> we rewrite <math>A_7</math> recursively:<br />
<cmath>\begin{align*}<br />
A_7 &= A_5+2S_5 \\<br />
&=\left(A_3+2S_3\right)+2S_5 \\<br />
&=\left(\left(A_1+2S_1\right)+2S_3\right)+2S_5 \\<br />
&=2S_1+2S_3+2S_5 \\<br />
&=2(3)+2\left(3^3\right)+2\left(3^5\right).<br />
\end{align*}</cmath><br />
<u><b>Answer</b></u><br />
<br />
The requested probability is <cmath>p=\frac{A_7}{3^7}=\frac{2(3)+2\left(3^3\right)+2\left(3^5\right)}{3^7}=\frac{2(1)+2\left(3^2\right)+2\left(3^4\right)}{3^6}=\frac{182}{729},</cmath> from which <math>n=\boxed{182}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
== Solution 3 (Multivariable Recursion by Table) ==<br />
Define notation <math>V_k</math> as Solution 2 does.<br />
<br />
In fact, we can generalize the following relationships for all <i><b>nonnegative</b></i> integers <math>k:</math><br />
<cmath>\begin{align*}<br />
A_0&=1, \\<br />
B_0&=0, \\<br />
C_0&=0, \\<br />
D_0&=0, \\<br />
A_{k+1}&=B_k+C_k+D_k, \\<br />
B_{k+1}&=A_k+C_k+D_k, \\<br />
C_{k+1}&=A_k+B_k+D_k, \\<br />
D_{k+1}&=A_k+B_k+C_k. \\<br />
\end{align*}</cmath><br />
Using these equations, we recursively fill out the table below:<br />
<cmath>\begin{array}{c||c|c|c|c|c|c|c|c} <br />
\hspace{7mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&& \\ [-2.5ex]<br />
\boldsymbol{k} & \boldsymbol{0} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} \\ <br />
\hline \hline<br />
&&&&&&&& \\ [-2.25ex]<br />
\boldsymbol{A_k} &1&0&3&6&21&60&183&546 \\ \hline <br />
&&&&&&&& \\ [-2.25ex]<br />
\boldsymbol{B_k} &0&1&2&7&20&61&182&547 \\ \hline <br />
&&&&&&&& \\ [-2.25ex]<br />
\boldsymbol{C_k} &0&1&2&7&20&61&182&547 \\ \hline <br />
&&&&&&&& \\ [-2.25ex]<br />
\boldsymbol{D_k} &0&1&2&7&20&61&182&547 \\<br />
\end{array}</cmath><br />
Note that the paths from <math>V</math> to <math>A</math> and the paths from <math>A</math> to <math>V</math> have one-to-one correspondence. So, we must get <cmath>A_k+B_k+C_k+D_k=3^k</cmath> for all values of <math>k.</math><br />
<br />
The requested probability is <cmath>p=\frac{A_7}{3^7}=\frac{546}{2187}=\frac{182}{729},</cmath> from which <math>n=\boxed{182}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
== Solution 4 (Single Variable Version of Solution 2) ==<br />
Let <math>a_n</math> denotes the number of ways that the bug arrives at <math>A</math> after crawling <math>n</math> meters, then we have <math>a_1=0</math>.<br />
<br />
Notice that there is respectively <math>1</math> way to arrive at <math>A</math> for each of the different routes after the previous <math>n-1</math> crawls, excluding the possibility that the bug ends up at <math>A</math> after the <math>(n-1)</math>th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation <cmath>a_n=3^{n-1}-a_{n-1}.</cmath><br />
Quick calculations yield<br />
<cmath>\begin{align*}<br />
a_7&=3^6-a_6\\<br />
&=3^6-\left(3^5-3^4+3^3-3^2+3-a_1\right)\\<br />
&=546.<br />
\end{align*}</cmath><br />
Thus, <math>p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}</math>.<br />
<br />
~Nafer<br />
<br />
== Solution 5 (Generating Functions and Roots of Unity Filter) ==<br />
The generating function for a problem with this general form (<math>4</math> states, <math>n</math> steps) is <math>(x+x^2+x^3)^n</math>, so the generating function of interest for this problem is <math>(x+x^2+x^3)^7</math>. Our goal is to find the coefficients of every <math>x^{4n}</math> and add them up before dividing by <math>3^7</math>. We can do this by applying a roots of unity filter. <br />
<br />
Let <math>\omega = e^{i\pi / 2}</math>. We have that if <math>G(x) = (x+x^2+x^3)^7</math>, then <cmath>\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)}{4} = \frac{2187-1-1-1}{4} = 546.</cmath> From here, the desired probability is <math>\frac{546}{2187} = \frac{182}{729}</math>. Therefore, the answer is <math>n=\boxed{182}</math>.<br />
<br />
~RedFlame2112<br />
<br />
== Solution 6 (Partitions) ==<br />
We can find the number of different times the bug reaches vertex <math>A</math> before the <math>7</math>th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at <math>A.</math><br />
<br />
Define <math>f(x)</math> to be the number of paths of length <math>x</math> which start and end at <math>A</math> but do not pass through <math>A</math> otherwise. Obviously <math>f(1) = 0.</math> In general for <math>f(x),</math> the bug has three initial edges to pick from. From there, since the bug cannot return to <math>A</math> by definition, the bug has exactly two choices. This continues from the <math>2</math>nd move up to the <math>(x-1)</math>th move. The last move must be a return to <math>A,</math> so this move is determined. So <math>f(x) = 2^{x-2}3.</math><br />
<br />
Now we need to find the number of cycles by which the bug can reach <math>A</math> at the end. Since <math>f(1) = 0,</math> we know that <math>f(6)</math> cannot be used, as on the <math>7</math>th move the bug cannot move from <math>A</math> to <math>A.</math> So we need to find the number of [[partition]]s of <math>7</math> using only <math>2,3,4,5,</math> and <math>7.</math> These are <math>f(2)f(2)f(3),f(2)f(5),f(3)f(4),</math> and <math>f(7).</math> We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition:<br />
<cmath>{3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7) = 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53) = 546.</cmath><br />
Finally, this is a probability question, so we divide by <math>3^7,</math> or <math>\frac{546}{3^7} = \frac{182}{3^6}.</math> The answer is <math>n=\boxed{182}.</math><br />
<br />
== Remark ==<br />
Here is a similar problem from another AIME test: [[2003 AIME II Problems/Problem 13|2003 AIME II Problem 13]], in which we have an equilateral triangle instead.<br />
<br />
== See also ==<br />
{{AIME box|year=1985|num-b=11|num-a=13}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics Competition Resources]]<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_25&diff=1710112018 AMC 10B Problems/Problem 252022-02-06T04:44:41Z<p>Junche: </p>
<hr />
<div>{{duplicate|[[2018 AMC 10B Problems#Problem 25 | 2018 AMC 10B #25]] and [[2018 AMC 12B Problems#Problem 24|2018 AMC 12B #24]]}}<br />
<br />
== Problem ==<br />
Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many real numbers <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>?<br />
<br />
<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math><br />
<br />
==Solution 1==<br />
This rewrites itself to <math>x^2=10,000\{x\}</math>.<br />
<br />
Graphing <math>y=10,000\{x\}</math> and <math>y=x^2</math> we see that the former is a set of line segments with slope <math>10,000</math> from <math>0</math> to <math>1</math> with a hole at <math>x=1</math>, then <math>1</math> to <math>2</math> with a hole at <math>x=2</math> etc.<br />
Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization.<br />
<br />
<asy><br />
import graph;<br />
size(400);<br />
xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5}));<br />
yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}));<br />
real y(real x) {return x^2;}<br />
draw(circle((-4,16), 0.1));<br />
draw(circle((-3,16), 0.1));<br />
draw(circle((-2,16), 0.1));<br />
draw(circle((-1,16), 0.1));<br />
draw(circle((0,16), 0.1));<br />
draw(circle((1,16), 0.1));<br />
draw(circle((2,16), 0.1));<br />
draw(circle((3,16), 0.1));<br />
draw(circle((4,16), 0.1));<br />
draw((-5,0)--(-4,16), black);<br />
draw((-4,0)--(-3,16), black);<br />
draw((-3,0)--(-2,16), black);<br />
draw((-2,0)--(-1,16), black);<br />
draw((-1,0)--(-0,16), black);<br />
draw((0,0)--(1,16), black);<br />
draw((1,0)--(2,16), black);<br />
draw((2,0)--(3,16), black);<br />
draw((3,0)--(4,16), black);<br />
draw(graph(y,-4.2,4.2),green);<br />
</asy><br />
<br />
Now notice that when <math>x=\pm 100</math> then graph has a hole at <math>(\pm 100,10,000)</math> which the equation <math>y=x^2</math> passes through and then continues upwards. Thus our set of possible solutions is bounded by <math>(-100,100)</math>. We can see that <math>y=x^2</math> intersects each of the lines once and there are <math>99-(-99)+1=199</math> lines for an answer of <math>\boxed{\text{(C)}~199}</math>.<br />
<br />
==Solution 2==<br />
<br />
Same as the first solution, <math>x^2=10,000\{x\} </math>.<br />
<br />
<br />
We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>\{x\}</math> in terms of <math>\lfloor x \rfloor</math>:<br />
<cmath> \{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0</cmath><br />
<br />
<br />
We use the quadratic formula to solve for <math>\{x\}</math> :<br />
<cmath> \{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -40,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2} </cmath><br />
<br />
<br />
Since <math> 0 \leq \{x\} < 1 </math>, we get an inequality which we can then solve. After simplifying a lot, we get that <math>\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0</math>.<br />
<br />
<br />
Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of <math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.<br />
<br />
==Solution 3==<br />
<br />
Let <math>x = a+k</math> where <math>a</math> is the integer part of <math>x</math> and <math>k</math> is the fractional part of <math>x</math>.<br />
We can then rewrite the problem below:<br />
<br />
<math>(a+k)^2 + 10000a = 10000(a+k)</math><br />
<br />
From here, we get<br />
<br />
<math>(a+k)^2 + 10000a = 10000a + 10000k</math><br />
<br />
Solving for <math>a+k = x</math><br />
<br />
<math>(a+k)^2 = 10000k</math><br />
<br />
<math>x = a+k = \pm100\sqrt{k}</math><br />
<br />
Because <math>0 \leq k < 1</math>, we know that <math>a+k</math> cannot be less than or equal to <math>-100</math> nor greater than or equal to <math>100</math>. Therefore:<br />
<br />
<math>-99 \leq x \leq 99</math><br />
<br />
There are <math>199</math> elements in this range, so the answer is <math>\boxed{\textbf{(C)} \text{ 199}}</math>.<br />
<br />
Note (not by author): this solution seems to be invalid at first, because one can not determine whether <math>x</math> is an integer or not. However, it actually works because although <math>x</math> itself might not be an integer, it is very close to one, so there are 199 potential <math>x</math>.<br />
<br />
==Solution 4==<br />
<br />
Notice the given equation is equivilent to <math>(\lfloor x \rfloor+\{x\})^2=10,000\{x\} </math><br />
<br />
Now we now that <math>\{x\} < 1</math> so plugging in <math>1</math> for <math>\{x\}</math> we can find the upper and lower bounds for the values.<br />
<br />
<math>(\lfloor x \rfloor +1)^2 = 10,000(1)</math><br />
<br />
<math>(\lfloor x \rfloor +1) = \pm 100</math><br />
<br />
<math>\lfloor x \rfloor = 99, -101</math><br />
<br />
And just like <math>\textbf{Solution 2}</math>, we see that <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of <math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.<br />
<br />
==Solution 5==<br />
<br />
Firstly, if <math>x</math> is an integer, then <math>10,000\lfloor x \rfloor=10,000x</math>, so <math>x</math> must be <math>0</math>.<br />
<br />
If <math>0<x<1</math>, then we know the following:<br />
<br />
<math>0<x^2<1</math><br />
<br />
<math>10,000\lfloor x \rfloor =0</math><br />
<br />
<math>0<10,000x<10,000</math><br />
<br />
Therefore, <math>0<x^2+10,000\lfloor x \rfloor <1</math>, which overlaps with <math>0<10,000x<10,000</math>. This means that there is at least one real solution between <math>0</math> and <math>1</math>. Since <math>x^2+10,000\lfloor x \rfloor </math> increases quadratically and <math>10,000x</math> increases linearly, there is only one solution for this case. <br />
<br />
Similarly, if <math>1<x<2</math>, then we know the following:<br />
<br />
<math>1<x^2<4</math><br />
<br />
<math>10,000\lfloor x \rfloor =10,000</math><br />
<br />
<math><10,000<10,000x<20,000</math><br />
<br />
By following similar logic, we can find that there is one solution between <math>1</math> ad <math>2</math>. <br />
<br />
We can also follow the same process to find that there are negative solutions for <math>x</math> as well.<br />
<br />
There are not an infinite amount of solutions, so at one point there will be no solutions when <math>n<x<n+1</math> for some integer <math>n</math>. For there to be no solutions in a given range means that the range of <math>10,000\lfloor x \rfloor + x^2</math> does not intersect the range of <math>10,000x</math>. <math>x^2</math> will always be positive, and <math>10,000\lfloor x \rfloor</math> is less than <math>10,000</math> less than <math>10,000x</math>, so when <math>x^2 >= 10,000</math>, the equation will have no solutions. This means that there are <math>99</math> positive solutions, <math>99</math> negative solutions, and <math>0</math> for a total of <math>\boxed{\text{(C)}~199}</math> solutions.<br />
<br />
~Owen1204<br />
<br />
==Solution 6 (General Equation)==<br />
<br />
General solution to this type of equation <math>f(x, \lfloor x \rfloor) = 0</math>:<br />
<br />
1. solve <math>f(x, \lfloor x \rfloor) = 0</math> for <math>x</math> to get <math>x = g(\lfloor x \rfloor )</math><br />
2. apply <math>\lfloor x \rfloor \le x < \lfloor x \rfloor+1</math>, solve <math>\lfloor x \rfloor \le g(\lfloor x \rfloor) < \lfloor x \rfloor+1</math> to get the domain of <math>\lfloor x \rfloor</math><br />
3. get <math> \lfloor x \rfloor</math> from the domain of <math> \lfloor x \rfloor</math> because <math> \lfloor x \rfloor</math> is integer, then get <math>x</math> from <math> \lfloor x \rfloor</math> by <math>x = g( \lfloor x \rfloor) </math><br />
Note: function <math>\lfloor x \rfloor</math> maps <math>x</math> to its floor. By solving <math>f(x, \lfloor x \rfloor) = 0</math>, we get function <math>x = g( \lfloor x \rfloor) </math>, mapping <math>x</math>'s floor to <math>x</math><br />
<br />
<math>x^2 - 10000x + 10000 \lfloor x \rfloor =0</math><br />
<br />
<math>x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}</math>, <math>\lfloor x \rfloor \le 2500</math><br />
<br />
<math>\lfloor x \rfloor \le x < \lfloor x \rfloor + 1</math><br />
<br />
If <math>x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}</math>, <math>x \ge 5000</math>, it contradicts <math>x < \lfloor x \rfloor + 1 \le 2501</math><br />
<br />
So <math>x= 5000 - 100 \sqrt{2500 - \lfloor x \rfloor}</math><br />
<br />
Let <math>k = \lfloor x \rfloor</math> , <math>x= 5000 - 100 \sqrt{2500 - k}</math><br />
<br />
<math>k \le 5000 - 100 \sqrt{2500 - k} < k + 1</math><br />
<br />
<math>0 \le 5000 - k - 100 \sqrt{2500 - k} < 1</math><br />
<br />
<math>0 \le 2500 - k - 100 \sqrt{2500 - k} + 2500 < 1</math><br />
<br />
<math>0 \le (\sqrt{2500 - k} - 50)^2 < 1</math><br />
<br />
<math>-1 < \sqrt{2500 - k} - 50 < 1</math><br />
<br />
<math>49 < \sqrt{2500 - k} < 51</math><br />
<br />
<math>-101 < k < 99</math><br />
<br />
So the number of <math>k</math>'s values is <math>99-(-101)-1=199</math>. Because <math>x=5000-100\sqrt{2500-k}</math>, for each value of <math>k</math>, there is a value for <math>x</math>. The answer is <math>\boxed{\textbf{(C)} 199}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Solution 7==<br />
Subtracting <math>10000\lfloor x\rfloor</math> from both sides gives <math>x^2=10000(x-\lfloor x\rfloor)=10000\{x\}</math>. Dividing both sides by <math>10000</math> gives <math>\left(\frac{x}{100}\right)^2=\{x\}<1</math>. <math>\left(\frac{x}{100}\right)^2<1</math> when <math>-100<x<100</math> so the answer is <math>\boxed{199}</math><br />
<br />
~randomdude10807<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=vHKPbaXwJUE<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=1709742019 AMC 10B Problems/Problem 252022-02-05T15:25:45Z<p>Junche: </p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems#Problem 25|2019 AMC 10B #25]] and [[2019 AMC 12B Problems#Problem 23|2019 AMC 12B #23]]}}<br />
<br />
==Problem==<br />
<br />
