https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Brudder&feedformat=atomAoPS Wiki - User contributions [en]2024-03-30T02:19:14ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1118452019 AMC 8 Problems/Problem 242019-11-21T01:37:23Z<p>Brudder: /* Solution 2 (Mass Points) */</p>
<hr />
<div><br />
<br />
<br />
==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <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 />
Draw <math>X</math> so that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE</math>=<math>ED</math>. <math>FC</math>=3<math>XD</math> so <math>BC</math>=4<math>BC</math>. Since <math>AF</math>=3<math>EF</math>( <math>XE</math>=<math>EF</math> and <math>AX</math>=1/3<math>AF</math>, so <math>XE</math>=<math>EF</math>=1/3<math>AF</math>), the altitude of triangle <math>BEF</math> is equal to 1/3 of the altitude of <math>ABC</math>. The area of <math>ABC</math> is 360, so the area of <math>BEF</math>=1/3*1/4*360=<math>\boxed{(B) 30}</math>~heeeeeeeheeeee<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 3. By equality, point <math>B</math> has a mass of 3 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{B}(30)</math><br />
-Brudder<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{B}</math><br />
-Brudder<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1118442019 AMC 8 Problems/Problem 242019-11-21T01:37:03Z<p>Brudder: /* Solution 1 */</p>
<hr />
<div><br />
<br />
<br />
==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <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 />
Draw <math>X</math> so that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE</math>=<math>ED</math>. <math>FC</math>=3<math>XD</math> so <math>BC</math>=4<math>BC</math>. Since <math>AF</math>=3<math>EF</math>( <math>XE</math>=<math>EF</math> and <math>AX</math>=1/3<math>AF</math>, so <math>XE</math>=<math>EF</math>=1/3<math>AF</math>), the altitude of triangle <math>BEF</math> is equal to 1/3 of the altitude of <math>ABC</math>. The area of <math>ABC</math> is 360, so the area of <math>BEF</math>=1/3*1/4*360=<math>\boxed{(B) 30}</math>~heeeeeeeheeeee<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 3. By equality, point <math>B</math> has a mass of 3 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{B}30)</math><br />
-Brudder<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{B}</math><br />
-Brudder<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_15&diff=1114762005 AMC 12A Problems/Problem 152019-11-16T22:28:45Z<p>Brudder: /* Solution 6 (Mass Points) */</p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{AB}</math> be a [[diameter]] of a [[circle]] and <math>C</math> be a point on <math>\overline{AB}</math> with <math>2 \cdot AC = BC</math>. Let <math>D</math> and <math>E</math> be [[point]]s on the circle such that <math>\overline{DC} \perp \overline{AB}</math> and <math>\overline{DE}</math> is a second diameter. What is the [[ratio]] of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math><br />
<br />
== Solution ==<br />
<br />
===Solution 1===<br />
<br />
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or <math>\frac{CF}{CD}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>).<br />
<br />
Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. <br />
<br />
Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]:<br />
<br />
<div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div><br />
<br />
By the [[Pythagorean Theorem]] in <math>\triangle OFC</math>, we have <math>\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r</math>.<br />
<br />
Our answer is <math>\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 2===<br />
<br />
Let the center of the circle be <math>O</math>.<br />
<br />
Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>DE \Rightarrow</math> Area of <math>\triangle DCE = 2 \cdot</math> Area of <math>\triangle DCO = 2 \cdot (\frac{1}{6} \cdot</math> Area of <math>\triangle ABD) = \frac{1}{3} \cdot</math> Area of <math>\triangle ABD \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 3===<br />
<br />
Let <math>r</math> be the radius of the circle. Note that <math>AC+BC = 2r</math> so <math>AC = \frac{2}{3}r</math>.<br />
<br />
By [[Power of a Point Theorem]], <math>CD^2= AC \cdot BC = 2\cdot AC^2</math>, and thus <math>CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r</math><br />
<br />
Then the area of <math>\triangle ABD</math> is <math>\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2</math>. Similarly, the area of <math>\triangle DCE</math> is <math>\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2</math>, so the desired ratio is <math>\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}</math><br />
<br />
===Solution 4===<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), e=(-D.x,-D.y);<br />
