https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Panda08&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T02:06:02ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=1407202018 AMC 8 Problems/Problem 222020-12-27T04:32:20Z<p>Panda08: /* Solution 5 */</p>
<hr />
<div>==Problem 22==<br />
Point <math>E</math> is the midpoint of side <math>\overline{CD}</math> in square <math>ABCD,</math> and <math>\overline{BE}</math> meets diagonal <math>\overline{AC}</math> at <math>F.</math> The area of quadrilateral <math>AFED</math> is <math>45.</math> What is the area of <math>ABCD?</math><br />
<br />
<asy><br />
size(5cm);<br />
draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br />
draw((0,6)--(6,0)); draw((3,0)--(6,6));<br />
label("$A$",(0,6),NW);<br />
label("$B$",(6,6),NE);<br />
label("$C$",(6,0),SE);<br />
label("$D$",(0,0),SW);<br />
label("$E$",(3,0),S);<br />
label("$F$",(4,2),E);<br />
</asy><br />
<br />
<math>\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144</math><br />
<br />
==Solution 1==<br />
Let the area of <math>\triangle CEF</math> be <math>x</math>. Thus, the area of triangle <math>\triangle ACD</math> is <math>45+x</math> and the area of the square is <math>2(45+x) = 90+2x</math>.<br />
<br />
By AA similarity, <math>\triangle CEF \sim \triangle ABF</math> with a 1:2 ratio, so the area of triangle <math>\triangle ABF</math> is <math>4x</math>. Now consider trapezoid <math>ABED</math>. Its area is <math>45+4x</math>, which is three-fourths the area of the square. We set up an equation in <math>x</math>:<br />
<br />
<cmath> 45+4x = \frac{3}{4}\left(90+2x\right) </cmath><br />
Solving, we get <math>x = 9</math>. The area of square <math>ABCD</math> is <math>90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}</math>.<br />
<br />
==Solution 2==<br />
We can use analytic geometry for this problem.<br />
<br />
Let us start by giving <math>D</math> the coordinate <math>(0,0)</math>, <math>A</math> the coordinate <math>(0,1)</math>, and so forth. <math>\overline{AC}</math> and <math>\overline{EB}</math> can be represented by the equations <math>y=-x+1</math> and <math>y=2x-1</math>, respectively. Solving for their intersection gives point <math>F</math> coordinates <math>\left(\frac{2}{3},\frac{1}{3}\right)</math>. <br />
<br />
Now, <math>\triangle</math><math>EFC</math>’s area is simply <math>\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}</math> or <math>\frac{1}{12}</math>. This means that pentagon <math>ABCEF</math>’s area is <math>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}</math> of the entire square, and it follows that quadrilateral <math>AFED</math>’s area is <math>\frac{5}{12}</math> of the square. <br />
<br />
The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>.<br />
==Solution 3==<br />
Note that triangle <math>ABC</math> has half the area of the square and triangle <math>FEC</math> has <math>\dfrac1{12}</math>th. Thus the area of the quadrilateral is <math>1-1/2-1/12=5/12</math> th the area of the square. The area of the square is then <math>45\cdot\dfrac{12}{5}=\boxed{\textbf{(B.)}108}</math>.<br />
<br />
==Solution 4==<br />
Extend <math>\overline{AD}</math> and <math>\overline{BE}</math> to meet at <math>X</math>. Drop an altitude from <math>F</math> to <math>\overline{CE}</math> and call it <math>h</math>. Also, call <math>\overline{CE}</math> <math>x</math>. As stated before, we have <math>\triangle ABF \sim \triangle CEF</math>, so the ratio of their heights is in a <math>1:2</math> ratio, making the altitude from <math>F</math> to <math>\overline{AB}</math> <math>2h</math>. Note that this means that the side of the square is <math>3h</math>. In addition, <math>\triangle XDE \sim \triangle XAB</math> by AA Similarity in a <math>1:2</math> ratio. This means that the side length of the square is <math>2x</math>, making <math>3h=2x</math>.<br />
<br />
Now, note that <math>[ADEF]=[XAB]-[XDE]-[ABF]</math>. We have <math>[\triangle XAB]=(4x)(2x)/2=4x^2,</math> <math>[\triangle XDE]=(x)(2x)/2=x^2,</math> and <math>[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.</math> Subtracting makes <math>[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.</math> We are given that <math>[ADEF]=45,</math> so <math>5x^2/3=45 \Rightarrow x^2=27.</math> Therefore, <math>x= 3 \sqrt{3},</math> so our answer is <math>(2x)^2=4x^2=4(27)=\boxed{\textbf{(B) }108}.</math> - moony_eyed<br />
<br />
==Solution 5==<br />
<br />
Solution with Cartesian and Barycentric Coordinates:<br />
<br />
We start with the following<br />
<br />
Claim: Given a square <math>ABCD</math>, let <math>E</math> be the midpoint of <math>\overline{DC}</math> and let <math>BE\cap AC = F</math>. Then <math>\frac {AF}{FC}=2</math>.<br />
<br />
Proof. We use Cartesian coordinates. Let <math>D</math> be the origin, <math>A=(0,1),C=(0,1),B=(1,1)</math>. We have that <math>\overline{AC}</math> and <math>\overline{EB}</math> are governed by the equations <math>y=-x+1</math> and <math>y=2x-1</math>, respectively. Solving, <math>F=\left(\frac{2}{3},\frac{1}{3}\right)</math>. The result follows. <math>\square</math><br />
<br />
Now we apply Barycentric Coordinates w.r.t. <math>\triangle ACD</math>. We let <math>A=(1,0,0),D=(0,1,0),C=(0,0,1)</math>. Then <math>E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)</math>. <br />
<br />
In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.</cmath> Let <math>[FEC]=x</math> so that <math>[ACD]=45+x</math>. Then we have <math>\frac{x}{x+45}=\frac 16 \Rightarrow x=9</math> so the answer is <math>2(45+9)=108</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/veaDx64aX0g<br />
<br />
=See Also=<br />
{{AMC8 box|year=2018|num-b=21|num-a=23}}<br />
Set s to be the bottom left triangle.<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1407172019 AMC 8 Problems/Problem 242020-12-27T03:41:20Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div>==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> so that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and let <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?<br />
<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
<br />
Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}</math><br />
<br />
==Solution 2 (Mass Points)==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(7cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br />
draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br />
draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br />
draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br />
draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((0.28,2.39),dotstyle); <br />
label("$A$", (0.36,2.59), NE * labelscalefactor); <br />
dot((-2.8,-1.17),dotstyle); <br />
label("$B$", (-2.72,-0.97), NE * labelscalefactor); <br />
dot((3.78,-1.05),dotstyle); <br />
label("$C$", (3.86,-0.85), NE * labelscalefactor); <br />
dot((1.2887445398528459,1.3985482236874887),dotstyle); <br />
label("$D$", (1.36,1.59), NE * labelscalefactor); <br />
dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br />
label("$F$", (-0.64,-0.93), NE * labelscalefactor); <br />
dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br />
label("$E$", (-0.2,0.57), NE * labelscalefactor); <br />
label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br />
label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); <br />
label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br />
label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); <br />
label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br />
<br />
First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of <math>3</math>. By equality, point <math>B</math> has a mass of <math>3</math> also. <br />
<br />
Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>.<br />
<br />
Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases.<br />
<br />
So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}</math><br />
-Brudder<br />
<br />
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{\textbf{(B) }30}</math><br />
-Brudder<br />
<br />
==Solution 3==<br />
<math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{\textbf{(B) }30}</math>. ~~SmileKat32<br />
<br />
==Solution 4 (Similar Triangles)==<br />
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown:<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
</asy><br />
Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>.<br />
<br />
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>.<br />
<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$60$", (A+G+D)/3);<br />
label("$30$", (B+E+F)/3);<br />
</asy><br />
(Credit to MP8148 for the idea)<br />
<br />
==Solution 5 (Area Ratios)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. <br />
(Credit to scrabbler94 for the idea)<br />
<br />
==Solution 6 (Coordinate Bashing)==<br />
Let <math>ADB</math> be a right triangle, and <math>BD=CD</math><br />
<br />
Let <math>A=(-2\sqrt{30}, 0)</math><br />
<br />
<math>B=(0, 4\sqrt{30})</math><br />
<br />
<math>C=(4\sqrt{30}, 0)</math><br />
<br />
<math>D=(0, 0)</math><br />
<br />
<math>E=(0, 2\sqrt{30})</math><br />
<br />
<math>F=(\sqrt{30}, 3\sqrt{30})</math><br />
<br />
The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math><br />
