https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sakshamsethi&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:24:41ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Talk:2000_AIME_II_Problems/Problem_3&diff=156611Talk:2000 AIME II Problems/Problem 32021-06-22T13:14:33Z<p>Sakshamsethi: Created page with "no"</p>
<hr />
<div>no</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_16&diff=1544312014 AMC 10A Problems/Problem 162021-05-29T19:34:53Z<p>Sakshamsethi: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=1</math>, <math>BC=2</math>, and points <math>E</math>, <math>F</math>, and <math>G</math> are midpoints of <math>\overline{BC}</math>, <math>\overline{CD}</math>, and <math>\overline{AD}</math>, respectively. Point <math>H</math> is the midpoint of <math>\overline{GE}</math>. What is the area of the shaded region?<br />
<br />
<asy><br />
import graph;<br />
size(9cm);<br />
pen dps = fontsize(10); defaultpen(dps);<br />
pair D = (0,0);<br />
pair F = (1/2,0);<br />
pair C = (1,0);<br />
pair G = (0,1);<br />
pair E = (1,1);<br />
pair A = (0,2);<br />
pair B = (1,2);<br />
pair H = (1/2,1);<br />
<br />
// do not look<br />
pair X = (1/3,2/3);<br />
pair Y = (2/3,2/3);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(G--E);<br />
draw(A--F--B);<br />
draw(D--H--C);<br />
filldraw(H--X--F--Y--cycle,grey);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E,E);<br />
label("$F$",F,S);<br />
label("$G$",G,W);<br />
label("$H$",H,N);<br />
<br />
label("$\frac12$",(0.25,0),S);<br />
label("$\frac12$",(0.75,0),S);<br />
label("$1$",(1,0.5),E);<br />
label("$1$",(1,1.5),E);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1==<br />
Denote <math>D=(0,0)</math>. Then <math>A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)</math>. Let the intersection of <math>AF</math> and <math>DH</math> be <math>X</math>, and the intersection of <math>BF</math> and <math>CH</math> be <math>Y</math>. Then we want to find the coordinates of <math>X</math> so we can find <math>XY</math>. From our points, the slope of <math>AF</math> is <math>\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4</math>, and its <math>y</math>-intercept is just <math>2</math>. Thus the equation for <math>AF</math> is <math>y = -4x + 2</math>. We can also quickly find that the equation of <math>DH</math> is <math>y = 2x</math>. Setting the equations equal, we have <math>2x = -4x +2 \implies x = \frac13</math>. Because of symmetry, we can see that the distance from <math>Y</math> to <math>BC</math> is also <math>\frac13</math>, so <math>XY = 1 - 2 \cdot \frac13 = \frac13</math>. Now the area of the kite is simply the product of the two diagonals over <math>2</math>. Since the length <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.<br />
<br />
<asy><br />
import graph;<br />
size(9cm);<br />
pen dps = fontsize(10); defaultpen(dps);<br />
pair D = (0,0);<br />
pair F = (1/2,0);<br />
pair C = (1,0);<br />
pair G = (0,1);<br />
pair E = (1,1);<br />
pair A = (0,2);<br />
pair B = (1,2);<br />
pair H = (1/2,1);<br />
<br />
// do not look<br />
pair X = (1/3,2/3);<br />
pair Y = (2/3,2/3);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(G--E);<br />
draw(A--F--B);<br />
draw(D--H--C);<br />
filldraw(H--X--F--Y--cycle,grey);<br />
draw(X--Y,dashed);<br />
<br />
<br />
<br />
label("$A\: (0,2)$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D \: (0,0)$",D,SW);<br />
label("$E$",E,E);<br />
label("$F\: (\frac12,0)$",F,S);<br />
label("$G$",G,W);<br />
label("$H \: (\frac12,1)$",H,N);<br />
label("$Y$",Y,E);<br />
label("$X$",X,W);<br />
<br />
<br />
label("$\frac12$",(0.25,0),S);<br />
label("$\frac12$",(0.75,0),S);<br />
label("$1$",(1,0.5),E);<br />
label("$1$",(1,1.5),E);<br />
</asy><br />
<br />
==Solution 2==<br />
<br />
Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2}* IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = \boxed{\textbf{(E)} \: \frac{1}{6}}</math><br />
<br />
==Solution 3==<br />
<br />
From the diagram in Solution 1, let <math>e</math> be the height of <math>XHY</math> and <math>f</math> be the height of <math>XFY</math>. It is clear that their sum is <math>1</math> as they are parallel to <math>GD</math>. Let <math>k</math> be the ratio of the sides of the similar triangles <math>XFY</math> and <math>AFB</math>, which are similar because <math>XY</math> is parallel to <math>AB</math> and the triangles share angle <math>F</math>. Then <math>k = f/2</math>, as 2 is the height of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height of <math>XHY</math> will be equal to the base, like in <math>DHC</math>, making <math>XY = e</math>. However, <math>XY</math> is also the base of <math>XFY</math>, so <math>k = e / AB</math> where <math>AB = 1</math> so <math>k = e</math>. Subbing into <math>k = f/2</math> gives a system of linear equations, <math>e + f = 1</math> and <math>e = f/2</math>. Solving yields <math>e = XY = 1/3</math> and <math>f = 2/3</math>, and since the area of the kite is simply the product of the two diagonals over <math>2</math> and <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.<br />
<br />
==Solution 4==<br />
Let the unmarked vertices of the shaded area be labeled <math>I</math> and <math>J</math>, with <math>I</math> being closer to <math>GD</math> than <math>J</math>. Noting that kite <math>HJFI</math> can be split into triangles <math>HJI</math> and <math>JIF</math>. <br />
<br />
Lemma: The distance from line segment <math>JI</math> to <math>H</math> is half the distance from <math>JI</math> to <math>F</math> <br />
<br />
Proof: Drop perpendiculars of triangles <math>HJI</math> and <math>JIF</math> to line <math>JI</math>, and let the point of intersection be <math>Q</math>. Note that <math>HJI</math> and <math>JIF</math> are similar to <math>HDC</math> and <math>ABF</math>, respectively. Now, the ratio of <math>DC</math> to <math>HF</math> is <math>1:1</math>, which shows that the ratio of <math>JI</math> to <math>HQ</math> is <math>1:1</math>, because of similar triangles as described above. Similarly, the ratio of <math>JI</math> to <math>FQ</math> is <math>1:2</math>. Since these two triangles contain the same base, <math>JI</math>, the ratio of <math>HQ:FQ = 1:2</math>. <math>\square</math><br />
<br />
Because kite <math>HJFI</math> is orthodiagonal, we multiply <math>(1*(1/3))/2 = \boxed{\textbf{(E)} \: \frac{1}{6}}</math><br />
<br />
~Lemma proof by sakshamsethi<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_16&diff=1544302014 AMC 10A Problems/Problem 162021-05-29T19:34:21Z<p>Sakshamsethi: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=1</math>, <math>BC=2</math>, and points <math>E</math>, <math>F</math>, and <math>G</math> are midpoints of <math>\overline{BC}</math>, <math>\overline{CD}</math>, and <math>\overline{AD}</math>, respectively. Point <math>H</math> is the midpoint of <math>\overline{GE}</math>. What is the area of the shaded region?<br />
<br />
<asy><br />
import graph;<br />
size(9cm);<br />
pen dps = fontsize(10); defaultpen(dps);<br />
pair D = (0,0);<br />
pair F = (1/2,0);<br />
pair C = (1,0);<br />
pair G = (0,1);<br />
pair E = (1,1);<br />
pair A = (0,2);<br />
pair B = (1,2);<br />
pair H = (1/2,1);<br />
<br />
// do not look<br />
pair X = (1/3,2/3);<br />
pair Y = (2/3,2/3);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(G--E);<br />
draw(A--F--B);<br />
draw(D--H--C);<br />
filldraw(H--X--F--Y--cycle,grey);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E,E);<br />
label("$F$",F,S);<br />
label("$G$",G,W);<br />
label("$H$",H,N);<br />
<br />
label("$\frac12$",(0.25,0),S);<br />
label("$\frac12$",(0.75,0),S);<br />
label("$1$",(1,0.5),E);<br />
label("$1$",(1,1.5),E);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1==<br />
Denote <math>D=(0,0)</math>. Then <math>A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)</math>. Let the intersection of <math>AF</math> and <math>DH</math> be <math>X</math>, and the intersection of <math>BF</math> and <math>CH</math> be <math>Y</math>. Then we want to find the coordinates of <math>X</math> so we can find <math>XY</math>. From our points, the slope of <math>AF</math> is <math>\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4</math>, and its <math>y</math>-intercept is just <math>2</math>. Thus the equation for <math>AF</math> is <math>y = -4x + 2</math>. We can also quickly find that the equation of <math>DH</math> is <math>y = 2x</math>. Setting the equations equal, we have <math>2x = -4x +2 \implies x = \frac13</math>. Because of symmetry, we can see that the distance from <math>Y</math> to <math>BC</math> is also <math>\frac13</math>, so <math>XY = 1 - 2 \cdot \frac13 = \frac13</math>. Now the area of the kite is simply the product of the two diagonals over <math>2</math>. Since the length <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.<br />
<br />
<asy><br />
import graph;<br />
size(9cm);<br />
pen dps = fontsize(10); defaultpen(dps);<br />
pair D = (0,0);<br />
pair F = (1/2,0);<br />
pair C = (1,0);<br />
pair G = (0,1);<br />
pair E = (1,1);<br />
pair A = (0,2);<br />
pair B = (1,2);<br />
pair H = (1/2,1);<br />
<br />
// do not look<br />
pair X = (1/3,2/3);<br />
pair Y = (2/3,2/3);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(G--E);<br />
draw(A--F--B);<br />
draw(D--H--C);<br />
filldraw(H--X--F--Y--cycle,grey);<br />
draw(X--Y,dashed);<br />
<br />
<br />
<br />
label("$A\: (0,2)$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D \: (0,0)$",D,SW);<br />
label("$E$",E,E);<br />
label("$F\: (\frac12,0)$",F,S);<br />
label("$G$",G,W);<br />
label("$H \: (\frac12,1)$",H,N);<br />
label("$Y$",Y,E);<br />
label("$X$",X,W);<br />
<br />
<br />
label("$\frac12$",(0.25,0),S);<br />
label("$\frac12$",(0.75,0),S);<br />
label("$1$",(1,0.5),E);<br />
label("$1$",(1,1.5),E);<br />
</asy><br />
<br />
==Solution 2==<br />
<br />
Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2}* IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = \boxed{\textbf{(E)} \: \frac{1}{6}}</math><br />
<br />
==Solution 3==<br />
<br />
From the diagram in Solution 1, let <math>e</math> be the height of <math>XHY</math> and <math>f</math> be the height of <math>XFY</math>. It is clear that their sum is <math>1</math> as they are parallel to <math>GD</math>. Let <math>k</math> be the ratio of the sides of the similar triangles <math>XFY</math> and <math>AFB</math>, which are similar because <math>XY</math> is parallel to <math>AB</math> and the triangles share angle <math>F</math>. Then <math>k = f/2</math>, as 2 is the height of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height of <math>XHY</math> will be equal to the base, like in <math>DHC</math>, making <math>XY = e</math>. However, <math>XY</math> is also the base of <math>XFY</math>, so <math>k = e / AB</math> where <math>AB = 1</math> so <math>k = e</math>. Subbing into <math>k = f/2</math> gives a system of linear equations, <math>e + f = 1</math> and <math>e = f/2</math>. Solving yields <math>e = XY = 1/3</math> and <math>f = 2/3</math>, and since the area of the kite is simply the product of the two diagonals over <math>2</math> and <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.<br />
<br />
==Solution 4==<br />
Let the unmarked vertices of the shaded area be labeled <math>I</math> and <math>J</math>, with <math>I</math> being closer to <math>GD</math> than <math>J</math>. Noting that kite <math>HJFI</math> can be split into triangles <math>HJI</math> and <math>JIF</math>. <br />
<br />
Lemma: The distance from line segment <math>JI</math> to <math>H</math> is half the distance from <math>JI</math> to <math>F</math> <br />
<br />
Proof: Drop perpendiculars of triangles <math>HJI</math> and <math>JIF</math> to line <math>JI</math>, and let the point of intersection be <math>Q</math>. Note that <math>HJI</math> and <math>JIF</math> are similar to <math>HDC</math> and <math>ABF</math>, respectively. Now, the ratio of <math>DC</math> to <math>HF</math> is <math>1:1</math>, which shows that the ratio of <math>JI</math> to <math>HQ</math> is <math>1:1</math>, because of similar triangles as described above. Similarly, the ratio of <math>JI</math> to <math>FQ</math> is <math>1:2</math>. Since these two triangles contain the same base, <math>JI</math>, the ratio of <math>HQ:FQ = 1:2</math>. <math>\square</math><br />
<br />
<br />
<br />
<br />
<br />
<br />
, we know that the distance from line segment <math>JI</math> to <math>H</math> is half the distance from <math>JI</math> to <math>F</math>. <br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
Because kite <math>HJFI</math> is orthodiagonal, we multiply <math>(1*(1/3))/2 = \boxed{\textbf{(E)} \: \frac{1}{6}}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=1534942017 USAJMO Problems/Problem 32021-05-10T19:14:20Z<p>Sakshamsethi: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
(<math>*</math>) Let <math>ABC</math> be an equilateral triangle and let <math>P</math> be a point on its circumcircle. Let lines <math>PA</math> and <math>BC</math> intersect at <math>D</math>; let lines <math>PB</math> and <math>CA</math> intersect at <math>E</math>; and let lines <math>PC</math> and <math>AB</math> intersect at <math>F</math>. Prove that the area of triangle <math>DEF</math> is twice that of triangle <math>ABC</math>.<br />
<br />
<asy><br />
size(3inch);<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br />
draw(Circle(O, 2sqrt(3)), black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
</asy><br />
<br />
==Solution (No Trig/Bash)==<br />
<br />
Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math><br />
<br />
<br />
<br />
Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> Now let <math>E'=DE\cap AF</math> and <math>F'=DF\cap AE</math>. Then by an area/concurrence theorem, we have that <math>\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,</math> or <math>\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math><br />
<br />
<br />
<br />
Manipulating these gives <math>\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.</math> Thus <math>\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,</math> and we are done.<br />
<br />
~cocohearts<br />
<br />
==Solution 1==<br />
WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 1: If <math>BF = k</math>, then <math>CE = \frac 1k</math>.<br />
This lemma results directly from the fact that <math>\triangle FBC</math> ~ <math>\triangle BCE</math>; <math>\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}</math>, or <math>CE = \frac{1}{BF}</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 2: <math>[AEF] = (k+\frac 1k + 2)(X+Y)</math>.<br />
We see that <math>[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)</math>, as desired. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 3: <math>\frac{X}{Y} = k</math>.<br />
We see that <br />
<cmath>\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.</cmath><br />
However, after some angle chasing and by the Law of Sines in <math>\triangle BCF</math>, we have <math>\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}</math>, or <math>k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}</math>, which implies the result. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
By the area lemma, we have <math>[BDF] = kX</math> and <math>[CDF] = \frac{Y}{k}</math>. <br />
<br />
We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven.<br />
<br />
==Solution 2==<br />
<br />
We will use barycentric coordinates and vectors. Let <math>\vec{X}</math> be the position vector of a point <math>X.</math> The point <math>(x, y, z)</math> in barycentric coordinates denotes the point <math>x\vec{A} + y\vec{B} + z\vec{C}.</math> For all points in the plane of <math>\triangle{ABC},</math> we have <math>x + y + z = 1.</math> It is clear that <math>A = (1, 0, 0)</math>; <math>B = (0, 1, 0)</math>; and <math>C = (0, 0, 1).</math><br />
<br />
Define the point <math>P</math> as <math>P = \left(x_P, y_P, z_P\right).</math> The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with the condition <math>x_P + y_P + z_P = 1</math> inherent to barycentric coordinates, gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math><br />
<br />
We can write the equations of the following lines:<br />
<cmath>BC: x = 0</cmath><br />
<cmath>CA: y = 0</cmath><br />
<cmath>AB: z = 0</cmath><br />
<cmath>PA: \frac{y}{y_P} = \frac{z}{z_P}</cmath><br />
<cmath>PB: \frac{x}{x_P} = \frac{z}{z_P}</cmath><br />
<cmath>PC: \frac{x}{x_P} = \frac{y}{y_P}.</cmath><br />
<br />
We can then solve for the points <math>D, E, F</math>:<br />
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath><br />
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath><br />
<cmath>F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).</cmath><br />
<br />
The area of an arbitrary triangle <math>XYZ</math> is:<br />
<cmath>[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|</cmath><br />
<cmath>[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.</cmath><br />
<br />
To calculate <math>[DEF],</math> we wish to compute <math>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).</math> After a lot of computation, we obtain the following:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
Evaluating the denominator,<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)</cmath><br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.</cmath><br />
<br />
Since <math>x_P + y_P + z_P = 1</math> and <math>x_Py_P + y_Pz_P + z_Px_P = 0,</math> it follows that:<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.</cmath><br />
<br />
We thus conclude that:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
From this, it follows that <math>[DEF] = 2[ABC],</math> and we are done.<br />
<br />
<br />
==Solution 3==<br />
<center><asy><br />
import cse5;<br />
import graph;<br />
import olympiad;<br />
<br />
size(3inch);<br />
<br />
<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));<br />
pair P = (-1, -sqrt(11)+sqrt(3));<br />
path circle = Circle(O, 2sqrt(3));<br />
pair D = extension(A,P,B,C);<br />
pair E1 = extension(A,C,B,P);<br />
pair F=extension(A,B,C,P);<br />
draw(circle, black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
pair G = extension(O,D,F,E1);<br />
draw(O--G,dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
dot("O", O, SE);<br />
</asy></center><br />
We'll use coordinates and shoelace. Let the origin be the midpoint of <math>BC</math>. Let <math>AB=2</math>, and <math>BF = 2x</math>, then <math>F=(-x-1,-x\sqrt{3})</math>. Using the facts <math>\triangle{CBP} \sim \triangle{CFB}</math> and <math>\triangle{BCP} \sim \triangle{BEC}</math>, we have <math>BF * CE = BC^2</math>, so <math>CE = \frac{1}{2x}</math>, and <math> E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})</math>.<br />
<br />
The slope of <math>FE</math> is <br />
<cmath>k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}</cmath><br />
It is well-known that <math>\triangle{DFE}</math> is self-polar, so <math>FE</math> is the polar of <math>D</math>, i.e., <math>OD</math> is perpendicular to <math>FE</math>. Therefore, the slope of <math>OD</math> is <math>-\frac{1}{k}</math>. Since <math>O=(0,\frac{1}{\sqrt{3}})</math>, we get the x-coordinate of <math>D</math>, <math>x_D = \frac{k}{\sqrt{3}}</math>, i.e., <math>D = (\frac{k}{\sqrt{3}},0)</math>. Using shoelace,<br />
<cmath>2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})-<br />
(-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}} </cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2}<br />
{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 4\sqrt{3}</cmath><br />
So <math>[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]</math>. Q.E.D<br />
<br />
By Mathdummy.<br />
==Solution 4 Without the nasty computations ==<br />
Note that <math>\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60</math>. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. <br />
<br />
Let <math>BP = x</math> and <math>CP = y</math>. Then,<br />
From Law of Cosine, <math>BC^2 = x^2 + y^2 + xy</math>. <br />
<br />
From Ptolemy's theorem, <math>AP BC = x AC + y AB</math>, so <math>AP = x + y</math>.<br />
<br />
Lemma 1: In Triangle ABC with side lengths <math>a,b,c</math> and <math>\angle A =120^o</math>, the length of the angle bisector of <math>A</math> is <br />
<cmath> d = \frac{bc}{b+c}</cmath><br />
This can be easily proved with Stewart's and Law of Cosine.<br />
<br />
Using Lemma 1, we have <br />
<cmath>PD = \frac{xy}{x+y}</cmath><br />
<cmath> x = \frac{FP AP}{FP + AP}</cmath><br />
<cmath> y = \frac{EP AP}{EP + AP}</cmath><br />
Plug in <math>AP=x+y</math>, we get: <br />
<cmath> PD = \frac{xy}{x+y}</cmath> <br />
<cmath> FP = \frac{x(x+y)}{y}</cmath><br />
<cmath> EP = \frac{y(x+y)}{x}</cmath><br />
Then<br />
<cmath> [\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)</cmath><br />
<cmath>= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2}(x^2 + xy + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2} BC^2</cmath><br />
<cmath> = 2 [\triangle{ABC}]</cmath><br />
<br />
By Mathdummy.<br />
<br />
==See also==<br />
{{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=1534932017 USAJMO Problems/Problem 32021-05-10T19:14:09Z<p>Sakshamsethi: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
(<math>*</math>) Let <math>ABC</math> be an equilateral triangle and let <math>P</math> be a point on its circumcircle. Let lines <math>PA</math> and <math>BC</math> intersect at <math>D</math>; let lines <math>PB</math> and <math>CA</math> intersect at <math>E</math>; and let lines <math>PC</math> and <math>AB</math> intersect at <math>F</math>. Prove that the area of triangle <math>DEF</math> is twice that of triangle <math>ABC</math>.<br />
<br />
<asy><br />
size(3inch);<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br />
draw(Circle(O, 2sqrt(3)), black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
</asy><br />
<br />
==Solution (No Trig/Bash)==<br />
<br />
Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math><br />
<br />
<br />
<br />
Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> Now let <math>E'=DE\cap AF</math> and <math>F'=DF\cap AE</math>. Then by an area/concurrence theorem, we have that <math>\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,</math> or <math>\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math><br />
<br />
<br />
<br />
Manipulating these gives <math>\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.</math> Thus <math>\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,</math> and we are done.<br />
<br />
~cocohearts<br />
<br />
==Solution 1==<br />
WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 1: If <math>BF = k</math>, then <math>CE = \frac 1k</math>.<br />
This lemma results directly from the fact that <math>\triangle FBC</math> ~ <math>\triangle BCE</math>; <math>\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}</math>, or <math>CE = \frac{1}{BF}</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 2: <math>[AEF] = (k+\frac 1k + 2)(X+Y)</math>.<br />
We see that <math>[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)</math>, as desired. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 3: <math>\frac{X}{Y} = k</math>.<br />
We see that <br />
<cmath>\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.</cmath><br />
However, after some angle chasing and by the Law of Sines in <math>\triangle BCF</math>, we have <math>\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}</math>, or <math>k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}</math>, which implies the result. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
By the area lemma, we have <math>[BDF] = kX</math> and <math>[CDF] = \frac{Y}{k}</math>. <br />
<br />
We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven.<br />
<br />
==Solution 2==<br />
<br />
We will use barycentric coordinates and vectors. Let <math>\vec{X}</math> be the position vector of a point <math>X.</math> The point <math>(x, y, z)</math> in barycentric coordinates denotes the point <math>x\vec{A} + y\vec{B} + z\vec{C}.</math> For all points in the plane of <math>\triangle{ABC},</math> we have <math>x + y + z = 1.</math> It is clear that <math>A = (1, 0, 0)</math>; <math>B = (0, 1, 0)</math>; and <math>C = (0, 0, 1).</math><br />
<br />