pair H=(e.x,0);<br />
draw(A--B--D--cycle);<br />
draw(D--e--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",e,SSE);<br />
label("$B$",B,NE);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
label("$H$",H,SE);<br />
draw(e--H,dashed);<br />
label("O",(0,0),NE);<br />
label("1",(C--O),N);<br />
label("1",(H--O),N);<br />
label("2",(A--C),N);<br />
label("2",(H--B),N);<br />
label("3",(O--D),NE);<br />
label("3",(O--e),NE);<br />
label("$2\sqrt{2}$",(D--C),W);<br />
label("$2\sqrt{2}$",(H--e),E);<br />
draw(rightanglemark(e,(e.x,0),A,2));<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
Let the center of the circle be <math>O</math>.<br />
Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>.<br />
===Solution 5===<br />
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A</math>, <math>B</math>, and <math>C</math> which are <math>(-3,0)</math>, <math>(-1,0)</math>, and <math>(3,0)</math> respectively.<br />
By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>.<br />
===Solution 6 (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 />
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/* end of picture */<br />
</asy><br />
<br />
We set point <math>A</math> as a mass of 2. This means that point <math>B</math> has a mass of <math>1</math> since <math>2\times{AC} = 1\times{BC}</math>. This implies that point <math>C</math> has a mass of <math>2+1 = 3</math> and the center of the circle has a mass of <math>3+1 = 4</math>. After this, we notice that points <math>D</math> and <math>E</math> both must have a mass of <math>2</math> since <math>2+2 = 4</math> and they are both radii of the circle.<br />
<br />
To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply <math>\frac{3\times2\times2}{2\times2\times1}</math> which is <math>\boxed{\frac{1}{3}}</math> (the reciprocal of 3)<br />
<br />
-Brudder<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_15&diff=1114752005 AMC 12A Problems/Problem 152019-11-16T22:28:32Z<p>Brudder: /* Solution 6 (Mass Points) */</p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{AB}</math> be a [[diameter]] of a [[circle]] and <math>C</math> be a point on <math>\overline{AB}</math> with <math>2 \cdot AC = BC</math>. Let <math>D</math> and <math>E</math> be [[point]]s on the circle such that <math>\overline{DC} \perp \overline{AB}</math> and <math>\overline{DE}</math> is a second diameter. What is the [[ratio]] of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math><br />
<br />
== Solution ==<br />
<br />
===Solution 1===<br />
<br />
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or <math>\frac{CF}{CD}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>).<br />
<br />
Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. <br />
<br />
Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]:<br />
<br />
<div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div><br />
<br />
By the [[Pythagorean Theorem]] in <math>\triangle OFC</math>, we have <math>\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r</math>.<br />
<br />
Our answer is <math>\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 2===<br />
<br />
Let the center of the circle be <math>O</math>.<br />
<br />
Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>DE \Rightarrow</math> Area of <math>\triangle DCE = 2 \cdot</math> Area of <math>\triangle DCO = 2 \cdot (\frac{1}{6} \cdot</math> Area of <math>\triangle ABD) = \frac{1}{3} \cdot</math> Area of <math>\triangle ABD \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 3===<br />
<br />
Let <math>r</math> be the radius of the circle. Note that <math>AC+BC = 2r</math> so <math>AC = \frac{2}{3}r</math>.<br />
<br />
By [[Power of a Point Theorem]], <math>CD^2= AC \cdot BC = 2\cdot AC^2</math>, and thus <math>CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r</math><br />
<br />
Then the area of <math>\triangle ABD</math> is <math>\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2</math>. Similarly, the area of <math>\triangle DCE</math> is <math>\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2</math>, so the desired ratio is <math>\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}</math><br />
<br />
===Solution 4===<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), e=(-D.x,-D.y);<br />
pair H=(e.x,0);<br />
draw(A--B--D--cycle);<br />
draw(D--e--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",e,SSE);<br />
label("$B$",B,NE);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
label("$H$",H,SE);<br />
draw(e--H,dashed);<br />
label("O",(0,0),NE);<br />
label("1",(C--O),N);<br />
label("1",(H--O),N);<br />
label("2",(A--C),N);<br />
label("2",(H--B),N);<br />
label("3",(O--D),NE);<br />
label("3",(O--e),NE);<br />
label("$2\sqrt{2}$",(D--C),W);<br />
label("$2\sqrt{2}$",(H--e),E);<br />