<br />
The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math><br />
<br />
Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math><br />
<br />
Now we can find <math>EF=BF=2\sqrt{15}</math><br />
<br />
<math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare</math><br />
<br />
-Trex4days<br />
<br />
== Solution 7 ==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(15cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br />
<br />
/* draw figures */<br />
draw(circle((0,0), 5), linewidth(2)); <br />
draw((-4,-3)--(4,3), linewidth(2)); <br />
draw((-4,-3)--(0,5), linewidth(2)); <br />
draw((0,5)--(4,3), linewidth(2)); <br />
draw((12,-1)--(-4,-3), linewidth(2)); <br />
draw((0,5)--(0,-5), linewidth(2)); <br />
draw((-4,-3)--(0,-5), linewidth(2)); <br />
draw((4,3)--(0,2.48), linewidth(2)); <br />
draw((4,3)--(12,-1), linewidth(2)); <br />
draw((-4,-3)--(4,3), linewidth(2)); <br />
/* dots and labels */<br />
dot((0,0),dotstyle); <br />
label("E", (0.27,-0.24), NE * labelscalefactor); <br />
dot((-5,0),dotstyle); <br />
dot((-4,-3),dotstyle); <br />
label("B", (-4.45,-3.38), NE * labelscalefactor); <br />
dot((4,3),dotstyle); <br />
label("$D$", (4.15,3.2), NE * labelscalefactor); <br />
dot((0,5),dotstyle); <br />
label("A", (-0.09,5.26), NE * labelscalefactor); <br />
dot((12,-1),dotstyle); <br />
label("C", (12.23,-1.24), NE * labelscalefactor); <br />
dot((0,-5),dotstyle); <br />
label("$G$", (0.19,-4.82), NE * labelscalefactor); <br />
dot((0,2.48),dotstyle); <br />
label("I", (-0.33,2.2), NE * labelscalefactor); <br />
dot((0,0),dotstyle); <br />
label("E", (0.27,-0.24), NE * labelscalefactor); <br />
dot((0,-2.5),dotstyle); <br />
label("F", (0.23,-2.2), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
Let <math>A[\Delta XYZ]</math> = <math>\text{Area of Triangle XYZ}</math> <br />
<br />
<br />
<math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math><br />
<br />
<br />
<math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts)<br />
<br />
<br />
Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>.<br />
<br />
<br />
<math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic)<br />
<br />
<br />
But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>.<br />
<br />
\therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).<br />
<br />
In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>.<br />
<br />
Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math><br />
<br />
<br />
Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>.<br />
<br />
<math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>.<br />
<br />
Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math><br />
<br />
Thus <math>A[\Delta BEF] = \boxed{\textbf{(B) }30}</math><br />
<br />
<br />
~phoenixfire & flamewavelight<br />
<br />
== Solution 8 ==<br />
<asy><br />
import geometry;<br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF, M;<br />
B = (0,0); C = (3,0); M = (1.45,0);<br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
draw(EE--M,StickIntervalMarker(1,1));<br />
label("$M$",M,S);<br />
draw(A--DD,invisible,StickIntervalMarker(1,1));<br />
dot((DD+C)/2);<br />
draw(DD--C,invisible,StickIntervalMarker(2,1));<br />
</asy><br />
Using the ratio of <math>\overline{AD}</math> and <math>\overline{CD}</math>, we find the area of <math>\triangle ADB</math> is <math>120</math> and the area of <math>\triangle BDC</math> is <math>240</math>. Also using the fact that <math>E</math> is the midpoint of <math>\overline{BD}</math>, we know <math>\triangle ADE = \triangle ABE = 60</math>.<br />
Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. We immediatley know that <math>\triangle BEM \sim BDC</math> by <math>2</math>. Using that we can conclude <math>EM</math> has ratio <math>1</math>. Using <math>\triangle EFM \sim \triangle AFC</math>, we get <math>EF:AE = 1:2</math>. Therefore using the fact that <math>\triangle EBF</math> is in <math>\triangle ABF</math>, the area has ratio <math>\triangle BEF : \triangle ABE=1:2</math> and we know <math>\triangle ABE</math> has area <math>60</math> so <math>\triangle BEF</math> is <math>\boxed{\textbf{(B) }30}</math>. - fath2012<br />
<br />
==Solution 9 (Menelaus's Theorem)==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
By Menelaus's Theorem on triangle <math>BCD</math>, we have <cmath>\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.</cmath> Therefore, <cmath>[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.</cmath><br />
<br />
==Solution 10 (Graph Paper)==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,D,E,F,a,b,c,d,e,f;<br />
A = (2,3);<br />
B = (0,2); <br />
C = (2,0);<br />
D = (2/3)*A+(1/3)*C;<br />
E = (B+D)/2;<br />
F = intersectionpoint(B--C,A--A+2*(E-A));<br />
a = (0,0);<br />
b = (1,0);<br />
c = (2,1);<br />
d = (1,3);<br />
e = (0,3);<br />
f = (0,1);<br />
draw(a--C,dashed);<br />
draw(f--c,dashed);<br />
draw(e--A,dashed);<br />
draw(a--e,dashed);<br />
draw(b--d,dashed);<br />
draw(A--B--C--cycle);<br />
draw(A--F); <br />
draw(B--D);<br />
dot(A); <br />
label("$A$",A,NE);<br />
dot(B); <br />
label("$B$",B,dir(180));<br />
dot(C); <br />
label("$C$",C,SE);<br />
dot(D); <br />
label("$D$",D,dir(0));<br />
dot(E); <br />
label("$E$",E,SE);<br />
dot(F); <br />
label("$F$",F,SW);<br />
</asy><br />
<b>Note:</b> If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.<br><br />
<br><br />
As triangle <math>ABC</math> is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.<br><br />
<br><br />
As point <math>D</math> splits line segment <math>\overline{AC}</math> in a <math>1:2</math> ratio, we draw <math>\overline{AC}</math> as a vertical line segment <math>3</math> units long. Point <math>D</math> is thus <math>1</math> unit below point <math>A</math> and <math>2</math> units above point <math>C</math>. By definition, Point <math>E</math> splits line segment <math>\overline{BD}</math> in a <math>1:1</math> ratio, so we draw <math>\overline{BD}</math> <math>2</math> units long directly left of <math>D</math> and draw <math>E</math> directly between <math>B</math> and <math>D</math>, <math>1</math> unit away from both.<br><br />
<br><br />
We then draw line segments <math>\overline{AB}</math> and <math>\overline{BC}</math>. We can easily tell that triangle <math>ABC</math> occupies <math>3</math> square units of space. Constructing line <math>AE</math> and drawing <math>F</math> at the intersection of <math>AE</math> and <math>BC</math>, we can easily see that triangle <math>EBF</math> forms a right triangle occupying <math>\frac{1}{4}</math> of a square unit of space.<br><br />
<br><br />
The ratio of the areas of triangle <math>EBF</math> and triangle <math>ABC</math> is thus <math>\frac{1}{4}\div3=\frac{1}{12}</math>, and since the area of triangle <math>ABC</math> is <math>360</math>, this means that the area of triangle <math>EBF</math> is <math>\frac{1}{12}\times360=\boxed{\textbf{(B) }30}</math>. ~[[User:emerald_block|emerald_block]]<br><br />
<br><br />
<b>Additional note:</b> There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br />
<br />
==Solution 11==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF,G;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
G = (1.5,0);<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD); <br />
draw(G--DD);<br />
label("$A$",A,N);<br />
label("$B$",<br />
B,SW); <br />
label("$C$",C,SE);<br />
label("$D$",DD,NE); <br />
label("$E$",EE,NW);<br />
label("$F$",FF,S);<br />
label("$G$",G,S);<br />
</asy><br />
We know that <math>AD = \dfrac{1}{3} AC</math>, so <math>[ABD] = \dfrac{1}{3} [ABC] = 120</math>. Using the same method, since <math>BE = \dfrac{1}{2} BD</math>, <math>[ABE] = \dfrac{1}{2} [ABD] = 60</math>. Next, we draw <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{DG}</math> is parallel to <math>\overline{AF}</math> and create segment <math>DG</math>. We then observe that <math>\triangle AFC \sim \triangle DGC</math>, and since <math>AD:DC = 1:2</math>, <math>FG:GC</math> is also equal to <math>1:2</math>. Similarly (no pun intended), <math>\triangle DBG \sim \triangle EBF</math>, and since <math>BE:ED = 1:1</math>, <math>BF:FG</math> is also equal to <math>1:1</math>. Combining the information in these two ratios, we find that <math>BF:FG:GC = 1:1:2</math>, or equivalently, <math>BF = \dfrac{1}{4} BC</math>. Thus, <math>[BFA] = \dfrac{1}{4} [BCA] = 90</math>. We already know that <math>[ABE] = 60</math>, so the area of <math>\triangle EBF</math> is <math>[BFA] - [ABE] = \boxed{\textbf{(B) }30}</math>. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br />
<br />
==Solution 12 (Fastest Solution if you have no time)==<br />
The picture is misleading. Assume that the triangle ABC is right. <br />
<br />
Then find two factors of <math>720</math> that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near <math>720</math> to use difference of squares, we find <math>24</math> and <math>30</math> as our numbers. Then the coordinates of D are <math>(10,16)</math>(note, A=0,0). E is then <math>(5,8)</math>. Then the equation of the line AE is <math>-16x/5+24=y</math>. Plugging in <math>y=0</math>, we have <math>x=\dfrac{15}{2}</math>. Now notice that we have both the height and the base of EBF. <br />
<br />