Define the point <math>P</math> as <math>P = \left(x_P, y_P, z_P\right).</math> The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with the condition <math>x_P + y_P + z_P = 1</math> inherent to barycentric coordinates, gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math><br />
<br />
We can write the equations of the following lines:<br />
<cmath>BC: x = 0</cmath><br />
<cmath>CA: y = 0</cmath><br />
<cmath>AB: z = 0</cmath><br />
<cmath>PA: \frac{y}{y_P} = \frac{z}{z_P}</cmath><br />
<cmath>PB: \frac{x}{x_P} = \frac{z}{z_P}</cmath><br />
<cmath>PC: \frac{x}{x_P} = \frac{y}{y_P}.</cmath><br />
<br />
We can then solve for the points <math>D, E, F</math>:<br />
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath><br />
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath><br />
<cmath>F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).</cmath><br />
<br />
The area of an arbitrary triangle <math>XYZ</math> is:<br />
<cmath>[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|</cmath><br />
<cmath>[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.</cmath><br />
<br />
To calculate <math>[DEF],</math> we wish to compute <math>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).</math> After a lot of computation, we obtain the following:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
Evaluating the denominator,<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)</cmath><br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.</cmath><br />
<br />
Since <math>x_P + y_P + z_P = 1</math> and <math>x_Py_P + y_Pz_P + z_Px_P = 0,</math> it follows that:<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.</cmath><br />
<br />
We thus conclude that:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
From this, it follows that <math>[DEF] = 2[ABC],</math> and we are done.<br />
<br />
<br />
==Solution 3==<br />
<center><asy><br />
import cse5;<br />
import graph;<br />
import olympiad;<br />
<br />
size(3inch);<br />
<br />
<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));<br />
pair P = (-1, -sqrt(11)+sqrt(3));<br />
path circle = Circle(O, 2sqrt(3));<br />
pair D = extension(A,P,B,C);<br />
pair E1 = extension(A,C,B,P);<br />
pair F=extension(A,B,C,P);<br />
draw(circle, black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
pair G = extension(O,D,F,E1);<br />
draw(O--G,dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
dot("O", O, SE);<br />
</asy></center><br />
We'll use coordinates and shoelace. Let the origin be the midpoint of <math>BC</math>. Let <math>AB=2</math>, and <math>BF = 2x</math>, then <math>F=(-x-1,-x\sqrt{3})</math>. Using the facts <math>\triangle{CBP} \sim \triangle{CFB}</math> and <math>\triangle{BCP} \sim \triangle{BEC}</math>, we have <math>BF * CE = BC^2</math>, so <math>CE = \frac{1}{2x}</math>, and <math> E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})</math>.<br />
<br />
The slope of <math>FE</math> is <br />
<cmath>k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}</cmath><br />
It is well-known that <math>\triangle{DFE}</math> is self-polar, so <math>FE</math> is the polar of <math>D</math>, i.e., <math>OD</math> is perpendicular to <math>FE</math>. Therefore, the slope of <math>OD</math> is <math>-\frac{1}{k}</math>. Since <math>O=(0,\frac{1}{\sqrt{3}})</math>, we get the x-coordinate of <math>D</math>, <math>x_D = \frac{k}{\sqrt{3}}</math>, i.e., <math>D = (\frac{k}{\sqrt{3}},0)</math>. Using shoelace,<br />
<cmath>2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})-<br />
(-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}} </cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2}<br />
{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 4\sqrt{3}</cmath><br />
So <math>[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]</math>. Q.E.D<br />
<br />
By Mathdummy.<br />
==Solution 4 Without the nasty computations ==<br />
Note that <math>\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60</math>. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. <br />
<br />
Let <math>BP = x</math> and <math>CP = y</math>. Then,<br />
From Law of Cosine, <math>BC^2 = x^2 + y^2 + xy</math>. <br />
<br />
From Ptolemy's theorem, <math>AP BC = x AC + y AB</math>, so <math>AP = x + y</math>.<br />
<br />
Lemma 1: In Triangle ABC with side lengths <math>a,b,c</math> and <math>\angle A =120^o</math>, the length of the angle bisector of <math>A</math> is <br />
<cmath> d = \frac{bc}{b+c}</cmath><br />
This can be easily proved with Stewart's and Law of Cosine.<br />
<br />
Using Lemma 1, we have <br />
<cmath>PD = \frac{xy}{x+y}</cmath><br />
<cmath> x = \frac{FP AP}{FP + AP}</cmath><br />
<cmath> y = \frac{EP AP}{EP + AP}</cmath><br />
Plug in <math>AP=x+y</math>, we get: <br />
<cmath> PD = \frac{xy}{x+y}</cmath> <br />
<cmath> FP = \frac{x(x+y)}{y}</cmath><br />
<cmath> EP = \frac{y(x+y)}{x}</cmath><br />
Then<br />
<cmath> [\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)</cmath><br />
<cmath>= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2}(x^2 + xy + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2} BC^2</cmath><br />
<cmath> = 2 [\triangle{ABC}]</cmath><br />
<br />
By Mathdummy.<br />
<br />
==Solution 5==<br />
<br />
<br />
Extend AP to meet FE at point X. Note that there is symmetry of triangle <math>\triangle{AFE}</math> across line AX. Therefore, if we can prove that the area of <math>\triangle{FDX}</math> is twice the area of <math>\triangle{BAD}</math>, then we will be done. For simplicity, let the side length of equilateral triangle <math>\triangle{ABC}</math> be <math>1</math>. <br />
<br />
We can easily find the area of <math>\triangle{BAD}</math>. Since the area formula for an equilateral triangle is <math>\frac{s^2 \sqrt{3}}{4}</math> , where <math>s</math> is the side length, the area of <math>\triangle{ABC}</math> is <math>\frac{\sqrt{3}}{4}</math>. By symmetry, the area of <math>\triangle{BAD}</math> is exactly half the area of <math>\triangle{ABC}</math>. Thus the area of <math>\triangle{BAD} = \frac{\frac{\sqrt{3}}{4}}{2} = \boxed{\frac{\sqrt{3}}{8}}</math>.<br />
<br />
Now we just need to find the area of <math>\triangle{FDX}</math>. We have to find <math>FX</math> and <math>DX</math> for the area. Note that <math>FX = \frac{FE}{2}</math>. It looks by eye that <math>\triangle{AFE}</math> is equilateral. Let's try to prove it. <br />
<br />
<br />
<br />
<br />
<math>\textbf{\textit{Lemma 1: </math>\triangle{AFE}<math> is equilateral.}}</math><br />
<br />
<math>\textbf{\textit{Proof of Lemma 1:}}</math><br />
By symmetry, across lines <math>AX, FC, BE</math>, we arrive to some conclusions: <br />
\begin{center}<br />
<math>AF=AE</math><br />
<br />
<math>AF = FE</math><br />
<br />
<math>FE = AE</math>. <br />
\end{center}<br />
<br />
Substituting the first equation into the second, we see that <math>AF=AE=FE</math>, which is necessary and sufficient to prove that <math>\triangle{AFE}</math> is equilateral. <math>\square</math> <br />
<br />
<br />
<br />
Since <math>\triangle{AFE}</math> is equilateral, and so is <math>\triangle{ABC}</math>, we find that <math>\triangle{ABC} \sim \triangle{AFE}</math>. Because <math>AB = \frac{AF}{2}</math> by symmetry across <math>BE</math>, we know that the similarity ratio is <math>\frac{1}{2}</math>.<br />
Thus the side length of <math>\triangle{AFE}</math> is <math>2</math>, and <math>FX = \frac{FE}{2} = 1</math>.<br />
<br />
<br />
<br />
Now we need to find <math>DX</math>. We can do <math>AX-AD</math>. Well, that's easy! Since <math>\triangle{BAD}</math> is a 30-60-90 triangle by symmetry across line <math>AX</math>, and <math>BD = \frac{BC}{2} = \frac{1}{2}</math>, we know that <math>AD = \frac{\sqrt{3}}{2}</math>. To find <math>AX</math>, we do pretty much the same thing: <math>FAX</math> is a 30-60-90 triangle, and <math>FX = \frac{2}{2} = 1</math>, so <math>AX = \sqrt{3}</math>. We can now find <math>DX</math>: <br />
\begin{center}<br />
<math>DX = AX - AD = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}</math>. <br />
\end{center}<br />
<br />
<br />
<br />
<br />
Recalling why we were doing all this, we find the area of <math>\triangle{FDX}</math>. It is <math>\frac{FX \times DX}{2} = \frac{1 \times \frac{\sqrt{3}}{2}}{2} = \boxed{\frac{\sqrt{3}}{4}}</math>. <br />
<br />
<br />
<br />
<br />
Let's summarize: <br />
<br />
\begin{center}<br />
The area of <math>\triangle{ABC}</math> is <math>\boxed{\frac{\sqrt{3}}{8}}</math><br />
<br />
The area of <math>\triangle{FDX}</math> is <math>\boxed{\frac{\sqrt{3}}{4}}</math>, <br />
<br />
which proves our initial claim. <br />
<br />
<br />
<br />
\end{center}<br />
<br />
<math>\square{}</math><br />
<br />
==See also==<br />
{{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=1534922017 USAJMO Problems/Problem 32021-05-10T19:11:06Z<p>Sakshamsethi: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
(<math>*</math>) Let <math>ABC</math> be an equilateral triangle and let <math>P</math> be a point on its circumcircle. Let lines <math>PA</math> and <math>BC</math> intersect at <math>D</math>; let lines <math>PB</math> and <math>CA</math> intersect at <math>E</math>; and let lines <math>PC</math> and <math>AB</math> intersect at <math>F</math>. Prove that the area of triangle <math>DEF</math> is twice that of triangle <math>ABC</math>.<br />
<br />
<asy><br />
size(3inch);<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br />
draw(Circle(O, 2sqrt(3)), black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
</asy><br />
<br />
==Solution (No Trig/Bash)==<br />
<br />
Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math><br />
<br />
<br />
<br />
Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> Now let <math>E'=DE\cap AF</math> and <math>F'=DF\cap AE</math>. Then by an area/concurrence theorem, we have that <math>\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,</math> or <math>\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math><br />
<br />
<br />
<br />
Manipulating these gives <math>\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.</math> Thus <math>\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,</math> and we are done.<br />
<br />
~cocohearts<br />
<br />
==Solution 1==<br />
WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 1: If <math>BF = k</math>, then <math>CE = \frac 1k</math>.<br />
This lemma results directly from the fact that <math>\triangle FBC</math> ~ <math>\triangle BCE</math>; <math>\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}</math>, or <math>CE = \frac{1}{BF}</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 2: <math>[AEF] = (k+\frac 1k + 2)(X+Y)</math>.<br />
We see that <math>[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)</math>, as desired. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 3: <math>\frac{X}{Y} = k</math>.<br />
We see that <br />
<cmath>\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.</cmath><br />
However, after some angle chasing and by the Law of Sines in <math>\triangle BCF</math>, we have <math>\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}</math>, or <math>k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}</math>, which implies the result. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
By the area lemma, we have <math>[BDF] = kX</math> and <math>[CDF] = \frac{Y}{k}</math>. <br />
<br />
We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven.<br />
<br />
==Solution 2==<br />
<br />
We will use barycentric coordinates and vectors. Let <math>\vec{X}</math> be the position vector of a point <math>X.</math> The point <math>(x, y, z)</math> in barycentric coordinates denotes the point <math>x\vec{A} + y\vec{B} + z\vec{C}.</math> For all points in the plane of <math>\triangle{ABC},</math> we have <math>x + y + z = 1.</math> It is clear that <math>A = (1, 0, 0)</math>; <math>B = (0, 1, 0)</math>; and <math>C = (0, 0, 1).</math><br />
<br />
Define the point <math>P</math> as <math>P = \left(x_P, y_P, z_P\right).</math> The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with the condition <math>x_P + y_P + z_P = 1</math> inherent to barycentric coordinates, gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math><br />
<br />
We can write the equations of the following lines:<br />
<cmath>BC: x = 0</cmath><br />
<cmath>CA: y = 0</cmath><br />
<cmath>AB: z = 0</cmath><br />
<cmath>PA: \frac{y}{y_P} = \frac{z}{z_P}</cmath><br />
<cmath>PB: \frac{x}{x_P} = \frac{z}{z_P}</cmath><br />
<cmath>PC: \frac{x}{x_P} = \frac{y}{y_P}.</cmath><br />
<br />
We can then solve for the points <math>D, E, F</math>:<br />
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath><br />
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath><br />
<cmath>F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).</cmath><br />
<br />
The area of an arbitrary triangle <math>XYZ</math> is:<br />
<cmath>[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|</cmath><br />
<cmath>[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.</cmath><br />
<br />
To calculate <math>[DEF],</math> we wish to compute <math>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).</math> After a lot of computation, we obtain the following:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
Evaluating the denominator,<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)</cmath><br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.</cmath><br />
<br />
Since <math>x_P + y_P + z_P = 1</math> and <math>x_Py_P + y_Pz_P + z_Px_P = 0,</math> it follows that:<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.</cmath><br />
<br />
We thus conclude that:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
From this, it follows that <math>[DEF] = 2[ABC],</math> and we are done.<br />
<br />
<br />
==Solution 3==<br />
<center><asy><br />
import cse5;<br />
import graph;<br />
import olympiad;<br />
<br />
size(3inch);<br />
<br />
<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));<br />
pair P = (-1, -sqrt(11)+sqrt(3));<br />
path circle = Circle(O, 2sqrt(3));<br />
pair D = extension(A,P,B,C);<br />
pair E1 = extension(A,C,B,P);<br />
pair F=extension(A,B,C,P);<br />
draw(circle, black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
pair G = extension(O,D,F,E1);<br />
draw(O--G,dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
dot("O", O, SE);<br />
</asy></center><br />
We'll use coordinates and shoelace. Let the origin be the midpoint of <math>BC</math>. Let <math>AB=2</math>, and <math>BF = 2x</math>, then <math>F=(-x-1,-x\sqrt{3})</math>. Using the facts <math>\triangle{CBP} \sim \triangle{CFB}</math> and <math>\triangle{BCP} \sim \triangle{BEC}</math>, we have <math>BF * CE = BC^2</math>, so <math>CE = \frac{1}{2x}</math>, and <math> E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})</math>.<br />
<br />
The slope of <math>FE</math> is <br />
<cmath>k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}</cmath><br />
It is well-known that <math>\triangle{DFE}</math> is self-polar, so <math>FE</math> is the polar of <math>D</math>, i.e., <math>OD</math> is perpendicular to <math>FE</math>. Therefore, the slope of <math>OD</math> is <math>-\frac{1}{k}</math>. Since <math>O=(0,\frac{1}{\sqrt{3}})</math>, we get the x-coordinate of <math>D</math>, <math>x_D = \frac{k}{\sqrt{3}}</math>, i.e., <math>D = (\frac{k}{\sqrt{3}},0)</math>. Using shoelace,<br />
<cmath>2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})-<br />
(-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}} </cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2}<br />
{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 4\sqrt{3}</cmath><br />
So <math>[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]</math>. Q.E.D<br />
<br />
By Mathdummy.<br />
==Solution 4 Without the nasty computations ==<br />
Note that <math>\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60</math>. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. <br />
<br />
Let <math>BP = x</math> and <math>CP = y</math>. Then,<br />
From Law of Cosine, <math>BC^2 = x^2 + y^2 + xy</math>. <br />
<br />
From Ptolemy's theorem, <math>AP BC = x AC + y AB</math>, so <math>AP = x + y</math>.<br />
<br />
Lemma 1: In Triangle ABC with side lengths <math>a,b,c</math> and <math>\angle A =120^o</math>, the length of the angle bisector of <math>A</math> is <br />
<cmath> d = \frac{bc}{b+c}</cmath><br />
This can be easily proved with Stewart's and Law of Cosine.<br />
<br />
Using Lemma 1, we have <br />
<cmath>PD = \frac{xy}{x+y}</cmath><br />
<cmath> x = \frac{FP AP}{FP + AP}</cmath><br />
<cmath> y = \frac{EP AP}{EP + AP}</cmath><br />
Plug in <math>AP=x+y</math>, we get: <br />
<cmath> PD = \frac{xy}{x+y}</cmath> <br />
<cmath> FP = \frac{x(x+y)}{y}</cmath><br />
<cmath> EP = \frac{y(x+y)}{x}</cmath><br />
Then<br />
<cmath> [\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)</cmath><br />
<cmath>= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2}(x^2 + xy + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2} BC^2</cmath><br />
<cmath> = 2 [\triangle{ABC}]</cmath><br />
<br />
By Mathdummy.<br />
<br />
==Solution 5==<br />
<br />
<br />
Extend AP to meet FE at point X. Note that there is symmetry of triangle <math>\triangle{AFE}</math> across line AX. Therefore, if we can prove that the area of <math>\triangle{FDX}</math> is twice the area of <math>\triangle{BAD}</math>, then we will be done. For simplicity, let the side length of equilateral triangle <math>\triangle{ABC}</math> be <math>1</math>. <br />
<br />
We can easily find the area of <math>\triangle{BAD}</math>. Since the area formula for an equilateral triangle is <math>\frac{s^2 \sqrt{3}}{4}</math> , where <math>s</math> is the side length, the area of <math>\triangle{ABC}</math> is <math>\frac{\sqrt{3}}{4}</math>. By symmetry, the area of <math>\triangle{BAD}</math> is exactly half the area of <math>\triangle{ABC}</math>. Thus the area of <math>\triangle{BAD} = \frac{\frac{\sqrt{3}}{4}}{2} = \boxed{\frac{\sqrt{3}}{8}}</math>.<br />
<br />
Now we just need to find the area of <math>\triangle{FDX}</math>. We have to find <math>FX</math> and <math>DX</math> for the area. Note that <math>FX = \frac{FE}{2}</math>. It looks by eye that <math>\triangle{AFE}</math> is equilateral. Let's try to prove it. <br />
<br />
<br />
<br />
<br />
\textbf{\textit{Lemma 1: <math>\triangle{AFE}</math> is equilateral.}} <br />
<br />
\textbf{\textit{Proof of Lemma 1:}} <br />
By symmetry, across lines <math>AX, FC, BE</math>, we arrive to some conclusions: <br />
\begin{center}<br />
<math>AF=AE</math><br />
<br />
<math>AF = FE</math><br />
<br />
<math>FE = AE</math>. <br />
\end{center}<br />
<br />
Substituting the first equation into the second, we see that <math>AF=AE=FE</math>, which is necessary and sufficient to prove that <math>\triangle{AFE}</math> is equilateral. <math>\square</math> <br />
<br />
<br />
<br />
Since <math>\triangle{AFE}</math> is equilateral, and so is <math>\triangle{ABC}</math>, we find that <math>\triangle{ABC} \sim \triangle{AFE}</math>. Because <math>AB = \frac{AF}{2}</math> by symmetry across <math>BE</math>, we know that the similarity ratio is <math>\frac{1}{2}</math>.<br />
Thus the side length of <math>\triangle{AFE}</math> is <math>2</math>, and <math>FX = \frac{FE}{2} = 1</math>.<br />
<br />
<br />
<br />
Now we need to find <math>DX</math>. We can do <math>AX-AD</math>. Well, that's easy! Since <math>\triangle{BAD}</math> is a 30-60-90 triangle by symmetry across line <math>AX</math>, and <math>BD = \frac{BC}{2} = \frac{1}{2}</math>, we know that <math>AD = \frac{\sqrt{3}}{2}</math>. To find <math>AX</math>, we do pretty much the same thing: <math>FAX</math> is a 30-60-90 triangle, and <math>FX = \frac{2}{2} = 1</math>, so <math>AX = \sqrt{3}</math>. We can now find <math>DX</math>: <br />
\begin{center}<br />
<math>DX = AX - AD = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}</math>. <br />
\end{center}<br />
<br />
<br />
<br />
<br />
Recalling why we were doing all this, we find the area of <math>\triangle{FDX}</math>. It is <math>\frac{FX \times DX}{2} = \frac{1 \times \frac{\sqrt{3}}{2}}{2} = \boxed{\frac{\sqrt{3}}{4}}</math>. <br />
<br />
<br />
<br />
<br />
Let's summarize: <br />
<br />
\begin{center}<br />
The area of <math>\triangle{ABC}</math> is <math>\boxed{\frac{\sqrt{3}}{8}}</math><br />
<br />
The area of <math>\triangle{FDX}</math> is <math>\boxed{\frac{\sqrt{3}}{4}}</math>, <br />
<br />
which proves our initial claim. <br />
<br />
<br />
<br />
\end{center}<br />
<br />
<math>\square{}</math><br />
<br />
==See also==<br />
{{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=1534912017 USAJMO Problems/Problem 32021-05-10T19:10:25Z<p>Sakshamsethi: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
(<math>*</math>) Let <math>ABC</math> be an equilateral triangle and let <math>P</math> be a point on its circumcircle. Let lines <math>PA</math> and <math>BC</math> intersect at <math>D</math>; let lines <math>PB</math> and <math>CA</math> intersect at <math>E</math>; and let lines <math>PC</math> and <math>AB</math> intersect at <math>F</math>. Prove that the area of triangle <math>DEF</math> is twice that of triangle <math>ABC</math>.<br />
<br />
<asy><br />
size(3inch);<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br />
draw(Circle(O, 2sqrt(3)), black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
</asy><br />
<br />
==Solution (No Trig/Bash)==<br />
<br />
Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math><br />
<br />
<br />
<br />
Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> Now let <math>E'=DE\cap AF</math> and <math>F'=DF\cap AE</math>. Then by an area/concurrence theorem, we have that <math>\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,</math> or <math>\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math><br />
<br />
<br />
<br />
Manipulating these gives <math>\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.</math> Thus <math>\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,</math> and we are done.<br />
<br />
~cocohearts<br />
<br />
==Solution 1==<br />
WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 1: If <math>BF = k</math>, then <math>CE = \frac 1k</math>.<br />
This lemma results directly from the fact that <math>\triangle FBC</math> ~ <math>\triangle BCE</math>; <math>\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}</math>, or <math>CE = \frac{1}{BF}</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 2: <math>[AEF] = (k+\frac 1k + 2)(X+Y)</math>.<br />
We see that <math>[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)</math>, as desired. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 3: <math>\frac{X}{Y} = k</math>.<br />
We see that <br />
<cmath>\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.</cmath><br />
However, after some angle chasing and by the Law of Sines in <math>\triangle BCF</math>, we have <math>\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}</math>, or <math>k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}</math>, which implies the result. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
By the area lemma, we have <math>[BDF] = kX</math> and <math>[CDF] = \frac{Y}{k}</math>. <br />
<br />
We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven.<br />
<br />
==Solution 2==<br />
<br />
We will use barycentric coordinates and vectors. Let <math>\vec{X}</math> be the position vector of a point <math>X.</math> The point <math>(x, y, z)</math> in barycentric coordinates denotes the point <math>x\vec{A} + y\vec{B} + z\vec{C}.</math> For all points in the plane of <math>\triangle{ABC},</math> we have <math>x + y + z = 1.</math> It is clear that <math>A = (1, 0, 0)</math>; <math>B = (0, 1, 0)</math>; and <math>C = (0, 0, 1).</math><br />
<br />
Define the point <math>P</math> as <math>P = \left(x_P, y_P, z_P\right).</math> The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with the condition <math>x_P + y_P + z_P = 1</math> inherent to barycentric coordinates, gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math><br />
<br />
We can write the equations of the following lines:<br />
<cmath>BC: x = 0</cmath><br />
<cmath>CA: y = 0</cmath><br />
<cmath>AB: z = 0</cmath><br />
<cmath>PA: \frac{y}{y_P} = \frac{z}{z_P}</cmath><br />
<cmath>PB: \frac{x}{x_P} = \frac{z}{z_P}</cmath><br />
<cmath>PC: \frac{x}{x_P} = \frac{y}{y_P}.</cmath><br />
<br />