How many sequences of <math>0</math>s and <math>1</math>s of length <math>19</math> are there that begin with a <math>0</math>, end with a <math>0</math>, contain no two consecutive <math>0</math>s, and contain no three consecutive <math>1</math>s?<br />
<br />
<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math><br />
<br />
==Solution 1 (Recursion)==<br />
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.<br />
Let <math>f(n)</math> be the number of valid (meaning: the sequence contains 0s and 1s, starts and ends with both 0, and there are no two consecutive 0s and no three consecutive 1s) sequences of length <math>n</math>. <br />
<br />
Then we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, with <math>n \ge 3</math> (because otherwise, the sequence would contain only 0s and this is not allowed due to the give conditions). <br />
<br />
We derived the recursive function, since for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> to return to the starting position, 0, and the resulting sequence will still satisfy the given conditions.<br />
<br />
It is easy to find <math>f(3) = 1</math> since the only possible valid sequence is <math>010</math>. <math>f(4)=1</math> since the only possible valid sequence is <math>0110</math>. <math>f(5)=1</math> since the only possible valid sequence is <math>01010</math>. <br />
<br />
The recursive sequence is then as follows:<br />
<br />
<cmath>f(3)=1</cmath><br />
<cmath>f(4)=1</cmath><br />
<cmath>f(5) = 1</cmath><br />
<cmath>f(6) = 1 + 1 = 2</cmath> <br />
<cmath>f(7) = 1 + 1 = 2</cmath><br />
<cmath>f(8) = 1 + 2 = 3</cmath> <br />
<cmath>f(9) = 2 + 2 = 4</cmath><br />
<cmath>f(10) = 2 + 3 = 5</cmath><br />
<cmath>f(11) = 3 + 4 = 7</cmath> <br />
<cmath>f(12) = 4 + 5 = 9</cmath> <br />
<cmath>f(13) = 5 + 7 = 12</cmath><br />
<cmath>f(14) = 7 + 9 = 16</cmath> <br />
<cmath>f(15) = 9 + 12 = 21</cmath><br />
<cmath>f(16) = 12 + 16 = 28</cmath><br />
<cmath>f(17) = 16 + 21 = 37</cmath> <br />
<cmath>f(18) = 21 + 28 = 49</cmath><br />
<cmath>f(19) = 28 + 37 = 65</cmath><br />
<br />
So, our answer is <math>\boxed{\text{\bf{(C)} } 65}</math>.<br />
<br />
<br />
'''Contributors:'''<br />
<br />
~Original Author<br />
<br />
~solasky<br />
<br />
~BakedPotato66<br />
<br />
==Solution 2 (casework)==<br />
After any particular <math>0</math>, the next <math>0</math> in the sequence must appear exactly <math>2</math> or <math>3</math> positions down the line. In this case, we start at position <math>1</math> and end at position <math>19</math>, i.e. we move a total of <math>18</math> positions down the line. Therefore, we must add a series of <math>2</math>s and <math>3</math>s to get <math>18</math>. There are a number of ways to do this:<br />
<br />
'''Case 1''': nine <math>2</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.<br />
<br />
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose4} = 35</math> ways to arrange them.<br />
<br />
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.<br />
<br />
==Solution 3 (casework and blocks)==<br />
We can simplify the original problem into a problem where there are <math>2^{17}</math> binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of <math>0</math>s, <math>1</math>s, and <math>11</math>s. Now, we use casework: <br />
<br />
'''Case 1''': Alternating 1s and 0s. There is simply 1 way to do this: <math>0101010101010101010</math>. <br />
Now, we note that there cannot be only one block of <math>11</math> in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of <math>11</math>s this cannot be satisfied. This is true for all odd numbers of <math>11</math> blocks. <br />
<br />
'''Case 2''': There are 2 <math>11</math> blocks. Using the zeroes in the sequence as dividers, we have a sample as <math>0110110101010101010</math>. We know there are 8 places for <math>11</math>s, which will be filled by <math>1</math>s if the <math>11</math>s don't fill them. This is <math>{8\choose2} = 28</math> ways. <br />
<br />
'''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math>11</math> blocks. <br />
<br />
'''Case 4''': No single <math>1</math> blocks, only <math>11</math> blocks. There is simply one case for this, which is <math>0110110110110110110</math>. <br />
<br />
Adding these four cases, we have <math>1+28+35+1=\boxed{\textbf{(C) }65}</math> as our final answer. <br />
<br />
~Equinox8<br />
<br />
==Solution 4 (similar to #3)==<br />
Any valid sequence must start with a <math>0</math>. We can then think of constructing a sequence as adding groups of terms to this <math>0</math>, each ending in <math>0</math>. (This is always possible because every valid string ends in <math>0</math>.) For example, we can represent the string <math>01011010110110</math> as: <math>0-10-110-10-110-110</math>.<br />
To not have any consecutive 0s, we must have at least one <math>1</math> before the next <math>0</math>. However, we cannot have three or more <math>1</math>s before the next <math>0</math> because we cannot have three consecutive <math>1</math>s. Consequently, we can only have one or two <math>1</math>s. <br />
<br />
So we can have the groups: <math>10</math> and <math>110</math>.<br />
<br />
After the initial <math>0</math>, we have <math>18</math> digits left to fill in the string. Let the number of <math>10</math> blocks be <math>x</math>, and <math>110</math> be <math>y</math>. Then <math>x</math> and <math>y</math> must satisfy <math>2x+3y=18</math>. We recognize this as a Diophantine equation. Taking <math>\pmod{2}</math> yields <math>y=0 \pmod{2}</math>. Since <math>x</math> and <math>y</math> must both be nonnegative, we get the solutions <math>(9, 0)</math>, <math>(6, 2)</math>, <math>(3, 4)</math>, and <math>(0, 6)</math>. We now handle each of these cases separately.<br />
<br />
<math>(9, 0)</math>: Only one arrangement, namely all <math>10</math>s.<br />
<br />
<math>(6, 2)</math>: We have 6 groups of <math>11</math>, and <math>2</math> groups of <math>110</math>. This has <math>\binom{6+2}{2}=28</math> cases.<br />
<br />
<math>(3, 4)</math>: This means we have 3 groups of <math>10</math>, and 4 groups of <math>110</math>. This has <math>\binom{3+4}{3}=35</math> cases.<br />
<br />
<math>(0, 6)</math>: Only one arrangement, namely all <math>110</math>.<br />
<br />
Adding these, we have <math>1+28+35+1=65 \longrightarrow \boxed{(C)}</math>.<br />
~Math4Life2020<br />
<br />
~edited by alpha_2 for spelling and and typos<br />
<br />
==Solution 5 (Constructive Counting)==<br />
Suppose the number of <math>0</math>s is <math>n</math>. We can construct the sequence in two steps:<br />
<br />
Step 1: put <math>n-1</math> of <math>1</math>s between the <math>0</math>s;<br />
<br />
Step 2: put the rest <math>19-n-(n-1)=20-2n</math> of <math>1</math>s in the <math>n-1</math> spots where there is a <math>1</math>. There are <math>\binom{n-1}{20-2n}</math> ways of doing this.<br />
<br />
Now we find the possible values of <math>n</math>:<br />
<br />
First of all <math>n+(n-1) \leq 19 \Rightarrow n\leq 10</math> (otherwise there will be two consecutive <math>0</math>s); <br />
<br />
And secondly <math>20-2n \leq n-1\Rightarrow n\geq 7</math> (otherwise there will be three consecutive <math>1</math>s). <br />
<br />
Therefore the answer is<br />
<cmath><br />
\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{\textbf{(C) }65}.<br />
</cmath><br />
<br />
~ asops<br />
<br />
== Solution 6 (Recursion)==<br />
<br />
For a valid sequence of length <math>n</math>, the sequence must be in the form of <math>01xx...xx10</math>. By removing the <math>01</math> at the start of the sequence and the <math>10</math> at the end of the sequence, there are <math>n-4</math> bits left. The <math>n-4</math> bits left can be in the form of:<br />
<math>0yy...yy0</math>, the whole <math>(n-4)</math> bits are valid sequence, which is <math>f(n-4)</math><br />
<math>0yy...y01</math>, the <math>(n-5)</math> bits before the last <math>1</math> are valid sequence, which is <math>f(n-5)</math><br />
<math>10y...yy0</math>, the <math>(n-5)</math> bits after the first <math>1</math> are valid sequence, which is <math>f(n-5)</math><br />
<math>10y...y01</math>, the <math>(n-6)</math> bits between the first and last <math>1</math> are valid sequence, which is <math>f(n-6)</math><br />
<br />
So, <math>f(n) = f(n-4) + 2f(n-5) + f(n-6)</math><br />
<br />
We will calculate <math>f(19)</math> by [https://en.wikipedia.org/wiki/Dynamic_programming Dynamic Programming].<br />
<br />
<math>f(3) = 1</math><br />
<br />
<math>f(4) = 1</math><br />
<br />
<math>f(5) = 1</math><br />
<br />
<math>f(6) = 2</math><br />
<br />
<math>f(7) = 2</math><br />
<br />
<math>f(8) = 3</math><br />
<br />
<math>f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4</math><br />
<br />
<math>f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5</math><br />
<br />
<math>f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7</math><br />
<br />
<math>f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9</math><br />
<br />
<math>f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12</math><br />
<br />
<math>f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16</math><br />
<br />
<math>f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21</math><br />
<br />
<math>f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28</math><br />
<br />
<math>f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37</math><br />
<br />
<math>f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49</math><br />
<br />
<math>f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{\text{\bf{(C)} } 65}</math><br />
<br />
We can further prove <math>f(n) = f(n-4) + 2f(n-5) + f(n-6)</math> is equivalent to <math>f(n) = f(n-2) + f(n-3)</math><br />
<br />
Let <math>k(n) = f(n-2) + f(n-3)</math><br />
<br />
<math>k(n-2) = f(n-4) + f(n-5)</math><br />
<br />
<math>k(n-3) = f(n-5) + f(n-6)</math><br />
<br />
<math>k(n-2)+ k(n-3) = f(n-4) + 2f(n-5) + f(n-6) = f(n)</math><br />
<br />
<math>f(n) = k(n-2)+ k(n-3)</math><br />
<br />
<math>f(n-2) = k(n-4)+ k(n-5)</math><br />
<br />
<math>f(n-3) = k(n-5)+ k(n-6)</math><br />
<br />
<math>k(n) = f(n-2) + f(n-3) = k(n-4)+ k(n-5) + k(n-5)+ k(n-6) = k(n-4) + 2k(n-5) + k(n-6)</math><br />
<br />
So <math>k(n)</math> is the same as <math>f(n)</math>.<br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Video Solution==<br />
For those who want a video solution: https://youtu.be/VamT49PjmdI<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1709732019 AMC 10B Problems/Problem 232022-02-05T15:24:35Z<p>Junche: </p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems#Problem 23|2019 AMC 10B #23]] and [[2019 AMC 12B Problems#Problem 20|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A=(6,13)</math> and <math>B=(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is <math>(x, 0)</math>, the Pythagorean Theorem gives <math>\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}</math>. This simplifies to <math>x = 5</math>.<br />
<br />
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) <math>AOBX</math> is cyclic. <br />
<br />
Therefore, we can apply [[Ptolemy's Theorem]] to give:<br />
<br />
<math>2\sqrt{170}r = d \sqrt{40}</math>, where <math>r</math> is the radius of the circle and <math>d</math> is the distance between the circle's center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}r</math>. <br />
<br />
Using the Pythagorean Theorem on the right triangle <math>OAX</math> (or <math>OBX</math>), we find that <math>170 + r^2 = 17r^2</math>, so <math>r^2 = \frac{85}{8}</math>, and thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
===Diagram for Solution 1===<br />
[[File:Desmos-graph (1).png|900px|caption]]<br />
<br />
~BakedPotato66<br />
<br />
==Solution 2 (coordinate bash)==<br />
We firstly obtain <math>x=5</math> as in Solution 1. Label the point <math>(5,0)</math> as <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line passing through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. Therefore, the slope of the line perpendicular to <math>AC</math> is <math>-\frac{1}{13}</math>, so its equation is <math>y=-\frac{x}{13}+\frac{175}{13}</math>. <br />
<br />
But notice that this line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. Hence <math>3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}</math>. So the center of the circle is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. <br />
<br />
Finally, the distance between the center, <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>, and point <math>A</math> is <math>\frac{\sqrt{170}}{4}</math>. Thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The midpoint of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>x</math>-axis. Then <math>CD</math> is the perpendicular bisector of <math>AB</math>. Let the center of the circle be <math>O</math>. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, so <math>\frac{OA}{AC} = \frac{AD}{DC}</math>.<br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, so the slope of <math>CD</math> is <math>3</math>. Hence, the equation of <math>CD</math> is <math>y-12=3(x-9) \Rightarrow y=3x-15</math>. Letting <math>y=0</math>, we have <math>x=5</math>, so <math>C = (5,0)</math>.<br />
<br />
Now, we compute <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math>,<br />
<math>AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}</math>, and<br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math>.<br />
<br />
Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</math>,<br />
and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
<br />
==Solution 4 (how fast can you multiply two-digit numbers?)==<br />
Let <math>(x,0)</math> be the intersection on the x-axis. By Power of a Point Theorem, <math>(x-6)^2+13^2=(x-12)^2+11^2\implies x=5</math>. Then the equations for the tangent lines passing <math>A</math> and <math>B</math>, respectively, are <math>13(x-6)+13=y</math> and <math>\frac{11}{7}(x-12)+11=y</math>. Then the lines normal (perpendicular) to them are <math>-\frac{1}{13}(x-6)+13=y</math> and <math>-\frac{7}{11}(x-12)+11=y</math>. Solving for <math>x</math>, we have<br />
<br />
<br />
<br />
<cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath><br />
<cmath>\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}</cmath><br />
<cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath><br />
<br />
After condensing, <math>x=\frac{37}{4}</math>. Then, the center of <math>\omega</math> is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. Apply distance formula. WLOG, assume you use <math>A</math>. Then, the area of <math>\omega</math> is <cmath>\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.</cmath><br />
<br />
==Solution 5 (power of a point)==<br />
<br />
Firstly, the point of intersection of the two tangent lines has an equal distance to points <math>A</math> and <math>B</math> due to power of a point theorem. This means we can easily find the point, which is <math>(5, 0)</math>. Label this point <math>X</math>. <math>\triangle{XAB}</math> is an isosceles triangle with lengths, <math>\sqrt{170}</math>, <math>\sqrt{170}</math>, and <math>2\sqrt{10}</math>. Label the midpoint of segment <math>AB</math> as <math>M</math>. The height of this triangle, or <math>\overline{XM}</math>, is <math>4\sqrt{10}</math>. Since <math>\overline{XM}</math> bisects <math>\overline{AB}</math>, <math>\overleftrightarrow{XM}</math> contains the diameter of circle <math>\omega</math>. Let the two points on circle <math>\omega</math> where <math>\overleftrightarrow{XM}</math> intersects be <math>P</math> and <math>Q</math> with <math>\overline{XP}</math> being the shorter of the two. Now let <math>\overline{MP}</math> be <math>x</math> and <math>\overline{MQ}</math> be <math>y</math>. By Power of a Point on <math>\overline{PQ}</math> and <math>\overline{AB}</math>, <math>xy = (\sqrt{10})^2 = 10</math>. Applying Power of a Point again on <math>\overline{XQ}</math> and <math>\overline{XA}</math>, <math>(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170</math>. Expanding while using the fact that <math>xy = 10</math>, <math>y=x+\frac{\sqrt{10}}{2}</math>. Plugging this into <math>xy=10</math>, <math>2x^2+\sqrt{10}x-20=0</math>. Using the quadratic formula, <math>x = \frac{\sqrt{170}-\sqrt{10}}{4}</math>, and since <math>x+y=2x+\frac{\sqrt{10}}{2}</math>, <math>x+y=\frac{\sqrt{170}}{2}</math>. Since this is the diameter, the radius of circle <math>\omega</math> is <math>\frac{\sqrt{170}}{4}</math>, and so the area of circle <math>\omega</math> is <math>\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 6 (Similar to #3)==<br />