Solving for the area, we have <math>(8)(15/2)(1/2)=30</math>.<br />
<br />
== Solution 13 ==<br />
<math>AD : DC = 1:2</math>, so <math>ADB</math> has area <math>120</math> and <math>CDB</math> has area <math>240</math>. <math>BE = ED</math> so the area of <math>ABE</math> is equal to the area of <math>ADE = 60</math>.<br />
Draw <math>\overline{DG}</math> parallel to <math>\overline{AF}</math>.<br> <br />
Set area of BEF = <math>x</math>. BEF is similar to BDG in ratio of 1:2<br> <br />
so area of BDG = <math>4x</math>, area of EFDG=<math>3x</math>, and area of CDG<math>=240-4x</math>.<br><br />
CDG is similar to CAF in ratio of 2:3 so area CDG = <math>4/9</math> area CAF, and area AFDG=<math>5/4</math> area CDG.<br> <br />
Thus <math>60+3x=5/4(240-4x)</math> and <math>x=30</math>.<br />
~EFrame<br />
<br />
== Solution 14 - Geometry & Algebra==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(DD--FF,blue);<br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
We draw line <math>FD</math> so that we can define a variable <math>x</math> for the area of <math> \triangle BEF = \triangle DEF</math>. Knowing that <math> \triangle ABE</math> and <math> \triangle ADE</math> share both their height and base, we get that <math>ABE = ADE = 60</math>.<br />
<br />
Since we have a rule where 2 triangles, (<math>\triangle A</math> which has base <math>a</math> and vertex <math>c</math>), and (<math>\triangle B</math> which has Base <math>b</math> and vertex <math>c</math>)who share the same vertex (which is vertex <math>c</math> in this case), and share a common height, their relationship is : Area of <math>A : B = a : b</math> (the length of the two bases), we can list the equation where <math>\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}</math>. Substituting <math>x</math> into the equation we get: <br />
<br />
<cmath>\frac{x+60}{300-x} = \frac{2x}{240-2x}</cmath>. <cmath>(2x)(300-x) = (60+x)(240-x).</cmath> <cmath>600-2x^2 = 14400 - 120x + 240x - 2x^2.</cmath> <cmath>480x = 14400.</cmath> and we now have that <math> \triangle BEF=30.</math> <br />
~<math>\bold{\color{blue}{onionheadjr}}</math><br />
<br />
==Video Solutions==<br />
Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br />
<br />
https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br />
<br />
https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br />
<br />
https://www.youtube.com/watch?v=nyevg9w-CCI&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=6 ~ MathEx <br />
<br />
https://www.youtube.com/watch?v=m04K0Q2SNXY&t=1s<br />
<br />
==Solution 16 (Straightforward Solution)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
Since <math>AD:DC=1:2</math> thus <math>\triangle ABD=\frac{1}{3} \cdot 360 = 120.</math><br />
Similarly, <math>\triangle DBC = \frac{2}{3} \cdot 360 = 240.</math><br />
Now, since <math>E</math> is a midpoint of <math>BD</math>, <math>\triangle ABE = \triangle AED = 120 \div 2 = 60.</math><br />
We can use the fact that <math>E</math> is a midpoint of <math>BD</math> even further. Connect lines <math>E</math> and <math>C</math> so that <math>\triangle BEC</math> and <math>\triangle DEC</math> share 2 sides. We know that <math>\triangle BEC=\triangle DEC=240 \div 2 = 120</math> since <math>E</math> is a midpoint of <math>BD.</math><br />
Let's label <math>\triangle BEF</math> <math>x</math>. We know that <math>\triangle EFC</math> is <math>120-x</math> since <math>\triangle BEC = 120.</math><br />
Note that with this information now, we can deduct more things that are needed to finish the solution.<br />
Note that <math>\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.</math> because of triangles <math>EBF, ABE, AEC,</math> and <math>EFC.</math> We want to find <math>x.</math><br />
This is a simple equation, and solving we get <math>x=\boxed{\textbf{(B)}30}.</math><br />
<br />
By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=1407162019 AMC 8 Problems/Problem 232020-12-27T03:39:26Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div>==Problem 23==<br />
After Euclid High School's last basketball game, it was determined that <math>\frac{1}{4}</math> of the team's points were scored by Alexa and <math>\frac{2}{7}</math> were scored by Brittany. Chelsea scored <math>15</math> points. None of the other <math>7</math> team members scored more than <math>2</math> points. What was the total number of points scored by the other <math>7</math> team members?<br />
<br />
<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math><br />
<br />
==Solution 1==<br />
Since <math>\frac{\text{total points}}{4}</math> and <math>\frac{2(\text{total points})}{7}</math> are integers, we have <math>28 | \text{total points}</math>. We see that the number of points scored by the other team members is less than or equal to <math>14</math> and greater than or equal to <math>0</math>. We let the total number of points be <math>t</math> and the total number of points scored by the other team members be <math>x</math>, which means that <math>\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14</math>, which means <math>15 \le \frac{13t}{28} \le 29</math>. The only value of <math>t</math> that satisfies all conditions listed is <math>56</math>, so <math>x=\boxed{\textbf{(B)} 11}</math>.<br />
<br />
==Solution 2==<br />
Starting from the above equation <math>\frac{t}{4}+\frac{2t}{7} + 15 + x = t</math> where <math>t</math> is the total number of points scored and <math>x\le 14</math> is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation <math>x+15 = \frac{13}{28}t</math>, or <math>28x+28\cdot 15=13t</math>. Since <math>t</math> is necessarily divisible by 28, let <math>t=28u</math> where <math>u \ge 0</math> and divide by 28 to obtain <math>x + 15 = 13u</math>. Then it is easy to see <math>u=2</math> (<math>t=56</math>) is the only candidate, giving <math>x=\boxed{\textbf{(B)} 11}</math>. -scrabbler94<br />
<br />
==Solution 3==<br />
We first start by setting the total number of points as <math>28</math>, since <math>\text{LCM}(4,7) = 28</math>. However, we see that this does not work since we surpass the number of points just with the information given (<math>28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30</math> <math>(> 28)</math> ). Next, we can see that the total number of points scored is <math>56</math> as, if it is more than or equal to <math>84</math>, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: <math>56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45</math>, and thus, the other seven players would have scored a total of <math>56-45 = \boxed{\textbf{(B)} 11}</math> (We see that this works since we could have <math>4</math> of them score <math>2</math> points, and the other <math>3</math> of them score <math>1</math> point) -aops5234 -Edited by [[User: Penguin_Spellcaster|Penguin_Spellcaster]]<br />
<br />
==Solution 4 — Modular Arithmetic ==<br />
<br />
Adding together Alexa's and Brittany's fractions, we get <math>\frac{15}{28}</math> as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: <math>\frac{15x}{28x}</math> where <math>x</math> is the common ratio. Let <math>y</math> and <math>z</math> and <math>w</math> be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have <math>\frac{13x}{28x} = 15 + y + 2z + 0w.</math> We want all of our variables to be integers. Thus, we want <math>15 + y + 2z = 0 \pmod {13}.</math> Simplifying, <math>y+2z = 11 \pmod {13}.</math> The only possible value, as this integer sum has to be less than <math>7 \cdot 2 + 1 = 15,</math> must be 11. Therefore <math>y+2z = 11,</math> and the answer is <math>\boxed{ \textbf{(B) 11}}</math> - ab2024<br />
<br />
==Video explaining solution== <br />
<br />
Associated video - https://www.youtube.com/watch?v=jE-7Se7ay1c<br />
<br />
https://www.youtube.com/watch?v=3Mae_6qFxoU&t=204s ~ hi_im_bob<br />
<br />
https://youtu.be/wsYCn2FqZJE<br />
<br />
https://www.youtube.com/watch?v=fKjmw_zzCUU<br />
<br />
https://www.youtube.com/watch?v=o2mcnLOVFBA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=5 ~ MathEx<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=22|num-a=24}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_22&diff=1407152019 AMC 8 Problems/Problem 222020-12-27T03:35:38Z<p>Panda08: /* Solution 2 */</p>
<hr />
<div>==Problem 22==<br />
A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was <math>84\%</math> of the original price, by what percent was the price increased and decreased?<br />
<br />
<math>\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0.84</math>; so <math>1-x^{2}=0.84</math>, and <math>(x^{2})=0.16</math>, obtaining <math>x=0.4</math>. Therefore, the price was increased and decreased by <math>40</math>%, or <math>\boxed{\textbf{(E)}\ 40}</math>.<br />
<br />
==Solution 2 (Answer options)==<br />
We can try out every option and see which one works out. By this method, we get <math>\boxed{\textbf{(E)}\ 40}</math>.<br />
<br />
==Solution 3==<br />
Let x be the discount. We can also work in reverse such as (<math>84</math>)<math>(\frac{100}{100-x})</math><math>(\frac{100}{100+x})</math> = <math>100</math>.<br />
<br />
Thus <math>8400</math> = <math>(100+x)(100-x)</math>. Solving for <math>x</math> gives us <math>x = 40, -40</math>. But <math>x</math> has to be positive. Thus <math>x</math> = <math>40</math>.<br />