We can then solve for the points <math>D, E, F</math>:<br />
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath><br />
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath><br />
<cmath>F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).</cmath><br />
<br />
The area of an arbitrary triangle <math>XYZ</math> is:<br />
<cmath>[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|</cmath><br />
<cmath>[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.</cmath><br />
<br />
To calculate <math>[DEF],</math> we wish to compute <math>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).</math> After a lot of computation, we obtain the following:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
Evaluating the denominator,<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)</cmath><br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.</cmath><br />
<br />
Since <math>x_P + y_P + z_P = 1</math> and <math>x_Py_P + y_Pz_P + z_Px_P = 0,</math> it follows that:<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.</cmath><br />
<br />
We thus conclude that:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
From this, it follows that <math>[DEF] = 2[ABC],</math> and we are done.<br />
<br />
<br />
==Solution 3==<br />
<center><asy><br />
import cse5;<br />
import graph;<br />
import olympiad;<br />
<br />
size(3inch);<br />
<br />
<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));<br />
pair P = (-1, -sqrt(11)+sqrt(3));<br />
path circle = Circle(O, 2sqrt(3));<br />
pair D = extension(A,P,B,C);<br />
pair E1 = extension(A,C,B,P);<br />
pair F=extension(A,B,C,P);<br />
draw(circle, black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
pair G = extension(O,D,F,E1);<br />
draw(O--G,dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
dot("O", O, SE);<br />
</asy></center><br />
We'll use coordinates and shoelace. Let the origin be the midpoint of <math>BC</math>. Let <math>AB=2</math>, and <math>BF = 2x</math>, then <math>F=(-x-1,-x\sqrt{3})</math>. Using the facts <math>\triangle{CBP} \sim \triangle{CFB}</math> and <math>\triangle{BCP} \sim \triangle{BEC}</math>, we have <math>BF * CE = BC^2</math>, so <math>CE = \frac{1}{2x}</math>, and <math> E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})</math>.<br />
<br />
The slope of <math>FE</math> is <br />
<cmath>k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}</cmath><br />
It is well-known that <math>\triangle{DFE}</math> is self-polar, so <math>FE</math> is the polar of <math>D</math>, i.e., <math>OD</math> is perpendicular to <math>FE</math>. Therefore, the slope of <math>OD</math> is <math>-\frac{1}{k}</math>. Since <math>O=(0,\frac{1}{\sqrt{3}})</math>, we get the x-coordinate of <math>D</math>, <math>x_D = \frac{k}{\sqrt{3}}</math>, i.e., <math>D = (\frac{k}{\sqrt{3}},0)</math>. Using shoelace,<br />
<cmath>2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})-<br />
(-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}} </cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2}<br />
{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 4\sqrt{3}</cmath><br />
So <math>[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]</math>. Q.E.D<br />
<br />
By Mathdummy.<br />
==Solution 4 Without the nasty computations ==<br />
Note that <math>\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60</math>. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. <br />
<br />
Let <math>BP = x</math> and <math>CP = y</math>. Then,<br />
From Law of Cosine, <math>BC^2 = x^2 + y^2 + xy</math>. <br />
<br />
From Ptolemy's theorem, <math>AP BC = x AC + y AB</math>, so <math>AP = x + y</math>.<br />
<br />
Lemma 1: In Triangle ABC with side lengths <math>a,b,c</math> and <math>\angle A =120^o</math>, the length of the angle bisector of <math>A</math> is <br />
<cmath> d = \frac{bc}{b+c}</cmath><br />
This can be easily proved with Stewart's and Law of Cosine.<br />
<br />
Using Lemma 1, we have <br />
<cmath>PD = \frac{xy}{x+y}</cmath><br />
<cmath> x = \frac{FP AP}{FP + AP}</cmath><br />
<cmath> y = \frac{EP AP}{EP + AP}</cmath><br />
Plug in <math>AP=x+y</math>, we get: <br />
<cmath> PD = \frac{xy}{x+y}</cmath> <br />
<cmath> FP = \frac{x(x+y)}{y}</cmath><br />
<cmath> EP = \frac{y(x+y)}{x}</cmath><br />
Then<br />
<cmath> [\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)</cmath><br />
<cmath>= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2}(x^2 + xy + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2} BC^2</cmath><br />
<cmath> = 2 [\triangle{ABC}]</cmath><br />
<br />
By Mathdummy.<br />
<br />
==Solution 5==<br />
<br />
<br />
Extend AP to meet FE at point X. Note that there is symmetry of triangle <math>\triangle{AFE}</math> across line AX. Therefore, if we can prove that the area of <math>\triangle{FDX}</math> is twice the area of <math>\triangle{BAD}</math>, then we will be done. For simplicity, let the side length of equilateral triangle <math>\triangle{ABC}</math> be <math>1</math>. \\ <br />
<br />
We can easily find the area of <math>\triangle{BAD}</math>. Since the area formula for an equilateral triangle is <math>\frac{s^2 \sqrt{3}}{4}</math> , where <math>s</math> is the side length, the area of <math>\triangle{ABC}</math> is <math>\frac{\sqrt{3}}{4}</math>. By symmetry, the area of <math>\triangle{BAD}</math> is exactly half the area of <math>\triangle{ABC}</math>. Thus the area of <math>\triangle{BAD} = \frac{\frac{\sqrt{3}}{4}}{2} = \boxed{\frac{\sqrt{3}}{8}}</math>.<br />
<br />
Now we just need to find the area of <math>\triangle{FDX}</math>. We have to find <math>FX</math> and <math>DX</math> for the area. Note that <math>FX = \frac{FE}{2}</math>. It looks by eye that <math>\triangle{AFE}</math> is equilateral. Let's try to prove it. <br />
<br />
<br />
<br />
<br />
\textbf{\textit{Lemma 1: <math>\triangle{AFE}</math> is equilateral.}} <br />
<br />
\textbf{\textit{Proof of Lemma 1:}} <br />
By symmetry, across lines <math>AX, FC, BE</math>, we arrive to some conclusions: <br />
\begin{center}<br />
<math>AF=AE</math><br />
<br />
<math>AF = FE</math><br />
<br />
<math>FE = AE</math>. <br />
\end{center}<br />
<br />
Substituting the first equation into the second, we see that <math>AF=AE=FE</math>, which is necessary and sufficient to prove that <math>\triangle{AFE}</math> is equilateral. <math>\square</math> <br />
<br />
<br />
<br />
Since <math>\triangle{AFE}</math> is equilateral, and so is <math>\triangle{ABC}</math>, we find that <math>\triangle{ABC} \sim \triangle{AFE}</math>. Because <math>AB = \frac{AF}{2}</math> by symmetry across <math>BE</math>, we know that the similarity ratio is <math>\frac{1}{2}</math>.<br />
Thus the side length of <math>\triangle{AFE}</math> is <math>2</math>, and <math>FX = \frac{FE}{2} = 1</math>.<br />
<br />
<br />
<br />
Now we need to find <math>DX</math>. We can do <math>AX-AD</math>. Well, that's easy! Since <math>\triangle{BAD}</math> is a 30-60-90 triangle by symmetry across line <math>AX</math>, and <math>BD = \frac{BC}{2} = \frac{1}{2}</math>, we know that <math>AD = \frac{\sqrt{3}}{2}</math>. To find <math>AX</math>, we do pretty much the same thing: <math>FAX</math> is a 30-60-90 triangle, and <math>FX = \frac{2}{2} = 1</math>, so <math>AX = \sqrt{3}</math>. We can now find <math>DX</math>: <br />
\begin{center}<br />
<math>DX = AX - AD = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}</math>. <br />
\end{center}<br />
<br />
<br />
<br />
<br />
Recalling why we were doing all this, we find the area of <math>\triangle{FDX}</math>. It is <math>\frac{FX \times DX}{2} = \frac{1 \times \frac{\sqrt{3}}{2}}{2} = \boxed{\frac{\sqrt{3}}{4}}</math>. <br />
<br />
<br />
<br />
<br />
Let's summarize: <br />
<br />
\begin{center}<br />
The area of <math>\triangle{ABC}</math> is <math>\boxed{\frac{\sqrt{3}}{8}}</math>\\<br />
<br />
The area of <math>\triangle{FDX}</math> is <math>\boxed{\frac{\sqrt{3}}{4}}</math>, \\<br />
<br />
which proves our initial claim. <br />
<br />
<br />
<br />
\end{center}<br />
<br />
<math>\square{}</math><br />
<br />
==See also==<br />
{{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=1534902017 USAJMO Problems/Problem 32021-05-10T18:43:21Z<p>Sakshamsethi: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
(<math>*</math>) Let <math>ABC</math> be an equilateral triangle and let <math>P</math> be a point on its circumcircle. Let lines <math>PA</math> and <math>BC</math> intersect at <math>D</math>; let lines <math>PB</math> and <math>CA</math> intersect at <math>E</math>; and let lines <math>PC</math> and <math>AB</math> intersect at <math>F</math>. Prove that the area of triangle <math>DEF</math> is twice that of triangle <math>ABC</math>.<br />
<br />
<asy><br />
size(3inch);<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br />
draw(Circle(O, 2sqrt(3)), black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
</asy><br />
<br />
==Solution (No Trig/Bash)==<br />
<br />
Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math><br />
<br />
<br />
<br />
Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> Now let <math>E'=DE\cap AF</math> and <math>F'=DF\cap AE</math>. Then by an area/concurrence theorem, we have that <math>\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,</math> or <math>\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math><br />
<br />
<br />
<br />
Manipulating these gives <math>\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.</math> Thus <math>\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,</math> and we are done.<br />
<br />
~cocohearts<br />
<br />
==Solution 1==<br />
WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 1: If <math>BF = k</math>, then <math>CE = \frac 1k</math>.<br />
This lemma results directly from the fact that <math>\triangle FBC</math> ~ <math>\triangle BCE</math>; <math>\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}</math>, or <math>CE = \frac{1}{BF}</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 2: <math>[AEF] = (k+\frac 1k + 2)(X+Y)</math>.<br />
We see that <math>[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)</math>, as desired. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 3: <math>\frac{X}{Y} = k</math>.<br />
We see that <br />
<cmath>\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.</cmath><br />
However, after some angle chasing and by the Law of Sines in <math>\triangle BCF</math>, we have <math>\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}</math>, or <math>k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}</math>, which implies the result. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
By the area lemma, we have <math>[BDF] = kX</math> and <math>[CDF] = \frac{Y}{k}</math>. <br />
<br />
We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven.<br />
<br />
==Solution 2==<br />
<br />
We will use barycentric coordinates and vectors. Let <math>\vec{X}</math> be the position vector of a point <math>X.</math> The point <math>(x, y, z)</math> in barycentric coordinates denotes the point <math>x\vec{A} + y\vec{B} + z\vec{C}.</math> For all points in the plane of <math>\triangle{ABC},</math> we have <math>x + y + z = 1.</math> It is clear that <math>A = (1, 0, 0)</math>; <math>B = (0, 1, 0)</math>; and <math>C = (0, 0, 1).</math><br />
<br />
Define the point <math>P</math> as <math>P = \left(x_P, y_P, z_P\right).</math> The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with the condition <math>x_P + y_P + z_P = 1</math> inherent to barycentric coordinates, gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math><br />
<br />
We can write the equations of the following lines:<br />
<cmath>BC: x = 0</cmath><br />
<cmath>CA: y = 0</cmath><br />
<cmath>AB: z = 0</cmath><br />
<cmath>PA: \frac{y}{y_P} = \frac{z}{z_P}</cmath><br />
<cmath>PB: \frac{x}{x_P} = \frac{z}{z_P}</cmath><br />
<cmath>PC: \frac{x}{x_P} = \frac{y}{y_P}.</cmath><br />
<br />
We can then solve for the points <math>D, E, F</math>:<br />
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath><br />
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath><br />
<cmath>F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).</cmath><br />
<br />
The area of an arbitrary triangle <math>XYZ</math> is:<br />
<cmath>[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|</cmath><br />
<cmath>[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.</cmath><br />
<br />
To calculate <math>[DEF],</math> we wish to compute <math>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).</math> After a lot of computation, we obtain the following:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
Evaluating the denominator,<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)</cmath><br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.</cmath><br />
<br />
Since <math>x_P + y_P + z_P = 1</math> and <math>x_Py_P + y_Pz_P + z_Px_P = 0,</math> it follows that:<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.</cmath><br />
<br />
We thus conclude that:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
From this, it follows that <math>[DEF] = 2[ABC],</math> and we are done.<br />
<br />
<br />
==Solution 3==<br />
<center><asy><br />
import cse5;<br />
import graph;<br />
import olympiad;<br />
<br />
size(3inch);<br />
<br />
<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));<br />
pair P = (-1, -sqrt(11)+sqrt(3));<br />
path circle = Circle(O, 2sqrt(3));<br />
pair D = extension(A,P,B,C);<br />
pair E1 = extension(A,C,B,P);<br />
pair F=extension(A,B,C,P);<br />
draw(circle, black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
pair G = extension(O,D,F,E1);<br />
draw(O--G,dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
dot("O", O, SE);<br />
</asy></center><br />
We'll use coordinates and shoelace. Let the origin be the midpoint of <math>BC</math>. Let <math>AB=2</math>, and <math>BF = 2x</math>, then <math>F=(-x-1,-x\sqrt{3})</math>. Using the facts <math>\triangle{CBP} \sim \triangle{CFB}</math> and <math>\triangle{BCP} \sim \triangle{BEC}</math>, we have <math>BF * CE = BC^2</math>, so <math>CE = \frac{1}{2x}</math>, and <math> E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})</math>.<br />
<br />
The slope of <math>FE</math> is <br />
<cmath>k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}</cmath><br />
It is well-known that <math>\triangle{DFE}</math> is self-polar, so <math>FE</math> is the polar of <math>D</math>, i.e., <math>OD</math> is perpendicular to <math>FE</math>. Therefore, the slope of <math>OD</math> is <math>-\frac{1}{k}</math>. Since <math>O=(0,\frac{1}{\sqrt{3}})</math>, we get the x-coordinate of <math>D</math>, <math>x_D = \frac{k}{\sqrt{3}}</math>, i.e., <math>D = (\frac{k}{\sqrt{3}},0)</math>. Using shoelace,<br />
<cmath>2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})-<br />
(-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}} </cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2}<br />
{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 4\sqrt{3}</cmath><br />
So <math>[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]</math>. Q.E.D<br />
<br />
By Mathdummy.<br />
==Solution 4 Without the nasty computations ==<br />
Note that <math>\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60</math>. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. <br />
<br />
Let <math>BP = x</math> and <math>CP = y</math>. Then,<br />
From Law of Cosine, <math>BC^2 = x^2 + y^2 + xy</math>. <br />
<br />
From Ptolemy's theorem, <math>AP BC = x AC + y AB</math>, so <math>AP = x + y</math>.<br />
<br />
Lemma 1: In Triangle ABC with side lengths <math>a,b,c</math> and <math>\angle A =120^o</math>, the length of the angle bisector of <math>A</math> is <br />
<cmath> d = \frac{bc}{b+c}</cmath><br />
This can be easily proved with Stewart's and Law of Cosine.<br />
<br />
Using Lemma 1, we have <br />
<cmath>PD = \frac{xy}{x+y}</cmath><br />
<cmath> x = \frac{FP AP}{FP + AP}</cmath><br />
<cmath> y = \frac{EP AP}{EP + AP}</cmath><br />
Plug in <math>AP=x+y</math>, we get: <br />
<cmath> PD = \frac{xy}{x+y}</cmath> <br />
<cmath> FP = \frac{x(x+y)}{y}</cmath><br />
<cmath> EP = \frac{y(x+y)}{x}</cmath><br />
Then<br />
<cmath> [\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)</cmath><br />
<cmath>= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2}(x^2 + xy + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2} BC^2</cmath><br />
<cmath> = 2 [\triangle{ABC}]</cmath><br />
<br />
By Mathdummy.<br />
<br />
==Solution 5==<br />
<br />
<br />
Extend AP to meet FE at point X. Note that there is symmetry of triangle <math>\triangle{AFE}</math> across line AX. Therefore, if we can prove that the area of <math>\triangle{FDX}</math> is twice the area of <math>\triangle{BAD}</math>, then we will be done. For simplicity, let the side length of equilateral triangle <math>\triangle{ABC}</math> be <math>1</math>. \\ <br />
<br />
We can easily find the area of <math>\triangle{BAD}</math>. Since the area formula for an equilateral triangle is <math>\frac{s^2 \sqrt{3}}{4}</math> , where <math>s</math> is the side length, the area of <math>\triangle{ABC}</math> is <math>\frac{\sqrt{3}}{4}</math>. By symmetry, the area of <math>\triangle{BAD}</math> is exactly half the area of <math>\triangle{ABC}</math>. Thus the area of <math>\triangle{BAD} = \frac{\frac{\sqrt{3}}{4}}{2} = \boxed{\frac{\sqrt{3}}{8}}</math>.<br />
<br />
Now we just need to find the area of <math>\triangle{FDX}</math>. We have to find <math>FX</math> and <math>DX</math> for the area. Note that <math>FX = \frac{FE}{2}</math>. It looks by eye that <math>\triangle{AFE}</math> is equilateral. Let's try to prove it. \\<br />
<br />
<br />
<br />
<br />
\textbf{\textit{Lemma 1: <math>\triangle{AFE}</math> is equilateral.}} \\<br />
<br />
\textbf{\textit{Proof of Lemma 1:}} <br />
By symmetry, across lines <math>AX, FC, BE</math>, we arrive to some conclusions: <br />
\begin{center}<br />
<math>AF=AE</math><br />
<br />
<math>AF = FE</math><br />
<br />
<math>FE = AE</math>. <br />
\end{center}<br />
<br />
Substituting the first equation into the second, we see that <math>AF=AE=FE</math>, which is necessary and sufficient to prove that <math>\triangle{AFE}</math> is equilateral. <math>\square</math> <br />
\\<br />
<br />
<br />
Since <math>\triangle{AFE}</math> is equilateral, and so is <math>\triangle{ABC}</math>, we find that <math>\triangle{ABC} \sim \triangle{AFE}</math>. Because <math>AB = \frac{AF}{2}</math> by symmetry across <math>BE</math>, we know that the similarity ratio is <math>\frac{1}{2}</math>.<br />
Thus the side length of <math>\triangle{AFE}</math> is <math>2</math>, and <math>FX = \frac{FE}{2} = 1</math>.<br />
\\<br />
<br />
<br />
Now we need to find <math>DX</math>. We can do <math>AX-AD</math>. Well, that's easy! Since <math>\triangle{BAD}</math> is a 30-60-90 triangle by symmetry across line <math>AX</math>, and <math>BD = \frac{BC}{2} = \frac{1}{2}</math>, we know that <math>AD = \frac{\sqrt{3}}{2}</math>. To find <math>AX</math>, we do pretty much the same thing: <math>FAX</math> is a 30-60-90 triangle, and <math>FX = \frac{2}{2} = 1</math>, so <math>AX = \sqrt{3}</math>. We can now find <math>DX</math>: <br />
\begin{center}<br />
<math>DX = AX - AD = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}</math>. <br />
\end{center}<br />
<br />
<br />
\\<br />
<br />
Recalling why we were doing all this, we find the area of <math>\triangle{FDX}</math>. It is <math>\frac{FX \times DX}{2} = \frac{1 \times \frac{\sqrt{3}}{2}}{2} = \boxed{\frac{\sqrt{3}}{4}}</math>. \\<br />
<br />
<br />
\\<br />
<br />
Let's summarize: \\<br />
<br />
\begin{center}<br />
The area of <math>\triangle{ABC}</math> is <math>\boxed{\frac{\sqrt{3}}{8}}</math>\\<br />
<br />
The area of <math>\triangle{FDX}</math> is <math>\boxed{\frac{\sqrt{3}}{4}}</math>, \\<br />
<br />
which proves our initial claim. <br />
<br />
<br />
<br />
\end{center}<br />
<br />
<math>\square{}</math><br />
<br />
==See also==<br />
{{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=1534892017 USAJMO Problems/Problem 32021-05-10T18:42:26Z<p>Sakshamsethi: /* Solution 4 Without the nasty computations */</p>
<hr />
<div>==Problem==<br />
(<math>*</math>) Let <math>ABC</math> be an equilateral triangle and let <math>P</math> be a point on its circumcircle. Let lines <math>PA</math> and <math>BC</math> intersect at <math>D</math>; let lines <math>PB</math> and <math>CA</math> intersect at <math>E</math>; and let lines <math>PC</math> and <math>AB</math> intersect at <math>F</math>. Prove that the area of triangle <math>DEF</math> is twice that of triangle <math>ABC</math>.<br />
<br />
<asy><br />
size(3inch);<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br />
draw(Circle(O, 2sqrt(3)), black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
</asy><br />
<br />
==Solution (No Trig/Bash)==<br />
<br />
Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math><br />
<br />
<br />
<br />
Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> Now let <math>E'=DE\cap AF</math> and <math>F'=DF\cap AE</math>. Then by an area/concurrence theorem, we have that <math>\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,</math> or <math>\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math><br />
<br />
<br />
<br />
Manipulating these gives <math>\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.</math> Thus <math>\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,</math> and we are done.<br />
<br />
~cocohearts<br />
<br />
==Solution 1==<br />
WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 1: If <math>BF = k</math>, then <math>CE = \frac 1k</math>.<br />
This lemma results directly from the fact that <math>\triangle FBC</math> ~ <math>\triangle BCE</math>; <math>\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}</math>, or <math>CE = \frac{1}{BF}</math>. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 2: <math>[AEF] = (k+\frac 1k + 2)(X+Y)</math>.<br />
We see that <math>[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)</math>, as desired. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
Lemma 3: <math>\frac{X}{Y} = k</math>.<br />
We see that <br />
<cmath>\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.</cmath><br />
However, after some angle chasing and by the Law of Sines in <math>\triangle BCF</math>, we have <math>\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}</math>, or <math>k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}</math>, which implies the result. <br />
<br />
------------------------------------------------------------------------------------------------------<br />
<br />
By the area lemma, we have <math>[BDF] = kX</math> and <math>[CDF] = \frac{Y}{k}</math>. <br />
<br />
We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven.<br />
<br />
==Solution 2==<br />
<br />
We will use barycentric coordinates and vectors. Let <math>\vec{X}</math> be the position vector of a point <math>X.</math> The point <math>(x, y, z)</math> in barycentric coordinates denotes the point <math>x\vec{A} + y\vec{B} + z\vec{C}.</math> For all points in the plane of <math>\triangle{ABC},</math> we have <math>x + y + z = 1.</math> It is clear that <math>A = (1, 0, 0)</math>; <math>B = (0, 1, 0)</math>; and <math>C = (0, 0, 1).</math><br />
<br />
Define the point <math>P</math> as <math>P = \left(x_P, y_P, z_P\right).</math> The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with the condition <math>x_P + y_P + z_P = 1</math> inherent to barycentric coordinates, gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math><br />
<br />
We can write the equations of the following lines:<br />
<cmath>BC: x = 0</cmath><br />
<cmath>CA: y = 0</cmath><br />
<cmath>AB: z = 0</cmath><br />
<cmath>PA: \frac{y}{y_P} = \frac{z}{z_P}</cmath><br />
<cmath>PB: \frac{x}{x_P} = \frac{z}{z_P}</cmath><br />
<cmath>PC: \frac{x}{x_P} = \frac{y}{y_P}.</cmath><br />
<br />
We can then solve for the points <math>D, E, F</math>:<br />
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath><br />
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath><br />
<cmath>F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).</cmath><br />
<br />
The area of an arbitrary triangle <math>XYZ</math> is:<br />
<cmath>[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|</cmath><br />
<cmath>[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.</cmath><br />