draw(rightanglemark(e,(e.x,0),A,2));<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
Let the center of the circle be <math>O</math>.<br />
Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>.<br />
===Solution 5===<br />
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A</math>, <math>B</math>, and <math>C</math> which are <math>(-3,0)</math>, <math>(-1,0)</math>, and <math>(3,0)</math> respectively.<br />
By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>.<br />
===Solution 6 (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 />
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dot((2.1561002471522333,-3.887757016355488),dotstyle); <br />
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label("2", (-0.8831167323620728,2.2919841753236083), NE * labelscalefactor,wrwrwr); <br />
label("2", (2.5444446389790625,-4.5969076449088275), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
We set point <math>A</math> as a mass of 2. This means that point <math>B</math> has a mass of <math>1</math> since <math>2\times{AC} = 1\times{BC}</math>. This implies that point <math>C</math> has a mass of <math>2+1 = 3</math> and the center of the circle has a mass of <math>3+1 = 4</math>. After this, we notice that points <math>D</math> and <math>E</math> both must have a mass of <math>2</math> since <math>2+2 = 4</math> and they are both radii of the circle.<br />
<br />
To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply <math>\frac{3\times2\times2}{2\times2\times1}</math> which is <math>\boxed{\frac{1}{3}}</math> (the reciprocal of 3)<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_15&diff=1114742005 AMC 12A Problems/Problem 152019-11-16T22:27:53Z<p>Brudder: /* Solution 6 (Mass Points) */</p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{AB}</math> be a [[diameter]] of a [[circle]] and <math>C</math> be a point on <math>\overline{AB}</math> with <math>2 \cdot AC = BC</math>. Let <math>D</math> and <math>E</math> be [[point]]s on the circle such that <math>\overline{DC} \perp \overline{AB}</math> and <math>\overline{DE}</math> is a second diameter. What is the [[ratio]] of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math><br />
<br />
== Solution ==<br />
<br />
===Solution 1===<br />
<br />
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or <math>\frac{CF}{CD}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>).<br />
<br />
Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. <br />
<br />
Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]:<br />
<br />
<div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div><br />
<br />
By the [[Pythagorean Theorem]] in <math>\triangle OFC</math>, we have <math>\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r</math>.<br />
<br />
Our answer is <math>\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 2===<br />
<br />
Let the center of the circle be <math>O</math>.<br />
<br />
Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>DE \Rightarrow</math> Area of <math>\triangle DCE = 2 \cdot</math> Area of <math>\triangle DCO = 2 \cdot (\frac{1}{6} \cdot</math> Area of <math>\triangle ABD) = \frac{1}{3} \cdot</math> Area of <math>\triangle ABD \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 3===<br />
<br />
Let <math>r</math> be the radius of the circle. Note that <math>AC+BC = 2r</math> so <math>AC = \frac{2}{3}r</math>.<br />
<br />
By [[Power of a Point Theorem]], <math>CD^2= AC \cdot BC = 2\cdot AC^2</math>, and thus <math>CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r</math><br />
<br />
Then the area of <math>\triangle ABD</math> is <math>\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2</math>. Similarly, the area of <math>\triangle DCE</math> is <math>\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2</math>, so the desired ratio is <math>\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}</math><br />
<br />
===Solution 4===<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), e=(-D.x,-D.y);<br />
pair H=(e.x,0);<br />
draw(A--B--D--cycle);<br />
draw(D--e--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",e,SSE);<br />
label("$B$",B,NE);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
label("$H$",H,SE);<br />
draw(e--H,dashed);<br />
label("O",(0,0),NE);<br />
label("1",(C--O),N);<br />
label("1",(H--O),N);<br />
label("2",(A--C),N);<br />
label("2",(H--B),N);<br />
label("3",(O--D),NE);<br />
label("3",(O--e),NE);<br />
label("$2\sqrt{2}$",(D--C),W);<br />
label("$2\sqrt{2}$",(H--e),E);<br />
draw(rightanglemark(e,(e.x,0),A,2));<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
Let the center of the circle be <math>O</math>.<br />
Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>.<br />
===Solution 5===<br />