<br />
==Solution 4 ~ using the answer choices==<br />
<br />
Let our original cost be <math>\$ 100.</math> We are looking for a result of <math>\$ 84,</math> then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try <math>\boxed{40\%}</math>, and we have the answer; it worked.<br />
<br />
==Video explaining solution== <br />
<br />
Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc<br />
<br />
https://youtu.be/gX_l0PGsQao<br />
<br />
https://www.youtube.com/watch?v=_TheVi-6LWE<br />
<br />
https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 ~ MathEx<br />
<br />
https://youtu.be/h2GlK7itc2g<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_22&diff=1407142019 AMC 8 Problems/Problem 222020-12-27T03:35:05Z<p>Panda08: /* Solution 3 */</p>
<hr />
<div>==Problem 22==<br />
A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was <math>84\%</math> of the original price, by what percent was the price increased and decreased?<br />
<br />
<math>\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0.84</math>; so <math>1-x^{2}=0.84</math>, and <math>(x^{2})=0.16</math>, obtaining <math>x=0.4</math>. Therefore, the price was increased and decreased by <math>40</math>%, or <math>\boxed{\textbf{(E)}\ 40}</math>.<br />
<br />
==Solution 2 (Answer options)==<br />
We can try out every option and see which one works out. By this method, we get <math>\boxed{\textbf{(E)}\ 40}</math>. ~avamarora<br />
<br />
==Solution 3==<br />
Let x be the discount. We can also work in reverse such as (<math>84</math>)<math>(\frac{100}{100-x})</math><math>(\frac{100}{100+x})</math> = <math>100</math>.<br />
<br />
Thus <math>8400</math> = <math>(100+x)(100-x)</math>. Solving for <math>x</math> gives us <math>x = 40, -40</math>. But <math>x</math> has to be positive. Thus <math>x</math> = <math>40</math>.<br />
<br />
==Solution 4 ~ using the answer choices==<br />
<br />
Let our original cost be <math>\$ 100.</math> We are looking for a result of <math>\$ 84,</math> then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try <math>\boxed{40\%}</math>, and we have the answer; it worked.<br />
<br />
==Video explaining solution== <br />
<br />
Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc<br />
<br />
https://youtu.be/gX_l0PGsQao<br />
<br />
https://www.youtube.com/watch?v=_TheVi-6LWE<br />
<br />
https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 ~ MathEx<br />
<br />
https://youtu.be/h2GlK7itc2g<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_21&diff=1407132019 AMC 8 Problems/Problem 212020-12-27T03:31:52Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div><br />
==Problem 21==<br />
What is the area of the triangle formed by the lines <math>y=5</math>, <math>y=1+x</math>, and <math>y=1-x</math>?<br />
<br />
<math>\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16</math><br />
<br />
==Solution 1==<br />
First we need to find the coordinates where the graphs intersect. <br />
<br />
We want the points x and y to be the same. Thus, we set <math>5=x+1,</math> and get <math>x=4.</math> Plugging this into the equation, <math>y=1-x,</math> <br />
<math>y=5</math>, and <math>y=1+x</math> intersect at <math>(4,5)</math>, we call this line x.<br />
<br />
Doing the same thing, we get <math>x=-4.</math> Thus <math>y=5</math> also.<br />
<math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>, we call this line y.<br />
<br />
It's apparent the only solution to <math>1-x=1+x</math> is <math>0.</math> Thus, <math>y=1.</math><br />
<math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>, we call this line z.<br />
<br />
Using the [[Shoelace Theorem]] we get: <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}</cmath> <math>=</math> So our answer is <math>\boxed{\textbf{(E)}\ 16.}</math><br />
<br />
We might also see that the lines <math>y</math> and <math>z</math> are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by <math>-1</math> to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is <math>5-1=4,</math> so <math>\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}</math><br />
<br />
Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!<br />
<br />
==Solution 2==<br />
Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>.<br />
<br />
==Solution 3==<br />
<br />
<math>y = x + 1</math> and <math>y = -x + 1</math> have <math>y</math>-intercepts at <math>(1, 0)</math> and slopes of <math>1</math> and <math>-1</math>, respectively. Since the product of these slopes is <math>-1</math>, the two lines are perpendicular. From <math>y = 5</math>, we see that <math>(-4, 5)</math> and <math>(4, 5)</math> are the other two intersection points, and they are <math>8</math> units apart. By symmetry, this triangle is a <math>45-45-90</math> triangle, so the legs are <math>4\sqrt{2}</math> each and the area is <math>\frac{(4\sqrt{2})^2}{2} = \boxed{\textbf{(E)}\ 16}</math>. <br />
==Video Solutions==<br />
Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY<br />
<br />
https://www.youtube.com/watch?v=9nlX9VCisQc<br />
<br />
https://www.youtube.com/watch?v=mz3DY1rc5ao<br />
<br />
https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx<br />
<br />
https://youtu.be/RvtOX17DemY<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=20|num-a=22}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_21&diff=1407112019 AMC 8 Problems/Problem 212020-12-27T03:31:21Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div><br />
==Problem 21==<br />
What is the area of the triangle formed by the lines <math>y=5</math>, <math>y=1+x</math>, and <math>y=1-x</math>?<br />
<br />
<math>\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16</math><br />
<br />
==Solution 1==<br />
First we need to find the coordinates where the graphs intersect. <br />
<br />
We want the points x and y to be the same. Thus, we set <math>5=x+1,</math> and get <math>x=4.</math> Plugging this into the equation, <math>y=1-x,</math> <br />
<math>y=5</math>, and <math>y=1+x</math> intersect at <math>(4,5)</math>, we call this line x.<br />
<br />
Doing the same thing, we get <math>x=-4.</math> Thus <math>y=5</math> also.<br />
<math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>, we call this line y.<br />
<br />
It's apparent the only solution to <math>1-x=1+x</math> is <math>0.</math> Thus, <math>y=1.</math><br />
<math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>, we call this line z.<br />
<br />
Using the [[hoelace Theorem]] we get: <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}</cmath> <math>=</math> So our answer is <math>\boxed{\textbf{(E)}\ 16.}</math><br />
<br />
We might also see that the lines <math>y</math> and <math>z</math> are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by <math>-1</math> to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is <math>5-1=4,</math> so <math>\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}</math><br />
<br />
Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!<br />
<br />
==Solution 2==<br />
Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>.<br />
<br />
==Solution 3==<br />
<br />
<math>y = x + 1</math> and <math>y = -x + 1</math> have <math>y</math>-intercepts at <math>(1, 0)</math> and slopes of <math>1</math> and <math>-1</math>, respectively. Since the product of these slopes is <math>-1</math>, the two lines are perpendicular. From <math>y = 5</math>, we see that <math>(-4, 5)</math> and <math>(4, 5)</math> are the other two intersection points, and they are <math>8</math> units apart. By symmetry, this triangle is a <math>45-45-90</math> triangle, so the legs are <math>4\sqrt{2}</math> each and the area is <math>\frac{(4\sqrt{2})^2}{2} = \boxed{\textbf{(E)}\ 16}</math>. <br />
==Video Solutions==<br />
Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY<br />
<br />
https://www.youtube.com/watch?v=9nlX9VCisQc<br />
<br />
https://www.youtube.com/watch?v=mz3DY1rc5ao<br />
<br />
https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx<br />
<br />
https://youtu.be/RvtOX17DemY<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=20|num-a=22}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_19&diff=1407102019 AMC 8 Problems/Problem 192020-12-27T03:27:14Z<p>Panda08: /* Solution 3 */</p>
<hr />
<div>==Problem 19==<br />
In a tournament there are six teams that play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br />
<br />
<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math><br />
<br />
==Solution 1==<br />
<br />
After fully understanding the problem, we immediately know that the three top teams, say team <math>A</math>, team <math>B</math>, and team <math>C</math>, must beat the other three teams <math>D</math>, <math>E</math>, <math>F</math>. Therefore, <math>A</math>,<math>B</math>,<math>C</math> must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against <math>D</math>, <math>E</math>, and <math>F</math>. For games between <math>A</math>, <math>B</math>, <math>C</math>, we have 2 cases. In both cases, there is an equality of points between <math>A</math>, <math>B</math>, and <math>C</math>.<br />
<br />
Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have <math>(1+1)*2=4</math> points (they play twice). Therefore, this case brings a total of <math>4+18=22</math> points.<br />