<br />
To calculate <math>[DEF],</math> we wish to compute <math>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).</math> After a lot of computation, we obtain the following:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
Evaluating the denominator,<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)</cmath><br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.</cmath><br />
<br />
Since <math>x_P + y_P + z_P = 1</math> and <math>x_Py_P + y_Pz_P + z_Px_P = 0,</math> it follows that:<br />
<cmath>(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.</cmath><br />
<br />
We thus conclude that:<br />
<cmath>(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].</cmath><br />
<br />
From this, it follows that <math>[DEF] = 2[ABC],</math> and we are done.<br />
<br />
<br />
==Solution 3==<br />
<center><asy><br />
import cse5;<br />
import graph;<br />
import olympiad;<br />
<br />
size(3inch);<br />
<br />
<br />
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));<br />
pair P = (-1, -sqrt(11)+sqrt(3));<br />
path circle = Circle(O, 2sqrt(3));<br />
pair D = extension(A,P,B,C);<br />
pair E1 = extension(A,C,B,P);<br />
pair F=extension(A,B,C,P);<br />
draw(circle, black);<br />
draw(A--B--C--cycle);<br />
draw(B--E1--C);<br />
draw(C--F--B);<br />
draw(A--P);<br />
draw(D--E1--F--cycle, dashed);<br />
pair G = extension(O,D,F,E1);<br />
draw(O--G,dashed);<br />
label("A", A, N);<br />
label("B", B, W);<br />
label("C", C, E);<br />
label("P", P, S);<br />
label("D", D, NW);<br />
label("E", E1, SE);<br />
label("F", F, SW);<br />
dot("O", O, SE);<br />
</asy></center><br />
We'll use coordinates and shoelace. Let the origin be the midpoint of <math>BC</math>. Let <math>AB=2</math>, and <math>BF = 2x</math>, then <math>F=(-x-1,-x\sqrt{3})</math>. Using the facts <math>\triangle{CBP} \sim \triangle{CFB}</math> and <math>\triangle{BCP} \sim \triangle{BEC}</math>, we have <math>BF * CE = BC^2</math>, so <math>CE = \frac{1}{2x}</math>, and <math> E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})</math>.<br />
<br />
The slope of <math>FE</math> is <br />
<cmath>k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}</cmath><br />
It is well-known that <math>\triangle{DFE}</math> is self-polar, so <math>FE</math> is the polar of <math>D</math>, i.e., <math>OD</math> is perpendicular to <math>FE</math>. Therefore, the slope of <math>OD</math> is <math>-\frac{1}{k}</math>. Since <math>O=(0,\frac{1}{\sqrt{3}})</math>, we get the x-coordinate of <math>D</math>, <math>x_D = \frac{k}{\sqrt{3}}</math>, i.e., <math>D = (\frac{k}{\sqrt{3}},0)</math>. Using shoelace,<br />
<cmath>2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})-<br />
(-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}} </cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2}<br />
{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})</cmath><br />
<cmath> = 4\sqrt{3}</cmath><br />
So <math>[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]</math>. Q.E.D<br />
<br />
By Mathdummy.<br />
==Solution 4 Without the nasty computations ==<br />
Note that <math>\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60</math>. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. <br />
<br />
Let <math>BP = x</math> and <math>CP = y</math>. Then,<br />
From Law of Cosine, <math>BC^2 = x^2 + y^2 + xy</math>. <br />
<br />
From Ptolemy's theorem, <math>AP BC = x AC + y AB</math>, so <math>AP = x + y</math>.<br />
<br />
Lemma 1: In Triangle ABC with side lengths <math>a,b,c</math> and <math>\angle A =120^o</math>, the length of the angle bisector of <math>A</math> is <br />
<cmath> d = \frac{bc}{b+c}</cmath><br />
This can be easily proved with Stewart's and Law of Cosine.<br />
<br />
Using Lemma 1, we have <br />
<cmath>PD = \frac{xy}{x+y}</cmath><br />
<cmath> x = \frac{FP AP}{FP + AP}</cmath><br />
<cmath> y = \frac{EP AP}{EP + AP}</cmath><br />
Plug in <math>AP=x+y</math>, we get: <br />
<cmath> PD = \frac{xy}{x+y}</cmath> <br />
<cmath> FP = \frac{x(x+y)}{y}</cmath><br />
<cmath> EP = \frac{y(x+y)}{x}</cmath><br />
Then<br />
<cmath> [\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)</cmath><br />
<cmath>= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2}(x^2 + xy + y^2) </cmath><br />
<cmath> = \frac{\sqrt{3}}{2} BC^2</cmath><br />
<cmath> = 2 [\triangle{ABC}]</cmath><br />
<br />
By Mathdummy.<br />
<br />
==Solution 5==<br />
<br />
==See also==<br />
{{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_19&diff=1530472012 AMC 10A Problems/Problem 192021-05-01T20:26:29Z<p>Sakshamsethi: /* Solution 2 (Modular Arithmetic) */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #13]] and [[2012 AMC 10A Problems|2012 AMC 10A #19]]}}<br />
<br />
==Problem 19==<br />
<br />
Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?<br />
<br />
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60 </math><br />
<br />
==Solution==<br />
Let Paula work at a rate of <math>p</math>, the two helpers work at a combined rate of <math>h</math>, and the time it takes to eat lunch be <math>L</math>, where <math>p</math> and <math>h</math> are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:<br />
<br />
<cmath>(8-L)(p+h)=50</cmath><br />
<br />
<cmath>(6.2-L)h=24</cmath><br />
<br />
<cmath>(11.2-L)p=26</cmath><br />
<br />
With three equations and three variables, we need to find the value of <math>L</math>.<br />
Adding the second and third equations together gives us <math>6.2h+11.2p-L(p+h)=50</math>. Subtracting the first equation from this new one gives us <math>-1.8h+3.2p=0</math>, so we get <math>h=\frac{16}{9}p</math>. <br />
Plugging into the second equation:<br />
<br />
<cmath>(6.2-L)\frac{16}{9}p=24</cmath><br />
<cmath>(6.2-L)p=\frac{27}{2}</cmath><br />
<br />
We can then subtract this from the third equation:<br />
<br />
<cmath>5p=26-\frac{27}{2}</cmath><br />
<cmath>p=\frac{5}{2}</cmath><br />
Plugging <math>p</math> into our third equation gives: <cmath>L=\frac{4}{5}</cmath><br />
<br />
Converting <math>L</math> from hours to minutes gives us <math>L=48</math> minutes, which is <math>\boxed{\textbf{(D)}\ 48}</math>.<br />
<br />
==Solution 2 (Modular Arithmetic)==<br />
Because Paula worked from <cmath>8:00 \text{A.M.}</cmath> to <cmath>7:12 \text{P.M.}</cmath>, she worked for 11 hours and 12 minutes = 672 minutes. Since there is <math>100-50-24=26</math>% of the house left, we get the equation <math>26a=672</math>. Because <math>672</math> is <math>22</math> mod <math>26</math>, looking at out answer choices, the only answer that is <math>22</math> <math>\text{mod}</math> <math>26</math> is <math>48</math>. So the answer is <math>\boxed{\textbf{(D)}\ 48}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2012|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_19&diff=1530462012 AMC 10A Problems/Problem 192021-05-01T20:25:59Z<p>Sakshamsethi: /* Solution 2 (Easy process of elimination) */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #13]] and [[2012 AMC 10A Problems|2012 AMC 10A #19]]}}<br />
<br />
==Problem 19==<br />
<br />
Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?<br />
<br />
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60 </math><br />
<br />
==Solution==<br />
Let Paula work at a rate of <math>p</math>, the two helpers work at a combined rate of <math>h</math>, and the time it takes to eat lunch be <math>L</math>, where <math>p</math> and <math>h</math> are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:<br />
<br />
<cmath>(8-L)(p+h)=50</cmath><br />
<br />
<cmath>(6.2-L)h=24</cmath><br />
<br />
<cmath>(11.2-L)p=26</cmath><br />
<br />
With three equations and three variables, we need to find the value of <math>L</math>.<br />
Adding the second and third equations together gives us <math>6.2h+11.2p-L(p+h)=50</math>. Subtracting the first equation from this new one gives us <math>-1.8h+3.2p=0</math>, so we get <math>h=\frac{16}{9}p</math>. <br />
Plugging into the second equation:<br />
<br />
<cmath>(6.2-L)\frac{16}{9}p=24</cmath><br />
<cmath>(6.2-L)p=\frac{27}{2}</cmath><br />
<br />
We can then subtract this from the third equation:<br />
<br />
<cmath>5p=26-\frac{27}{2}</cmath><br />
<cmath>p=\frac{5}{2}</cmath><br />
Plugging <math>p</math> into our third equation gives: <cmath>L=\frac{4}{5}</cmath><br />
<br />
Converting <math>L</math> from hours to minutes gives us <math>L=48</math> minutes, which is <math>\boxed{\textbf{(D)}\ 48}</math>.<br />
<br />
==Solution 2 (Modular Arithmetic)==<br />
Because Paula worked from <cmath>8:00 A.M.</cmath> to <cmath>7:12 P.M.</cmath>, she worked for 11 hours and 12 minutes = 672 minutes. Since there is <math>100-50-24=26</math>% of the house left, we get the equation <math>26a=672</math>. Because <math>672</math> is <math>22</math> mod <math>26</math>, looking at out answer choices, the only answer that is <math>22</math> <math>\text{mod}</math> <math>26</math> is <math>48</math>. So the answer is <math>\boxed{\textbf{(D)}\ 48}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2012|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_25&diff=1517972017 AMC 10A Problems/Problem 252021-04-17T21:07:47Z<p>Sakshamsethi: /* Solution 5: A Slightly Adjusted Version of Solution 2 */</p>
<hr />
<div>==Problem==<br />
How many integers between <math>100</math> and <math>999</math>, inclusive, have the property that some permutation of its digits is a multiple of <math>11</math> between <math>100</math> and <math>999?</math> For example, both <math>121</math> and <math>211</math> have this property.<br />
<br />
<math> \mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486</math><br />
<br />
==Solution 1==<br />
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.<br />
<br />
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.<br />
<br />
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each. <br />
<br />
There are 110, 220, 330 ... 990, yielding 9 extra permutations<br />
<br />
Also, there are 209, 308, 407...902, yielding 8 more permutations.<br />
<br />
Now, just subtract these 17 from the total (243), getting 226. <math>\boxed{\textbf{(A) } 226}</math><br />
<br />
*Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer. <br />
<br />
==Solution 3==<br />
<br />
We note that we only have to consider multiples of <math>11</math> and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of <math>11</math> has:<br />
<br />
<math>\textbf{Case 1:}</math> All three digits are the same. <br />
By inspection, we find that there are no multiples of <math>11</math> here.<br />
<br />
<math>\textbf{Case 2:}</math> Two of the digits are the same, and the third is different.<br />
<br />
<math>\textbf{Case 2a:}</math><br />
There are <math>8</math> multiples of <math>11</math> without a zero that have this property:<br />
<math>121</math>, <math>242</math>, <math>363</math>, <math>484</math>, <math>616</math>, <math>737</math>, <math>858</math>, <math>979</math>.<br />
Each contributes <math>3</math> valid permutations, so there are <math>8 \cdot 3 = 24</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 2b:}</math><br />
There are <math>9</math> multiples of <math>11</math> with a zero that have this property:<br />
<math>110</math>, <math>220</math>, <math>330</math>, <math>440</math>, <math>550</math>, <math>660</math>, <math>770</math>, <math>880</math>, <math>990</math>.<br />
Each one contributes <math>2</math> valid permutations (the first digit can't be zero), so there are <math>9 \cdot 2 = 18</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 3:}</math> All the digits are different.<br />
Since there are <math>\frac{990-110}{11}+1 = 81</math> multiples of <math>11</math> between <math>100</math> and <math>999</math>, there are <math>81-8-9 = 64</math> multiples of <math>11</math> remaining in this case. However, <math>8</math> of them contain a zero, namely <math>209</math>, <math>308</math>, <math>407</math>, <math>506</math>, <math>605</math>, <math>704</math>, <math>803</math>, and <math>902</math>. Each of those multiples of <math>11</math> contributes <math>2 \cdot 2=4</math> valid permutations, but we overcounted by a factor of <math>2</math>; every permutation of <math>209</math>, for example, is also a permutation of <math>902</math>. Therefore, there are <math>8 \cdot 4 / 2 = 16</math>. Therefore, there are <math>64-8=56</math> remaining multiples of <math>11</math> without a <math>0</math> in this case. Each one contributes <math>3! = 6</math> valid permutations, but once again, we overcounted by a factor of <math>2</math> (note that if a number ABC is a multiple of <math>11</math>, then so is CBA). Therefore, there are <math>56 \cdot 6 / 2 = 168</math> valid permutations in this subcase.<br />
<br />
Adding up all the permutations from all the cases, we have <math>24+18+16+168 = \boxed{\textbf{(A) } 226}</math>.<br />
<br />
==Solution 4 ==<br />
<br />
We can first overcount and then subtract.<br />
We know that there are <math>81</math> multiples of <math>11</math>.<br />
<br />
We can then multiply by <math>6</math> for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)<br />
<br />
Now divide by <math>2</math>, because if a number <math>abc</math> with digits <math>a</math>, <math>b</math>, and <math>c</math> is a multiple of <math>11</math>, then <math>cba</math> is also a multiple of <math>11</math> so we have counted the same permutations twice. <br />
<br />
Basically, each multiple of <math>11</math> has its own <math>3</math> permutations (say <math>abc</math> has <math>abc</math> <math>acb</math> and <math>bac</math> whereas <math>cba</math> has <math>cba</math> <math>cab</math> and <math>bca</math>). We know that each multiple of <math>11</math> has at least <math>3</math> permutations because it cannot have <math>3</math> repeating digits.<br />
<br />
Hence we have <math>243</math> permutations without subtracting for overcounting.<br />
Now note that we overcounted cases in which we have <math>0</math>'s at the start of each number. So, in theory, we could just answer <math>A</math> and then move on.<br />
<br />
If we want to solve it, then we continue.<br />
<br />
We overcounted cases where the middle digit of the number is <math>0</math> and the last digit is <math>0</math>.<br />
<br />
Note that we assigned each multiple of <math>11</math> three permutations.<br />
<br />
The last digit is <math>0</math> gives <math>9</math> possibilities where we overcounted by <math>1</math> permutation for each of <math>110, 220, ... , 990</math>.<br />
<br />
The middle digit is <math>0</math> gives <math>8</math> possibilities where we overcount by <math>1</math>.<br />
<math>605, 704, 803, 902</math> and <math>506, 407, 308, 209</math><br />
<br />
Subtracting <math>17</math> gives <math>\boxed{\textbf{(A) } 226}</math>.<br />
<br />
Now, we may ask if there is further overlap (i.e if two of <math>abc</math> and <math>bac</math> and <math>acb</math> were multiples of <math>11</math>). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod <math>11</math> and adding, we get that <math>2a</math>, <math>2b</math>, or <math>2c</math> is congruent to <math>0\ (mod\ 11)</math>. Since <math>a, b, c</math> are digits, this can never happen as none of them can equal <math>11</math> and they can't equal <math>0</math> as they are the leading digit of a three-digit number in each of the cases.<br />
<br />
== Solution 6 (1 but quicker) ==<br />
<br />
The smallest multiple of <math>11</math> above <math>100</math> is <math>110 = 11 \cdot 10</math>, while the largest multiple of <math>11</math> less than <math>999</math> is <math>990 = 11 \cdot 90</math>. This means there are <math>90 - 10 + 1 = 81</math> multiples of <math>11</math> between <math>100</math> and <math>999</math>.<br />
<br />
As there are <math>3</math> permutations for each multiple, we have <math>81 * 3 = 243</math>. However, we have overcounted, as numbers like <math>099</math> shouldn't be counted. Looking at the answer choices, we notice there is only one that is less than <math>243</math>, and so we have our answer as <math>\boxed{\textbf{(A) } 226}</math>.<br />
<br />
-¢<br />
<br />
==Video Solution==<br />
Two different variations on solving it.<br />
https://youtu.be/z5KNZEwmrWM<br />
<br />
https://youtu.be/MBcHwu30MX4<br />
-Video Solution by Richard Rusczyk<br />
<br />
https://youtu.be/Ly69GHOq9Yw<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_25&diff=1517962017 AMC 10A Problems/Problem 252021-04-17T21:07:24Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
How many integers between <math>100</math> and <math>999</math>, inclusive, have the property that some permutation of its digits is a multiple of <math>11</math> between <math>100</math> and <math>999?</math> For example, both <math>121</math> and <math>211</math> have this property.<br />
<br />
<math> \mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486</math><br />
<br />
==Solution 1==<br />
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.<br />
<br />
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.<br />
<br />
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each. <br />
<br />
There are 110, 220, 330 ... 990, yielding 9 extra permutations<br />
<br />
Also, there are 209, 308, 407...902, yielding 8 more permutations.<br />
<br />
Now, just subtract these 17 from the total (243), getting 226. <math>\boxed{\textbf{(A) } 226}</math><br />
<br />
*Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer. <br />
<br />
==Solution 3==<br />
<br />
We note that we only have to consider multiples of <math>11</math> and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of <math>11</math> has:<br />
<br />
<math>\textbf{Case 1:}</math> All three digits are the same. <br />
By inspection, we find that there are no multiples of <math>11</math> here.<br />
<br />
<math>\textbf{Case 2:}</math> Two of the digits are the same, and the third is different.<br />
<br />
<math>\textbf{Case 2a:}</math><br />
There are <math>8</math> multiples of <math>11</math> without a zero that have this property:<br />
<math>121</math>, <math>242</math>, <math>363</math>, <math>484</math>, <math>616</math>, <math>737</math>, <math>858</math>, <math>979</math>.<br />
Each contributes <math>3</math> valid permutations, so there are <math>8 \cdot 3 = 24</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 2b:}</math><br />
There are <math>9</math> multiples of <math>11</math> with a zero that have this property:<br />
<math>110</math>, <math>220</math>, <math>330</math>, <math>440</math>, <math>550</math>, <math>660</math>, <math>770</math>, <math>880</math>, <math>990</math>.<br />
Each one contributes <math>2</math> valid permutations (the first digit can't be zero), so there are <math>9 \cdot 2 = 18</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 3:}</math> All the digits are different.<br />
Since there are <math>\frac{990-110}{11}+1 = 81</math> multiples of <math>11</math> between <math>100</math> and <math>999</math>, there are <math>81-8-9 = 64</math> multiples of <math>11</math> remaining in this case. However, <math>8</math> of them contain a zero, namely <math>209</math>, <math>308</math>, <math>407</math>, <math>506</math>, <math>605</math>, <math>704</math>, <math>803</math>, and <math>902</math>. Each of those multiples of <math>11</math> contributes <math>2 \cdot 2=4</math> valid permutations, but we overcounted by a factor of <math>2</math>; every permutation of <math>209</math>, for example, is also a permutation of <math>902</math>. Therefore, there are <math>8 \cdot 4 / 2 = 16</math>. Therefore, there are <math>64-8=56</math> remaining multiples of <math>11</math> without a <math>0</math> in this case. Each one contributes <math>3! = 6</math> valid permutations, but once again, we overcounted by a factor of <math>2</math> (note that if a number ABC is a multiple of <math>11</math>, then so is CBA). Therefore, there are <math>56 \cdot 6 / 2 = 168</math> valid permutations in this subcase.<br />
<br />
Adding up all the permutations from all the cases, we have <math>24+18+16+168 = \boxed{\textbf{(A) } 226}</math>.<br />
<br />
==Solution 4 ==<br />
<br />
We can first overcount and then subtract.<br />
We know that there are <math>81</math> multiples of <math>11</math>.<br />
<br />
We can then multiply by <math>6</math> for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)<br />
<br />
Now divide by <math>2</math>, because if a number <math>abc</math> with digits <math>a</math>, <math>b</math>, and <math>c</math> is a multiple of <math>11</math>, then <math>cba</math> is also a multiple of <math>11</math> so we have counted the same permutations twice. <br />
<br />
Basically, each multiple of <math>11</math> has its own <math>3</math> permutations (say <math>abc</math> has <math>abc</math> <math>acb</math> and <math>bac</math> whereas <math>cba</math> has <math>cba</math> <math>cab</math> and <math>bca</math>). We know that each multiple of <math>11</math> has at least <math>3</math> permutations because it cannot have <math>3</math> repeating digits.<br />
<br />
Hence we have <math>243</math> permutations without subtracting for overcounting.<br />
Now note that we overcounted cases in which we have <math>0</math>'s at the start of each number. So, in theory, we could just answer <math>A</math> and then move on.<br />
<br />
If we want to solve it, then we continue.<br />
<br />
We overcounted cases where the middle digit of the number is <math>0</math> and the last digit is <math>0</math>.<br />
<br />
Note that we assigned each multiple of <math>11</math> three permutations.<br />
<br />
The last digit is <math>0</math> gives <math>9</math> possibilities where we overcounted by <math>1</math> permutation for each of <math>110, 220, ... , 990</math>.<br />
<br />
The middle digit is <math>0</math> gives <math>8</math> possibilities where we overcount by <math>1</math>.<br />
<math>605, 704, 803, 902</math> and <math>506, 407, 308, 209</math><br />
<br />
Subtracting <math>17</math> gives <math>\boxed{\textbf{(A) } 226}</math>.<br />
<br />
Now, we may ask if there is further overlap (i.e if two of <math>abc</math> and <math>bac</math> and <math>acb</math> were multiples of <math>11</math>). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod <math>11</math> and adding, we get that <math>2a</math>, <math>2b</math>, or <math>2c</math> is congruent to <math>0\ (mod\ 11)</math>. Since <math>a, b, c</math> are digits, this can never happen as none of them can equal <math>11</math> and they can't equal <math>0</math> as they are the leading digit of a three-digit number in each of the cases.<br />
<br />
== Solution 5: A Slightly Adjusted Version of Solution 2 ==<br />
<br />
<br />
<math>\textbf{WARNING:}</math> If you do not feel comfortable looking at a massive amount of casework, please skip the solution.<br />
<br />
<br />
Recalling the divisibility rule for <math>11</math>, if we have a number <math>ABC</math> where <math>A</math>, <math>B</math>, <math>C</math> are digits, then <math>11\mid -A+B-C</math>.<br />
<br />
Notice that for any three-digit positive integer <math>ABC</math>, <math>-A+B-C<11</math>, thus we have 2 possibilities: <math>-A+B-C=0</math> and <math>-A+B-C=-11</math>.<br />
<br />
<math>\textbf{Case 1:}</math> <math>-A+B-C=0\Longrightarrow A+C=B</math><br />
<br />
Subcase <math>a</math>: <math>A\neq B\neq C\neq0</math><br />
<br />
We have these values for <math>A+C=B</math><br />
<cmath>1+2=3,1+3=4,1+4=5,...,1+8=9</cmath><br />
<cmath>2+3=5,2+4=6,...,2+7=9</cmath><br />
<cmath>3+4=7,3+5=8,3+6=9</cmath><br />
<cmath>4+5=9</cmath><br />
From which we get <math>7+5+3+1=16</math> triples <math>(A,B,C)</math>. Counting every permutation, we have <math>16\cdot3!=96</math> possibilities.<br />
<br />
Subcase <math>b</math>: <math>A=C</math>, <math>A,B,C\neq0</math><br />
<br />
We have<br />
<cmath>1+1=2,2+2=4,3+3=6,4+4=8</cmath><br />
From which we get <math>4\cdot3=12</math> possibilities.<br />
<br />
Subcase <math>c</math>: <math>A=B,C=0</math><br />
<br />
We have<br />
<cmath>1+0=1,2+0=2,...,9+0=9</cmath><br />
Since <math>0</math> can't be the hundreds digit, from here we get <math>9\cdot2=18</math> possibilities. Summing up case <math>1</math>, we have <math>96+12+18=126</math> possibilities.<br />
<br />
<math>\textbf{Case 2:}</math> <math>-A+B-C=-11\Longrightarrow A+C-B=11</math><br />