Let the tangent lines from A and B intersect at X. Let the center of <math>\omega</math> be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is <math>(5, 0)</math>. However, notice that CMX is the perpendicular bisector of AB. Thus, AM is the altitude from A to CX. Using the distance formula on AX, we get that the length of <math>AX=\sqrt{170}=\sqrt{17}\sqrt{10}</math>. Using the distance formula on AM, we get that <math>AM=\sqrt{10}</math>. Using the distance formula on MX, we get that <math>MX=4\sqrt{10}</math>. To get AC (the radius of <math>\omega</math>), we use either of these methods:<br />
<br />
Method 1: Since CAX is a right angle, the altitude AM is the geometric mean of XM and MC. We get that <math>MC=\frac{\sqrt{17}}{4}</math>. Thus, XC has length <math>XC=\frac{17\sqrt{10}}{4}</math>. Using the Pythagorean Theorem on CAX yields <math>CA=\frac{\sqrt{170}}{4}</math>. <br />
<br />
Method 2: Note that CAX and AMX are similar. Thus, <math>\frac{AM}{MX}=\frac{AC}{AX}</math>. Solving for AC yields <math>\frac{AX \cdot AM}{MX}=\frac{\sqrt{170}}{4}</math>. <br />
<br />
Using the area formula for a circle yields that the area is <math>\frac{85\pi}{8} \longrightarrow \boxed{(C)}</math>.<br />
~Math4Life2020<br />
<br />
==Video Solution==<br />
For those who want a video solution: (Is similar to Solution 1)<br />
https://youtu.be/WI2NVuIp1Ik<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/W1zuqrTlBtU<br />
<br />
~IceMatrix<br />
==Video Solution by The Power of Logic==<br />
https://www.youtube.com/watch?v=sQIWSrio_Hc<br />
<br />
~The Power of Logic<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_20&diff=1709712019 AMC 12B Problems/Problem 202022-02-05T15:16:36Z<p>Junche: Redirected page to 2019 AMC 10B Problems/Problem 23</p>
<hr />
<div>#REDIRECT[[2019_AMC_10B_Problems/Problem_23]]</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1709702019 AMC 10B Problems/Problem 232022-02-05T15:10:56Z<p>Junche: </p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A=(6,13)</math> and <math>B=(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is <math>(x, 0)</math>, the Pythagorean Theorem gives <math>\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}</math>. This simplifies to <math>x = 5</math>.<br />
<br />
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) <math>AOBX</math> is cyclic. <br />
<br />
Therefore, we can apply [[Ptolemy's Theorem]] to give:<br />
<br />
<math>2\sqrt{170}r = d \sqrt{40}</math>, where <math>r</math> is the radius of the circle and <math>d</math> is the distance between the circle's center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}r</math>. <br />
<br />
Using the Pythagorean Theorem on the right triangle <math>OAX</math> (or <math>OBX</math>), we find that <math>170 + r^2 = 17r^2</math>, so <math>r^2 = \frac{85}{8}</math>, and thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
===Diagram for Solution 1===<br />
[[File:Desmos-graph (1).png|900px|caption]]<br />
<br />
~BakedPotato66<br />
<br />
==Solution 2 (coordinate bash)==<br />
We firstly obtain <math>x=5</math> as in Solution 1. Label the point <math>(5,0)</math> as <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line passing through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. Therefore, the slope of the line perpendicular to <math>AC</math> is <math>-\frac{1}{13}</math>, so its equation is <math>y=-\frac{x}{13}+\frac{175}{13}</math>. <br />
<br />
But notice that this line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. Hence <math>3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}</math>. So the center of the circle is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. <br />
<br />
Finally, the distance between the center, <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>, and point <math>A</math> is <math>\frac{\sqrt{170}}{4}</math>. Thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The midpoint of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>x</math>-axis. Then <math>CD</math> is the perpendicular bisector of <math>AB</math>. Let the center of the circle be <math>O</math>. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, so <math>\frac{OA}{AC} = \frac{AD}{DC}</math>.<br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, so the slope of <math>CD</math> is <math>3</math>. Hence, the equation of <math>CD</math> is <math>y-12=3(x-9) \Rightarrow y=3x-15</math>. Letting <math>y=0</math>, we have <math>x=5</math>, so <math>C = (5,0)</math>.<br />
<br />
Now, we compute <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math>,<br />
<math>AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}</math>, and<br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math>.<br />
<br />
Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</math>,<br />
and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
<br />
==Solution 4 (how fast can you multiply two-digit numbers?)==<br />
Let <math>(x,0)</math> be the intersection on the x-axis. By Power of a Point Theorem, <math>(x-6)^2+13^2=(x-12)^2+11^2\implies x=5</math>. Then the equations for the tangent lines passing <math>A</math> and <math>B</math>, respectively, are <math>13(x-6)+13=y</math> and <math>\frac{11}{7}(x-12)+11=y</math>. Then the lines normal (perpendicular) to them are <math>-\frac{1}{13}(x-6)+13=y</math> and <math>-\frac{7}{11}(x-12)+11=y</math>. Solving for <math>x</math>, we have<br />
<br />
<br />
<br />
<cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath><br />
<cmath>\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}</cmath><br />
<cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath><br />
<br />
After condensing, <math>x=\frac{37}{4}</math>. Then, the center of <math>\omega</math> is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. Apply distance formula. WLOG, assume you use <math>A</math>. Then, the area of <math>\omega</math> is <cmath>\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.</cmath><br />
<br />
==Solution 5 (power of a point)==<br />
<br />
Firstly, the point of intersection of the two tangent lines has an equal distance to points <math>A</math> and <math>B</math> due to power of a point theorem. This means we can easily find the point, which is <math>(5, 0)</math>. Label this point <math>X</math>. <math>\triangle{XAB}</math> is an isosceles triangle with lengths, <math>\sqrt{170}</math>, <math>\sqrt{170}</math>, and <math>2\sqrt{10}</math>. Label the midpoint of segment <math>AB</math> as <math>M</math>. The height of this triangle, or <math>\overline{XM}</math>, is <math>4\sqrt{10}</math>. Since <math>\overline{XM}</math> bisects <math>\overline{AB}</math>, <math>\overleftrightarrow{XM}</math> contains the diameter of circle <math>\omega</math>. Let the two points on circle <math>\omega</math> where <math>\overleftrightarrow{XM}</math> intersects be <math>P</math> and <math>Q</math> with <math>\overline{XP}</math> being the shorter of the two. Now let <math>\overline{MP}</math> be <math>x</math> and <math>\overline{MQ}</math> be <math>y</math>. By Power of a Point on <math>\overline{PQ}</math> and <math>\overline{AB}</math>, <math>xy = (\sqrt{10})^2 = 10</math>. Applying Power of a Point again on <math>\overline{XQ}</math> and <math>\overline{XA}</math>, <math>(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170</math>. Expanding while using the fact that <math>xy = 10</math>, <math>y=x+\frac{\sqrt{10}}{2}</math>. Plugging this into <math>xy=10</math>, <math>2x^2+\sqrt{10}x-20=0</math>. Using the quadratic formula, <math>x = \frac{\sqrt{170}-\sqrt{10}}{4}</math>, and since <math>x+y=2x+\frac{\sqrt{10}}{2}</math>, <math>x+y=\frac{\sqrt{170}}{2}</math>. Since this is the diameter, the radius of circle <math>\omega</math> is <math>\frac{\sqrt{170}}{4}</math>, and so the area of circle <math>\omega</math> is <math>\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 6 (Similar to #3)==<br />
Let the tangent lines from A and B intersect at X. Let the center of <math>\omega</math> be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is <math>(5, 0)</math>. However, notice that CMX is the perpendicular bisector of AB. Thus, AM is the altitude from A to CX. Using the distance formula on AX, we get that the length of <math>AX=\sqrt{170}=\sqrt{17}\sqrt{10}</math>. Using the distance formula on AM, we get that <math>AM=\sqrt{10}</math>. Using the distance formula on MX, we get that <math>MX=4\sqrt{10}</math>. To get AC (the radius of <math>\omega</math>), we use either of these methods:<br />
<br />
Method 1: Since CAX is a right angle, the altitude AM is the geometric mean of XM and MC. We get that <math>MC=\frac{\sqrt{17}}{4}</math>. Thus, XC has length <math>XC=\frac{17\sqrt{10}}{4}</math>. Using the Pythagorean Theorem on CAX yields <math>CA=\frac{\sqrt{170}}{4}</math>. <br />
<br />
Method 2: Note that CAX and AMX are similar. Thus, <math>\frac{AM}{MX}=\frac{AC}{AX}</math>. Solving for AC yields <math>\frac{AX \cdot AM}{MX}=\frac{\sqrt{170}}{4}</math>. <br />
<br />
Using the area formula for a circle yields that the area is <math>\frac{85\pi}{8} \longrightarrow \boxed{(C)}</math>.<br />
~Math4Life2020<br />
<br />
==Video Solution==<br />
For those who want a video solution: (Is similar to Solution 1)<br />
https://youtu.be/WI2NVuIp1Ik<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/W1zuqrTlBtU<br />
<br />
~IceMatrix<br />
==Video Solution by The Power of Logic==<br />
https://www.youtube.com/watch?v=sQIWSrio_Hc<br />
<br />
~The Power of Logic<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_24&diff=1708692016 AMC 10A Problems/Problem 242022-02-03T04:13:43Z<p>Junche: </p>
<hr />
<div>{{duplicate|[[2016 AMC 10A Problems#Problem 24|2016 AMC 10A #24]] and [[2016 AMC 12A Problems#Problem 21|2016 AMC 12A #21]]}}<br />
<br />
==Problem==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
<br />
==Solution 1 (Algebra)==<br />
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by <math>200</math> for now, then multiply it back at the end of our solution.<br />
<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(B,D,O,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$E$",E,WSW);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(B--D);<br />
draw(rightanglemark(C,E,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite.<br />
<br />
We set lengths <math>BE=ED</math> equal to <math>x</math> (Solution 1.1 begins from here). By the Pythagorean Theorem,<br />
<cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath><br />
<br />
We solve for <math>x</math>:<br />
<cmath>1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2</cmath><br />
<cmath>2\sqrt{(1-x^2)(2-x^2)}=2x^2-1</cmath><br />
<cmath>4(1-x^2)(2-x^2)=(2x^2-1)^2</cmath><br />
<cmath>8-12x^2+4x^4=4x^4-4x^2+1</cmath><br />
<cmath>8x^2=7</cmath><br />
<cmath>x=\frac{\sqrt{14}}{4}</cmath><br />
<br />
By Ptolemy's Theorem,<br />
<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath><br />
<br />
Substituting values,<br />
<cmath>1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2</cmath><br />
<cmath>1+AD=\frac{7}{2}</cmath><br />
<cmath>AD=\frac{5}{2}</cmath><br />
<br />
Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>.<br />
<br />
==Solution 2 (HARD Algebra)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
</asy><br />
<br />
Let quadrilateral <math>ABCD</math> be inscribed in circle <math>O</math>, where <math>AD</math> is the side of unknown length. Draw the radii from center <math>O</math> to all four vertices of the quadrilateral, and draw the altitude of <math>\triangle BOC</math> such that it passes through side <math>AD</math> at the point <math>G</math> and meets side <math>BC</math> at the point <math>H</math>.<br />
<br />
By the Pythagorean Theorem, the length of <math>OH</math> is<br />
<cmath>\begin{align*}<br />
\sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2}<br />
\\ &= \sqrt{80000 - 10000}<br />
\\ &= \sqrt{70000}<br />
\\ &= 100\sqrt{7}.<br />
\end{align*}</cmath><br />
<br />
Note that <math>[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].</math> Let the length of <math>OG</math> be <math>h</math> and the length of <math>AD</math> be <math>x</math>; then we have that <br />
<br />
<math> [AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].</math> <br />
<br />
Furthermore,<br />
<cmath>\begin{align*}<br />
h &= \sqrt{OD^2 - GD^2}<br />
\\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2}<br />
\\ &= \sqrt{80000 - \frac{x^2}{4}}<br />
\end{align*}</cmath><br />
<br />
Substituting this value of <math>h</math> into the previous equation and evaluating for <math>x</math>, we get:<br />
<cmath>\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}</cmath><br />
<cmath>\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}</cmath><br />
<cmath>60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)</cmath><br />
<cmath>40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)</cmath><br />
<cmath>7x^2 - 5600x + 1120000 = 320000 - x^2</cmath><br />
<cmath>8x^2 - 5600x + 800000 = 0</cmath><br />
<cmath>x^2 - 700x + 100000 = 0</cmath><br />
<br />
The roots of this quadratic are found by using the quadratic formula:<br />
<cmath>\begin{align*}<br />
x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1}<br />
\\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2}<br />
\\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2}<br />
\\ &= 350 \pm \frac{300}{2}<br />
\\ &= 200, 500<br />
\end{align*}</cmath><br />
<br />
If the length of <math>AD</math> is <math>200</math>, then quadrilateral <math>ABCD</math> would be a square and thus, the radius of the circle would be<br />
<cmath>\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}</cmath><br />
Which is a contradiction. Therefore, our answer is <math>\boxed{500}.</math><br />
<br />
==Solution 3 (Trigonometry Bash)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Angle mark for BOC<br />
draw(anglemark(C,O,B));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on <math>\Delta BOC</math>; let <math> \theta = \angle BOC</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath><br />
Because <math>\angle AOB</math>, <math>\angle BOC</math>, and <math>\angle COD</math> are congruent, <math>\angle AOD = 3\theta</math>. To find the remaining side (<math>AD</math>), we simply have to apply the law of cosines to <math>\Delta AOD</math> . Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath><br />
<br />
==Solution 4 (Easier Trigonometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=foot(A,B,C);<br />
F=foot(D,B,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(A--E);<br />
draw(E--B);<br />
draw(C--F);<br />
draw(F--D);<br />
label("$E$",E,NW);<br />
label("$F$",F,NE);<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
draw(rightanglemark(A,E,B));<br />
draw(rightanglemark(C,F,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle <math>O</math> with <math>AD</math> being the desired side. Then, drop perpendiculars from <math>A</math> and <math>D</math> to the extended line of <math>\overline{BC}</math> and let these points be <math>E</math> and <math>F</math>, respectively. Also, let <math>\theta = \angle BOC</math>. From the [[Law of Cosines]] on <math>\triangle BOC</math>, we have <math>\cos \theta = \frac{3}{4}</math>.<br />