<br />
Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have <math>3</math> points, so a team gets <math>3\times2=6</math> points if they each win a game and lose a game. This case brings a total of <math>18+6=24</math> points. <br />
<br />
Therefore, we use Case2 since it brings the greater amount of points, or <math>\boxed{(C)24}</math>.<br />
<br />
<br />
--------------------------<br />
Note that case 2 can be easily seen to be better as follows. Let <math>x_A</math> be the number of points <math>A</math> gets, <math>x_B</math> be the number of points <math>B</math> gets, and <math>x_C</math> be the number of points <math>C</math> gets. Since <math>x_A = x_B = x_C</math>, to maximize <math>x_A</math>, we can just maximize <math>x_A + x_B + x_C</math>. But in each match, if one team wins then the total sum increases by <math>3</math> points, whereas if they tie, the total sum increases by <math>2</math> points. So it is best if there are the fewest ties possible.<br />
<br />
==Solution 2==<br />
<br />
(1st match(3) + 2nd match(1)) * number of teams(6) = 24, <math>\boxed{C}</math>.<br />
<br />
Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So that means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won & drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is <math>\boxed{C}</math>.<br />
<br />
==Solution 3==<br />
<br />
<br />
We can name the top three teams as <math>A, B,</math> and <math>C</math>. We can see that <math>A=B=C</math>, because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: <math>AB, BC,</math> and <math>AC</math> come twice. In order to even out the scores and get the maximum score, we can say that in match <math>AB, A</math> and <math>B</math> each win once out of the two games that they play. We can say the same thing for <math>AC</math> and <math>BC</math>. This tells us that each team <math>A, B,</math> and <math>C</math> win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as <math>D, E,</math> and <math>F</math>. We can write down every match that <math>A, B,</math> or <math>C</math> plays in that we haven't counted yet: <math>AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,</math> and <math>CF</math>. We can say <math>A, B,</math> and <math>C</math> win each of these in order to obtain the maximum score that <math>A, B,</math> and <math>C</math> can have. If <math>A, B,</math> and <math>C</math> win all six of their matches, <math>A, B,</math> and <math>C</math> will have a score of <math>18</math>. <math>18 + 6</math> results in a maximum score of <math>\boxed{24}</math>. This tells us that the correct answer choice is <math>\boxed{C}</math>.<br />
<br />
== Video Solutions ==<br />
<br />
Associated Video - https://youtu.be/s0O3_uXZrOI<br />
<br />
https://youtu.be/hM4sHJSMNDs<br />
<br />
-Happpytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_19&diff=1407092019 AMC 8 Problems/Problem 192020-12-27T03:26:42Z<p>Panda08: /* Solution 2 */</p>
<hr />
<div>==Problem 19==<br />
In a tournament there are six teams that play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br />
<br />
<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math><br />
<br />
==Solution 1==<br />
<br />
After fully understanding the problem, we immediately know that the three top teams, say team <math>A</math>, team <math>B</math>, and team <math>C</math>, must beat the other three teams <math>D</math>, <math>E</math>, <math>F</math>. Therefore, <math>A</math>,<math>B</math>,<math>C</math> must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against <math>D</math>, <math>E</math>, and <math>F</math>. For games between <math>A</math>, <math>B</math>, <math>C</math>, we have 2 cases. In both cases, there is an equality of points between <math>A</math>, <math>B</math>, and <math>C</math>.<br />
<br />
Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have <math>(1+1)*2=4</math> points (they play twice). Therefore, this case brings a total of <math>4+18=22</math> points.<br />
<br />
Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have <math>3</math> points, so a team gets <math>3\times2=6</math> points if they each win a game and lose a game. This case brings a total of <math>18+6=24</math> points. <br />
<br />
Therefore, we use Case2 since it brings the greater amount of points, or <math>\boxed{(C)24}</math>.<br />
<br />
<br />
--------------------------<br />
Note that case 2 can be easily seen to be better as follows. Let <math>x_A</math> be the number of points <math>A</math> gets, <math>x_B</math> be the number of points <math>B</math> gets, and <math>x_C</math> be the number of points <math>C</math> gets. Since <math>x_A = x_B = x_C</math>, to maximize <math>x_A</math>, we can just maximize <math>x_A + x_B + x_C</math>. But in each match, if one team wins then the total sum increases by <math>3</math> points, whereas if they tie, the total sum increases by <math>2</math> points. So it is best if there are the fewest ties possible.<br />
<br />
==Solution 2==<br />
<br />
(1st match(3) + 2nd match(1)) * number of teams(6) = 24, <math>\boxed{C}</math>.<br />
<br />
Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So that means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won & drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is <math>\boxed{C}</math>.<br />
<br />
==Solution 3==<br />
<br />
<br />
We can name the top three teams as <math>A, B,</math> and <math>C</math>. We can see that <math>A=B=C</math>, because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: <math>AB, BC,</math> and <math>AC</math> come twice. In order to even out the scores and get the maximum score, we can say that in match <math>AB, A</math> and <math>B</math> each win once out of the two games that they play. We can say the same thing for <math>AC</math> and <math>BC</math>. This tells us that each team <math>A, B,</math> and <math>C</math> win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as <math>D, E,</math> and <math>F</math>. We can write down every match that <math>A, B,</math> or <math>C</math> plays in that we haven't counted yet: <math>AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,</math> and <math>CF</math>. We can say <math>A, B,</math> and <math>C</math> win each of these in order to obtain the maximum score that <math>A, B,</math> and <math>C</math> can have. If <math>A, B,</math> and <math>C</math> win all six of their matches, <math>A, B,</math> and <math>C</math> will have a score of <math>18</math>. <math>18 + 6</math> results in a maximum score of <math>\boxed{24}</math>. This tells us that the correct answer choice is <math>\boxed{C}</math>.<br />
<br />
~Champion1234<br />
<br />
<br />
== Video Solutions ==<br />
<br />
Associated Video - https://youtu.be/s0O3_uXZrOI<br />
<br />
https://youtu.be/hM4sHJSMNDs<br />
<br />
-Happpytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_19&diff=1407082019 AMC 8 Problems/Problem 192020-12-27T03:26:12Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div>==Problem 19==<br />
In a tournament there are six teams that play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br />
<br />
<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math><br />
<br />
==Solution 1==<br />
<br />
After fully understanding the problem, we immediately know that the three top teams, say team <math>A</math>, team <math>B</math>, and team <math>C</math>, must beat the other three teams <math>D</math>, <math>E</math>, <math>F</math>. Therefore, <math>A</math>,<math>B</math>,<math>C</math> must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against <math>D</math>, <math>E</math>, and <math>F</math>. For games between <math>A</math>, <math>B</math>, <math>C</math>, we have 2 cases. In both cases, there is an equality of points between <math>A</math>, <math>B</math>, and <math>C</math>.<br />
<br />
Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have <math>(1+1)*2=4</math> points (they play twice). Therefore, this case brings a total of <math>4+18=22</math> points.<br />
<br />
Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have <math>3</math> points, so a team gets <math>3\times2=6</math> points if they each win a game and lose a game. This case brings a total of <math>18+6=24</math> points. <br />
<br />
Therefore, we use Case2 since it brings the greater amount of points, or <math>\boxed{(C)24}</math>.<br />
<br />
<br />
--------------------------<br />
Note that case 2 can be easily seen to be better as follows. Let <math>x_A</math> be the number of points <math>A</math> gets, <math>x_B</math> be the number of points <math>B</math> gets, and <math>x_C</math> be the number of points <math>C</math> gets. Since <math>x_A = x_B = x_C</math>, to maximize <math>x_A</math>, we can just maximize <math>x_A + x_B + x_C</math>. But in each match, if one team wins then the total sum increases by <math>3</math> points, whereas if they tie, the total sum increases by <math>2</math> points. So it is best if there are the fewest ties possible.<br />
<br />
==Solution 2==<br />
<br />
(1st match(3) + 2nd match(1)) * number of teams(6) = 24, <math>\boxed{C}</math>.<br />
<br />
Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So that means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won & drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is <math>\boxed{C}</math>.<br />