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A</math>, <math>B</math>, and <math>C</math> which are <math>(-3,0)</math>, <math>(-1,0)</math>, and <math>(3,0)</math> respectively.<br />
By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>.<br />
===Solution 6 (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 />
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pen dotstyle = black; /* point style */ <br />
real xmin = -4.24313994860289, xmax = 6.360350402147026, ymin = -8.17642986522568, ymax = 4.1323989018072735; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
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draw(circle((0.9223776980185863,-1.084225871478444), 3.171161249925393), linewidth(2) + wrwrwr); <br />
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draw((-0.4813673187299407,-1.0971666418223938)--(2.1729847859273126,-3.904641555130568), linewidth(2) + wrwrwr); <br />
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draw((-2.2487343499798245,-1.1018908670262892)--(-0.5623104956355617,1.717909856970905), linewidth(2) + wrwrwr); <br />
draw((-0.5623104956355617,1.717909856970905)--(4.0935388680341624,-1.0849377800575102), linewidth(2) + wrwrwr); <br />
draw(circle((-2.858607769046372,-1.1524617347926092), 0.45463011998128017), linewidth(2) + wrwrwr); <br />
draw(circle((4.790088296064635,-1.144019465405069), 0.45463011998128006), linewidth(2) + wrwrwr); <br />
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draw(circle((1.4976032349241348,-1.3128648531558642), 0.4546301199812797), linewidth(2) + wrwrwr); <br />
draw(circle((-0.815578577261755,2.334195522261307), 0.4546301199812809), linewidth(2) + wrwrwr); <br />
draw(circle((2.6119827940793803,-4.5546962979711285), 0.45463011998128033), linewidth(2) + wrwrwr); <br />
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dot((0.9223776980185863,-1.084225871478444),dotstyle); <br />
dot((-0.5623104956355617,1.717909856970905),dotstyle); <br />
label("$D$", (-0.49477234053524466,1.8867552447217006), NE * labelscalefactor); <br />
dot((-0.39884850438732933,-3.9670413347199647),dotstyle); <br />
dot((-2.2487343499798245,-1.1018908670262892),dotstyle); <br />
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dot((4.0935388680341624,-1.0849377800575102),dotstyle); <br />
label("B", (4.165360361386693,-0.9160781919414961), NE * labelscalefactor); <br />
dot((-0.4813673187299407,-1.0971666418223938),linewidth(4pt) + dotstyle); <br />
label("C", (-0.41034964665984724,-0.9667318082667347), NE * labelscalefactor); <br />
dot((2.1729847859273126,-3.904641555130568),dotstyle); <br />
dot((2.1561002471522333,-3.887757016355488),dotstyle); <br />
label("E", (2.2236384022525524,-3.7189116286046926), NE * labelscalefactor); <br />
label("2", (-2.9261459241466903,-1.2031153511178476), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.722550140964316,-1.186230812342768), NE * labelscalefactor,wrwrwr); <br />
label("3", (-0.9844239650125497,-1.6758824368200735), NE * labelscalefactor,wrwrwr); <br />
label("4", (1.4300650798238166,-1.355076200093563), NE * labelscalefactor,wrwrwr); <br />
label("2", (-0.8831167323620728,2.2919841753236083), NE * labelscalefactor,wrwrwr); <br />
label("2", (2.5444446389790625,-4.5969076449088275), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
We set point <math>A</math> as a mass of 2. This means that point <math>B</math> has a mass of <math>1</math> since <math>2\times{AC} = 1\times{BC}</math>. This implies that point <math>C</math> has a mass of <math>2+1 = 3</math> and the center of the circle has a mass of <math>3+1 = 4</math>. After this, we notice that points <math>D</math> and <math>E</math> both must have a mass of <math>2</math> since <math>2+2 = 4</math> and they are both radii of the circle.<br />
<br />
To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply <math>\frac{3\times2\times2}{2\times2\times1}</math> which is <math>\boxed{\frac{1}{3}}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_15&diff=1114722005 AMC 12A Problems/Problem 152019-11-16T22:27:05Z<p>Brudder: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{AB}</math> be a [[diameter]] of a [[circle]] and <math>C</math> be a point on <math>\overline{AB}</math> with <math>2 \cdot AC = BC</math>. Let <math>D</math> and <math>E</math> be [[point]]s on the circle such that <math>\overline{DC} \perp \overline{AB}</math> and <math>\overline{DE}</math> is a second diameter. What is the [[ratio]] of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math><br />
<br />
== Solution ==<br />
<br />
===Solution 1===<br />
<br />
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or <math>\frac{CF}{CD}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>).<br />
<br />
Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. <br />
<br />
Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]:<br />
<br />