<br />
Now, since <math>\triangle BOC</math> is isosceles with <math>\overline{OB} \cong \overline{OC}</math>, we have that <math>\angle BCO = \angle CBO = 90 - \frac{\theta}{2}</math>. In addition, we know that <math>\overline{BC} \cong \overline{CD}</math> as they are both equal to <math>200</math> and <math>\overline{OB} \cong \overline{OC} \cong \overline{OD}</math> as they are both radii of the same circle. By SSS Congruence, we have that <math>\triangle OBC \cong \triangle OCD</math>, so we have that <math>\angle OCD = \angle BCO = 90 - \frac{\theta}{2}</math>, so <math>\angle DCF = \theta</math>.<br />
<br />
Thus, we have <math>\frac{FC}{DC} = \cos \theta = \frac{3}{4}</math>, so <math>FC = 150</math>. Similarly, <math>BE = 150</math>, and <math>AD = 150 + 200 + 150 = \boxed{500}</math>.<br />
<br />
==Solution 5 (Just Geometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
E=extension(B,O,A,D);<br />
<br />
label("$E$",E,NE);<br />
<br />
F=extension(C,O,A,D);<br />
<br />
label("$F$",F,NE);<br />
<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
<br />
</asy><br />
<br />
Let AD intersect OB at E and OC at F.<br />
<br />
<br />
<math>\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta</math><br />
<br />
<math>\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}</math><br />
<br />
<br />
From there, <math>\triangle{OAB} \sim \triangle{ABE}</math>, thus:<br />
<br />
<math>\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}</math><br />
<br />
<math>OA = OB</math> because they are both radii of <math>\odot{O}</math>. Since <math>\frac{OA}{AB} = \frac{OB}{AE}</math>, we have that <math>AB = AE</math>. Similarly, <math>CD = DF</math>.<br />
<br />
<math>OE = 100\sqrt{2} = \frac{OB}{2}</math> and <math>EF=\frac{BC}{2}=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}</math><br />
<br />
==Solution 6 (Ptolemy's Theorem)==<br />
<br />
<asy><br />
pathpen = black; pointpen = black;<br />
size(6cm);<br />
draw(unitcircle);<br />
pair A = D("A", dir(50), dir(50));<br />
pair B = D("B", dir(90), dir(90));<br />
pair C = D("C", dir(130), dir(130));<br />
pair D = D("D", dir(170), dir(170));<br />
pair O = D("O", (0,0), dir(-90));<br />
draw(A--C, red);<br />
draw(B--D, blue+dashed);<br />
draw(A--B--C--D--cycle);<br />
draw(A--O--C);<br />
draw(O--B);<br />
</asy><br />
<br />
Let <math>s = 200</math>. Let <math>O</math> be the center of the circle. Then <math>AC</math> is twice the altitude of <math>\triangle OBC</math> to <math>\overline{OB}</math>. Since <math>\triangle OBC</math> is isosceles we can compute its area to be <math>\frac{s^2 \sqrt{7}}{4}</math>, hence <math>CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}</math>.<br />
<br />
Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.</math> This gives us: <cmath>\boxed{\textbf{(E) } 500.}</cmath><br />
<br />
==Solution 7 (Trigonometry)==<br />
Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths <math>100,100\sqrt{7},200\sqrt{2}</math> by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is <math>\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}</math>. Similarly, the cosine is <math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}</math>.<br />
Since there are three sides, and since <math>\sin\theta=\sin\left(180-\theta\right)</math>,we seek to find <math>2r\sin 3\theta</math>.<br />
First, <math>\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}</math> and <math>\cos 2\theta=\frac{3}{4}</math> by Pythagorean.<br />
<cmath>\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}</cmath><br />
<cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}</cmath><br />
<br />
==Solution 8 (Area By Brahmagupta's Formula)==<br />
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=2</math> and <math>DA</math> is the missing side length. Let <math>DA=2x</math>. If <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>AD</math>, respectively, the height of the trapezoid is <math>OM-ON</math>. By the pythagorean theorem, <math>OM=\sqrt{OB^2-BM^2}=\sqrt7</math> and <math>ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}</math>. Thus the height of the trapezoid is <math>\sqrt7-\sqrt{8-x^2}</math>, so the area is <math>\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})</math>. By Brahmagupta's formula, the area is <math>\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Setting these two equal, we get <math>(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Dividing both sides by <math>x+1</math> and then squaring, we get <math>7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)</math>. Expanding the right hand side and canceling the <math>x^2</math> terms gives us <math>15-2(\sqrt7)(\sqrt{8-x^2})=2x+3</math>. Rearranging and dividing by two, we get <math>(\sqrt7)(\sqrt{8-x^2})=6-x</math>. Squaring both sides, we get <math>56-7x^2=x^2-12x+36</math>. Rearranging, we get <math>8x^2-12x-20=0</math>. Dividing by 4 we get <math>2x^2-3x-5=0</math>. Factoring we get, <math>(2x-5)(x+1)=0</math>, and since <math>x</math> cannot be negative, we get <math>x=2.5</math>. Since <math>DA=2x</math>, <math>DA=5</math>. Scaling up by 100, we get <math>\boxed{\textbf{(E)}\text{ 500}}</math>.<br />
<br />
==Solution 9 (Similar Triangles)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations, L is used to write alpha= statement<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O, L;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(A,D,O,B);<br />
F=extension(A,D,O,C);<br />
L=midpoint(C--D);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad = A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,NW);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,NE);<br />
label("$E$",E,SW);<br />
label("$F$",F,SE);<br />
label("$O$",O,SE);<br />
dot(O,linewidth(5));<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0));<br />
draw(anglemark(C,O,B));<br />
label("$\theta$",O,3N);<br />
draw(anglemark(E,F,O));<br />
label("$\alpha$",F,3SW);<br />
draw(anglemark(D,F,C));<br />
label("$\alpha$",F,3NE);<br />
draw(anglemark(F,C,D));<br />
label("$\alpha$",C,3SSE);<br />
draw(anglemark(C,D,F));<br />
label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW);<br />
</asy><br />
Label the points as shown, and let <math>\angle{EOF} = \theta</math>. Since <math>\overline{OB} = \overline{OC}</math>, and <math>\triangle{OFE} \sim \triangle{OCB}</math>, we get that <math>\angle{EFO} = 90-\frac{\theta}{2}</math>. We assign <math>\alpha</math> to <math>90-\frac{\theta}{2}</math> for simplicity. <br />
From here, by vertical angles <math>\angle{CFD} = \alpha</math>. Also, since <math>\triangle{OCB} \cong \triangle{ODC}</math>, <math>\angle{OCD} = \alpha</math>. This means that <math>\angle{CDF} = 180-2\alpha = \theta</math>, which leads to <math>\triangle{OCB} \sim \triangle{DCF}</math>. <br />
Since we know that <math>\overline{CD} = 200</math>, <math>\overline{DF} = 200</math>, and by similar reasoning <math>\overline{AE} = 200</math>. <br />
Finally, again using similar triangles, we get that <math>\overline{CF} = 100\sqrt{2}</math>, which means that <math>\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}</math>. We can again apply similar triangles (or use Power of a Point) to get <math>\overline{EF} = 100</math>, and finally <math>\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}</math> - ColtsFan10<br />
<br />
==Solution 10 (Parameshwara’s Formula for Circumradius) ==<br />
<br />
Scale down by <math>100</math>. We know that the semiperimeter of the quadrilateral is <math>\frac{(2 + 2 + 2 + x)}{2}</math> where <math>x = \overline {AD}</math>. Simplifying we get <math>\frac{6 + x}{2}</math>. Now, the radius is <math>2\sqrt {2}</math>, so <br />
<math>2\sqrt {2} = \frac{1}{4} \sqrt \frac {(4 + 2x)^{3}}{(\frac{2 + x}{2})^{3} (\frac {6 - x}{2})^{3}}</math>.<br />
<br />
Simplifying we get <math>x = 5</math>. So the answer is <math>500</math>.<br />
<br />
==Solution 11 (Complex Numbers)==<br />
<br />
We first scale down by a factor of <math>200\sqrt{2}</math>. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, so that <math>AD</math> is the length of the fourth side. We draw this in the complex plane so that <math>D</math> corresponds to the complex number <math>1</math>, and we let <math>C</math> correspond to the complex number <math>z</math>. Then, <math>A</math> corresponds to <math>z^3</math> and <math>B</math> corresponds to <math>z^2</math>. We are given that <math>\lvert z \rvert = 1</math> and <math>\lvert z-1 \rvert = 1/\sqrt{2}</math>, and we wish to find <math>\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert</math>. Let <math>z=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers. Then, <math>a^2+b^2=1</math> and <math>a^2-2a+1+b^2=1/2</math>; solving for <math>a</math> and <math>b</math> yields <math>a=3/4</math> and <math>b=\sqrt{7}/4</math>. Thus, <math>AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}</math>. Scaling back up gives us a final answer of <math>\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{\textbf{(E)} 500}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
==Video Solution by AoPS (Deven Ware)==<br />
https://www.youtube.com/watch?v=hpSyHZwsteM<br />
<br />
==Video Solution by Walt S.==<br />
https://www.youtube.com/watch?v=3iDqR9YNNkU<br />
== Video Solution (Ptolemy’s Theorem) ==<br />
https://youtu.be/NsQbhYfGh1Q?t=5094<br />
<br />
~ pi_is_3.14<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/gCmQlaiEG5A<br />
<br />
~IceMatrix<br />
==See Also==<br />
<br />
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_24&diff=1708522016 AMC 10A Problems/Problem 242022-02-02T18:02:57Z<p>Junche: </p>
<hr />
<div>==Problem==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
<br />
==Solution 1 (Algebra)==<br />
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by <math>200</math> for now, then multiply it back at the end of our solution.<br />
<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(B,D,O,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$E$",E,WSW);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(B--D);<br />
draw(rightanglemark(C,E,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite.<br />
<br />
We set lengths <math>BE=ED</math> equal to <math>x</math> (Solution 1.1 begins from here). By the Pythagorean Theorem,<br />
<cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath><br />
<br />
We solve for <math>x</math>:<br />
<cmath>1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2</cmath><br />
<cmath>2\sqrt{(1-x^2)(2-x^2)}=2x^2-1</cmath><br />
<cmath>4(1-x^2)(2-x^2)=(2x^2-1)^2</cmath><br />
<cmath>8-12x^2+4x^4=4x^4-4x^2+1</cmath><br />
<cmath>8x^2=7</cmath><br />
<cmath>x=\frac{\sqrt{14}}{4}</cmath><br />
<br />
By Ptolemy's Theorem,<br />
<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath><br />
<br />
Substituting values,<br />
<cmath>1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2</cmath><br />
<cmath>1+AD=\frac{7}{2}</cmath><br />
<cmath>AD=\frac{5}{2}</cmath><br />
<br />
Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>.<br />
<br />
==Solution 2 (HARD Algebra)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
</asy><br />
<br />
Let quadrilateral <math>ABCD</math> be inscribed in circle <math>O</math>, where <math>AD</math> is the side of unknown length. Draw the radii from center <math>O</math> to all four vertices of the quadrilateral, and draw the altitude of <math>\triangle BOC</math> such that it passes through side <math>AD</math> at the point <math>G</math> and meets side <math>BC</math> at the point <math>H</math>.<br />
<br />
By the Pythagorean Theorem, the length of <math>OH</math> is<br />
<cmath>\begin{align*}<br />
\sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2}<br />
\\ &= \sqrt{80000 - 10000}<br />
\\ &= \sqrt{70000}<br />
\\ &= 100\sqrt{7}.<br />
\end{align*}</cmath><br />
<br />
Note that <math>[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].</math> Let the length of <math>OG</math> be <math>h</math> and the length of <math>AD</math> be <math>x</math>; then we have that <br />
<br />
<math> [AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].</math> <br />
<br />
Furthermore,<br />
<cmath>\begin{align*}<br />
h &= \sqrt{OD^2 - GD^2}<br />
\\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2}<br />
\\ &= \sqrt{80000 - \frac{x^2}{4}}<br />
\end{align*}</cmath><br />
<br />
Substituting this value of <math>h</math> into the previous equation and evaluating for <math>x</math>, we get:<br />
<cmath>\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}</cmath><br />
<cmath>\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}</cmath><br />
<cmath>60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)</cmath><br />
<cmath>40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)</cmath><br />
<cmath>7x^2 - 5600x + 1120000 = 320000 - x^2</cmath><br />
<cmath>8x^2 - 5600x + 800000 = 0</cmath><br />
<cmath>x^2 - 700x + 100000 = 0</cmath><br />
<br />
The roots of this quadratic are found by using the quadratic formula:<br />
<cmath>\begin{align*}<br />
x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1}<br />
\\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2}<br />
\\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2}<br />
\\ &= 350 \pm \frac{300}{2}<br />
\\ &= 200, 500<br />
\end{align*}</cmath><br />
<br />
If the length of <math>AD</math> is <math>200</math>, then quadrilateral <math>ABCD</math> would be a square and thus, the radius of the circle would be<br />
<cmath>\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}</cmath><br />
Which is a contradiction. Therefore, our answer is <math>\boxed{500}.</math><br />
<br />
==Solution 3 (Trigonometry Bash)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Angle mark for BOC<br />
draw(anglemark(C,O,B));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on <math>\Delta BOC</math>; let <math> \theta = \angle BOC</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath><br />
Because <math>\angle AOB</math>, <math>\angle BOC</math>, and <math>\angle COD</math> are congruent, <math>\angle AOD = 3\theta</math>. To find the remaining side (<math>AD</math>), we simply have to apply the law of cosines to <math>\Delta AOD</math> . Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath><br />
<br />
==Solution 4 (Easier Trigonometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=foot(A,B,C);<br />
F=foot(D,B,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(A--E);<br />
draw(E--B);<br />
draw(C--F);<br />
draw(F--D);<br />
label("$E$",E,NW);<br />
label("$F$",F,NE);<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
draw(rightanglemark(A,E,B));<br />
draw(rightanglemark(C,F,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle <math>O</math> with <math>AD</math> being the desired side. Then, drop perpendiculars from <math>A</math> and <math>D</math> to the extended line of <math>\overline{BC}</math> and let these points be <math>E</math> and <math>F</math>, respectively. Also, let <math>\theta = \angle BOC</math>. From the [[Law of Cosines]] on <math>\triangle BOC</math>, we have <math>\cos \theta = \frac{3}{4}</math>.<br />
<br />
Now, since <math>\triangle BOC</math> is isosceles with <math>\overline{OB} \cong \overline{OC}</math>, we have that <math>\angle BCO = \angle CBO = 90 - \frac{\theta}{2}</math>. In addition, we know that <math>\overline{BC} \cong \overline{CD}</math> as they are both equal to <math>200</math> and <math>\overline{OB} \cong \overline{OC} \cong \overline{OD}</math> as they are both radii of the same circle. By SSS Congruence, we have that <math>\triangle OBC \cong \triangle OCD</math>, so we have that <math>\angle OCD = \angle BCO = 90 - \frac{\theta}{2}</math>, so <math>\angle DCF = \theta</math>.<br />
<br />
Thus, we have <math>\frac{FC}{DC} = \cos \theta = \frac{3}{4}</math>, so <math>FC = 150</math>. Similarly, <math>BE = 150</math>, and <math>AD = 150 + 200 + 150 = \boxed{500}</math>.<br />