<br />
~MRLUIGI<br />
<br />
<br />
==Solution 3==<br />
<br />
<br />
We can name the top three teams as <math>A, B,</math> and <math>C</math>. We can see that <math>A=B=C</math>, because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: <math>AB, BC,</math> and <math>AC</math> come twice. In order to even out the scores and get the maximum score, we can say that in match <math>AB, A</math> and <math>B</math> each win once out of the two games that they play. We can say the same thing for <math>AC</math> and <math>BC</math>. This tells us that each team <math>A, B,</math> and <math>C</math> win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as <math>D, E,</math> and <math>F</math>. We can write down every match that <math>A, B,</math> or <math>C</math> plays in that we haven't counted yet: <math>AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,</math> and <math>CF</math>. We can say <math>A, B,</math> and <math>C</math> win each of these in order to obtain the maximum score that <math>A, B,</math> and <math>C</math> can have. If <math>A, B,</math> and <math>C</math> win all six of their matches, <math>A, B,</math> and <math>C</math> will have a score of <math>18</math>. <math>18 + 6</math> results in a maximum score of <math>\boxed{24}</math>. This tells us that the correct answer choice is <math>\boxed{C}</math>.<br />
<br />
~Champion1234<br />
<br />
<br />
== Video Solutions ==<br />
<br />
Associated Video - https://youtu.be/s0O3_uXZrOI<br />
<br />
https://youtu.be/hM4sHJSMNDs<br />
<br />
-Happpytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_19&diff=1407072019 AMC 8 Problems/Problem 192020-12-27T03:25:27Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div>==Problem 19==<br />
In a tournament there are six teams that play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br />
<br />
<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math><br />
<br />
==Solution 1==<br />
<br />
After fully understanding the problem, we immediately know that the three top teams, say team <math>A</math>, team <math>B</math>, and team <math>C</math>, must beat the other three teams <math>D</math>, <math>E</math>, <math>F</math>. Therefore, <math>A</math>,<math>B</math>,<math>C</math> must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against <math>D</math>, <math>E</math>, and <math>F</math>. For games between <math>A</math>, <math>B</math>, <math>C</math>, we have 2 cases. In both cases, there is an equality of points between <math>A</math>, <math>B</math>, and <math>C</math>.<br />
<br />
Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have <math>(1+1)*2=4</math> points (they play twice). Therefore, this case brings a total of <math>4+18=22</math> points.<br />
<br />
Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have <math>3</math> points, so a team gets <math>3\times2=6</math> points if they each win a game and lose a game. This case brings a total of <math>18+6=24</math> points. <br />
<br />
Therefore, we use Case2 since it brings the greater amount of points, or <math>\boxed{24}</math>, so the answer is <math>\boxed{C}</math>.<br />
<br />
<br />
--------------------------<br />
Note that case 2 can be easily seen to be better as follows. Let <math>x_A</math> be the number of points <math>A</math> gets, <math>x_B</math> be the number of points <math>B</math> gets, and <math>x_C</math> be the number of points <math>C</math> gets. Since <math>x_A = x_B = x_C</math>, to maximize <math>x_A</math>, we can just maximize <math>x_A + x_B + x_C</math>. But in each match, if one team wins then the total sum increases by <math>3</math> points, whereas if they tie, the total sum increases by <math>2</math> points. So it is best if there are the fewest ties possible.<br />
<br />
==Solution 2==<br />
<br />
(1st match(3) + 2nd match(1)) * number of teams(6) = 24, <math>\boxed{C}</math>.<br />
<br />
Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So that means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won & drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is <math>\boxed{C}</math>.<br />
<br />
~MRLUIGI<br />
<br />
<br />
==Solution 3==<br />
<br />
<br />
We can name the top three teams as <math>A, B,</math> and <math>C</math>. We can see that <math>A=B=C</math>, because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: <math>AB, BC,</math> and <math>AC</math> come twice. In order to even out the scores and get the maximum score, we can say that in match <math>AB, A</math> and <math>B</math> each win once out of the two games that they play. We can say the same thing for <math>AC</math> and <math>BC</math>. This tells us that each team <math>A, B,</math> and <math>C</math> win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as <math>D, E,</math> and <math>F</math>. We can write down every match that <math>A, B,</math> or <math>C</math> plays in that we haven't counted yet: <math>AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,</math> and <math>CF</math>. We can say <math>A, B,</math> and <math>C</math> win each of these in order to obtain the maximum score that <math>A, B,</math> and <math>C</math> can have. If <math>A, B,</math> and <math>C</math> win all six of their matches, <math>A, B,</math> and <math>C</math> will have a score of <math>18</math>. <math>18 + 6</math> results in a maximum score of <math>\boxed{24}</math>. This tells us that the correct answer choice is <math>\boxed{C}</math>.<br />
<br />
~Champion1234<br />
<br />
<br />
== Video Solutions ==<br />
<br />
Associated Video - https://youtu.be/s0O3_uXZrOI<br />
<br />
https://youtu.be/hM4sHJSMNDs<br />
<br />
-Happpytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=1407062019 AMC 8 Problems/Problem 182020-12-27T03:23:34Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div>==Problem 18==<br />
The faces of each of two fair dice are numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>8</math>. When the two dice are tossed, what is the probability that their sum will be an even number?<br />
<br />
<math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math><br />
<br />
==Solution 1==<br />
The approach to this problem:<br />
There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the <math>2</math> numbers add up to an odd number and subtract the answer from <math>1</math>. <br />
<br />
How to solve the problem:<br />
The probability of getting an odd number first is <math>\frac{4}{6}=\frac{2}{3}</math>. In order to make the sum odd, we must select an even number next. The probability of getting an even number is <math>\frac{2}{6}=\frac{1}{3}</math>. Now we multiply the two fractions: <math>\frac{2}{3}\times\frac{1}{3}=2/9</math>. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do <math>\frac{2}{9}\times2=\frac{4}{9}</math>. This is the probability of getting an odd number. In order to get the probability of getting an even number we do <math>1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}</math><br />
<br />
==Solution 2==<br />
We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the rolls must consist of <math>2</math> odds or <math>2</math> evens. <br />
<br />
Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to roll <math>2</math> odds is <math>4*4=16</math>, as there are <math>4</math> choices for the first odd on the first roll and <math>4</math> choices for the second odd on the second roll.<br />
<br />
Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.<br />
<br />
==Solution 3 (Complementary Counting)==<br />
We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is <math>\frac{1}{3}</math>, and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>.<br />
<br />
==Solution 4==<br />
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.<br />
<br />
==Solution 5 (Casework)==<br />
To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.<br />
<br />
==Video Solution==<br />
<br />
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw<br />
<br />
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=17|num-a=19}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=1407052019 AMC 8 Problems/Problem 182020-12-27T03:23:09Z<p>Panda08: /* Solution 3 (Complementary Counting) */</p>
<hr />
<div>==Problem 18==<br />
The faces of each of two fair dice are numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>8</math>. When the two dice are tossed, what is the probability that their sum will be an even number?<br />
<br />
<math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math><br />
<br />
==Solution 1==<br />
The approach to this problem:<br />
There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the <math>2</math> numbers add up to an odd number and subtract the answer from <math>1</math>. <br />
<br />
How to solve the problem:<br />
The probability of getting an odd number first is <math>\frac{4}{6}=\frac{2}{3}</math>. In order to make the sum odd, we must select an even number next. The probability of getting an even number is <math>\frac{2}{6}=\frac{1}{3}</math>. Now we multiply the two fractions: <math>\frac{2}{3}\times\frac{1}{3}=2/9</math>. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do <math>\frac{2}{9}\times2=\frac{4}{9}</math>. This is the probability of getting an odd number. In order to get the probability of getting an even number we do <math>1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}</math><br />