<div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div><br />
<br />
By the [[Pythagorean Theorem]] in <math>\triangle OFC</math>, we have <math>\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r</math>.<br />
<br />
Our answer is <math>\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 2===<br />
<br />
Let the center of the circle be <math>O</math>.<br />
<br />
Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB</math>.<br />
<br />
<math>O</math> is midpoint of <math>DE \Rightarrow</math> Area of <math>\triangle DCE = 2 \cdot</math> Area of <math>\triangle DCO = 2 \cdot (\frac{1}{6} \cdot</math> Area of <math>\triangle ABD) = \frac{1}{3} \cdot</math> Area of <math>\triangle ABD \Longrightarrow \mathrm{(C)}</math>.<br />
<br />
===Solution 3===<br />
<br />
Let <math>r</math> be the radius of the circle. Note that <math>AC+BC = 2r</math> so <math>AC = \frac{2}{3}r</math>.<br />
<br />
By [[Power of a Point Theorem]], <math>CD^2= AC \cdot BC = 2\cdot AC^2</math>, and thus <math>CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r</math><br />
<br />
Then the area of <math>\triangle ABD</math> is <math>\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2</math>. Similarly, the area of <math>\triangle DCE</math> is <math>\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2</math>, so the desired ratio is <math>\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}</math><br />
<br />
===Solution 4===<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), e=(-D.x,-D.y);<br />
pair H=(e.x,0);<br />
draw(A--B--D--cycle);<br />
draw(D--e--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",e,SSE);<br />
label("$B$",B,NE);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
label("$H$",H,SE);<br />
draw(e--H,dashed);<br />
label("O",(0,0),NE);<br />
label("1",(C--O),N);<br />
label("1",(H--O),N);<br />
label("2",(A--C),N);<br />
label("2",(H--B),N);<br />
label("3",(O--D),NE);<br />
label("3",(O--e),NE);<br />
label("$2\sqrt{2}$",(D--C),W);<br />
label("$2\sqrt{2}$",(H--e),E);<br />
draw(rightanglemark(e,(e.x,0),A,2));<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
Let the center of the circle be <math>O</math>.<br />
Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>.<br />
===Solution 5===<br />
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A</math>, <math>B</math>, and <math>C</math> which are <math>(-3,0)</math>, <math>(-1,0)</math>, and <math>(3,0)</math> respectively.<br />
By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>.<br />
===Solution 6 (Mass Points)===<br />
<asy><br />
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import graph; size(0cm); <br />
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<br />
We set point <math>A</math> as a mass of 2. This means that point <math>B</math> has a mass of <math>1</math> since <math>2\timesAC = 1\timesBC</math>. This implies that point <math>C</math> has a mass of <math>2+1 = 3</math> and the center of the circle has a mass of <math>3+1 = 4</math>. After this, we notice that points <math>D</math> and <math>E</math> both must have a mass of <math>2</math> since <math>2+2 = 4</math> and they are both radii of the circle.<br />
<br />
To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD devided by each other. Which is simply <math>\frac{3\times2\times2}{2\times2\times1}</math> which is <math>\boxed{\frac{1}{3}}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_4&diff=1095402007 AIME I Problems/Problem 42019-08-30T02:57:10Z<p>Brudder: </p>
<hr />
<div>== Problem ==<br />
Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at [[constant]] speed. Their periods are 60, 84, and 140. The three planets and the star are currently [[collinear]]. What is the fewest number of years from now that they will all be collinear again?<br />
<br />
== Solution ==<br />
<br />
Denote the planets <math>A, B, C </math> respectively. Let <math>a(t), b(t), c(t) </math> denote the angle which each of the respective planets makes with its initial position after <math>t </math> years. These are given by <math> a(t) = \frac{t \pi}{30} </math>, <math> b(t) = \frac{t \pi}{42} </math>, <math>c(t) = \frac{t \pi}{70}</math>.<br />
<br />
In order for the planets and the central star to be collinear, <math>a(t)</math>, <math>b(t) </math>, and <math>c(t) </math> must differ by a multiple of <math>\pi </math>. Note that <math> a(t) - b(t) = \frac{t \pi}{105}</math> and <math> b(t) - c(t) = \frac{t \pi}{105}</math>, so <math> a(t) - c(t) = \frac{ 2 t \pi}{105} </math>. These are simultaneously multiples of <math>\pi </math> exactly when <math>t </math> is a multiple of <math>105</math>, so the planets and the star will next be collinear in <math>\boxed{105}</math> years.<br />
<br />
== Solution 2 ==<br />
This problem is trivialized since the answer is the LCM/GCF of <math>60,84,140</math>. This is just <math>420/4</math> which is <math>\boxed{105}</math><br />