<br />
==Solution 5 (Just Geometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
E=extension(B,O,A,D);<br />
<br />
label("$E$",E,NE);<br />
<br />
F=extension(C,O,A,D);<br />
<br />
label("$F$",F,NE);<br />
<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
<br />
</asy><br />
<br />
Let AD intersect OB at E and OC at F.<br />
<br />
<br />
<math>\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta</math><br />
<br />
<math>\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}</math><br />
<br />
<br />
From there, <math>\triangle{OAB} \sim \triangle{ABE}</math>, thus:<br />
<br />
<math>\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}</math><br />
<br />
<math>OA = OB</math> because they are both radii of <math>\odot{O}</math>. Since <math>\frac{OA}{AB} = \frac{OB}{AE}</math>, we have that <math>AB = AE</math>. Similarly, <math>CD = DF</math>.<br />
<br />
<math>OE = 100\sqrt{2} = \frac{OB}{2}</math> and <math>EF=\frac{BC}{2}=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}</math><br />
<br />
==Solution 6 (Ptolemy's Theorem)==<br />
<br />
<asy><br />
pathpen = black; pointpen = black;<br />
size(6cm);<br />
draw(unitcircle);<br />
pair A = D("A", dir(50), dir(50));<br />
pair B = D("B", dir(90), dir(90));<br />
pair C = D("C", dir(130), dir(130));<br />
pair D = D("D", dir(170), dir(170));<br />
pair O = D("O", (0,0), dir(-90));<br />
draw(A--C, red);<br />
draw(B--D, blue+dashed);<br />
draw(A--B--C--D--cycle);<br />
draw(A--O--C);<br />
draw(O--B);<br />
</asy><br />
<br />
Let <math>s = 200</math>. Let <math>O</math> be the center of the circle. Then <math>AC</math> is twice the altitude of <math>\triangle OBC</math> to <math>\overline{OB}</math>. Since <math>\triangle OBC</math> is isosceles we can compute its area to be <math>\frac{s^2 \sqrt{7}}{4}</math>, hence <math>CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}</math>.<br />
<br />
Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.</math> This gives us: <cmath>\boxed{\textbf{(E) } 500.}</cmath><br />
<br />
==Solution 7 (Trigonometry)==<br />
Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths <math>100,100\sqrt{7},200\sqrt{2}</math> by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is <math>\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}</math>. Similarly, the cosine is <math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}</math>.<br />
Since there are three sides, and since <math>\sin\theta=\sin\left(180-\theta\right)</math>,we seek to find <math>2r\sin 3\theta</math>.<br />
First, <math>\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}</math> and <math>\cos 2\theta=\frac{3}{4}</math> by Pythagorean.<br />
<cmath>\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}</cmath><br />
<cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}</cmath><br />
<br />
==Solution 8 (Area By Brahmagupta's Formula)==<br />
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=2</math> and <math>DA</math> is the missing side length. Let <math>DA=2x</math>. If <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>AD</math>, respectively, the height of the trapezoid is <math>OM-ON</math>. By the pythagorean theorem, <math>OM=\sqrt{OB^2-BM^2}=\sqrt7</math> and <math>ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}</math>. Thus the height of the trapezoid is <math>\sqrt7-\sqrt{8-x^2}</math>, so the area is <math>\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})</math>. By Brahmagupta's formula, the area is <math>\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Setting these two equal, we get <math>(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Dividing both sides by <math>x+1</math> and then squaring, we get <math>7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)</math>. Expanding the right hand side and canceling the <math>x^2</math> terms gives us <math>15-2(\sqrt7)(\sqrt{8-x^2})=2x+3</math>. Rearranging and dividing by two, we get <math>(\sqrt7)(\sqrt{8-x^2})=6-x</math>. Squaring both sides, we get <math>56-7x^2=x^2-12x+36</math>. Rearranging, we get <math>8x^2-12x-20=0</math>. Dividing by 4 we get <math>2x^2-3x-5=0</math>. Factoring we get, <math>(2x-5)(x+1)=0</math>, and since <math>x</math> cannot be negative, we get <math>x=2.5</math>. Since <math>DA=2x</math>, <math>DA=5</math>. Scaling up by 100, we get <math>\boxed{\textbf{(E)}\text{ 500}}</math>.<br />
<br />
==Solution 9 (Similar Triangles)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations, L is used to write alpha= statement<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O, L;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(A,D,O,B);<br />
F=extension(A,D,O,C);<br />
L=midpoint(C--D);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad = A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,NW);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,NE);<br />
label("$E$",E,SW);<br />
label("$F$",F,SE);<br />
label("$O$",O,SE);<br />
dot(O,linewidth(5));<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0));<br />
draw(anglemark(C,O,B));<br />
label("$\theta$",O,3N);<br />
draw(anglemark(E,F,O));<br />
label("$\alpha$",F,3SW);<br />
draw(anglemark(D,F,C));<br />
label("$\alpha$",F,3NE);<br />
draw(anglemark(F,C,D));<br />
label("$\alpha$",C,3SSE);<br />
draw(anglemark(C,D,F));<br />
label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW);<br />
</asy><br />
Label the points as shown, and let <math>\angle{EOF} = \theta</math>. Since <math>\overline{OB} = \overline{OC}</math>, and <math>\triangle{OFE} \sim \triangle{OCB}</math>, we get that <math>\angle{EFO} = 90-\frac{\theta}{2}</math>. We assign <math>\alpha</math> to <math>90-\frac{\theta}{2}</math> for simplicity. <br />
From here, by vertical angles <math>\angle{CFD} = \alpha</math>. Also, since <math>\triangle{OCB} \cong \triangle{ODC}</math>, <math>\angle{OCD} = \alpha</math>. This means that <math>\angle{CDF} = 180-2\alpha = \theta</math>, which leads to <math>\triangle{OCB} \sim \triangle{DCF}</math>. <br />
Since we know that <math>\overline{CD} = 200</math>, <math>\overline{DF} = 200</math>, and by similar reasoning <math>\overline{AE} = 200</math>. <br />
Finally, again using similar triangles, we get that <math>\overline{CF} = 100\sqrt{2}</math>, which means that <math>\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}</math>. We can again apply similar triangles (or use Power of a Point) to get <math>\overline{EF} = 100</math>, and finally <math>\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}</math> - ColtsFan10<br />
<br />
==Solution 10 (Parameshwara’s Formula for Circumradius) ==<br />
<br />
Scale down by <math>100</math>. We know that the semiperimeter of the quadrilateral is <math>\frac{(2 + 2 + 2 + x)}{2}</math> where <math>x = \overline {AD}</math>. Simplifying we get <math>\frac{6 + x}{2}</math>. Now, the radius is <math>2\sqrt {2}</math>, so <br />
<math>2\sqrt {2} = \frac{1}{4} \sqrt \frac {(4 + 2x)^{3}}{(\frac{2 + x}{2})^{3} (\frac {6 - x}{2})^{3}}</math>.<br />
<br />
Simplifying we get <math>x = 5</math>. So the answer is <math>500</math>.<br />
<br />
==Solution 11 (Complex Numbers)==<br />
<br />
We first scale down by a factor of <math>200\sqrt{2}</math>. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, so that <math>AD</math> is the length of the fourth side. We draw this in the complex plane so that <math>D</math> corresponds to the complex number <math>1</math>, and we let <math>C</math> correspond to the complex number <math>z</math>. Then, <math>A</math> corresponds to <math>z^3</math> and <math>B</math> corresponds to <math>z^2</math>. We are given that <math>\lvert z \rvert = 1</math> and <math>\lvert z-1 \rvert = 1/\sqrt{2}</math>, and we wish to find <math>\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert</math>. Let <math>z=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers. Then, <math>a^2+b^2=1</math> and <math>a^2-2a+1+b^2=1/2</math>; solving for <math>a</math> and <math>b</math> yields <math>a=3/4</math> and <math>b=\sqrt{7}/4</math>. Thus, <math>AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}</math>. Scaling back up gives us a final answer of <math>\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{\textbf{(E)} 500}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
==Video Solution by AoPS (Deven Ware)==<br />
https://www.youtube.com/watch?v=hpSyHZwsteM<br />
<br />
==Video Solution by Walt S.==<br />
https://www.youtube.com/watch?v=3iDqR9YNNkU<br />
== Video Solution (Ptolemy’s Theorem) ==<br />
https://youtu.be/NsQbhYfGh1Q?t=5094<br />
<br />
~ pi_is_3.14<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/gCmQlaiEG5A<br />
<br />
~IceMatrix<br />
==See Also==<br />
<br />
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=User:Isabelchen&diff=169139User:Isabelchen2022-01-02T15:48:22Z<p>Junche: /* AOPS Contributions */</p>
<hr />
<div>==AOPS Contributions==<br />
<br />
===AMC 10===<br />
<br />
1. [https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_24#Graph_Theory_Insight 2013 AMC10A Problem 24 Graph Theory Insight] (Graph Theory)<br />
<br />
2. [https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25#Solution_5_.28Case_Work_with_Drawing.29 2013 AMC10A Problem 25 Solution 5] (Discrete Geometry)<br />
<br />
3. [https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_22#Side_Note_to_Solution_1_.26_2 2013 AMC10B Problem 22 Side Note] (Number Theory)<br />
<br />
4. [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18#Solution_2 2014 AMC10A Problem 18 Solution 2] (Analytic Geometry)<br />
<br />
5. [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18#Solution_3 2014 AMC10A Problem 18 Solution 3] (Analytic Geometry)<br />
<br />
6. [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22#Solution_6_.28Pure_Euclidian_Geometry.29 2014 AMC10A Problem 22 Solution 6] (Geometry)<br />
<br />
7. [https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22#Solution_7_.28Pure_Euclidian_Geometry.29 2014 AMC10A Problem 22 Solution 7] (Geometry)<br />
<br />
8. [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21#Solution_3 2015 AMC10A Problem 21 Solution 3] (3D Geometry)<br />
<br />
9. [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_22#Solution_4_.28Recursion.29 2015 AMC10A Problem 22 Solution 4] (Combinatorics)<br />
<br />
10. [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_24#Solution_3 2015 AMC10A Problem 24 Solution 3] (Geometry)<br />
<br />
11. [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_19#Solution_3 2015 AMC10B Problem 19 Solution 3] (Geometry)<br />
<br />
12. [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_20#Solution_4_.28Casework.29 2016 AMC10A Problem 20 Solution 4] (Combinatorics)<br />
<br />
13. [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_19#Solution_4_.28Area.29 2016 AMC10B Problem 19 Solution 4] (Geometry)<br />
<br />
14. [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_19#Solution_5_.28Area.29 2016 AMC10B Problem 19 Solution 5] (Geometry)<br />
<br />
15. [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22#Solution_4_.28Aggregate_Counting.29 2016 AMC10B Problem 22 Solution 4] (Graph Theory)<br />
<br />
16. [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_25#Supplement 2016 AMC10B Problem 25 Solution 1 Supplement] (Number Theory)<br />
<br />
17. [https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_25#Solution_3_.28Casework.29 2016 AMC10B Problem 25 Solution 3] (Number Theory)<br />
<br />
18. [https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17#Solution_4 2017 AMC10B Problem 17 Solution 4] (Combinatorics)<br />
<br />
19. [https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_25#Solution_4_.28Working_Backwards.29 2017 AMC10B Problem 25 Solution 4] (Number Theory)<br />
<br />
20. [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_10#Solution_8_.28Analytic_Geometry.29 2018 AMC10A Problem 10 Solution 8] (Analytic Geometry)<br />
<br />
21. [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22#Solution_6 2018 AMC10A Problem 22 Solution 6] (Number Theory)<br />
<br />
22. [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18#Solution_6_.28Derangement.29 2018 AMC10B Problem 18 Solution 6] (Combinatorics)<br />
<br />
23. [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_24#Solution_4_.28Area_Subtraction.29 2018 AMC10B Problem 24 Solution 4] (Geometry)<br />
<br />
24. [https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25#Solution_6 2018 AMC10B Problem 25 Solution 6] (Algebra)<br />
<br />
25. [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_22#Solution_3_.28with_Table.29 2019 AMC10A Problem 22 Solution 3] (Probability)<br />
<br />
26. [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_24#Solution_2_.28Pure_Elementary_Algebra.29 2019 AMC10A Problem 24 Solution 2] (Algebra)<br />
<br />
27. [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_22#Solution_4_.28Markov_Chain.29 2019 AMC10B Problem 22 Solution 4] (Probability, Markov Chain)<br />
<br />
28. [https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25#Solution_6_.28Recursion.29 2019 AMC10B Problem 25 Solution 6] (Combinatorics)<br />
<br />
29. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_20#Solution_8_.28Solving_Equations.29 2020 AMC10A Problem 20 Solution 8] (Geometry)<br />
<br />
30. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24#Solution_12_.28Diophantine_Equation.29 2020 AMC10A Problem 24 Solution 12] (Number Theory)<br />
<br />
31. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24#Solution_13_.28Euclidian_Algorithm.29 2020 AMC10A Problem 24 Solution 13] (Number Theory)<br />
<br />
32. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_13#Solution_3_.28Find_the_Pattern.29 2020 AMC10B Problem 13 Solution 3] (Combinatorics)<br />
<br />
33. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_21#Solution_6 2020 AMC10B Problem 21 Solution 6] (Geometry)<br />
<br />
34. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_23#Solution_7_.28Group_Theory.29 2020 AMC10B Problem 23 Solution 7] (Group Theory)<br />
<br />
35. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24#Solution_8 2020 AMC10B Problem 24 Solution 8] (Algebra)<br />
<br />
36. [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25#Solution_7_.28Integer_Partition.29 2020 AMC10B Problem 25 Solution 7] (Combinatorics)<br />
<br />
37. [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_23#Solution_8_.28Markov_Chain.29 2021 Spring AMC10A Problem 23 Solution 8] (Probability, Markov Chain)<br />
<br />
38. [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_24#Solution_7_.28Slope_and_Intercept.29 2021 Spring AMC10A Problem 24 Solution 7] (Analytic Geometry)<br />
<br />
39. [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25#Solution_4_.28Casework_and_Symmetry.29 2021 Spring AMC10A Problem 25 Solution 4] (Combinatorics)<br />
<br />
40. [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24#Solution_4_.28Graph_Coloring.29 2021 Fall AMC10A Problem 24 Solution 4] (Graph Theory)<br />
<br />
41. [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_7#Solution_4_.28Completing_the_Square.29 2021 Fall AMC10B Problem 12 Solution 4] (Algebra)<br />
<br />
42. [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_20#Side_Note_.28Bayes.27_Theorem.29 2021 Fall AMC10B Problem 20 Side Note] (Probability)<br />
<br />
43. [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_23#Solution_2_.28Ramsey.27s_Theorem.29 2021 Fall AMC10B Problem 23 Solution 2] (Graph Theory)<br />
<br />
44. [https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_20#Solution_4_.28Topology.29 2021 Fall AMC10B Problem 24 Solution 4] (Topology)<br />
<br />
===AMC 12===<br />
<br />
1. 2019 AMC12B Problem 20 Solution 6 (Trigonometry)</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_4&diff=1691382006 AMC 12A Problems/Problem 42022-01-02T15:20:49Z<p>Junche: </p>