<br />
Subcase <math>a</math>: <math>A\neq B\neq C\neq0</math><br />
<br />
We have these values for <math>A+C-B=11</math><br />
<cmath>9+8-6=11,9+7-5=11,...9+3-1=11</cmath><br />
<cmath>8+7-4=11,8+6-3=11,8+5-2=11,8+4-1</cmath><br />
<cmath>...</cmath><br />
From which we get <math>(6+4+2)\cdot3!=72</math> possibilities.<br />
<br />
Subcase <math>b</math>: <math>A=C</math>, <math>A,B,C\neq0</math><br />
<br />
We have <br />
<cmath>9+9=7,8+8-5,7+7-3=11,6+6-1=11</cmath><br />
From which we get <math>4\cdot3=12</math> possibilities.<br />
<br />
Subcase <math>c</math>: <math>B=0\Longrightarrow A+C+11</math><br />
<br />
We have<br />
<cmath>9+2=8+3=7+4=6+5=11</cmath><br />
From which we get <math>2\cdot2\cdot4=16</math> possibilities. Summing up case <math>2</math>, we have <math>72+12+16=100</math> possibilities.<br />
<br />
Adding the <math>2</math> cases, we get a total of <math>126+100=226</math> possibilities. <math>\boxed{\mathrm{(A)}}</math><br />
<br />
~ Nafer<br />
== Solution 6 (1 but quicker) ==<br />
<br />
The smallest multiple of <math>11</math> above <math>100</math> is <math>110 = 11 \cdot 10</math>, while the largest multiple of <math>11</math> less than <math>999</math> is <math>990 = 11 \cdot 90</math>. This means there are <math>90 - 10 + 1 = 81</math> multiples of <math>11</math> between <math>100</math> and <math>999</math>.<br />
<br />
As there are <math>3</math> permutations for each multiple, we have <math>81 * 3 = 243</math>. However, we have overcounted, as numbers like <math>099</math> shouldn't be counted. Looking at the answer choices, we notice there is only one that is less than <math>243</math>, and so we have our answer as <math>\boxed{\textbf{(A) } 226}</math>.<br />
<br />
-¢<br />
<br />
==Video Solution==<br />
Two different variations on solving it.<br />
https://youtu.be/z5KNZEwmrWM<br />
<br />
https://youtu.be/MBcHwu30MX4<br />
-Video Solution by Richard Rusczyk<br />
<br />
https://youtu.be/Ly69GHOq9Yw<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=Cube_(geometry)&diff=151304Cube (geometry)2021-04-10T16:07:06Z<p>Sakshamsethi: </p>
<hr />
<div>A '''cube''', or regular '''hexahedron''', is a solid composed of six [[Square (geometry) | square]] [[face]]s. A cube is [[Platonic solid #Duality | dual]] to the regular [[octahedron]] and has [[octahedral symmetry]]. A cube is a [[Platonic solid]]. All edges of cubes are equal to each other.<br />
<br />
==Formulas==<br />
A cube with [[edge]]-[[length]] <math>s</math> has:<br />
* Four space [[diagonal]]s of same lengths <math>s\sqrt{3}</math>(<math>\sqrt{s^2+s^2+s^2}=\sqrt{3s^2}=s\sqrt{3}</math>)<br />
* [[Surface area]] of <math>6s^2</math>. (6 sides of areas <math>s \cdot s</math>.)<br />
* [[Volume]] <math>s^3</math>(<math>s \cdot s \cdot s</math>)<br />
* A [[circumscribe]]d [[sphere]] of [[radius]] <math>\frac{s\sqrt{3}}{2}</math><br />
* An [[inscribe]]d sphere of radius <math>\frac{s}{2}</math><br />
* A sphere [[tangent]] to all of its edges of radius <math>\frac{s\sqrt{2}}{2}</math><br />
* A regular tetrahedron can fit in exactly two ways inside a cube<br />
<br />
==See also==<br />
* [[Hexahedron]]<br />
* [[Prism]]<br />
<br />
[[Category:Platonic solids]]<br />
{{stub}}<br />
[[Category:Geometry]]</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_11&diff=1362392014 AMC 10B Problems/Problem 112020-10-31T13:44:13Z<p>Sakshamsethi: /* Solution 2 (a bit easier) */</p>
<hr />
<div>==Problem==<br />
For the consumer, a single discount of <math>n\%</math> is more advantageous than any of the following discounts:<br />
<br />
(1) two successive <math>15\%</math> discounts<br />
<br />
(2) three successive <math>10\%</math> discounts<br />
<br />
(3) a <math>25\%</math> discount followed by a <math>5\%</math> discount<br />
<br />
What is the smallest possible positive integer value of <math>n</math>?<br />
<br />
<math> \textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33 </math><br />
<br />
==Solution 1==<br />
Let the original price be <math>x</math>. Then, for option <math>1</math>, the discounted price is <math>(1-.15)(1-.15)x = .7225x</math>. For option <math>2</math>, the discounted price is <math>(1-.1)(1-.1)(1-.1)x = .729x</math>. Finally, for option <math>3</math>, the discounted price is <math>(1-.25)(1-.05) = .7125x</math>. Therefore, <math>n</math> must be greater than <math>\max(x - .7225x, x-.729x, x-.7125x)</math>. It follows <math>n</math> must be greater than <math>.2875</math>. We multiply this by <math>100</math> to get the percent value, and then round up because <math>n</math> is the smallest integer that provides a greater discount than <math>28.75</math>, leaving us with the answer of <math>\boxed{\textbf{(C) } 29}</math><br />
<br />
<br />
==Solution 2 (a bit easier)==<br />
Assume the original price was <math>100</math> dollars. Thus, after a discount of <math>n\%</math>, the price will be <math>100-n</math> dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: <math>100-n</math>, <math>\frac{289}{4}</math>, and <math>\frac{9^3}{10}</math>. Simplify these to get <math>100-n</math>, <math>72</math> (rounded up), <math>72.9</math>, and <math>71</math> (rounded down). To have the greatest discount, we need the least price, which is <math>71</math> dollars. Now we get the original fractional value of this, and the discount of this is <math>100-\frac{285}{4}</math>, and we round this down to <math>28</math>. Now, it's pretty easy. The integer value greater than this is <math>\boxed{\textbf{(C) } 29}</math><br />
<br />
~sakshamsethi<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_11&diff=1362382014 AMC 10B Problems/Problem 112020-10-31T13:43:54Z<p>Sakshamsethi: /* Solution */</p>
<hr />
<div>==Problem==<br />
For the consumer, a single discount of <math>n\%</math> is more advantageous than any of the following discounts:<br />
<br />
(1) two successive <math>15\%</math> discounts<br />
<br />
(2) three successive <math>10\%</math> discounts<br />
<br />
(3) a <math>25\%</math> discount followed by a <math>5\%</math> discount<br />
<br />
What is the smallest possible positive integer value of <math>n</math>?<br />
<br />
<math> \textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33 </math><br />
<br />
==Solution 1==<br />
Let the original price be <math>x</math>. Then, for option <math>1</math>, the discounted price is <math>(1-.15)(1-.15)x = .7225x</math>. For option <math>2</math>, the discounted price is <math>(1-.1)(1-.1)(1-.1)x = .729x</math>. Finally, for option <math>3</math>, the discounted price is <math>(1-.25)(1-.05) = .7125x</math>. Therefore, <math>n</math> must be greater than <math>\max(x - .7225x, x-.729x, x-.7125x)</math>. It follows <math>n</math> must be greater than <math>.2875</math>. We multiply this by <math>100</math> to get the percent value, and then round up because <math>n</math> is the smallest integer that provides a greater discount than <math>28.75</math>, leaving us with the answer of <math>\boxed{\textbf{(C) } 29}</math><br />
<br />
<br />
==Solution 2 (a bit easier)==<br />
Assume the original price was <math>100</math> dollars. Thus, after a discount of <math>n\%</math>, the price will be <math>100-n</math> dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: <math>100-n</math>, <math>\frac{289}{4}</math>, and <math>\frac{9^3}{10}</math>. Simplify these to get <math>100-n</math>, <math>72</math> (rounded up), <math>72.9</math>, and <math>71</math> (rounded down). To have the greatest discount, we need the least price, which is <math>71</math> dollars. Now we get the original fractional value of this, and the discount of this is <math>100-\frac{285}{4}</math>, and we round this down to <math>28</math>. Now, it's pretty easy. The integer value greater than this is <math>\boxed{\textbf{(C) } 29}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_8&diff=1354282010 AMC 12A Problems/Problem 82020-10-19T23:07:10Z<p>Sakshamsethi: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?<br />
<br />
<math>\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ</math><br />
<br />
== Solution ==<br />
<br />
<center>[[File:AMC 2010 12A Problem 8.png]]</center><br />
<br />
<br />
Let <math>\angle BAE = \angle ACD = x</math>.<br />
<br />
<cmath>\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\<br />
\angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\<br />
\angle EAC &= 60^\circ - x\\<br />
\angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath><br />
<br />
Since <math>\frac{AC}{AB} = \frac{1}{2}</math> and the angle between the hypotenuse and the shorter side is <math>60^\circ</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>.<br />
<br />
== Solution 2(Trig and Angle Chasing) ==<br />
Let <math>AB=2a, AC=a</math>. Let <math>\angle BAE=\angle ACD=x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle FCE=60</math>, so <math>\angle ACB=60+x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle CFE=60</math>. Angles <math>AFD</math> and <math>CFE</math> are vertical, so <math>\angle AFD=60</math>. By triangle <math>ADF</math>, we have <math>\angle ADF=120-x</math>, and because of line <math>AB</math>, we have <math>\angle BDC=60+x</math>. Because Of line <math>BC</math>, we have <math>\angle AEB=120</math>, and by line <math>CD</math>, we have <math>\angle DFE=120</math>. By quadrilateral <math>BDFE</math>, we have <math>\angle ABC=60-x</math>.<br />
<br />
By the Law of Sines, we have <math>\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies \sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)</math>. By the sine addition formula(which states <math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)</math> by the way), we have <math>2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)</math>. Because cosine is an even function, and sine is an odd function, we have <math>2\sin(60)\cos(x)-2\cos(60)\sin(x)=\sin(60)\cos(x)+\cos(60)\sin(x) \implies \sin(60)\cos(x)=3\cos(60)\sin(x)</math>. We know that <math>\sin(60)=\frac{\sqrt{3}}{2}</math>, and <math>\cos(60)=\frac{1}{2}</math>, hence <math>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</math>. The only value of <math>x</math> that satisfies <math>60+x<180</math>(because <math>60+x</math> is an angle of the triangle) is <math>x=30^{\circ}</math>. We seek to find <math>\angle ACB</math>, which as we found before is <math>60+x</math>, which is <math>90</math>. The answer is <math>90, \text{or} \textbf{(C)}</math><br />
<br />
-vsamc<br />
<br />
==Video Solution==<br />
https://youtu.be/kU70k1-ONgM?t=785<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=1341372019 AMC 8 Problems/Problem 252020-09-27T00:49:59Z<p>Sakshamsethi: /* Solution 5 (EASIEST AND FASTEST SOLUTION) */</p>
<hr />
<div>==Problem 25==<br />
Alice has <math>24</math> apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?<br />
<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math><br />
<br />
==Solution 1==<br />
We use [[stars and bars]]. Let Alice get <math>k</math> apples, let Becky get <math>r</math> apples, let Chris get <math>y</math> apples.<br />
<cmath>\implies k + r + y = 24</cmath>We can manipulate this into an equation which can be solved using stars and bars.<br />
<br />
All of them get at least <math>2</math> apples, so we can subtract <math>2</math> from <math>k</math>, <math>2</math> from <math>r</math>, and <math>2</math> from <math>y</math>.<br />
<cmath>\implies (k - 2) + (r - 2) + (y - 2) = 18</cmath>Let <math>k' = k - 2</math>, let <math>r' = r - 2</math>, let <math>y' = y - 2</math>.<br />
<cmath>\implies k' + r' + y' = 18</cmath>We can allow either of them to equal to <math>0</math>, hence this can be solved by stars and bars.<br />
<br />
<br />
By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{190}</math>.<br />
<br />
==Solution 2==<br />
[[Without loss of generality]], let's assume that Alice has <math>2</math> apples. There are <math>19</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>3</math> apples, there are <math>18</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>4</math> apples, there are <math>17</math> ways to split the rest. So the total number of ways to split <math>24</math> apples between the three friends is equal to <math>19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}</math><br />
<br />
==Solution 3==<br />
Let's assume that the three of them have <math>x, y, z</math> apples. Since each of them has to have at least <math>2</math> apples, we say that <math>a+2=x, b+2=y</math> and <math>c+2=z</math>. Thus, <math>a+b+c+6=24 \implies a+b+c=18</math>, and so by stars and bars, the number of solutions for this is <math>{n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}</math> - aops5234 <br />
<br />
==Solution 4 (EASIEST AND FASTEST SOLUTION)==<br />
<br />
Since we have to give each of the <math>3</math> friends at least <math>2</math> apples, we need to spend a total of <math>2+2+2=6</math> apples to solve the restriction. Now we have <math>24-6=18</math> apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have <math>18</math> stones and <math>2</math> sticks, which have a total of <math>\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}</math> ways to arrange. <br />
<br />
~by sakshamsethi<br />
<br />
==Videos explaining solution==<br />
<br />
https://www.youtube.com/watch?v=2dBUklyUaNI<br />
<br />
https://www.youtube.com/watch?v=EJzSOPXULBc<br />
<br />
https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu<br />
<br />
https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 ~ MathEx<br />
<br />
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=24|after=Last Problem}}<br />
<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=1341362019 AMC 8 Problems/Problem 252020-09-27T00:49:44Z<p>Sakshamsethi: /* Solution 4 */</p>
<hr />
<div>==Problem 25==<br />
Alice has <math>24</math> apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?<br />
<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math><br />
<br />
==Solution 1==<br />
We use [[stars and bars]]. Let Alice get <math>k</math> apples, let Becky get <math>r</math> apples, let Chris get <math>y</math> apples.<br />
<cmath>\implies k + r + y = 24</cmath>We can manipulate this into an equation which can be solved using stars and bars.<br />
<br />
All of them get at least <math>2</math> apples, so we can subtract <math>2</math> from <math>k</math>, <math>2</math> from <math>r</math>, and <math>2</math> from <math>y</math>.<br />
<cmath>\implies (k - 2) + (r - 2) + (y - 2) = 18</cmath>Let <math>k' = k - 2</math>, let <math>r' = r - 2</math>, let <math>y' = y - 2</math>.<br />
<cmath>\implies k' + r' + y' = 18</cmath>We can allow either of them to equal to <math>0</math>, hence this can be solved by stars and bars.<br />
<br />
<br />
By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{190}</math>.<br />
<br />
==Solution 2==<br />
[[Without loss of generality]], let's assume that Alice has <math>2</math> apples. There are <math>19</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>3</math> apples, there are <math>18</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>4</math> apples, there are <math>17</math> ways to split the rest. So the total number of ways to split <math>24</math> apples between the three friends is equal to <math>19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}</math><br />
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==Solution 3==<br />
Let's assume that the three of them have <math>x, y, z</math> apples. Since each of them has to have at least <math>2</math> apples, we say that <math>a+2=x, b+2=y</math> and <math>c+2=z</math>. Thus, <math>a+b+c+6=24 \implies a+b+c=18</math>, and so by stars and bars, the number of solutions for this is <math>{n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}</math> - aops5234 <br />
<br />
==Solution 5 (EASIEST AND FASTEST SOLUTION)==<br />
<br />
Since we have to give each of the <math>3</math> friends at least <math>2</math> apples, we need to spend a total of <math>2+2+2=6</math> apples to solve the restriction. Now we have <math>24-6=18</math> apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have <math>18</math> stones and <math>2</math> sticks, which have a total of <math>\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}</math> ways to arrange. <br />
<br />
~by sakshamsethi<br />
<br />
==Videos explaining solution==<br />
<br />
https://www.youtube.com/watch?v=2dBUklyUaNI<br />
<br />
https://www.youtube.com/watch?v=EJzSOPXULBc<br />
<br />
https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu<br />
<br />
https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 ~ MathEx<br />
<br />
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=24|after=Last Problem}}<br />
<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=1341352019 AMC 8 Problems/Problem 252020-09-27T00:49:16Z<p>Sakshamsethi: /* Solution 5 */</p>
<hr />
<div>==Problem 25==<br />
Alice has <math>24</math> apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?<br />
<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math><br />
<br />
==Solution 1==<br />
We use [[stars and bars]]. Let Alice get <math>k</math> apples, let Becky get <math>r</math> apples, let Chris get <math>y</math> apples.<br />
<cmath>\implies k + r + y = 24</cmath>We can manipulate this into an equation which can be solved using stars and bars.<br />
<br />
All of them get at least <math>2</math> apples, so we can subtract <math>2</math> from <math>k</math>, <math>2</math> from <math>r</math>, and <math>2</math> from <math>y</math>.<br />
<cmath>\implies (k - 2) + (r - 2) + (y - 2) = 18</cmath>Let <math>k' = k - 2</math>, let <math>r' = r - 2</math>, let <math>y' = y - 2</math>.<br />
<cmath>\implies k' + r' + y' = 18</cmath>We can allow either of them to equal to <math>0</math>, hence this can be solved by stars and bars.<br />
<br />
<br />
By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{190}</math>.<br />
<br />
==Solution 2==<br />
[[Without loss of generality]], let's assume that Alice has <math>2</math> apples. There are <math>19</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>3</math> apples, there are <math>18</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>4</math> apples, there are <math>17</math> ways to split the rest. So the total number of ways to split <math>24</math> apples between the three friends is equal to <math>19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}</math><br />
<br />
==Solution 3==<br />
Let's assume that the three of them have <math>x, y, z</math> apples. Since each of them has to have at least <math>2</math> apples, we say that <math>a+2=x, b+2=y</math> and <math>c+2=z</math>. Thus, <math>a+b+c+6=24 \implies a+b+c=18</math>, and so by stars and bars, the number of solutions for this is <math>{n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}</math> - aops5234 <br />
<br />
==Solution 4==<br />
We can give each person one apple first so that <math>21</math> apples are shared between the three people, where each person receives at least one apple. Using Stars and Bars, the number of ways to do this is <math>\binom{21-1}{3-1}=\binom{20}{2}=\boxed{\textbf{(C)}\ 190}</math>.<br />
<br />
==Solution 5 (EASIEST AND FASTEST SOLUTION)==<br />
<br />
Since we have to give each of the <math>3</math> friends at least <math>2</math> apples, we need to spend a total of <math>2+2+2=6</math> apples to solve the restriction. Now we have <math>24-6=18</math> apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have <math>18</math> stones and <math>2</math> sticks, which have a total of <math>\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}</math> ways to arrange. <br />
<br />
~by sakshamsethi<br />
<br />
==Videos explaining solution==<br />
<br />
https://www.youtube.com/watch?v=2dBUklyUaNI<br />
<br />
https://www.youtube.com/watch?v=EJzSOPXULBc<br />
<br />
https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu<br />
<br />
https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 ~ MathEx<br />
<br />
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=24|after=Last Problem}}<br />
<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=1341341985 AJHSME Problems/Problem 212020-09-27T00:16:54Z<p>Sakshamsethi: /* Solution */</p>
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<div>==Problem==<br />
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Mr. Green receives a <math>10\% </math> raise every year. His salary after four such raises has gone up by what percent?<br />
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<math>\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% </math><br />
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==Solution==<br />
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Assume his salary is originally <math>100</math> dollars. Then, in the next year, he would have <math>110</math> dollars, and in the next, he would have <math>121</math> dollars. The next year he would have <math>133.1</math> dollars and in the final year, he would have <math>146.41</math>. As the total increase is greater than <math>45\%</math>, the answer is <math>\boxed{\text{E}}</math>. <br />
<br />
~sakshamsethi<br />
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==See Also==<br />
<br />
{{AJHSME box|year=1985|num-b=20|num-a=22}}<br />
[[Category:Introductory Algebra Problems]]<br />
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<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems&diff=1307112012 AMC 10B Problems2020-08-05T23:20:24Z<p>Sakshamsethi: /* Problem 7 */</p>
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<div>{{AMC10 Problems|year=2012|ab=B}}<br />
== Problem 1 ==<br />
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Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?<br />
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<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 </math><br />
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[[2012 AMC 12B Problems/Problem 1|Solution]]<br />
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== Problem 2 ==<br />
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A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?<br />
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<asy><br />
draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); <br />
draw(circle((10,5),5));</asy><br />
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<math> \textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200 </math><br />
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[[2012 AMC 10B Problems/Problem 2|Solution]]<br />
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== Problem 3 ==<br />
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The point in the xy-plane with coordinates <math>(1000, 2012)</math> is reflected across the line <math>y = 2000</math>. What are the coordinates of the reflected point?<br />
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<math> \textbf{(A)}\ (998,2012)\qquad\textbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012) </math><br />
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[[2012 AMC 10B Problems/Problem 3|Solution]]<br />
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== Problem 4 ==<br />
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When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?<br />
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math><br />
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[[2012 AMC 10B Problems/Problem 4|Solution]]<br />
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== Problem 5 ==<br />
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Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in dollars?<br />
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<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math><br />
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[[2012 AMC 10B Problems/Problem 5|Solution]]<br />
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== Problem 6 ==<br />
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In order to estimate the value of <math>x-y</math> where <math>x</math> and <math>y</math> are real numbers with <math>x > y > 0</math>, Xiaoli rounded <math>x</math> up by a small amount, rounded <math>y</math> down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct? <br />
<br />