<br />
- ViratKohli2018 (VK18)<br />
<br />
==Solution 2==<br />
We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the rolls must consist of <math>2</math> odds or <math>2</math> evens. <br />
<br />
Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to roll <math>2</math> odds is <math>4*4=16</math>, as there are <math>4</math> choices for the first odd on the first roll and <math>4</math> choices for the second odd on the second roll.<br />
<br />
Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.<br />
<br />
==Solution 3 (Complementary Counting)==<br />
We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is <math>\frac{1}{3}</math>, and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>.<br />
<br />
==Solution 4==<br />
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.<br />
<br />
==Solution 5 (Casework)==<br />
To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.<br />
<br />
==Video Solution==<br />
<br />
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw<br />
<br />
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=17|num-a=19}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=1407042019 AMC 8 Problems/Problem 182020-12-27T03:22:41Z<p>Panda08: /* Solution 5 (Casework) */</p>
<hr />
<div>==Problem 18==<br />
The faces of each of two fair dice are numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>8</math>. When the two dice are tossed, what is the probability that their sum will be an even number?<br />
<br />
<math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math><br />
<br />
==Solution 1==<br />
The approach to this problem:<br />
There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the <math>2</math> numbers add up to an odd number and subtract the answer from <math>1</math>. <br />
<br />
How to solve the problem:<br />
The probability of getting an odd number first is <math>\frac{4}{6}=\frac{2}{3}</math>. In order to make the sum odd, we must select an even number next. The probability of getting an even number is <math>\frac{2}{6}=\frac{1}{3}</math>. Now we multiply the two fractions: <math>\frac{2}{3}\times\frac{1}{3}=2/9</math>. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do <math>\frac{2}{9}\times2=\frac{4}{9}</math>. This is the probability of getting an odd number. In order to get the probability of getting an even number we do <math>1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}</math><br />
<br />
- ViratKohli2018 (VK18)<br />
<br />
==Solution 2==<br />
We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the rolls must consist of <math>2</math> odds or <math>2</math> evens. <br />
<br />
Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to roll <math>2</math> odds is <math>4*4=16</math>, as there are <math>4</math> choices for the first odd on the first roll and <math>4</math> choices for the second odd on the second roll.<br />
<br />
Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.<br />
<br />
==Solution 3 (Complementary Counting)==<br />
We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is <math>\frac{1}{3}</math>, and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>. - juliankuang<br />
<br />
==Solution 4==<br />
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.<br />
<br />
==Solution 5 (Casework)==<br />
To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.<br />
<br />
==Video Solution==<br />
<br />
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw<br />
<br />
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=17|num-a=19}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=1407032019 AMC 8 Problems/Problem 182020-12-27T03:22:16Z<p>Panda08: /* Video Solution */</p>
<hr />
<div>==Problem 18==<br />
The faces of each of two fair dice are numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>8</math>. When the two dice are tossed, what is the probability that their sum will be an even number?<br />
<br />
<math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math><br />
<br />
==Solution 1==<br />
The approach to this problem:<br />
There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the <math>2</math> numbers add up to an odd number and subtract the answer from <math>1</math>. <br />
<br />
How to solve the problem:<br />
The probability of getting an odd number first is <math>\frac{4}{6}=\frac{2}{3}</math>. In order to make the sum odd, we must select an even number next. The probability of getting an even number is <math>\frac{2}{6}=\frac{1}{3}</math>. Now we multiply the two fractions: <math>\frac{2}{3}\times\frac{1}{3}=2/9</math>. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do <math>\frac{2}{9}\times2=\frac{4}{9}</math>. This is the probability of getting an odd number. In order to get the probability of getting an even number we do <math>1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}</math><br />
<br />
- ViratKohli2018 (VK18)<br />
<br />
==Solution 2==<br />
We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the rolls must consist of <math>2</math> odds or <math>2</math> evens. <br />
<br />
Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to roll <math>2</math> odds is <math>4*4=16</math>, as there are <math>4</math> choices for the first odd on the first roll and <math>4</math> choices for the second odd on the second roll.<br />
<br />
Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.<br />
<br />
==Solution 3 (Complementary Counting)==<br />
We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is <math>\frac{1}{3}</math>, and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>. - juliankuang<br />
<br />
==Solution 4==<br />
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.<br />
<br />
==Solution 5 (Casework)==<br />
To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.~binderclips1<br />
<br />
==Video Solution==<br />
<br />
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw<br />
<br />
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=17|num-a=19}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_17&diff=1407022019 AMC 8 Problems/Problem 172020-12-27T03:14:48Z<p>Panda08: /* Solution 3 */</p>
<hr />
<div>==Problem 17==<br />
What is the value of the product <br />
<cmath>\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?</cmath><br />
<br />
<math>\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50</math><br />
<br />
==Solution 1(Telescoping)==<br />
We rewrite: <cmath>\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}</cmath><br />
<br />
The middle terms cancel, leaving us with<br />
<br />
<cmath>\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}</cmath><br />
<br />
==Solution 2==<br />
If you calculate the first few values of the equation, all of the values tend to <math>\frac{1}{2}</math>, but are not equal to it. The answer closest to <math>\frac{1}{2}</math> but not equal to it is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==Solution 3==<br />
Rewriting the numerator and the denominator, we get <math>\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}</math>. We can simplify by canceling 99! on both sides, leaving us with: <math>\frac{100 \cdot 98!}{2 \cdot 99!}</math> We rewrite <math>99!</math> as <math>99 \cdot 98!</math> and cancel <math>98!</math>, which gets <math>\boxed{(B)\frac{50}{99}}</math>.<br />
<br />
==Video Solution==<br />
<br />
Associated video - https://www.youtube.com/watch?v=yPQmvyVyvaM<br />
<br />
https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1 ~ MathEx<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=16|num-a=18}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_17&diff=1407012019 AMC 8 Problems/Problem 172020-12-27T03:13:26Z<p>Panda08: /* Solution 2 */</p>
<hr />
<div>==Problem 17==<br />
What is the value of the product <br />
<cmath>\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?</cmath><br />
<br />
<math>\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50</math><br />
<br />
==Solution 1(Telescoping)==<br />
We rewrite: <cmath>\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}</cmath><br />
<br />
The middle terms cancel, leaving us with<br />
<br />
<cmath>\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}</cmath><br />
<br />
==Solution 2==<br />
If you calculate the first few values of the equation, all of the values tend to <math>\frac{1}{2}</math>, but are not equal to it. The answer closest to <math>\frac{1}{2}</math> but not equal to it is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==Solution 3==<br />
Rewriting the numerator and the denominator, we get <math>\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}</math>. We can simplify by canceling 99! on both sides, leaving us with: <math>\frac{100 \cdot 98!}{2 \cdot 99!}</math> We rewrite <math>99!</math> as <math>99 \cdot 98!</math> and cancel <math>98!</math>, which gets <math>\boxed{\frac{50}{99}}</math>. Answer B.<br />
<br />
==Video Solution==<br />
<br />
Associated video - https://www.youtube.com/watch?v=yPQmvyVyvaM<br />
<br />