<hr />
<div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2006 AMC 10A #4]]}}<br />
== Problem ==<br />
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?<br />
<br />
<math>\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23</math><br />
<br />
== Solution 1 ==<br />
From the [[greedy algorithm]], we have <math>9</math> in the hours section and <math>59</math> in the minutes section. <math>9+5+9=\boxed{\textbf{(E) }23}</math><br />
<br />
==Solution 2 ([[matrix]]) ==<br />
<br />
With a matrix, we can see<br />
<math><br />
\begin{bmatrix}<br />
1+2&9&6&3\\<br />
1+1&8&5&2\\<br />
1+0&7&4&1<br />
\end{bmatrix}<br />
</math><br />
The largest single digit sum we can get is <math>9</math>.<br />
For the minutes digits, we can combine the largest <math>2</math> digits, which are <math>9,5 \Rightarrow 9+5=14</math>, and finally <math>14+9=\boxed{\textbf{(E) }23}</math><br />
<br />
==Solution 3==<br />
<br />
We first note that since the watch displays time in AM and PM, the value for the hours section varies from <math>00-12</math>. Therefore, the maximum value of the digits for the hours is when the watch displays <math>09</math>, which gives us <math>0+9=9</math>.<br />
<br />
Next, we look at the value of the minutes section, which varies from <math>00-59</math>. Let this value be a number <math>ab</math>. We quickly find that the maximum value for <math>a</math> and <math>b</math> is respectively <math>5</math> and <math>9</math>.<br />
<br />
Adding these up, we get <math>9+5+9=\boxed{\textbf{(E) }23}</math>.<br />
<br />
~[[User:Dairyqueenxd|Dairyqueenxd]]<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{AMC10 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=User:Junche&diff=168723User:Junche2021-12-28T14:18:40Z<p>Junche: /* AOPS Contributions: */</p>
<hr />
<div>==AOPS Contributions:==<br />
<br />
1. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22#Solution_6 2012 AMC10B Problem 22 Solution 6] (Combinatorics)<br />
<br />
2. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_16#Solution_4 2012 AMC10B Problem 24 Solution 4] (Graph Theory)</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=User:Junche&diff=168722User:Junche2021-12-28T14:17:39Z<p>Junche: /* AOPS Contributions: */</p>
<hr />
<div>==AOPS Contributions:==<br />
<br />
1. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22#Solution_6 2012 AMC10B Problem 22 Solution 6] (Combinatorics)<br />
<br />
2. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_16#Solution_4 2012 AMC10B Problem 22 Solution 4] (Graph Theory)</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=User:Junche&diff=168721User:Junche2021-12-28T14:15:13Z<p>Junche: /* AOPS Contributions: */</p>
<hr />
<div>==AOPS Contributions:==<br />
<br />
1. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_16#Solution_4 2012 AMC10B Problem 22 Solution 4] (Combinatorics)<br />
<br />
2. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22#Solution_6 2012 AMC10B Problem 24 Solution 4] (Graph Theory)</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=1687202012 AMC 12B Problems/Problem 162021-12-28T14:12:15Z<p>Junche: /* Solution 4 */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br />
<br />
== Problem==<br />
Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br />
<br />
<math> \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 </math><br />
<br />
==Solutions==<br />
<br />
=== Solution 1===<br />
Let the ordered triple <math>(a,b,c)</math> denote that <math>a</math> songs are liked by Amy and Beth, <math>b</math> songs by Beth and Jo, and <math>c</math> songs by Jo and Amy. We claim that the only possible triples are <math>(1,1,1), (2,1,1), (1,2,1)(1,1,2)</math>. <br />
<br />
To show this, observe these are all valid conditions. Second, note that none of <math>a,b,c</math> can be bigger than 3. Suppose otherwise, that <math>a = 3</math>. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either <math>b</math> or <math>c</math> to be at least 1. In fact, we require either <math>b</math> or <math>c</math> to equal 1, otherwise there will be a song liked by all three. Suppose <math>b = 1</math>. Then we must have <math>c=0</math> since no song is liked by all three girls, a contradiction.<br />
<br />
'''Case 1''': How many ways are there for <math>(a,b,c)</math> to equal <math>(1,1,1)</math>? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So <math>(a,b,c)=(1,1,1)</math> in <math>4\cdot3\cdot2\cdot4 = 96</math> ways.<br />
<br />
'''Case 2''': To find the number of ways for <math>(a,b,c) = (2,1,1)</math>, observe there are <math>\binom{4}{2} = 6</math> choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are <math>6\cdot2\cdot3=36</math> ways for the girls to like the songs.<br />
<br />
That gives a total of <math>96 + 36 = 132</math> ways for the girls to like the songs, so the answer is <math>\boxed{(\textrm{\textbf{B}})}</math>.<br />
<br />
=== Solution 2===<br />
<br />
<br />
Let <math>AB, BJ</math>, and <math>AJ</math> denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let <math>A, B, J,</math> and <math>N</math> denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least <math>1\: AB, BJ</math>, and <math>AJ</math>, they must be <math>3</math> songs out of the <math>4</math> that Amy, Beth, and Jo listened to. The fourth song can be of any type <math>N, A, B, J, AB, BJ</math>, and <math>AJ</math> (there is no <math>ABJ</math> because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange <math>AB, BJ, AJ</math>, and a song from the set <math>\{N, A, B, J, AB, BJ, AJ\}</math>.<br />
<br />
Case 1: Fourth song = <math>N, A, B, J </math><br />
<br />
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!)</math> rearrangements for each choice <math>*\: 4</math> choices = <math>96</math>.<br />
<br />
Case 2: Fourth song = <math>AB, BJ, AJ</math><br />
<br />
Note that in Case <math>2</math>, all three of the choices for the fourth song repeat somewhere in the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!/2!)</math> rearrangements for each choice <math>*\: 3</math> choices = <math>36</math>.<br />
<br />
<math>96 + 36 = \boxed{\textbf{(B)} \: 132}</math>.<br />
<br />
=== Solution 3===<br />
<br />
<br />
There are <math>\binom{4}{3}</math> ways to choose the three songs that are liked by the three pairs of girls.<br />
<br />
There are <math>3!</math> ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br />
<br />
In total, there are <math>\binom{4}{3}\cdot3!</math> possibilities for the first <math>3</math> songs.<br />
<br />
There are <math>3</math> cases for the 4th song, call it song D.<br />
<br />
Case <math>1</math>: D is disliked by all <math>3</math> girls <math>\implies</math> there is only <math>1</math> possibility.<br />
<br />
Case <math>2</math>: D is liked by exactly <math>1</math> girl <math>\implies</math> there are <math>3</math> possibility.<br />
<br />
Case <math>3</math>: D is liked by exactly <math>2</math> girls <math>\implies</math> there are <math>3</math> pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first <math>3</math> songs liked by the same girls.<br />
<br />
Counting the overlaps, there are <math>3</math> ways to choose the pair with overlaps and <math>4\cdot3=12</math> ways to choose what the other <math>2</math> pairs like independently. In total, there are <math>3\cdot12=36</math> overlapped possibilities.<br />
<br />
Finally, there are <math>\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132</math> ways for the songs to be likely by the girls. <math>\boxed{\mathrm{(B)}}</math><br />
<br />
~ Nafer<br />
<br />
===Solution 4===<br />
This is a bipartite graph problem, with the girls as left vertices and songs as right vertices. An edge connecting left vertex and right vertex means that a girl like a song.<br />
<br />
Condition 1: "No song is liked by all three", means that the degree of right vertices is at most 2. <br />
<br />
Condition 2: "for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third", means that for any pair of left vertices, there is at least a right vertex connecting to them.<br />
<br />
To meet condition 2, there are at least 3 right vertices with 2 edges connecting to left vertices. There are 2 cases:<br />
<br />
Case 1: there are only 3 such right vertices. There are <math>\binom{4}{3}</math> such vertices, with <math>3!</math> ways of connections to the left vertices, total arrangements are <math>\binom{4}{3}\cdot3! = 24</math>. The fourth right vertex either has no edge to the 3 left vertices, or 1 edge to 1 of the 3 left vertices. So there are <math>24\cdot(1+3) = 96</math> ways.<br />
<br />
Case 2: there are 4 such right vertices, 2 of them have edges to the same pair of left vertices. There are <math>\binom{4}{2}</math> such vertices, with <math>3!</math> ways of connections. So there are <math>\binom{4}{2}\cdot3! = 36</math> ways.<br />
<br />
Total ways are <math>96+36=132</math>.<br />
<br />
Another way is to overcount then subtract overlap ways. Similar to previous case 1, the fourth right vertex could have all possible connection to the left vertices except connecting to all 3, so it is <math>2^3-1=7</math> ways, so the total ways are <math>\binom{4}{3}\cdot3!\cdot7 = 24\cdot7 = 168</math>. But this overcounts the case 2 with 36 ways. So total ways are <math>168-36=132</math>.<br />
<br />
-[https://artofproblemsolving.com/wiki/index.php/User:Junche junche]<br />
<br />
==Video Solutions:==<br />
==Video Solution by Richard Rusczyk==<br />
https://artofproblemsolving.com/videos/amc/2012amc10b/272<br />
<br />
~dolphin7<br />
<br />
<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_22&diff=1687192012 AMC 10B Problems/Problem 222021-12-28T14:11:23Z<p>Junche: /* Solution 6 */</p>
<hr />
<div>==Problem==<br />
Let (<math>a_1</math>, <math>a_2</math>, ... <math>a_{10}</math>) be a list of the first 10 positive integers such that for each <math>2\le</math> <math>i</math> <math>\le10</math> either <math>a_i + 1</math> or <math>a_i-1</math> or both appear somewhere before <math>a_i</math> in the list. How many such lists are there?<br />
<br />
<br />
<math>\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880</math><br />
<br />
==Solution 1==<br />
If we have 1 as the first number, then the only possible list is <math>(1,2,3,4,5,6,7,8,9,10)</math>. <br />
<br />
If we have 2 as the first number, then we have 9 ways to choose where the <math>1</math> goes, and the numbers ascend from the first number, <math>2</math>, with the exception of the <math>1</math>.<br />
For example, <math>(2,3,1,4,5,6,7,8,9,10)</math>, or <math>(2,3,4,1,5,6,7,8,9,10)</math>. There are <math>\dbinom{9}{1}</math> ways to do so.<br />
<br />
If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are <math>\dbinom{9}{2}</math> ways to do this.<br />
<br />
In the same way, the total number of lists is:<br />
<math>\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}</math><br />
<br />
By the binomial theorem, this is <math>2^{9}</math> = <math>512</math>, or <math>\boxed{\textbf{(B)}}</math><br />
<br />
==Solution 2==<br />
Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is <math>(1,2,3,4,5...10)</math>. There are 9 arrows, so the answer is <math>2^{9}</math> = <math>512</math> <math>\boxed{\textbf{(B)}}</math><br />
<br />
NOTE:<br />
Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.<br />
<br />
==Solution 3==<br />
Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, <math>(1, 2)</math> and <math>(2, 1)</math>. If we replace it with 3, there are four lists, <math>(1, 2, 3), (2, 1, 3), (2, 3, 1),</math> and <math>(3, 2, 1)</math>. Since 2 and 4 are both powers of 2, it is likely that the number of lists is <math>2^{n-1}</math>, where <math>n</math> is the length of the lists. <math>2^{10-1}=512=\boxed{\textbf{(B)}}</math><br />
<br />
==Solution 4 (Recursion)==<br />
If <math>a_1=10</math>, the sequence must be <math>10, 9, 8,7,6,5,4,3,2,1</math>. If <math>a_2=10</math>, then <math>a_1=9</math>, and the sequence is <math>9, 10, 8, 7, 6, 5,4,3,2,1</math>. If <math>a_3=10</math>, then the possible sequences are <cmath>9,8,10,7,6,5,4,3,2,1 \text{ and}</cmath><cmath>8,9,10,7,6,5,4,3,2,1.</cmath> In general, for an <math>n</math>-length sequence, if <math>a_i=n</math>, then <math>a_1</math> through <math>a_{i-1}</math> can be filled in <math>f(i-1)</math> ways with <math>n-i+1</math> through <math>n-1</math>, and <math>a_{i+1}</math> through <math>a_{n}</math> must be sorted in decreasing order with the remaining numbers (<math>1</math> through <math>n-i</math>), in one way. Thus <math>f(n) = \sum_{i=0}^{n-1} f(i)</math>, where <math>f(0)=1</math>.<br />
<br />
We can see (or prove by induction) that <math>f(n)=2^{n-1} ~\forall~ n \ge 1</math>. Hence, <math>f(10)=2^9=\boxed{\textbf{(D) }512}</math>.<br />
<br />
- ColtsFan10<br />
<br />
==Solution 5 ==<br />
Assume the same conditions to be held and let's look at several smaller cases to find a pattern. If we are only arranging <math>1,2</math> there are trivially only <math>2</math> ways. Now let us look at arranging <math>1,2,3</math>. You can arrange this in <math>4</math> ways. Looking at <math>1,2,3,4</math> you can arrange this in <math>8</math> ways. The pattern becomes evident now. If there are <math>n</math> numbers there are <math>2^{n-1}</math> ways. Hence our answer would be <math>2^{10-1} = 512</math> ways which is <math>\boxed{B}</math>.<br />
<br />
-srisainandan6<br />
<br />
==Solution 6==<br />
Solution 3 and 5 states that <math>f(n)=2^{n-1} ~\forall~ n \ge 1</math> without formal proof. Solution 4 gives a formal proof. Here is another formal proof:<br />
<br />
<math>f(1)=1</math>. When the list goes from <math>n-1</math> numbers to <math>n</math> numbers, there are <math>2</math> ways to make the new lists:<br />
<br />
Case 1: append <math>n</math> to the end of lists with <math>n-1</math> numbers to make a new list, the number of the new lists is <math>f(n-1)</math>;<br />
<br />
Case 2: put number <math>1</math> at the end of the new lists, the way to arrange <math>(2,3,...,n-1,n)</math> as the first <math>n-1</math> items is the same as to arrange <math>(1,2,...,n-2,n-1)</math>, by subtracting 1 from each of the elements, so the number of the new lists is also <math>f(n-1)</math>.<br />
<br />
So <math>f(n)=f(n-1)+f(n-1)=2f(n-1)=2^{n-1} ~\forall~ n \ge 1</math><br />
<br />
-[https://artofproblemsolving.com/wiki/index.php/User:Junche junche]<br />
<br />
==Video Solution by Richard Rusczyk==<br />
https://artofproblemsolving.com/videos/amc/2012amc10b/269<br />
<br />
~dolphin7<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/bXPSv93GVbg<br />
<br />
~IceMatrix<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=User:Junche&diff=168718User:Junche2021-12-28T14:10:39Z<p>Junche: Created page with "==AOPS Contributions:== 1. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_16#Solution_4 2012 AMC10B Problem 22 Solution 6] (Combinatorics) 2...."</p>
<hr />
<div>==AOPS Contributions:==<br />
<br />
1. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_16#Solution_4 2012 AMC10B Problem 22 Solution 6] (Combinatorics)<br />
<br />
2. [https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22#Solution_6 2012 AMC10B Problem 24 Solution 4] (Graph Theory)</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&diff=1635412016 AMC 10B Problems/Problem 252021-10-15T14:14:51Z<p>Junche: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)</math>, where <math>\lfloor r \rfloor</math> denotes the greatest integer less than or equal to <math>r</math>. How many distinct values does <math>f(x)</math> assume for <math>x \ge 0</math>?<br />
<br />
<math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math><br />
<br />
==Solution 1==<br />
<br />
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have <br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath><br />
<br />