<math> \textbf{(A)} </math> Her estimate is larger than <math>x-y</math> <br />
<math> \textbf{(B)} </math> Her estimate is smaller than <math>x-y</math><br />
<math> \textbf{(C)} </math> Her estimate equals <math>x-y</math><br />
<math> \textbf{(D)} </math> Her estimate equals <math>y - x</math><br />
<math> \textbf{(E)} </math> Her estimate is <math>0</math><br />
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[[2012 AMC 10B Problems/Problem 6|Solution]]<br />
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== Problem 7 ==<br />
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For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br />
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<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math><br />
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[[2012 AMC 10B Problems/Problem 7|Solution]]<br />
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== Problem 8 ==<br />
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What is the sum of all integer solutions to <math>1<(x-2)^2<25</math>?<br />
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<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25 </math><br />
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[[2012 AMC 10B Problems/Problem 8|Solution]]<br />
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== Problem 9 ==<br />
Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?<br />
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math><br />
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[[2012 AMC 10B Problems/Problem 9|Solution]]<br />
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== Problem 10 ==<br />
How many ordered pairs of positive integers <math>(M,N)</math> satisfy the equation <math>\frac {M}{6} = \frac{6}{N}?</math><br />
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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math><br />
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[[2012 AMC 10B Problems/Problem 10|Solution]]<br />
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== Problem 11 ==<br />
A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?<br />
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<math> \textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 </math><br />
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[[2012 AMC 10B Problems/Problem 11|Solution]]<br />
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== Problem 12 ==<br />
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Point B is due east of point A. Point C is due north of point B. The distance between points A and C is <math>10\sqrt 2</math>, and <math>\angle BAC = 45^\circ</math>. Point D is 20 meters due north of point C. The distance AD is between which two integers?<br />
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<math>\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35</math><br />
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[[2012 AMC 10B Problems/Problem 12|Solution]]<br />
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== Problem 13 ==<br />
It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?<br />
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<math> \textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52 </math><br />
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[[2012 AMC 10B Problems/Problem 13|Solution]]<br />
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== Problem 14 ==<br />
Two equilateral triangles are contained in a square whose side length is <math>2\sqrt 3</math>. The bases of these triangles are the opposite sides of the square, and their intersection is a rhombus. What is the area of the rhombus?<br />
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<math><br />
\text{(A) } \frac{3}{2}<br />
\qquad<br />
\text{(B) } \sqrt 3<br />
\qquad<br />
\text{(C) } 2\sqrt 2 - 1 <br />
\qquad<br />
\text{(D) } 8\sqrt 3 - 12<br />
\qquad<br />
\text{(E)} \frac{4\sqrt 3}{3} <br />
</math><br />
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[[2012 AMC 10B Problems/Problem 14|Solution]]<br />
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==Problem 15==<br />
In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end on the tournament?<br />
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<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math><br />
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[[2012 AMC 10B Problems/Problem 15|Solution]]<br />
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==Problem 16==<br />
Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?<br />
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<asy><br />
filldraw((0,0)--(2,0)--(1,sqrt(3))--cycle,gray,gray);<br />
filldraw(circle((1,sqrt(3)),1),gray);<br />
filldraw(circle((0,0),1),gray);<br />
filldraw(circle((2,0),1),grey);</asy><br />
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<math>\textbf{(A)}\ 10\pi+4\sqrt{3}\qquad\textbf{(B)}\ 13\pi-\sqrt{3}\qquad\textbf{(C)}\ 12\pi+\sqrt{3}\qquad\textbf{(D)}\ 10\pi+9\qquad\textbf{(E)}\ 13\pi</math><br />
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[[2012 AMC 10B Problems/Problem 16|Solution]]<br />
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==Problem 17==<br />
Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br />
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<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{\sqrt{10}}{10}\qquad\textbf{(D)}\ \frac{\sqrt{5}}{6}\qquad\textbf{(E)}\ \frac{\sqrt{5}}{5}</math><br />
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[[2012 AMC 10B Problems/Problem 17|Solution]]<br />
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==Problem 18==<br />
Suppose that one of every 500 people in a certain population has a particular disease, which displays no symptoms. A blood test is available for screening for this disease. For a person who has this disease, the test always turns out positive. For a person who does not have the disease, however, there is a <math>2\%</math> false positive rate--in other words, for such people, <math>98\%</math> of the time the test will turn out negative, but <math>2\%</math> of the time the test will turn out positive and will incorrectly indicate that the person has the disease. Let <math>p</math> be the probability that a person who is chosen at random from this population and gets a positive test result actually has the disease. Which of the following is closest to <math>p</math>?<br />
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<math>\textbf{(A)}\ \frac{1}{98}\qquad\textbf{(B)}\ \frac{1}{9}\qquad\textbf{(C)}\ \frac{1}{11}\qquad\textbf{(D)}\ \frac{49}{99}\qquad\textbf{(E)}\ \frac{98}{99}</math><br />
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[[2012 AMC 10B Problems/Problem 18|Solution]]<br />
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==Problem 19==<br />
In rectangle <math>ABCD</math>, <math>AB=6</math>, <math>AD=30</math>, and <math>G</math> is the midpoint of <math>\overline{AD}</math>. Segment <math>AB</math> is extended 2 units beyond <math>B</math> to point <math>E</math>, and <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>. What is the area of <math>BFDG</math>?<br />
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<math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math><br />
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[[2012 AMC 10B Problems/Problem 19|Solution]]<br />
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==Problem 20==<br />
Bernardo and Silvia play the following game. An integer between 0 and 999, inclusive, is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let <math>N</math> be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of <math>N</math>?<br />
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<math>\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11</math><br />
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[[2012 AMC 10B Problems/Problem 20|Solution]]<br />
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==Problem 21==<br />
Four distinct points are arranged on a plane so that the segments connecting them have lengths <math>a</math>, <math>a</math>, <math>a</math>, <math>a</math>, <math>2a</math>, and <math>b</math>. What is the ratio of <math>b</math> to <math>a</math>?<br />
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<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math><br />
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[[2012 AMC 10B Problems/Problem 21|Solution]]<br />
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==Problem 22==<br />
Let <math>(a_1,a_2, \dots ,a_{10})</math> be a list of the first 10 positive integers such that for each <math>2 \le i \le 10</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 />
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<math>\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880</math><br />
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[[2012 AMC 10B Problems/Problem 22|Solution]]<br />
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== Problem 23 ==<br />
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A solid tetrahedron is sliced off a wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} </math><br />
<br />
[[2012 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<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 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 />
[[2012 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
A bug travels from <math>A</math> to <math>B</math> along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br />
<br />
<asy><br />
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filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br />
filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br />
filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);</asy><br />
<br />
<math>\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400</math><br />
<br />
[[2012 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC10 box|year=2012|ab=B|before=[[2012 AMC 10A Problems]]|after=[[2013 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2012 AMC 10B]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_7&diff=1307102012 AMC 10B Problems/Problem 72020-08-05T23:19:53Z<p>Sakshamsethi: /* Problem 7 */</p>
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== Problem 7 ==<br />
<br />
For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br />
<br />
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math><br />
<br />
==Solution 1==<br />
Let <math>x</math> be the number of acorns that both animals had.<br />
<br />
So by the info in the problem:<br />
<br />
<math>\frac{x}{3}=\left( \frac{x}{4} \right)+4</math><br />
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Subtracting <math> \frac{x}{4}</math> from both sides leaves<br />
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<math>\frac{x}{12}=4</math><br />
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<math>\boxed{x=48}</math><br />
<br />
This is answer choice <math>\textbf{(D)}</math><br />
<br />
==Solution 2==<br />
Instead of an Algebraic Solution, we can just find a residue in the common multiples of <math>3</math> and <math>4</math>, so <math>lcm[3,4]=12</math>, the next largest is <math>12\cdot2=24</math>, the next is <math>36</math>, and so on, with all of them being multiples of <math>12</math>, now we can see that per every common multiple, we can see a pattern such as <br />
<br />
<math>12=4\cdot3=3\cdot4</math> so <math>4-3=1</math> hole less. <br />
<br />
<math>24=4\cdot6=3\cdot8</math> so <math>8-6=2</math> holes less. <br />
<br />
<math>36=4\cdot9=3\cdot12</math> so <math>12-9=3</math> holes less. <br />
<br />
<math>48=4\cdot12=3\cdot16</math> so <math>16-12=4</math> holes less. <br />
<br />
So we see that <math>48</math> is the number we need which is <math>\textbf{48(D)}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_7&diff=1307092012 AMC 10B Problems/Problem 72020-08-05T23:19:40Z<p>Sakshamsethi: /* Problem 7 */</p>
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== Problem 7 ==<br />
<br />
For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br />
<br />
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math><br />
<br />
==Solution 1==<br />
Let <math>x</math> be the number of acorns that both animals had.<br />
<br />
So by the info in the problem:<br />
<br />
<math>\frac{x}{3}=\left( \frac{x}{4} \right)+4</math><br />
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Subtracting <math> \frac{x}{4}</math> from both sides leaves<br />
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<math>\frac{x}{12}=4</math><br />
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<math>\boxed{x=48}</math><br />
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This is answer choice <math>\textbf{(D)}</math><br />
<br />
==Solution 2==<br />
Instead of an Algebraic Solution, we can just find a residue in the common multiples of <math>3</math> and <math>4</math>, so <math>lcm[3,4]=12</math>, the next largest is <math>12\cdot2=24</math>, the next is <math>36</math>, and so on, with all of them being multiples of <math>12</math>, now we can see that per every common multiple, we can see a pattern such as <br />
<br />
<math>12=4\cdot3=3\cdot4</math> so <math>4-3=1</math> hole less. <br />
<br />
<math>24=4\cdot6=3\cdot8</math> so <math>8-6=2</math> holes less. <br />
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<math>36=4\cdot9=3\cdot12</math> so <math>12-9=3</math> holes less. <br />
<br />
<math>48=4\cdot12=3\cdot16</math> so <math>16-12=4</math> holes less. <br />
<br />
So we see that <math>48</math> is the number we need which is <math>\textbf{48(D)}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_19&diff=1296522001 AMC 10 Problems/Problem 192020-07-28T16:22:35Z<p>Sakshamsethi: /* Solution 3 */</p>
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<div>== Problem ==<br />
<br />
Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?<br />
<br />
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math><br />
<br />
== Solution==<br />
<br />
Let's use [[stars and bars]].<br />
Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, <math> O|OO|O </math> represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the balls and urn (stars and bars) identity.<br />
<br />
==Solution 2==<br />
<br />
Simple casework works here as well:<br />
Set up the following ratios:<br />
<cmath>4:0:0</cmath><br />
<cmath>3:1:0</cmath><br />
<cmath>2:2:0</cmath><br />
<cmath>2:1:1</cmath><br />
<br />
In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get <math>\frac{4!}{3!2!}</math> for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is <math>3*3+6 = \boxed{15}</math><br />
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Solution by IronicNinja<br />
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== See Also ==<br />
<br />
{{AMC10 box|year=2001|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_6&diff=1296481998 AJHSME Problems/Problem 62020-07-28T16:06:33Z<p>Sakshamsethi: /* Solutions */</p>
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<div>==Problem==<br />
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Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is<br />
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for(int a=0; a<4; ++a)<br />
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draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle);<br />
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<math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
== Solutions ==<br />
===Solution 1===<br />
We could count the area contributed by each square on the <math>3 \times 3</math> grid:<br />
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Top-left: <math>0</math><br />
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Top: Triangle with area <math>\frac{1}{2}</math><br />
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Top-right: <math>0</math><br />
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Left: Square with area <math>1</math><br />
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Center: Square with area <math>1</math><br />
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Right: Square with area <math>1</math><br />
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Bottom-left: Square with area <math>1</math><br />
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Bottom: Triangle with area <math>\frac{1}{2}</math><br />
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Bottom-right: Square with area <math>1</math><br />
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Adding all of these together, we get <math>\boxed{6}</math> or <math>\boxed{B}</math><br />
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===Solution 2===<br />
By Pick's Theorem, we get the formula, <math>A=I+\frac{b}{2}-1</math> where <math>I</math> is the number of lattice points in the interior and <math>b</math> being the number of lattice points on the boundary. In this problem, we can see that <math>I=1</math> and <math>B=12</math>. Substituting gives us <math>A=1+\frac{12}{2}-1=6</math> Thus, the answer is <math>\boxed{\text{(B) 6}}</math><br />
<br />
===Solution 3===<br />
<br />
Notice that the extra triangle on the top with area <math>1</math> can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area <math>1</math>. This creates a <math>2*3</math> rectangle, with a area of <math>6</math>. The answer is <math>\boxed{\text{(B) 6}}</math><br />
~sakshamsethi<br />
<br />
== See Also ==<br />
{{AJHSME box|year=1998|num-b=5|num-a=7}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1988_AJHSME_Problems/Problem_6&diff=1296471988 AJHSME Problems/Problem 62020-07-28T16:04:22Z<p>Sakshamsethi: /* Solution 2 */</p>
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<div>==Problem==<br />
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<math>\frac{(.2)^3}{(.02)^2} =</math><br />
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<math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20</math><br />
<br />
==Solution==<br />
Converting the [[decimal|decimals]] to [[fraction|fractions]] gives us <math>\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}</math>.<br />
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==Solution 2==<br />
<br />
We expand <math>\frac{(0.2)^3}{(0.02)^2}</math>, and get <math>\frac{(0.2) \times (0.2) \times (0.2)}{(0.02) \times (0.02)}</math>. The two <math>0.02</math>'s "cancel" out with the two <math>0.2</math>'s, leaving the fraction as: <math>(10) \times (10) \times (0.2)</math>. Using basic calculations, we compute this expression to get <math>20\Rightarrow \mathrm{(E)}</math>.<br />
<br />
~sakshamsethi<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1988|num-b=5|num-a=7}}<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1988_AJHSME_Problems/Problem_4&diff=1296461988 AJHSME Problems/Problem 42020-07-28T16:02:57Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by<br />
<br />
<math>\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math><br />
<br />
<asy><br />
unitsize(12);<br />
for(int a=0; a<7; ++a)<br />
{<br />
fill((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black);<br />
draw((2a+1,0)--(2a+2,0));<br />
}<br />
for(int b=7; b<15; ++b)<br />
{<br />
fill((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black);<br />
}<br />
for(int c=1; c<7; ++c)<br />
{<br />
fill((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black);<br />
}<br />
for(int d=1; d<6; ++d)<br />
{<br />
draw((2d+1,1)--(2d+2,1));<br />
}<br />
fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); draw((5,4)--(6,4));<br />
fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); draw((7,4)--(8,4));<br />
fill((8,4)--(9,4)--(9,5)--(8,5)--cycle,black); draw((9,4)--(10,4));<br />
label("same",(6.3,2.45),N);<br />
label("pattern here",(7.5,1.4),N);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
If, for a moment, we disregard the white squares, we notice that the number of black squares in each row increases by 1 continuously as we go down the pyramid.<br />
Thus, the number of black squares is <math> 1 + 2 + \cdots + 8 </math>.<br />
<br />
Same goes for the white squares, except it starts a row later, making it <math> 1 + 2 + \cdots + 7</math>.<br />
<br />
Subtracting the number of white squares from the number of black squares...<br />
<cmath>1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8 \Rightarrow (B)</cmath><br />
<br />
<br />
==Solution 2==<br />
It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are <math>8</math> rows, there are <math>8</math> more black squares than the white squares. <math>8\rightarrow \boxed{\text{D}}</math><br />
<br />
~sakshamsethi<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1988|num-b=3|num-a=5}}<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_24&diff=1296451986 AJHSME Problems/Problem 242020-07-28T16:02:13Z<p>Sakshamsethi: /* Solution 3 (easiest) */</p>
<hr />
<div>==Problem==<br />
<br />
The <math>600</math> students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately<br />
<br />
<math>\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad \text{(E)}\ \frac{1}{3}</math><br />
<br />
==Solution 1==<br />
There are <math>\binom{3}{1}</math> ways to choose which group the three kids are in and the chance that all three are in the same group is <math>\frac{1}{27}</math>. Hence <math>\frac{1}{9}</math> or <math>\boxed {B}</math>.<br />
<br />
==Solution 2==<br />
One of the statements, that there are <math>600</math> students in the school is redundant. Taking that there are <math>3</math> students and there are <math>3</math> groups, we can easily deduce there are <math>81</math> ways to group the <math>3</math> students, and there are <math>3</math> ways to group them in the same <math>1</math> group, so we might think <math>\frac{3}{54}=\frac{1}{27}</math> is the answer but as there are 3 groups we do <math>\frac{1}{27} (3)=\frac{1}{9}</math> which is <math>\boxed{\text{(B)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=23|num-a=25}}<br />
[[Category:Introductory Probability Problems]]<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_19&diff=1296402001 AMC 10 Problems/Problem 192020-07-28T15:33:08Z<p>Sakshamsethi: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
<br />
Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?<br />
<br />
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math><br />
<br />
== Solution==<br />
<br />
Let's use [[stars and bars]].<br />
Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, <math> O|OO|O </math> represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the balls and urn (stars and bars) identity.<br />
<br />
==Solution 2==<br />
<br />
Simple casework works here as well:<br />
Set up the following ratios:<br />
<cmath>4:0:0</cmath><br />
<cmath>3:1:0</cmath><br />
<cmath>2:2:0</cmath><br />
<cmath>2:1:1</cmath><br />
<br />
In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get <math>\frac{4!}{3!2!}</math> for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is <math>3*3+6 = \boxed{15}</math><br />
<br />
Solution by IronicNinja<br />
<br />
<br />
==Solution 3==<br />
<br />
Since there are three choices to pick from, the value of <math>a</math> in the expression <math>{a \choose b}</math> has to be a multiple of <math>3</math>. <math>{3 \choose 4}</math> would make no sense, because that would mean: "How many ways are there to choose 4 donuts out of 3?". Next we could try <math>{6 \choose 4}</math>, which is equal to <math>15</math> ways. This works, but now let <math>a</math> be <math>9</math>. So if there were <math>9</math> donuts, (which is possible since it's a multiple of 3), the number of ways we could choose <math>4</math> donuts would be <math>{9 \choose 4} = 126</math>. Anything greater than <math>18</math> will be incorrect, since <math>18</math> is the largest number in the answer choices. Thus, our answer is <math>\boxed {15}</math>.<br />
<br />
~sakshamsethi<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2001|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_19&diff=1296392001 AMC 10 Problems/Problem 192020-07-28T15:32:53Z<p>Sakshamsethi: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
<br />
Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?<br />
<br />
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math><br />
<br />
== Solution==<br />
<br />
Let's use [[stars and bars]].<br />
Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, <math> O|OO|O </math> represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the balls and urn (stars and bars) identity.<br />
<br />
==Solution 2==<br />
<br />
Simple casework works here as well:<br />
Set up the following ratios:<br />
<cmath>4:0:0</cmath><br />
<cmath>3:1:0</cmath><br />
<cmath>2:2:0</cmath><br />
<cmath>2:1:1</cmath><br />
<br />
In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get <math>\frac{4!}{3!2!}</math> for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is <math>3*3+6 = \boxed{15}</math><br />
<br />
Solution by IronicNinja<br />
<br />
<br />
==Solution 3==<br />
<br />
Since there are three choices to pick from, the value of <math>a</math> in the expression <math>{a \choose b}</math> has to be a multiple of <math>3</math>. <math>{3 \choose 4}</math> would make no sense, because that would mean: "How many ways are there to choose 4 donuts out of 3?". Next we could try <math>{6 \choose 4}</math>, which is equal to <math>15</math> ways. This works, but now let <math>a</math> be <math>9</math>. So if there were <math>9</math> donuts, (which is possible since it's a multiple of 3), the number of ways we could choose <math>4</math> donuts would be <math>{9 \choose 4} = 126</math>. Anything greater than <math>18</math> will be incorrect, since <math>18</math> is the largest number in the answer choices. Thus, our answer is <math>\boxed {15}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2001|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_19&diff=1296382001 AMC 10 Problems/Problem 192020-07-28T15:20:50Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?<br />
<br />
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math><br />
<br />
== Solution==<br />
<br />
Let's use [[stars and bars]].<br />
Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, <math> O|OO|O </math> represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the balls and urn (stars and bars) identity.<br />
<br />
==Solution 2==<br />
<br />
Simple casework works here as well:<br />
Set up the following ratios:<br />
<cmath>4:0:0</cmath><br />
<cmath>3:1:0</cmath><br />
<cmath>2:2:0</cmath><br />
<cmath>2:1:1</cmath><br />
<br />
In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get <math>\frac{4!}{3!2!}</math> for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is <math>3*3+6 = \boxed{15}</math><br />
<br />
Solution by IronicNinja<br />
<br />
<br />
==Solution 3==<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2001|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_21&diff=1295202000 AMC 12 Problems/Problem 212020-07-27T18:06:42Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #21]] and [[2000 AMC 10 Problems|2000 AMC 10 #19]]}}<br />
<br />
== Problem ==<br />
Through a point on the [[hypotenuse]] of a [[right triangle]], lines are drawn [[parallel]] to the legs of the triangle so that the triangle is divided into a [[square]] and two smaller right triangles. The area of one of the two small right triangles is <math>m</math> times the area of the square. The [[ratio]] of the area of the other small right triangle to the area of the square is <br />
<br />
<math>\text {(A)}\ \frac{1}{2m+1} \qquad \text {(B)}\ m \qquad \text {(C)}\ 1-m \qquad \text {(D)}\ \frac{1}{4m} \qquad \text {(E)}\ \frac{1}{8m^2}</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
<center><asy><br />
unitsize(36);<br />
draw((0,0)--(6,0)--(0,3)--cycle);<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
label("$1$",(1,2),S);<br />
label("$1$",(2,1),W);<br />
label("$2m$",(4,0),S);<br />
label("$x$",(0,2.5),W);<br />
</asy></center><br />
<br />
WLOG, let a side of the square be <math>1</math>. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A = \frac{1}{2}bh = \frac{h}{2}</math>, the height of the triangle with area <math>m</math> is <math>2m</math>. Therefore <math>\frac{2m}{1} = \frac{1}{x}</math> where <math>x</math> is the base of the other triangle. <math>x = \frac{1}{2m}</math>, and the area of that triangle is <math>\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{(D)}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
<br />
<center><asy><br />
unitsize(36);<br />
draw((0,0)--(6,0)--(0,3)--cycle);<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
label("$b$",(2.5,0),S);<br />
label("$a$",(0,1.5),W);<br />
label("$c$",(2.5,1),W);<br />
label("$A$",(0.5,2.5),W);<br />
label("$B$",(3.5,0.75),W);<br />
label("$C$",(1,1),W);<br />
</asy></center><br />
<br />
From the diagram from the previous solution, we have <math>a</math>, <math>b</math> as the legs and <math>c</math> as the side length of the square. WLOG, let the area of triangle <math>A</math><br />
be <math>m</math> times the area of square <math>C</math>.<br />
<br />
Since triangle <math>A</math> is similar to the large triangle, it has <math>h_A = a(\frac{c}{b}) = \frac{ac}{b}</math>, <math>b_A = c</math> and <cmath>[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2</cmath><br />
Thus <math>\frac{a}{2b} = m</math><br />
<br />
Now since triangle <math>B</math> is similar to the large triangle, it has <math>h_B = c</math>, <math>b_B = b\frac{c}{a} = \frac{bc}{a}</math> and <cmath>[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]</cmath> <br />
<br />
Thus <math>n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}</math>. <math>\text{(D)}</math>.<br />
<br />
~ Nafer<br />
<br />
<br />
[I think there is a mistake. It says it is a problem 19 for AMC 10 and problem 21 for AMC 12. They are both the same year, 2000. Is this a mistake, or meant to be like this?]<br />
<br />
-Canadian<br />
<br />
<br />
<br />
<br />
Answer to the question asked by Canadian: <br />
Many problems overlap in the AMC 10 and AMC 12 for the same year. Yes, it is meant to be like this.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2000|num-b=20|num-a=22}}<br />
{{AMC10 box|year=2000|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_21&diff=1295192000 AMC 12 Problems/Problem 212020-07-27T18:06:14Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #21]] and [[2000 AMC 10 Problems|2000 AMC 10 #19]]}}<br />
<br />
== Problem ==<br />
Through a point on the [[hypotenuse]] of a [[right triangle]], lines are drawn [[parallel]] to the legs of the triangle so that the triangle is divided into a [[square]] and two smaller right triangles. The area of one of the two small right triangles is <math>m</math> times the area of the square. The [[ratio]] of the area of the other small right triangle to the area of the square is <br />
<br />
<math>\text {(A)}\ \frac{1}{2m+1} \qquad \text {(B)}\ m \qquad \text {(C)}\ 1-m \qquad \text {(D)}\ \frac{1}{4m} \qquad \text {(E)}\ \frac{1}{8m^2}</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
<center><asy><br />
unitsize(36);<br />
draw((0,0)--(6,0)--(0,3)--cycle);<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
label("$1$",(1,2),S);<br />
label("$1$",(2,1),W);<br />
label("$2m$",(4,0),S);<br />
label("$x$",(0,2.5),W);<br />
</asy></center><br />
<br />
WLOG, let a side of the square be <math>1</math>. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A = \frac{1}{2}bh = \frac{h}{2}</math>, the height of the triangle with area <math>m</math> is <math>2m</math>. Therefore <math>\frac{2m}{1} = \frac{1}{x}</math> where <math>x</math> is the base of the other triangle. <math>x = \frac{1}{2m}</math>, and the area of that triangle is <math>\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{(D)}</math>.<br />
<br />
<br />
=== Solution 2 ===<br />
<br />
<center><asy><br />
unitsize(36);<br />
draw((0,0)--(6,0)--(0,3)--cycle);<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
label("$b$",(2.5,0),S);<br />
label("$a$",(0,1.5),W);<br />
label("$c$",(2.5,1),W);<br />
label("$A$",(0.5,2.5),W);<br />
label("$B$",(3.5,0.75),W);<br />
label("$C$",(1,1),W);<br />
</asy></center><br />
<br />
From the diagram from the previous solution, we have <math>a</math>, <math>b</math> as the legs and <math>c</math> as the side length of the square. WLOG, let the area of triangle <math>A</math><br />
be <math>m</math> times the area of square <math>C</math>.<br />
<br />
Since triangle <math>A</math> is similar to the large triangle, it has <math>h_A = a(\frac{c}{b}) = \frac{ac}{b}</math>, <math>b_A = c</math> and <cmath>[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2</cmath><br />
Thus <math>\frac{a}{2b} = m</math><br />
<br />
Now since triangle <math>B</math> is similar to the large triangle, it has <math>h_B = c</math>, <math>b_B = b\frac{c}{a} = \frac{bc}{a}</math> and <cmath>[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]</cmath> <br />
<br />
Thus <math>n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}</math>. <math>\text{(D)}</math>.<br />
<br />
~ Nafer<br />
<br />
<br />
[I think there is a mistake. It says it is a problem 19 for AMC 10 and problem 21 for AMC 12. They are both the same year, 2000. Is this a mistake, or meant to be like this?]<br />
<br />
-Canadian<br />
Answer to the question asked above by Canadian: <br />
Many problems overlap in the AMC 10 and AMC 12 for the same year. Yes, it is meant to be like this.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2000|num-b=20|num-a=22}}<br />
{{AMC10 box|year=2000|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=Triangular_number&diff=128368Triangular number2020-07-15T23:46:37Z<p>Sakshamsethi: /* Definition */</p>
<hr />
<div>The '''triangular numbers''' are the numbers <math>T_n</math> which are the sum of the first <math>n</math> [[natural number]]s from <math>1</math> to <math>n</math>.<br />
<br />
==Definition==<br />
The <math>n^{th}</math> triangular number is the sum of all natural numbers from one to <math>n</math>.<br />
That is, the <math>n^{th}</math> triangle number is <br />
<math>1 +2+3 + 4............. +(n-1)+(n)</math>.<br />
<br />
For example, the first few triangular numbers can be calculated by adding <br />
1, 1+2, 1+2+3, ... etc. <br />
<math>}<br />
rowStart -= 0.5;<br />
}<br />
return 0;<br />
}<br />
<br />
for (int n=1; n<5; ++n)<br />
{<br />
real value= n*(n+1)/2;<br />
draw_triangle((value+5,n),n);<br />
label( (string) value, (value+5, -2));<br />
}<br />
</asy></math><br />
<br />
==Formula==<br />
<br />
Using the sum of an [[arithmetic series]] formula, a formula can be calculated for <math>T_n</math>:<br />
<br />
:<math>T_n =\sum_{k=1}^{n}k = 1 + 2 + \ldots + n = \frac{n(n+1)}2</math><br />
<br />
<br />
The formula for finding the <math>n^{th}</math> triangular number can be written as <math>\dfrac{n(n+1)}{2}</math>.<br />
<br />
It can also be expressed as the sum of the <math>n^{th}</math> row in [[Pascal's Triangle]] and all the rows above it. Keep in mind that the triangle starts at Row 0.<br />
<br />
<br />
<br />
<br />
<br />
{{stub}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_25&diff=1283282002 AMC 8 Problems/Problem 252020-07-14T22:59:23Z<p>Sakshamsethi: /* Solution 2 (easiest) */</p>
<hr />
<div>==Problem==<br />
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br />
<br />
<math> \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} </math><br />
<br />
==Solution==<br />
<br />
Since Ott gets equal amounts of money from each friend, we can say that he gets <math>x</math> dollars from each friend. This means that Moe has <math>5x</math> dollars, Loki has <math>4x</math> dollars, and Nick has <math>3x</math> dollars. The total amount is <math>12x</math> dollars, and since Ott gets <math>3x</math> dollars total, <math>\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}</math>.<br />
<math>\blacksquare</math><br />
<br />
<br />
==Solution 2 (easiest)==<br />
Assume Moe, Loki, and Nick each give Ott <math>\$ 1</math>. Therefore, Moe has <math>\$ 5</math>, Loki has <math>\$ 4</math>, and Nick has <math>\$ 3</math>. After everyone gives Ott some fraction of their money, the total money at the end situation will be the same as the original; which is <math>\$ 12</math>. Ott gets <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>. Thus, the answer is <math>\frac{3}{12}=\boxed{\text{(B)}\ \frac14}</math>. <br />
<br />
~sakshamsethi<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/ysNxyATCxzg - Happytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_25&diff=1283272002 AMC 8 Problems/Problem 252020-07-14T22:53:33Z<p>Sakshamsethi: /* Solution 2 (easiest) */</p>
<hr />
<div>==Problem==<br />
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br />
<br />
<math> \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} </math><br />
<br />
==Solution==<br />
<br />
Since Ott gets equal amounts of money from each friend, we can say that he gets <math>x</math> dollars from each friend. This means that Moe has <math>5x</math> dollars, Loki has <math>4x</math> dollars, and Nick has <math>3x</math> dollars. The total amount is <math>12x</math> dollars, and since Ott gets <math>3x</math> dollars total, <math>\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}</math>.<br />
<math>\blacksquare</math><br />
<br />
<br />
==Solution 2 (easiest)==<br />
Assume Moe, Loki, and Nick each give Ott <math>\$ 1</math>. Therefore, Moe has <math>\$ 5</math>, Loki has <math>\$ 4</math>, and Nick has <math>\$ 3</math>. After everyone gives Ott some fraction of their money, he gets <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>. reviewing the situation, Moe now has <math>\$ 5</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 4</math>, Loki has <math>\$ 4</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>, and Nick has <math>\$ 3</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 2</math>. Ott has <math>\frac{3}{4+3+2+3}</math> fraction of the group's money. Thus, the answer is <math>\frac{3}{12}=\boxed{\text{(B)}\ \frac14}</math>. <br />
<br />
~sakshamsethi<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/ysNxyATCxzg - Happytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_25&diff=1283262002 AMC 8 Problems/Problem 252020-07-14T22:46:47Z<p>Sakshamsethi: /* Solution 2 (easiest) */</p>
<hr />
<div>==Problem==<br />
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br />
<br />
<math> \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} </math><br />
<br />
==Solution==<br />
<br />
Since Ott gets equal amounts of money from each friend, we can say that he gets <math>x</math> dollars from each friend. This means that Moe has <math>5x</math> dollars, Loki has <math>4x</math> dollars, and Nick has <math>3x</math> dollars. The total amount is <math>12x</math> dollars, and since Ott gets <math>3x</math> dollars total, <math>\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}</math>.<br />
<math>\blacksquare</math><br />
<br />
<br />
==Solution 2 (easiest)==<br />
Assume Moe has <math>\$ 5</math>. Using the information given in the problem, we can conclude that Loki has <math>\$ 4</math> and Nick has <math>\$ 3</math>. After everyone gives Ott some fraction of their money, he gets <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>. reviewing the situation, Moe now has <math>\$ 5</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 4</math>, Loki has <math>\$ 4</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>, and Nick has <math>\$ 3</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 2</math>. Ott has <math>\frac{3}{4+3+2+3}</math> fraction of the group's money. Thus, the answer is <math>\frac{3}{12}=\boxed{\text{(B)}\ \frac14}</math>. <br />
<br />
~sakshamsethi<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/ysNxyATCxzg - Happytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_25&diff=1283252002 AMC 8 Problems/Problem 252020-07-14T22:46:15Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br />
<br />
<math> \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} </math><br />
<br />
==Solution==<br />
<br />
Since Ott gets equal amounts of money from each friend, we can say that he gets <math>x</math> dollars from each friend. This means that Moe has <math>5x</math> dollars, Loki has <math>4x</math> dollars, and Nick has <math>3x</math> dollars. The total amount is <math>12x</math> dollars, and since Ott gets <math>3x</math> dollars total, <math>\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}</math>.<br />
<math>\blacksquare</math><br />
<br />
<br />
==Solution 2 (easiest)==<br />
Assume Moe has <math>\$ 5</math>. Using the information given in the problem, we can conclude that Loki has <math>\$ 4</math> and Nick has <math>\$ 3</math>. After everyone gives Ott some fraction of their money, he gets <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>. reviewing the situation, Moe now has <math>\$ 5</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 4</math>, Loki has <math>\$ 4</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>, and Nick has <math>\$ 3</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 2</math>. Ott has <math>\frac{3}{</math>4+3+2+3<math>}</math> fraction of the group's money. Thus, the answer is <math>\frac{3}{12}=\boxed{\text{(B)}\ \frac14}</math>. <br />
<br />
~sakshamsethi<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/ysNxyATCxzg - Happytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_25&diff=1283242002 AMC 8 Problems/Problem 252020-07-14T22:24:54Z<p>Sakshamsethi: /* Solution */</p>
<hr />
<div>==Problem==<br />
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br />
<br />
<math> \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} </math><br />
<br />
==Solution==<br />
<br />
Since Ott gets equal amounts of money from each friend, we can say that he gets <math>x</math> dollars from each friend. This means that Moe has <math>5x</math> dollars, Loki has <math>4x</math> dollars, and Nick has <math>3x</math> dollars. The total amount is <math>12x</math> dollars, and since Ott gets <math>3x</math> dollars total, <math>\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}</math>.<br />
<math>\blacksquare</math><br />
<br />
<br />
==Solution 2==<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/ysNxyATCxzg - Happytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=Chicken_McNugget_Theorem&diff=128203Chicken McNugget Theorem2020-07-13T15:59:10Z<p>Sakshamsethi: /* Simple */</p>
<hr />
<div>The '''Chicken McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s <math>m,n</math>, the greatest integer that cannot be written in the form <math>am + bn</math> for [[nonnegative]] integers <math>a, b</math> is <math>mn-m-n</math>.<br />
<br />
A consequence of the theorem is that there are exactly <math>\frac{(m - 1)(n - 1)}{2}</math> positive integers which cannot be expressed in the form <math>am + bn</math>. The proof is based on the fact that in each pair of the form <math>(k, (m - 1)(n - 1) - k+1)</math>, exactly one element is expressible.<br />
<br />
== Origins ==<br />
There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins.<br />
<br />
<br />
<br />
<br />
<br />
==Proof Without Words==<br />
<math>\begin{array}{ccccccc}<br />
0\mod{m}&1\mod{m}&2\mod{m}&...&...&...&(m-1)\mod{m}\\<br />
\hline<br />
\cancel{0n}&1&2&&...&&m-1\\<br />
\cancel{0n+m}&...&&\vdots&&...&\\<br />
\cancel{0n+2m}&...&&\cancel{1n}&&...&\\<br />
\cancel{0n+3m}&&&\cancel{1n+m}&&\vdots&\\<br />
\cancel{0n+4m}&&&\cancel{1n+2m}&&\cancel{2n}&\\<br />
\cancel{0n+5m}&&&\cancel{1n+3m}&&\cancel{2n+m}&\\<br />
\vdots&&&\vdots&&\vdots&\\<br />
\cancel{\qquad}&\cancel{\qquad}&\cancel{ \qquad}&\cancel{ \qquad}&\mathbf{(m-1)n-m}&\cancel{\qquad }&\cancel{\qquad }\\<br />
\cancel{\qquad}&\cancel{\qquad}&\cancel{ \qquad}&\cancel{ \qquad}&\cancel{(m-1)n}&\cancel{\qquad }&\cancel{\qquad }<br />
\end{array}</math><br />
<br />
==Proof 1==<br />
<b>Definition</b>. An integer <math>N \in \mathbb{Z}</math> will be called <i>purchasable</i> if there exist nonnegative integers <math>a,b</math> such that <math>am+bn = N</math>.<br />
<br />
We would like to prove that <math>mn-m-n</math> is the largest non-purchasable integer. We are required to show that (1) <math>mn-m-n</math> is non-purchasable, and (2) every <math>N > mn-m-n</math> is purchasable. <br />
Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty.<br />
<br />
<b>Lemma</b>. Let <math>A_{N} \subset \mathbb{Z} \times \mathbb{Z}</math> be the set of solutions <math>(x,y)</math> to <math>xm+yn = N</math>. Then <math>A_{N} = \{(x+kn,y-km) \;:\; k \in \mathbb{Z}\}</math> for any <math>(x,y) \in A_{N}</math>.<br />
<br />
<i>Proof</i>: By [[Bezout's Lemma]], there exist integers <math>x',y'</math> such that <math>x'm+y'n = 1</math>. Then <math>(Nx')m+(Ny')n = N</math>. Hence <math>A_{N}</math> is nonempty. It is easy to check that <math>(Nx'+kn,Ny'-km) \in A_{N}</math> for all <math>k \in \mathbb{Z}</math>. We now prove that there are no others. Suppose <math>(x_{1},y_{1})</math> and <math>(x_{2},y_{2})</math> are solutions to <math>xm+yn=N</math>. Then <math>x_{1}m+y_{1}n = x_{2}m+y_{2}n</math> implies <math>m(x_{1}-x_{2}) = n(y_{2}-y_{1})</math>. Since <math>m</math> and <math>n</math> are coprime and <math>m</math> divides <math>n(y_{2}-y_{1})</math>, <math>m</math> divides <math>y_{2}-y_{1}</math> and <math>y_{2} \equiv y_{1} \pmod{m}</math>. Similarly <math>x_{2} \equiv x_{1} \pmod{n}</math>. Let <math>k_{1},k_{2}</math> be integers such that <math>x_{2}-x_{1} = k_{1}n</math> and <math>y_{2}-y_{1} = k_{2}m</math>. Then <math>m(-k_{1}n) = n(k_{2}m)</math> implies <math>k_{1} = -k_{2}.</math> We have the desired result. <math>\square</math><br />
<br />
<b>Lemma</b>. For any integer <math>N</math>, there exists unique <math>(a_{N},b_{N}) \in \mathbb{Z} \times \{0,1,\ldots,m-1\}</math> such that <math>a_{N}m + b_{N}n = N</math>.<br />
<br />
<i>Proof</i>: By the division algorithm, there exists one and only one <math>k</math> such that <math>0 \le y-km \le m-1</math>. <math>\square</math><br />
<br />
<b>Lemma</b>. <math>N</math> is purchasable if and only if <math>a_{N} \ge 0</math>.<br />
<br />
<i>Proof</i>: If <math>a_{N} \ge 0</math>, then we may simply pick <math>(a,b) = (a_{N},b_{N})</math> so <math>N</math> is purchasable. If <math>a_{N} < 0</math>, then <math>a_{N}+kn < 0</math> if <math>k \le 0</math> and <math>b_{N}-km < 0</math> if <math>k > 0</math>, hence at least one coordinate of <math>(a_{N}+kn,b_{N}-km)</math> is negative for all <math>k \in \mathbb{Z}</math>. Thus <math>N</math> is not purchasable. <math>\square</math><br />
<br />
Thus the set of non-purchasable integers is <math>\{xm+yn \;:\; x<0,0 \le y \le m-1\}</math>. We would like to find the maximum of this set. <br />
Since both <math>m,n</math> are positive, the maximum is achieved when <math>x = -1</math> and <math>y = m-1</math> so that <math>xm+yn = (-1)m+(m-1)n = mn-m-n</math>.<br />
<br />
==Proof 2==<br />
We start with this statement taken from [[Fermat%27s_Little_Theorem#Proof_2_.28Inverses.29|Proof 2 of Fermat's Little Theorem]]:<br />
<br />
"Let <math>S = \{1,2,3,\cdots, p-1\}</math>. Then, we claim that the set <math>a \cdot S</math>, consisting of the product of the elements of <math>S</math> with <math>a</math>, taken modulo <math>p</math>, is simply a permutation of <math>S</math>. In other words, <br />
<br />
<center><cmath>S \equiv \{1a, 2a, \cdots, (p-1)a\} \pmod{p}.</cmath></center><br><br />
<br />
Clearly none of the <math>ia</math> for <math>1 \le i \le p-1</math> are divisible by <math>p</math>, so it suffices to show that all of the elements in <math>a \cdot S</math> are distinct. Suppose that <math>ai \equiv aj \pmod{p}</math> for <math>i \neq j</math>. Since <math>\text{gcd}\, (a,p) = 1</math>, by the cancellation rule, that reduces to <math>i \equiv j \pmod{p}</math>, which is a contradiction."<br />
<br />
Because <math>m</math> and <math>n</math> are coprime, we know that multiplying the residues of <math>m</math> by <math>n</math> simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form <math>am+bn</math>, <math>a</math> is <math>0</math> and <math>b</math> is the original residue. We now prove the following lemma.<br />
<br />
<b>Lemma</b>: For any nonnegative integer <math>c < m</math>, <math>cn</math> is the least purchasable number <math>\equiv cn \bmod m</math>.<br />
<br />