https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1 ~ MathEx<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=16|num-a=18}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_17&diff=1407002019 AMC 8 Problems/Problem 172020-12-27T03:12:03Z<p>Panda08: /* Solution 1(Telescoping) */</p>
<hr />
<div>==Problem 17==<br />
What is the value of the product <br />
<cmath>\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?</cmath><br />
<br />
<math>\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50</math><br />
<br />
==Solution 1(Telescoping)==<br />
We rewrite: <cmath>\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}</cmath><br />
<br />
The middle terms cancel, leaving us with<br />
<br />
<cmath>\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}</cmath><br />
<br />
==Solution 2==<br />
If you calculate the first few values of the equation, all of the values tend to <math>\frac{1}{2}</math>, but are not equal to it. The answer closest to <math>\frac{1}{2}</math> but not equal to it is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.~hpotter0104<br />
<br />
==Solution 3==<br />
Rewriting the numerator and the denominator, we get <math>\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}</math>. We can simplify by canceling 99! on both sides, leaving us with: <math>\frac{100 \cdot 98!}{2 \cdot 99!}</math> We rewrite <math>99!</math> as <math>99 \cdot 98!</math> and cancel <math>98!</math>, which gets <math>\boxed{\frac{50}{99}}</math>. Answer B.<br />
<br />
==Video Solution==<br />
<br />
Associated video - https://www.youtube.com/watch?v=yPQmvyVyvaM<br />
<br />
https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1 ~ MathEx<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=16|num-a=18}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_16&diff=1406992019 AMC 8 Problems/Problem 162020-12-27T03:09:51Z<p>Panda08: /* Solution 2 */</p>
<hr />
<div>==Problem 16==<br />
Qiang drives <math>15</math> miles at an average speed of <math>30</math> miles per hour. How many additional miles will he have to drive at <math>55</math> miles per hour to average <math>50</math> miles per hour for the entire trip?<br />
<br />
<math>\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135</math><br />
<br />
==Solution 1==<br />
The only option that is easily divisible by <math>55</math> is <math>110</math>. Which gives 2 hours of travel. And by the formula <math>\frac{15}{30} + \frac{110}{55} = \frac{5}{2}</math><br />
<br />
And <math>\text{Average Speed}</math> = <math>\frac{\text{Total Distance}}{\text{Total Time}}</math><br />
<br />
Thus <math>\frac{125}{50} = \frac{5}{2}</math><br />
<br />
Both are equal and thus our answer is <math>\boxed{\textbf{(D)}\ 110}.</math><br />
<br />
==Solution 2==<br />
Note that the average speed is simply the total distance over the total time. Let the number of additional miles he has to drive be <math>x.</math> Therefore, the total distance is <math>15+x</math> and the total time (in hours) is <cmath>\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.</cmath> We can set up the following equation: <cmath>\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.</cmath> Simplifying the equation, we get <cmath>15+x=25+\frac{10x}{11}.</cmath> Solving the equation yields <math>x=110,</math> so our answer is <math>\boxed{\textbf{(D)}\ 110}</math>.<br />
<br />
==Solution 3==<br />
If he travels <math>15</math> miles at a speed of <math>30</math> miles per hour, he travels for 30 min. Average rate is total distance over total time so <math>(15+d)/(0.5 + t) = 50</math>, where d is the distance left to travel and t is the time to travel that distance. solve for <math>d</math> to get <math>d = 10+50t</math>. you also know that he has to travel <math>55</math> miles per hour for some time, so <math>d=55t</math> plug that in for d to get <math>55t = 10+50t</math> and <math>t=2</math> and since <math>d=55t</math>, <math>d = 2\cdot55 =110</math> the answer is <math>\boxed{\textbf{(D)}\ 110}</math>.<br />
<br />
==Video Solution==<br />
<br />
Associated Video - https://www.youtube.com/watch?v=OC1KdFeZFeE<br />
<br />
https://youtu.be/5K1AgeZ8rUQ - happytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=15|num-a=17}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_16&diff=1406982019 AMC 8 Problems/Problem 162020-12-27T03:09:25Z<p>Panda08: /* Solution 3 */</p>
<hr />
<div>==Problem 16==<br />
Qiang drives <math>15</math> miles at an average speed of <math>30</math> miles per hour. How many additional miles will he have to drive at <math>55</math> miles per hour to average <math>50</math> miles per hour for the entire trip?<br />
<br />
<math>\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135</math><br />
<br />
==Solution 1==<br />
The only option that is easily divisible by <math>55</math> is <math>110</math>. Which gives 2 hours of travel. And by the formula <math>\frac{15}{30} + \frac{110}{55} = \frac{5}{2}</math><br />
<br />
And <math>\text{Average Speed}</math> = <math>\frac{\text{Total Distance}}{\text{Total Time}}</math><br />
<br />
Thus <math>\frac{125}{50} = \frac{5}{2}</math><br />
<br />
Both are equal and thus our answer is <math>\boxed{\textbf{(D)}\ 110}.</math><br />
<br />
==Solution 2==<br />
Note that the average speed is simply the total distance over the total time. Let the number of additional miles he has to drive be <math>x.</math> Therefore, the total distance is <math>15+x</math> and the total time (in hours) is <cmath>\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.</cmath> We can set up the following equation: <cmath>\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.</cmath> Simplifying the equation, we get <cmath>15+x=25+\frac{10x}{11}.</cmath> Solving the equation yields <math>x=110,</math> so our answer is <math>\boxed{\textbf{(D)}\ 110}</math>. <br />
<br />
~twinemma<br />
<br />
==Solution 3==<br />
If he travels <math>15</math> miles at a speed of <math>30</math> miles per hour, he travels for 30 min. Average rate is total distance over total time so <math>(15+d)/(0.5 + t) = 50</math>, where d is the distance left to travel and t is the time to travel that distance. solve for <math>d</math> to get <math>d = 10+50t</math>. you also know that he has to travel <math>55</math> miles per hour for some time, so <math>d=55t</math> plug that in for d to get <math>55t = 10+50t</math> and <math>t=2</math> and since <math>d=55t</math>, <math>d = 2\cdot55 =110</math> the answer is <math>\boxed{\textbf{(D)}\ 110}</math>.<br />
<br />
==Video Solution==<br />
<br />
Associated Video - https://www.youtube.com/watch?v=OC1KdFeZFeE<br />
<br />
https://youtu.be/5K1AgeZ8rUQ - happytwin<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=15|num-a=17}}<br />
<br />
{{MAA Notice}}</div>Panda08https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_4&diff=1406962019 AMC 8 Problems/Problem 42020-12-27T02:47:53Z<p>Panda08: /* Solution 1 */</p>
<hr />
<div>== Problem 4 ==<br />
<br />
Quadrilateral <math>ABCD</math> is a rhombus with perimeter <math>52</math> meters. The length of diagonal <math>\overline{AC}</math> is <math>24</math> meters. What is the area in square meters of rhombus <math>ABCD</math>?<br />
<br />
<asy><br />
draw((-13,0)--(0,5));<br />
draw((0,5)--(13,0));<br />
draw((13,0)--(0,-5));<br />
draw((0,-5)--(-13,0));<br />
dot((-13,0));<br />
dot((0,5));<br />
dot((13,0));<br />
dot((0,-5));<br />
label("A",(-13,0),W);<br />
label("B",(0,5),N);<br />
label("C",(13,0),E);<br />
label("D",(0,-5),S);<br />
</asy><br />
<br />
<math>\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144</math><br />
<br />
<br />
== Solution 1 ==<br />
<asy><br />
draw((-12,0)--(0,5));<br />
draw((0,5)--(12,0));<br />
draw((12,0)--(0,-5));<br />
draw((0,-5)--(-12,0));<br />
draw((0,0)--(12,0));<br />
draw((0,0)--(0,5));<br />
draw((0,0)--(-12,0));<br />
draw((0,0)--(0,-5));<br />
dot((-12,0));<br />
dot((0,5));<br />
dot((12,0));<br />
dot((0,-5));<br />
label("A",(-12,0),W);<br />
label("B",(0,5),N);<br />
label("C",(12,0),E);<br />
label("D",(0,-5),S);<br />
label("E",(0,0),SW);<br />
</asy><br />
<br />
A rhombus has sides of equal length. Because the perimeter of the rhombus is <math>52</math>, each side is <math>\frac{52}{4}=13</math>. In a rhombus, diagonals are perpendicular and bisect each other, which means <math>\overline{AE}</math> = <math>12</math> = <math>\overline{EC}</math>.<br />
<br />
Consider one of the right triangles:<br />
<br />
<asy><br />
draw((-12,0)--(0,5));<br />
draw((0,0)--(-12,0));<br />
draw((0,0)--(0,5));<br />
dot((-12,0));<br />
dot((0,5));<br />
label("A",(-12,0),W);<br />
label("B",(0,5),N);<br />
label("E",(0,0),SE);<br />
</asy><br />
<br />
<math>\overline{AB}</math> = <math>13</math>, and <math>\overline{AE}</math> = <math>12</math>. Using Pythagorean theorem, we find that <math>\overline{BE}</math> = <math>5</math>.<br />
You may recall the famous Pythagorean triple, (5, 12, 13).<br />
<br />
Thus the values of the two diagonals are <math>\overline{AC}</math> = <math>24</math> and <math>\overline{BD}</math> = <math>10</math>.<br />
The area of a rhombus is = <math>\frac{d_1\cdot{d_2}}{2}</math> = <math>\frac{24\cdot{10}}{2}</math> = <math>120</math><br />
<br />
<math>\boxed{\textbf{(D)}\ 120}</math><br />
<br />
==See also==<br />
{{AMC8 box|year=2019|num-b=3|num-a=5}}<br />
<br />
{{MAA Notice}}</div>Panda08