The function can then be simplified into <br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath><br />
<br />
which becomes<br />
<br />
<cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor</cmath><br />
<br />
We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>. <br />
<br />
Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when <math>\{ x \}</math> is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>.<br />
<br />
So we want to count how many distinct fractions less than <math>1</math> have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. '''Explanation for this is provided below.''' We can find this easily by computing<br />
<br />
<cmath>\sum_{k=2}^{10} \phi(k)</cmath><br />
<br />
where <math>\phi(k)</math> is the [[Euler Totient Function]]. Basically <math>\phi(k)</math> counts the number of fractions with <math>k</math> as its denominator (after simplification). This comes out to be <math>31</math>.<br />
<br />
Because the value of <math>f(x)</math> is at least <math>0</math> and can increase <math>31</math> times, there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>.<br />
<br />
===Explanation:===<br />
<br />
Arrange all such fractions in increasing order and take a current <math>\frac{m}{n}</math> to study. Let <math>p</math> denote the previous fraction in the list and <math>x_\text{old}</math> (<math>0 \le x_\text{old} < k</math> for each <math>k</math>) be the largest so that <math>\frac{x_\text{old}}{k} \le p</math>. Since <math>\text{ }\text{ }\frac{m}{n} > p</math>, we clearly have all <math>x_\text{new} \ge x_\text{old}</math>. Therefore, the change must be nonnegative.<br />
<br />
But among all numerators coprime to <math>n</math> so far, <math>m</math> is the largest. Therefore, choosing <math>\frac{m}{n}</math> as <math>{x}</math> increases the value <math>\lfloor n \{ x \} \rfloor</math>. Since the overall change in <math>f(x)</math> is positive as fractions <math>m/n</math> increase, we deduce that all such fractions correspond to different values of the function. <br />
<br />
Minor Latex Edits made by MATHWIZARD2010.<br />
<br />
===Supplement===<br />
<br />
Here are all the distinct <math>\frac{m}{n}</math> and <math>\phi(k):</math><br />
<br />
When <math>n=2</math> , <math>\frac{m}{n}=\frac{1}{2}</math> . <math>\phi(2)=1</math><br />
<br />
When <math>n=3</math> , <math>\frac{m}{n}=\frac{1}{3}</math> , <math>\frac{2}{3}</math> . <math>\phi(3)=2</math><br />
<br />
When <math>n=4</math> , <math>\frac{m}{n}=\frac{1}{4}</math> , <math>\frac{3}{4}</math> . <math>\phi(4)=2</math><br />
<br />
When <math>n=5</math> , <math>\frac{m}{n}=\frac{1}{5}</math> , <math>\frac{2}{5}</math> , <math>\frac{3}{5}</math> , <math>\frac{4}{5}</math> . <math>\phi(5)=4</math><br />
<br />
When <math>n=6</math> , <math>\frac{m}{n}=\frac{1}{6}</math> , <math>\frac{5}{6}</math> . <math>\phi(6)=2</math><br />
<br />
When <math>n=7</math> , <math>\frac{m}{n}=\frac{1}{7}</math> , <math>\frac{2}{7}</math> , <math>\frac{3}{7}</math> , <math>\frac{4}{7}</math> , <math>\frac{5}{7}</math> , <math>\frac{6}{7}</math> . <math>\phi(7)=6</math><br />
<br />
When <math>n=8</math> , <math>\frac{m}{n}=\frac{1}{8}</math> , <math>\frac{3}{8}</math> , <math>\frac{5}{8}</math> , <math>\frac{7}{8}</math> . <math>\phi(8)=4</math><br />
<br />
When <math>n=9</math> , <math>\frac{m}{n}=\frac{1}{9}</math> , <math>\frac{2}{9}</math> , <math>\frac{4}{9}</math> , <math>\frac{5}{9}</math> , <math>\frac{7}{9}</math> , <math>\frac{8}{9}</math> . <math>\phi(9)=6</math><br />
<br />
When <math>n=10</math> , <math>\frac{m}{n}=\frac{1}{10}</math> , <math>\frac{3}{10}</math> , <math>\frac{7}{10}</math> , <math>\frac{9}{10}</math> . <math>\phi(10)=4</math><br />
<br />
<math>\sum_{k=2}^{10} \phi(k)=31</math><br />
<br />
<math>31+1=\fbox{\textbf{(A)}\ 32}</math><br />
<br />
~isabelchen<br />
<br />
==Solution 2==<br />
<br />
<math>x = \lfloor x \rfloor + \{ x \}</math> so we have <cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor.</cmath> Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when <math>x</math> is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>. To get all the fractions,graphing this function gives us <math>46</math> different fractions. But on average, <math>3</math> in each of the <math>5</math> intervals don’t work. This means there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>.<br />
<br />
==Solution 3 (Casework)==<br />
<br />
Solution <math>1</math> is abstract. In this solution I will give a concrete explanation.<br />
<br />
WLOG, for example, when <math>x</math> increases from <math>\frac{2}{3}-\epsilon</math> to <math>\frac{2}{3}</math>, <math>\lfloor 3 \{ x \} \rfloor</math> will increase from <math>1</math> to <math>2</math>, <math>\lfloor 6 \{ x \} \rfloor</math> will increase from <math>3</math> to <math>4</math>, <math>\lfloor 9 \{ x \} \rfloor</math> will increase from <math>5</math> to <math>6</math>. In total, <math>f(x)</math> will increase by <math>3</math>. Because <math>\frac{1}{3}=\frac{2}{6}=\frac{3}{9}</math>, these <math>3</math> numbers are actually <math>1</math> distinct number to cause <math>f(x)</math> to change. In general, when <math>x</math> increases from <math>\frac{m}{n}-\epsilon</math> to <math>\frac{m}{n}</math>, <math>\lfloor k \{ x \} \rfloor</math> will increse from <math>k \cdot \frac{m}{n} -1</math> to <math>k \cdot \frac{m}{n} </math> if <math>k \cdot \frac{m}{n} </math> is an integer, and the value of <math>f(x)</math> will change. So the total number of distinct values <math>f(x)</math> could take is equal to the number of distinct values of <math>\frac{m}{n}</math>, where <math>0 < \frac{m}{n}<1</math> and <math>2 \le n \le 10</math>. <br />
<br />
Solution <math>1</math> uses Euler Totient Function to count the distinct number of <math>\frac{m}{n}</math>, I am going to use casework to count the distinct values of <math>\frac{m}{n}</math> by not counting the duplicate ones.<br />
<br />
When <math>n=10</math> , <math>\frac{m}{n}=\frac{1}{10}</math> , <math>\frac{2}{10}</math> , <math>...</math> , <math>\frac{9}{10}</math> <math>\Longrightarrow 9</math><br />
<br />
When <math>n=9</math> , <math>\frac{m}{n}=\frac{1}{9}</math> , <math>\frac{2}{9}</math> , <math>...</math> , <math>\frac{8}{9}</math> <math>\Longrightarrow 8</math><br />
<br />
When <math>n=8</math> , <math>\frac{m}{n}=\frac{1}{8}</math> , <math>\frac{2}{8}</math> , <math>...</math> , <math>\frac{7}{8}</math> <math>\Longrightarrow 6</math> ( <math>\frac{4}{8}</math> is duplicate)<br />
<br />
When <math>n=7</math> , <math>\frac{m}{n}=\frac{1}{7}</math> , <math>\frac{2}{7}</math> , <math>...</math> , <math>\frac{6}{7}</math> <math>\Longrightarrow 6</math><br />
<br />
When <math>n=6</math> , <math>\frac{m}{n}=\frac{1}{6}</math> , <math>\frac{5}{6}</math> <math>\Longrightarrow 2</math> ( <math>\frac{2}{6}</math> , <math>\frac{3}{6}</math> , and <math>\frac{4}{6}</math> is duplicate)<br />
<br />
When <math>n=5</math>, <math>4</math>, <math>3</math>, <math>2</math>, all the <math>\frac{m}{n}</math> is duplicate.<br />
<br />
<math>9+8+6+6+2=31</math>, <math>31+1=\fbox{\textbf{(A)}\ 32}</math><br />
<br />
~isabelchen<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=zXJrdDtZNbw<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=1626802012 AMC 12B Problems/Problem 162021-09-24T07:21:19Z<p>Junche: /* Solution 4 */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br />
<br />
== Problem==<br />
Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br />
<br />
<math> \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 </math><br />
<br />
==Solutions==<br />
<br />
=== Solution 1===<br />
Let the ordered triple <math>(a,b,c)</math> denote that <math>a</math> songs are liked by Amy and Beth, <math>b</math> songs by Beth and Jo, and <math>c</math> songs by Jo and Amy. We claim that the only possible triples are <math>(1,1,1), (2,1,1), (1,2,1)(1,1,2)</math>. <br />
<br />
To show this, observe these are all valid conditions. Second, note that none of <math>a,b,c</math> can be bigger than 3. Suppose otherwise, that <math>a = 3</math>. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either <math>b</math> or <math>c</math> to be at least 1. In fact, we require either <math>b</math> or <math>c</math> to equal 1, otherwise there will be a song liked by all three. Suppose <math>b = 1</math>. Then we must have <math>c=0</math> since no song is liked by all three girls, a contradiction.<br />
<br />
'''Case 1''': How many ways are there for <math>(a,b,c)</math> to equal <math>(1,1,1)</math>? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So <math>(a,b,c)=(1,1,1)</math> in <math>4\cdot3\cdot2\cdot4 = 96</math> ways.<br />
<br />
'''Case 2''': To find the number of ways for <math>(a,b,c) = (2,1,1)</math>, observe there are <math>\binom{4}{2} = 6</math> choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are <math>6\cdot2\cdot3=36</math> ways for the girls to like the songs.<br />
<br />
That gives a total of <math>96 + 36 = 132</math> ways for the girls to like the songs, so the answer is <math>\boxed{(\textrm{\textbf{B}})}</math>.<br />
<br />
=== Solution 2===<br />
<br />
<br />
Let <math>AB, BJ</math>, and <math>AJ</math> denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let <math>A, B, J,</math> and <math>N</math> denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least <math>1\: AB, BJ</math>, and <math>AJ</math>, they must be <math>3</math> songs out of the <math>4</math> that Amy, Beth, and Jo listened to. The fourth song can be of any type <math>N, A, B, J, AB, BJ</math>, and <math>AJ</math> (there is no <math>ABJ</math> because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange <math>AB, BJ, AJ</math>, and a song from the set <math>\{N, A, B, J, AB, BJ, AJ\}</math>.<br />
<br />
Case 1: Fourth song = <math>N, A, B, J </math><br />
<br />
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!)</math> rearrangements for each choice <math>*\: 4</math> choices = <math>96</math>.<br />
<br />
Case 2: Fourth song = <math>AB, BJ, AJ</math><br />
<br />
Note that in Case <math>2</math>, all three of the choices for the fourth song repeat somewhere in the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!/2!)</math> rearrangements for each choice <math>*\: 3</math> choices = <math>36</math>.<br />
<br />
<math>96 + 36 = \boxed{\textbf{(B)} \: 132}</math>.<br />
<br />
=== Solution 3===<br />
<br />
<br />
There are <math>\binom{4}{3}</math> ways to choose the three songs that are liked by the three pairs of girls.<br />
<br />
There are <math>3!</math> ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br />
<br />
In total, there are <math>\binom{4}{3}\cdot3!</math> possibilities for the first <math>3</math> songs.<br />
<br />
There are <math>3</math> cases for the 4th song, call it song D.<br />
<br />
Case <math>1</math>: D is disliked by all <math>3</math> girls <math>\implies</math> there is only <math>1</math> possibility.<br />
<br />
Case <math>2</math>: D is liked by exactly <math>1</math> girl <math>\implies</math> there are <math>3</math> possibility.<br />
<br />
Case <math>3</math>: D is liked by exactly <math>2</math> girls <math>\implies</math> there are <math>3</math> pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first <math>3</math> songs liked by the same girls.<br />
<br />
Counting the overlaps, there are <math>3</math> ways to choose the pair with overlaps and <math>4\cdot3=12</math> ways to choose what the other <math>2</math> pairs like independently. In total, there are <math>3\cdot12=36</math> overlapped possibilities.<br />
<br />
Finally, there are <math>\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132</math> ways for the songs to be likely by the girls. <math>\boxed{\mathrm{(B)}}</math><br />
<br />
~ Nafer<br />
<br />
===Solution 4===<br />
This is a bipartite graph problem, with the girls as left vertices and songs as right vertices. An edge connecting left vertex and right vertex means that a girl like a song.<br />
<br />
Condition 1: "No song is liked by all three", means that the degree of right vertices is at most 2. <br />
<br />
Condition 2: "for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third", means that for any pair of left vertices, there is at least a right vertex connecting to them.<br />
<br />
To meet condition 2, there are at least 3 right vertices with 2 edges connecting to left vertices. There are 2 cases:<br />
<br />
Case 1: there are only 3 such right vertices. There are <math>\binom{4}{3}</math> such vertices, with <math>3!</math> ways of connections to the left vertices, total arrangements are <math>\binom{4}{3}\cdot3! = 24</math>. The fourth right vertex either has no edge to the 3 left vertices, or 1 edge to 1 of the 3 left vertices. So there are <math>24\cdot(1+3) = 96</math> ways.<br />
<br />
Case 2: there are 4 such right vertices, 2 of them have edges to the same pair of left vertices. There are <math>\binom{4}{2}</math> such vertices, with <math>3!</math> ways of connections. So there are <math>\binom{4}{2}\cdot3! = 36</math> ways.<br />
<br />
Total ways are <math>96+36=132</math>.<br />
<br />
Another way is to overcount then subtract overlap ways. Similar to previous case 1, the fourth right vertex could have all possible connection to the left vertices except connecting to all 3, so it is <math>2^3-1=7</math> ways, so the total ways are <math>\binom{4}{3}\cdot3!\cdot7 = 24\cdot7 = 168</math>. But this overcounts the case 2 with 36 ways. So total ways are <math>168-36=132</math>.<br />
<br />
-junche<br />
<br />
==Video Solutions:==<br />
==Video Solution by Richard Rusczyk==<br />
https://artofproblemsolving.com/videos/amc/2012amc10b/272<br />
<br />
~dolphin7<br />
<br />
<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=1625942012 AMC 12B Problems/Problem 162021-09-22T03:58:53Z<p>Junche: /* Solution 4 */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br />
<br />
== Problem==<br />
Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br />
<br />
<math> \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 </math><br />
<br />
==Solutions==<br />
<br />
=== Solution 1===<br />
Let the ordered triple <math>(a,b,c)</math> denote that <math>a</math> songs are liked by Amy and Beth, <math>b</math> songs by Beth and Jo, and <math>c</math> songs by Jo and Amy. We claim that the only possible triples are <math>(1,1,1), (2,1,1), (1,2,1)(1,1,2)</math>. <br />
<br />
To show this, observe these are all valid conditions. Second, note that none of <math>a,b,c</math> can be bigger than 3. Suppose otherwise, that <math>a = 3</math>. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either <math>b</math> or <math>c</math> to be at least 1. In fact, we require either <math>b</math> or <math>c</math> to equal 1, otherwise there will be a song liked by all three. Suppose <math>b = 1</math>. Then we must have <math>c=0</math> since no song is liked by all three girls, a contradiction.<br />
<br />
'''Case 1''': How many ways are there for <math>(a,b,c)</math> to equal <math>(1,1,1)</math>? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So <math>(a,b,c)=(1,1,1)</math> in <math>4\cdot3\cdot2\cdot4 = 96</math> ways.<br />
<br />
'''Case 2''': To find the number of ways for <math>(a,b,c) = (2,1,1)</math>, observe there are <math>\binom{4}{2} = 6</math> choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are <math>6\cdot2\cdot3=36</math> ways for the girls to like the songs.<br />
<br />
That gives a total of <math>96 + 36 = 132</math> ways for the girls to like the songs, so the answer is <math>\boxed{(\textrm{\textbf{B}})}</math>.<br />
<br />
=== Solution 2===<br />
<br />
<br />