<i>Proof</i>: Any number that is less than <math>cn</math> and congruent to it <math>\bmod m</math> can be represented in the form <math>cn-dm</math>, where <math>d</math> is a positive integer. If this is purchasable, we can say <math>cn-dm=am+bn</math> for some nonnegative integers <math>a, b</math>. This can be rearranged into <math>(a+d)m=(c-b)n</math>, which implies that <math>(a+d)</math> is a multiple of <math>n</math> (since <math>\gcd(m, n)=1</math>). We can say that <math>(a+d)=gn</math> for some positive integer <math>g</math>, and substitute to get <math>gmn=(c-b)n</math>. Because <math>c < m</math>, <math>(c-b)n < mn</math>, and <math>gmn < mn</math>. We divide by <math>mn</math> to get <math>g<1</math>. However, we defined <math>g</math> to be a positive integer, and all positive integers are greater than or equal to <math>1</math>. Therefore, we have a contradiction, and <math>cn</math> is the least purchasable number congruent to <math>cn \bmod m</math>. <math>\square</math><br />
<br />
This means that because <math>cn</math> is purchasable, every number that is greater than <math>cn</math> and congruent to it <math>\bmod m</math> is also purchasable (because these numbers are in the form <math>am+bn</math> where <math>b=c</math>). Another result of this Lemma is that <math>cn-m</math> is the greatest number <math>\equiv cn \bmod m</math> that is not purchasable. <math>c \leq m-1</math>, so <math>cn-m \leq (m-1)n-m=mn-m-n</math>, which shows that <math>mn-m-n</math> is the greatest number in the form <math>cn-m</math>. Any number greater than this and congruent to some <math>cn \bmod m</math> is purchasable, because that number is greater than <math>cn</math>. All numbers are congruent to some <math>cn</math>, and thus all numbers greater than <math>mn-m-n</math> are purchasable.<br />
<br />
Putting it all together, we can say that for any coprime <math>m</math> and <math>n</math>, <math>mn-m-n</math> is the greatest number not representable in the form <math>am + bn</math> for nonnegative integers <math>a, b</math>. <math>\square</math><br />
<br />
==Corollary==<br />
This corollary is based off of Proof 2, so it is necessary to read that proof before this corollary. We prove the following lemma.<br />
<br />
<b>Lemma</b> For any integer <math>k</math>, exactly one of the integers <math>k</math>, <math>mn-m-n-k</math> is not purchasable.<br />
<br />
<i>Proof</i>: Because every number is congruent to some residue of <math>m</math> permuted by <math>n</math>, we can set <math>k \equiv cn \bmod m</math> for some <math>c</math>. We can break this into two cases.<br />
<br />
<i>Case 1</i>: <math>k \leq cn-m</math>. This implies that <math>k</math> is not purchasable, and that <math>mn-m-n-k \geq mn-m-n-(cn-m) = n(m-1-c)</math>. <math>n(m-1-c)</math> is a permuted residue, and a result of the lemma in Proof 2 was that a permuted residue is the least number congruent to itself <math>\bmod m</math> that is purchasable. Therefore, <math>mn-m-n-k \equiv n(m-1-c) \bmod m</math> and <math>mn-m-n-k \geq n(m-1-c)</math>, so <math>mn-m-n-k</math> is purchasable.<br />
<br />
<i>Case 2</i>: <math>k > cn-m</math>. This implies that <math>k</math> is purchasable, and that <math>mn-m-n-k < mn-m-n-(cn-m) = n(m-1-c)</math>. Again, because <math>n(m-1-c)</math> is the least number congruent to itself <math>\bmod m</math> that is purchasable, and because <math>mn-m-n-k \equiv n(m-1-c) \bmod m</math> and <math>mn-m-n-k < n(m-1-c)</math>, <math>mn-m-n-k</math> is not purchasable.<br />
<br />
We now limit the values of <math>k</math> to all integers <math>0 \leq k \leq \frac{mn-m-n}{2}</math>, which limits the values of <math>mn-m-n-k</math> to <math>mn-m-n \geq mn-m-n-k \geq \frac{mn-m-n}{2}</math>. Because <math>m</math> and <math>n</math> are coprime, only one of them can be a multiple of <math>2</math>. Therefore, <math>mn-m-n \equiv (0)(1)-0-1 \equiv -1 \equiv 1 \bmod 2</math>, showing that <math>\frac{mn-m-n}{2}</math> is not an integer and that <math>\frac{mn-m-n-1}{2}</math> and <math>\frac{mn-m-n+1}{2}</math> are integers. We can now set limits that are equivalent to the previous on the values of <math>k</math> and <math>mn-m-n-k</math> so that they cover all integers form <math>0</math> to <math>mn-m-n</math> without overlap: <math>0 \leq k \leq \frac{mn-m-n-1}{2}</math> and <math>\frac{mn-m-n+1}{2} \leq mn-m-n-k \leq mn-m-n</math>. There are <math>\frac{mn-m-n-1}{2}+1=\frac{(m-1)(n-1)}{2}</math> values of <math>k</math>, and each is paired with a value of <math>mn-m-n-k</math>, so we can make <math>\frac{(m-1)(n-1)}{2}</math> different ordered pairs of the form <math>(k, mn-m-n-k)</math>. The coordinates of these ordered pairs cover all integers from <math>0</math> to <math>mn-m-n</math> inclusive, and each contains exactly one not-purchasable integer, so that means that there are <math>\frac{(m-1)(n-1)}{2}</math> different not-purchasable integers from <math>0</math> to <math>mn-m-n</math>. All integers greater than <math>mn-m-n</math> are purchasable, so that means there are a total of <math>\frac{(m-1)(n-1)}{2}</math> integers <math>\geq 0</math> that are not purchasable.<br />
<br />
In other words, for every pair of coprime integers <math>m, n</math>, there are exactly <math>\frac{(m-1)(n-1)}{2}</math> nonnegative integers that cannot be represented in the form <math>am + bn</math> for nonnegative integers <math>a, b</math>. <math>\square</math><br />
<br />
==Generalization==<br />
If <math>m</math> and <math>n</math> are not relatively prime, then we can simply rearrange <math>am+bn</math> into the form<br />
<cmath>\gcd(m,n) \left( a\frac{m}{\gcd(m,n)}+b\frac{n}{\gcd(m,n)} \right)</cmath><br />
<math>\frac{m}{\gcd(m,n)}</math> and <math>\frac{n}{\gcd(m,n)}</math> are relatively prime, so we apply Chicken McNugget to find a bound<br />
<cmath>\frac{mn}{\gcd(m,n)^{2}}-\frac{m}{\gcd(m,n)}-\frac{n}{\gcd(m,n)}</cmath><br />
We can simply multiply <math>\gcd(m,n)</math> back into the bound to get<br />
<cmath>\frac{mn}{\gcd(m,n)}-m-n=\textrm{lcm}(m, n)-m-n</cmath><br />
Therefore, all multiples of <math>\gcd(m, n)</math> greater than <math>\textrm{lcm}(m, n)-m-n</math> are representable in the form <math>am+bn</math> for some positive integers <math>a, b</math>.<br />
<br />
=Problems=<br />
<br />
===Simple===<br />
*Marcy buys paint jars in containers of <math>2</math> and <math>7</math>. What's the largest number of paint jars that Marcy can't obtain? <br />
<br />
Answer: <math>5</math> containers<br />
<br />
*Bay Area Rapid food sells chicken nuggets. You can buy packages of <math>11</math> or <math>7</math>. What is the largest integer <math>n</math> such that there is no way to buy exactly <math>n</math> nuggets? Can you Generalize ?(ACOPS) <br />
<br />
Answer: <math>n=59</math> <br />
<br />
*If a game of American Football has only scores of field goals (<math>3</math> points) and touchdowns with the extra point (<math>7</math> points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)?<br />
<br />
Answer: <math>11</math> points<br />
<br />
*The town of Hamlet has <math>3</math> people for each horse, <math>4</math> sheep for each cow, and <math>3</math> ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?<br />
<br />
<math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math> [[2015 AMC 10B Problems/Problem 15|AMC 10B 2015 Problem 15]]<br />
<br />
Answer: <math>47\qquad\textbf{(C) }</math><br />
<br />
===Intermediate===<br />
*Ninety-four bricks, each measuring <math>4''\times10''\times19'',</math> are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes <math>4''\,</math> or <math>10''\,</math> or <math>19''\,</math> to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? [[1994 AIME Problems/Problem 11|AIME]]<br />
<br />
*Find the sum of all positive integers <math>n</math> such that, given an unlimited supply of stamps of denominations <math>5,n,</math> and <math>n+1</math> cents, <math>91</math> cents is the greatest postage that cannot be formed. [[2019 AIME II Problems/Problem 14|AIME II 2019 Problem 14]]<br />
<br />
===Olympiad===<br />
*On the real number line, paint red all points that correspond to integers of the form <math>81x+100y</math>, where <math>x</math> and <math>y</math> are positive integers. Paint the remaining integer points blue. Find a point <math>P</math> on the line such that, for every integer point <math>T</math>, the reflection of <math>T</math> with respect to <math>P</math> is an integer point of a different colour than <math>T</math>. (India TST)<br />
<br />
*Let <math>S</math> be a set of integers (not necessarily positive) such that<br />
<br />
(a) there exist <math>a,b \in S</math> with <math>\gcd(a,b)=\gcd(a-2,b-2)=1</math>;<br />
<br />
(b) if <math>x</math> and <math>y</math> are elements of <math>S</math> (possibly equal), then <math>x^2-y</math> also belongs to <math>S</math>. <br />
<br />
Prove that <math>S</math> is the set of all integers. (USAMO)<br />
<br />
==See Also==<br />
*[[Theorem]]<br />
*[[Prime]]<br />
<br />
[[Category:Theorems]]<br />
[[Category:Number theory]]</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_15&diff=1282022015 AMC 10B Problems/Problem 152020-07-13T15:52:10Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
The town of Hamlet has <math>3</math> people for each horse, <math>4</math> sheep for each cow, and <math>3</math> ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?<br />
<br />
<math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math><br />
<br />
==Solution==<br />
===Solution 1===<br />
Let the amount of people be <math>p</math>, horses be <math>h</math>, sheep be <math>s</math>, cows be <math>c</math>, and ducks be <math>d</math>. <br />
We know <cmath>3h=p</cmath> <cmath>4c=s</cmath> <cmath>3p=d</cmath> Then the total amount of people, horses, sheep, cows, and ducks may be written as <math>p+h+s+c+d = 3h+h+4c+c+(3\times3h)</math>. This is equivalent to <math>13h+5c</math>. Looking through the options, we see <math>47</math> is impossible to make for integer values of <math>h</math> and <math>c</math>. So the answer is <math>\boxed{\textbf{(B)} 47}</math>.<br />
<br />
===Solution 2===<br />
As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as <math>13d+5s</math>. However, instead of going through each of the solutions and testing the the options, you can use the [[Chicken McNugget]] Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form <math>13d+5s</math>. <br />
<cmath>13*5-13-5=47,</cmath> so our answer is <math>\boxed{\textbf{(B)} 47}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_15&diff=1282012015 AMC 10B Problems/Problem 152020-07-13T15:51:46Z<p>Sakshamsethi: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
The town of Hamlet has <math>3</math> people for each horse, <math>4</math> sheep for each cow, and <math>3</math> ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?<br />
<br />
<math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math><br />
<br />
==Solution==<br />
===Solution 1===<br />
Let the amount of people be <math>p</math>, horses be <math>h</math>, sheep be <math>s</math>, cows be <math>c</math>, and ducks be <math>d</math>. <br />
We know <cmath>3h=p</cmath> <cmath>4c=s</cmath> <cmath>3p=d</cmath> Then the total amount of people, horses, sheep, cows, and ducks may be written as <math>p+h+s+c+d = 3h+h+4c+c+(3\times3h)</math>. This is equivalent to <math>13h+5c</math>. Looking through the options, we see <math>47</math> is impossible to make for integer values of <math>h</math> and <math>c</math>. So the answer is <math>\boxed{\textbf{(B)} 47}</math>.<br />
<br />
===Solution 2===<br />
As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as <math>13d+5s</math>. However, instead of going through each of the solutions and testing the the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form <math>13d+5s</math>. <br />
<cmath>13*5-13-5=47,</cmath> so our answer is <math>\boxed{\textbf{(B)} 47}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_15&diff=1280712012 AMC 10B Problems/Problem 152020-07-11T15:19:58Z<p>Sakshamsethi: /* Solution */</p>
<hr />
<div>==Problem==<br />
In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end on the tournament?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math><br />
==Solution==<br />
The total number of games (and wins) in the tournament is <math>\frac{6 \times 5}{2}= 15</math>. A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:<br />
| 1 2 3 4 5 6 |<br />
|1 X W L W L W |<br />
|2 L X W L W W |<br />
|3 W L X W L W |<br />
|4 L W L X W W |<br />
|5 W L W L X W |<br />
|6 L L L L L X |<br />
The "X's" are for when it is where a team is set against itself, which cannot happen. The chart says that Team 6 has lost all of its matches, which means that each of the other teams won against it. Then, alternating Wins and Losses were tried. It shows that it is possible for 5 teams to tie for the same amount of wins, which in this case is 3 wins.<br />
Thus, the answer is <math>\boxed{\textbf{(D)}\ 5}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_9&diff=1280662012 AMC 10B Problems/Problem 92020-07-11T14:44:24Z<p>Sakshamsethi: /* Solution */</p>
<hr />
<div>== Problem 9 ==<br />
Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math><br />
<br />
== Solution ==<br />
<br />
Out of the first two integers, it's possible for both to be even: for example, <math>10 + 16 = 26.</math> But the next two integers, when added, increase the sum by <math>15,</math> which is odd, so one of them must be odd and the other must be even: for example, <math>3 + 12 = 15.</math> Finally, the next two integers increase the sum by <math>16,</math> which is even, so we can have both be even: for example, <math>2 + 14 = 16.</math> Therefore, <math>\boxed{\textbf{(A) } 1}</math> is the minimum number of integers that must be odd.<br />
<br />
<br />
Sidenote by Williamgolly:<br />
You can use the sum of each two pairs of numbers and take mod 2. See [[modular arithmetic]]<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_6&diff=1279792010 AMC 12B Problems/Problem 62020-07-10T14:49:06Z<p>Sakshamsethi: /* Problem 6 */</p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #6]] and [[2010 AMC 10B Problems|2010 AMC 10B #12]]}}<br />
<br />
== Problem 6 ==<br />
At the beginning of the school year, <math>50\%</math> of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and <math>50\%</math> answered "No." At the end of the school year, <math>70\%</math> answered "Yes" and <math>30\%</math> answered "No." Altogether, <math>x\%</math> of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80</math><br />
<br />
== Solution ==<br />
Clearly, the minimum possible value would be <math>70 - 50 = 20\%</math>. The maximum possible value would be <math>30 + 50 = 80\%</math>. The difference is <math>80 - 20 = \boxed{60}</math> <math>(D)</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=5|num-a=7|ab=B}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_5&diff=1279412003 AMC 10B Problems/Problem 52020-07-09T22:45:09Z<p>Sakshamsethi: /* Solution 2 (very easy to understand) */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #4]] and [[2003 AMC 10B Problems|2003 AMC 10B #5]]}}<br />
<br />
==Problem==<br />
<br />
Moe uses a mower to cut his rectangular <math>90</math>-foot by <math>150</math>-foot lawn. The swath he cuts is <math>28</math> inches wide, but he overlaps each cut by <math>4</math> inches to make sure that no grass is missed. He walks at the rate of <math>5000</math> feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn.<br />
<br />
<math>\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3 </math><br />
<br />
==Solution==<br />
<br />
Since the swath Moe actually mows is <math>24</math> inches, or <math>2</math> feet wide, he mows <math>10000</math> square feet in one hour. His lawn has an area of <math>13500</math>, so it will take Moe <math>1.35</math> hours to finish mowing the lawn. Thus the answer is <math>\boxed{\textbf{(C) } 1.35}</math>.<br />
<br />
<br />
<br />
==Solution 2 (very easy to understand)==<br />
<br />
Let's assume that the swath moves back and forth; parallel to the <math>90</math> feet side. Thus, the length of one strip is <math>90</math> feet. Now we need to find out how many strips there are. In reality, the swath Moe mows is <math>24</math> inches wide, which can be easily translated into <math>2</math> feet. <math>\frac{150}{2}</math> is the number of strips Moe needs to mow, which is equal to <math>75</math>. <br />
Therefore, the total number of feet Moe mows is <math>75\times90</math>. Since Moe's mowing rate is <math>5000</math> feet per hour, <math>\frac{75\times90}{5000}</math> is the number of hours it takes him to do his job. Using basic calulations, we compute the answer. <math>\boxed{\textbf{(C) } 1.35}</math>.<br />
<br />
~sakshamsethi<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=3|num-a=5}}<br />
{{AMC10 box|year=2003|ab=B|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_17&diff=1279001998 AJHSME Problems/Problem 172020-07-09T17:46:39Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>==Don't Crowd the Isles==<br />
<br />
Problems 15, 16, and 17 all refer to the following:<br />
<br />
<center><br />
In the very center of the Irenic Sea lie the beautiful Nisos Isles. In 1998 the number of people on these islands is only 200, but the population triples every 25 years. Queen Irene has decreed that there must be at least 1.5 square miles for every person living in the Isles. The total area of the Nisos Isles is 24,900 square miles.<br />
</center><br />
<br />
===Problem 17===<br />
<br />
In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?<br />
<br />
<math>\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}</math><br />
<br />
==Solution 1==<br />
<br />
We can divide the total area by how much will be occupied per person:<br />
<br />
<math>\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}</math> can stay on the island at its maximum capacity.<br />
<br />
We can divide this by the current population the island in the year <math>1998</math> to see by what factor the population increases:<br />
<br />
<math>\frac{16600}{200}=83</math>-fold increase in population.<br />
<br />
Thus, the population increases by a factor <math>83</math>. This is very close to <math>3 \times 3 \times 3 \times 3 = 81</math>, and so there are about <math>4</math> triplings of the island's population.<br />
<br />
It takes <math>4\times25=100=\boxed{C}</math> years to triple the island's population four times in succession.<br />
<br />
==Solution 2==<br />
<br />
We can continue the pattern, and because the pattern increases numbers rapidly, it won't be hard. <math>\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}</math> can live on the island at its maximum capacity. <br />
<br />
<math>1998: 200 \text{ people}</math> <math>\\$<br />
</math>2023: 600 \text{ people}<math> </math>\\$ <br />
<math>2048: 1800 \text{ people}</math> <math>\\$<br />
</math>2073: 5400 \text{ people}<math> </math>\\$<br />
<math>2098: 16200 \text{ people}</math> <math>\\$<br />
After the year </math>2098<math>, it will not be possible for the next increase to occur, because when </math>16200<math> is tripled, it is way more than the maximum capacity.\\ <br />
<br />
Thus, the answer is </math>2098-1998 = 100<math>, or </math>\boxed{C}$<br />
<br />
== See also ==<br />
{{AJHSME box|year=1998|num-b=16|num-a=18}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_17&diff=1278991998 AJHSME Problems/Problem 172020-07-09T17:38:20Z<p>Sakshamsethi: /* Solution 2 */</p>
<hr />
<div>==Don't Crowd the Isles==<br />
<br />
Problems 15, 16, and 17 all refer to the following:<br />
<br />
<center><br />
In the very center of the Irenic Sea lie the beautiful Nisos Isles. In 1998 the number of people on these islands is only 200, but the population triples every 25 years. Queen Irene has decreed that there must be at least 1.5 square miles for every person living in the Isles. The total area of the Nisos Isles is 24,900 square miles.<br />
</center><br />
<br />
===Problem 17===<br />
<br />
In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?<br />
<br />
<math>\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}</math><br />
<br />
==Solution 1==<br />
<br />
We can divide the total area by how much will be occupied per person:<br />
<br />
<math>\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}</math> can stay on the island at its maximum capacity.<br />
<br />
We can divide this by the current population the island in the year <math>1998</math> to see by what factor the population increases:<br />
<br />
<math>\frac{16600}{200}=83</math>-fold increase in population.<br />
<br />
Thus, the population increases by a factor <math>83</math>. This is very close to <math>3 \times 3 \times 3 \times 3 = 81</math>, and so there are about <math>4</math> triplings of the island's population.<br />
<br />
It takes <math>4\times25=100=\boxed{C}</math> years to triple the island's population four times in succession.<br />
<br />
==Solution 2==<br />
<br />
We can continue the pattern, and because the pattern increases numbers rapidly, it won't be hard. <math>\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}</math> can live on the island at its maximum capacity. <br />
<br />
<math>1998</math>: <math>200</math> \text{ people}<br />
<math>2023</math>: <math>600</math> \text{ people}<br />
<math>2048</math>: <math>1800</math> \text{ people}<br />
<math>2073</math>: <math>5400</math> \text{ people}<br />
<math>2098</math>: <math>16200</math> \text{ people}<br />
After 2098, it will not be possible for the next increase to occur, because when <math>16200</math> is tripled, it is way more than the maximum capacity. <br />
<br />
Thus, the answer is <math>2098-1998 = 100</math>, or <math>\boxed{C}</math><br />
<br />
== See also ==<br />
{{AJHSME box|year=1998|num-b=16|num-a=18}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Sakshamsethihttps://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_17&diff=1278981998 AJHSME Problems/Problem 172020-07-09T17:29:18Z<p>Sakshamsethi: /* Solution */</p>
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<div>==Don't Crowd the Isles==<br />
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Problems 15, 16, and 17 all refer to the following:<br />
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In the very center of the Irenic Sea lie the beautiful Nisos Isles. In 1998 the number of people on these islands is only 200, but the population triples every 25 years. Queen Irene has decreed that there must be at least 1.5 square miles for every person living in the Isles. The total area of the Nisos Isles is 24,900 square miles.<br />
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===Problem 17===<br />
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In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?<br />
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<math>\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}</math><br />
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==Solution 1==<br />
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We can divide the total area by how much will be occupied per person:<br />
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<math>\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}</math> can stay on the island at its maximum capacity.<br />
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We can divide this by the current population the island in the year <math>1998</math> to see by what factor the population increases:<br />
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<math>\frac{16600}{200}=83</math>-fold increase in population.<br />
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Thus, the population increases by a factor <math>83</math>. This is very close to <math>3 \times 3 \times 3 \times 3 = 81</math>, and so there are about <math>4</math> triplings of the island's population.<br />
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It takes <math>4\times25=100=\boxed{C}</math> years to triple the island's population four times in succession.<br />
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==Solution 2==<br />
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== See also ==<br />
{{AJHSME box|year=1998|num-b=16|num-a=18}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Sakshamsethi