Let <math>AB, BJ</math>, and <math>AJ</math> denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let <math>A, B, J,</math> and <math>N</math> denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least <math>1\: AB, BJ</math>, and <math>AJ</math>, they must be <math>3</math> songs out of the <math>4</math> that Amy, Beth, and Jo listened to. The fourth song can be of any type <math>N, A, B, J, AB, BJ</math>, and <math>AJ</math> (there is no <math>ABJ</math> because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange <math>AB, BJ, AJ</math>, and a song from the set <math>\{N, A, B, J, AB, BJ, AJ\}</math>.<br />
<br />
Case 1: Fourth song = <math>N, A, B, J </math><br />
<br />
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!)</math> rearrangements for each choice <math>*\: 4</math> choices = <math>96</math>.<br />
<br />
Case 2: Fourth song = <math>AB, BJ, AJ</math><br />
<br />
Note that in Case <math>2</math>, all three of the choices for the fourth song repeat somewhere in the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!/2!)</math> rearrangements for each choice <math>*\: 3</math> choices = <math>36</math>.<br />
<br />
<math>96 + 36 = \boxed{\textbf{(B)} \: 132}</math>.<br />
<br />
=== Solution 3===<br />
<br />
<br />
There are <math>\binom{4}{3}</math> ways to choose the three songs that are liked by the three pairs of girls.<br />
<br />
There are <math>3!</math> ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br />
<br />
In total, there are <math>\binom{4}{3}\cdot3!</math> possibilities for the first <math>3</math> songs.<br />
<br />
There are <math>3</math> cases for the 4th song, call it song D.<br />
<br />
Case <math>1</math>: D is disliked by all <math>3</math> girls <math>\implies</math> there is only <math>1</math> possibility.<br />
<br />
Case <math>2</math>: D is liked by exactly <math>1</math> girl <math>\implies</math> there are <math>3</math> possibility.<br />
<br />
Case <math>3</math>: D is liked by exactly <math>2</math> girls <math>\implies</math> there are <math>3</math> pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first <math>3</math> songs liked by the same girls.<br />
<br />
Counting the overlaps, there are <math>3</math> ways to choose the pair with overlaps and <math>4\cdot3=12</math> ways to choose what the other <math>2</math> pairs like independently. In total, there are <math>3\cdot12=36</math> overlapped possibilities.<br />
<br />
Finally, there are <math>\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132</math> ways for the songs to be likely by the girls. <math>\boxed{\mathrm{(B)}}</math><br />
<br />
~ Nafer<br />
<br />
===Solution 4===<br />
This is a bipartite graph problem, with the girls as left vertices and songs as right vertices. An edge connecting left vertex and right vertex means that a girl like a song.<br />
<br />
Condition 1: "No song is liked by all three", means that the degree of right vertices is at most 2. <br />
<br />
Condition 2: "for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third", means that for any pair of left vertices, there is at least a right vertex connecting to them.<br />
<br />
To meet condition 2, there are at least 3 right vertices with 2 edges connecting to left vertices. There are 2 cases:<br />
<br />
Case 1: there are only 3 such right vertices. There are <math>\binom{4}{3}</math> such vertices, with <math>3!</math> ways of connections to the left vertices, total arrangements are <math>\binom{4}{3}\cdot3! = 24</math>. The fourth right vertex either has no edge to the 3 left vertices, or 1 edge to 1 of the 3 left vertices. So there are <math>24\cdot(1+3) = 96</math> ways.<br />
<br />
Case 2: there are 4 such right vertices, 2 of them have edges to the same pair of left vertices. There are <math>\binom{4}{2}</math> such vertices, with <math>3!</math> ways of connections. So there are <math>\binom{4}{2}\cdot3! = 36</math> ways.<br />
<br />
Total ways are <math>96+36=132</math>.<br />
<br />
Another way is to overcount then subtract overlap ways. Similar to previous case 1, the fourth right vertex could have all possible connection to the left vertices except connecting to all 3, so it is <math>2^3-1=7</math> ways, so the total ways are <math>\binom{4}{3}\cdot3!\cdot7 = 24\cdot7 = 168</math>. But this overcounts the case 2 with 36 ways. So final ways are <math>168-36=132</math>.<br />
<br />
-junche<br />
<br />
==Video Solutions:==<br />
==Video Solution by Richard Rusczyk==<br />
https://artofproblemsolving.com/videos/amc/2012amc10b/272<br />
<br />
~dolphin7<br />
<br />
<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=1625742012 AMC 12B Problems/Problem 162021-09-21T18:09:29Z<p>Junche: </p>
<hr />
<div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br />
<br />
== Problem==<br />
Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br />
<br />
<math> \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 </math><br />
<br />
==Solutions==<br />
<br />
=== Solution 1===<br />
Let the ordered triple <math>(a,b,c)</math> denote that <math>a</math> songs are liked by Amy and Beth, <math>b</math> songs by Beth and Jo, and <math>c</math> songs by Jo and Amy. We claim that the only possible triples are <math>(1,1,1), (2,1,1), (1,2,1)(1,1,2)</math>. <br />
<br />
To show this, observe these are all valid conditions. Second, note that none of <math>a,b,c</math> can be bigger than 3. Suppose otherwise, that <math>a = 3</math>. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either <math>b</math> or <math>c</math> to be at least 1. In fact, we require either <math>b</math> or <math>c</math> to equal 1, otherwise there will be a song liked by all three. Suppose <math>b = 1</math>. Then we must have <math>c=0</math> since no song is liked by all three girls, a contradiction.<br />
<br />
'''Case 1''': How many ways are there for <math>(a,b,c)</math> to equal <math>(1,1,1)</math>? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So <math>(a,b,c)=(1,1,1)</math> in <math>4\cdot3\cdot2\cdot4 = 96</math> ways.<br />
<br />
'''Case 2''': To find the number of ways for <math>(a,b,c) = (2,1,1)</math>, observe there are <math>\binom{4}{2} = 6</math> choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are <math>6\cdot2\cdot3=36</math> ways for the girls to like the songs.<br />
<br />
That gives a total of <math>96 + 36 = 132</math> ways for the girls to like the songs, so the answer is <math>\boxed{(\textrm{\textbf{B}})}</math>.<br />
<br />
=== Solution 2===<br />
<br />
<br />
Let <math>AB, BJ</math>, and <math>AJ</math> denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let <math>A, B, J,</math> and <math>N</math> denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least <math>1\: AB, BJ</math>, and <math>AJ</math>, they must be <math>3</math> songs out of the <math>4</math> that Amy, Beth, and Jo listened to. The fourth song can be of any type <math>N, A, B, J, AB, BJ</math>, and <math>AJ</math> (there is no <math>ABJ</math> because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange <math>AB, BJ, AJ</math>, and a song from the set <math>\{N, A, B, J, AB, BJ, AJ\}</math>.<br />
<br />
Case 1: Fourth song = <math>N, A, B, J </math><br />
<br />
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!)</math> rearrangements for each choice <math>*\: 4</math> choices = <math>96</math>.<br />
<br />
Case 2: Fourth song = <math>AB, BJ, AJ</math><br />
<br />
Note that in Case <math>2</math>, all three of the choices for the fourth song repeat somewhere in the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!/2!)</math> rearrangements for each choice <math>*\: 3</math> choices = <math>36</math>.<br />
<br />
<math>96 + 36 = \boxed{\textbf{(B)} \: 132}</math>.<br />
<br />
=== Solution 3===<br />
<br />
<br />
There are <math>\binom{4}{3}</math> ways to choose the three songs that are liked by the three pairs of girls.<br />
<br />
There are <math>3!</math> ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br />
<br />
In total, there are <math>\binom{4}{3}\cdot3!</math> possibilities for the first <math>3</math> songs.<br />
<br />
There are <math>3</math> cases for the 4th song, call it song D.<br />
<br />
Case <math>1</math>: D is disliked by all <math>3</math> girls <math>\implies</math> there is only <math>1</math> possibility.<br />
<br />
Case <math>2</math>: D is liked by exactly <math>1</math> girl <math>\implies</math> there are <math>3</math> possibility.<br />
<br />
Case <math>3</math>: D is liked by exactly <math>2</math> girls <math>\implies</math> there are <math>3</math> pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first <math>3</math> songs liked by the same girls.<br />
<br />
Counting the overlaps, there are <math>3</math> ways to choose the pair with overlaps and <math>4\cdot3=12</math> ways to choose what the other <math>2</math> pairs like independently. In total, there are <math>3\cdot12=36</math> overlapped possibilities.<br />
<br />
Finally, there are <math>\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132</math> ways for the songs to be likely by the girls. <math>\boxed{\mathrm{(B)}}</math><br />
<br />
~ Nafer<br />
<br />
===Solution 4===<br />
This is a bipartite graph problem, with the girls as left vertices and songs as right vertices. An edge connecting left vertex and right vertex means that a girl like a song.<br />
<br />
Condition 1: "No song is liked by all three", means that the degree of right vertices is at most 2. <br />
<br />
Condition 2: "for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third", means that for any pair of left vertices, there is at least a right vertex connecting to them.<br />
<br />
To meet condition 2, there are at least 3 right vertices with 2 edges connecting to left vertices. There are 2 cases:<br />
<br />
Case 1: only 3 such right vertices. There are <math>\binom{4}{3}</math> such vertices, with <math>3!</math> ways of connections, total arrangements are <math>\binom{4}{3}\cdot3! = 24</math>. The fourth right vertex either has no edge to the 3 left vertices, or 1 edge to 1 of the 3 left vertices. So there are <math>24\cdot(1+3) = 96</math> ways.<br />
<br />
Case 2: there 4 such right vertices, 2 of them have edges to the same pair of left vertices. There are <math>\binom{4}{2}</math> such vertices, with <math>3!</math> ways of connections. So there are <math>\binom{4}{2}\cdot3! = 36</math> ways.<br />
<br />
Total ways are <math>96+36=132</math>.<br />
<br />
Another way is to overcount then subtract overlap ways. Similar to previous case 1, the fourth right vertex could have all possible connection to the left vertices except connecting to all 3, so it is <math>2^3-1=7</math> ways, so the total ways are <math>\binom{4}{3}\cdot3!\cdot7 = 24\cdot7 = 168</math>. But this overcount the case 2 with 36 ways. So final ways are <math>168-36=132</math>.<br />
<br />
-junche<br />
<br />
==Video Solutions:==<br />
==Video Solution by Richard Rusczyk==<br />
https://artofproblemsolving.com/videos/amc/2012amc10b/272<br />
<br />
~dolphin7<br />
<br />
<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Junchehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_22&diff=1625712012 AMC 10B Problems/Problem 222021-09-21T17:12:46Z<p>Junche: </p>
<hr />
<div>==Problem==<br />
Let (<math>a_1</math>, <math>a_2</math>, ... <math>a_{10}</math>) be a list of the first 10 positive integers such that for each <math>2\le</math> <math>i</math> <math>\le10</math> either <math>a_i + 1</math> or <math>a_i-1</math> or both appear somewhere before <math>a_i</math> in the list. How many such lists are there?<br />
<br />
<br />
<math>\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880</math><br />
<br />
==Solution 1==<br />
If we have 1 as the first number, then the only possible list is <math>(1,2,3,4,5,6,7,8,9,10)</math>. <br />
<br />
If we have 2 as the first number, then we have 9 ways to choose where the <math>1</math> goes, and the numbers ascend from the first number, <math>2</math>, with the exception of the <math>1</math>.<br />
For example, <math>(2,3,1,4,5,6,7,8,9,10)</math>, or <math>(2,3,4,1,5,6,7,8,9,10)</math>. There are <math>\dbinom{9}{1}</math> ways to do so.<br />
<br />
If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are <math>\dbinom{9}{2}</math> ways to do this.<br />
<br />
In the same way, the total number of lists is:<br />
<math>\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}</math><br />
<br />
By the binomial theorem, this is <math>2^{9}</math> = <math>512</math>, or <math>\boxed{\textbf{(B)}}</math><br />
<br />
==Solution 2==<br />
Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is <math>(1,2,3,4,5...10)</math>. There are 9 arrows, so the answer is <math>2^{9}</math> = <math>512</math> <math>\boxed{\textbf{(B)}}</math><br />
<br />
NOTE:<br />
Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.<br />
<br />
==Solution 3==<br />
Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, <math>(1, 2)</math> and <math>(2, 1)</math>. If we replace it with 3, there are four lists, <math>(1, 2, 3), (2, 1, 3), (2, 3, 1),</math> and <math>(3, 2, 1)</math>. Since 2 and 4 are both powers of 2, it is likely that the number of lists is <math>2^{n-1}</math>, where <math>n</math> is the length of the lists. <math>2^{10-1}=512=\boxed{\textbf{(B)}}</math><br />
<br />
==Solution 4 (Recursion)==<br />
If <math>a_1=10</math>, the sequence must be <math>10, 9, 8,7,6,5,4,3,2,1</math>. If <math>a_2=10</math>, then <math>a_1=9</math>, and the sequence is <math>9, 10, 8, 7, 6, 5,4,3,2,1</math>. If <math>a_3=10</math>, then the possible sequences are <cmath>9,8,10,7,6,5,4,3,2,1 \text{ and}</cmath><cmath>8,9,10,7,6,5,4,3,2,1.</cmath> In general, for an <math>n</math>-length sequence, if <math>a_i=n</math>, then <math>a_1</math> through <math>a_{i-1}</math> can be filled in <math>f(i-1)</math> ways with <math>n-i+1</math> through <math>n-1</math>, and <math>a_{i+1}</math> through <math>a_{n}</math> must be sorted in decreasing order with the remaining numbers (<math>1</math> through <math>n-i</math>), in one way. Thus <math>f(n) = \sum_{i=0}^{n-1} f(i)</math>, where <math>f(0)=1</math>.<br />
<br />
We can see (or prove by induction) that <math>f(n)=2^{n-1} ~\forall~ n \ge 1</math>. Hence, <math>f(10)=2^9=\boxed{\textbf{(D) }512}</math>.<br />
<br />
- ColtsFan10<br />
<br />
==Solution 5 ==<br />
Assume the same conditions to be held and let's look at several smaller cases to find a pattern. If we are only arranging <math>1,2</math> there are trivially only <math>2</math> ways. Now let us look at arranging <math>1,2,3</math>. You can arrange this in <math>4</math> ways. Looking at <math>1,2,3,4</math> you can arrange this in <math>8</math> ways. The pattern becomes evident now. If there are <math>n</math> numbers there are <math>2^{n-1}</math> ways. Hence our answer would be <math>2^{10-1} = 512</math> ways which is <math>\boxed{B}</math>.<br />
<br />
-srisainandan6<br />
<br />
==Solution 6==<br />
Solution 3 and 5 states that <math>f(n)=2^{n-1} ~\forall~ n \ge 1</math> without formal proof. Solution 4 gives a formal proof. Here is another formal proof:<br />
<br />
<math>f(1)=1</math>. When the list goes from <math>n-1</math> numbers to <math>n</math> numbers, there are <math>2</math> ways to make the new lists:<br />
<br />
Case 1: append <math>n</math> to the end of lists with <math>n-1</math> numbers to make a new list, the number of the new lists is <math>f(n-1)</math>;<br />
<br />
Case 2: put number <math>1</math> at the end of the new lists, the way to arrange <math>(2,3,...,n-1,n)</math> as the first <math>n-1</math> items is the same as to arrange <math>(1,2,...,n-2,n-1)</math>, by subtracting 1 from each of the elements, so the number of the new lists is also <math>f(n-1)</math>.<br />
<br />
So <math>f(n)=f(n-1)+f(n-1)=2f(n-1)=2^{n-1} ~\forall~ n \ge 1</math><br />
<br />
-junche<br />
<br />
==Video Solution by Richard Rusczyk==<br />
https://artofproblemsolving.com/videos/amc/2012amc10b/269<br />
<br />
~dolphin7<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/bXPSv93GVbg<br />
<br />
~IceMatrix<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Junche