-Brudder<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_4&diff=1094732005 AIME I Problems/Problem 42019-08-29T20:56:22Z<p>Brudder: /* = */</p>
<hr />
<div>== Problem ==<br />
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are <math>5</math> members left over. The director realizes that if he arranges the group in a formation with <math>7</math> more rows than columns, there are no members left over. Find the maximum number of members this band can have.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
If <math>n > 14</math> then <math>n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21</math> and so <math>(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5</math>. If <math>n</math> is an [[integer]] there are no numbers which are 5 more than a [[perfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the number of students is <math>n(n + 7)</math> which must be 5 more than a perfect square, so <math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larger number can. Thus, the answer is <math>\boxed{294}</math>.<br />
<br />
=== Solution 2 ===<br />
Define the number of rows/columns of the square formation as <math>s</math>, and the number of rows of the rectangular formation <math>r</math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\sqrt{4s^2 + 69}</math> must be an [[integer]], say <math>x</math>. Then <math>4s^2 + 69 = x^2</math> and <math>(x + 2s)(x - 2s) = 69</math>. The factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The latter obviously can be discarded, so there are <math>21</math> rows and <math>21 - 7 = 14</math> columns, making the answer <math>294</math>.<br />
<br />
=== Solution 3 ===<br />
The number of members is <math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by <math>4</math> and [[completing the square|complete the square]] to get <math>4m^2+69=(2n+7)^2</math>. Thus, we have <math>69=((2n+7)+2m)((2n+7)-2m)</math>. Since we want to maximize <math>n</math>, set the first factor equal to <math>69</math> and the second equal to <math>1</math>. Solving gives <math>n=14</math>, so the answer is <math>14\cdot21=294</math>.<br />
<br />
<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=I|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_4&diff=1094722005 AIME I Problems/Problem 42019-08-29T20:56:12Z<p>Brudder: </p>
<hr />
<div>== Problem ==<br />
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are <math>5</math> members left over. The director realizes that if he arranges the group in a formation with <math>7</math> more rows than columns, there are no members left over. Find the maximum number of members this band can have.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
If <math>n > 14</math> then <math>n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21</math> and so <math>(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5</math>. If <math>n</math> is an [[integer]] there are no numbers which are 5 more than a [[perfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the number of students is <math>n(n + 7)</math> which must be 5 more than a perfect square, so <math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larger number can. Thus, the answer is <math>\boxed{294}</math>.<br />
<br />
=== Solution 2 ===<br />
Define the number of rows/columns of the square formation as <math>s</math>, and the number of rows of the rectangular formation <math>r</math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\sqrt{4s^2 + 69}</math> must be an [[integer]], say <math>x</math>. Then <math>4s^2 + 69 = x^2</math> and <math>(x + 2s)(x - 2s) = 69</math>. The factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The latter obviously can be discarded, so there are <math>21</math> rows and <math>21 - 7 = 14</math> columns, making the answer <math>294</math>.<br />
<br />
=== Solution 3 ===<br />
The number of members is <math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by <math>4</math> and [[completing the square|complete the square]] to get <math>4m^2+69=(2n+7)^2</math>. Thus, we have <math>69=((2n+7)+2m)((2n+7)-2m)</math>. Since we want to maximize <math>n</math>, set the first factor equal to <math>69</math> and the second equal to <math>1</math>. Solving gives <math>n=14</math>, so the answer is <math>14\cdot21=294</math>.<br />
<br />
=== <br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=I|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_4&diff=1094712005 AIME I Problems/Problem 42019-08-29T20:42:35Z<p>Brudder: </p>
<hr />
<div>== Problem ==<br />
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are <math>5</math> members left over. The director realizes that if he arranges the group in a formation with <math>7</math> more rows than columns, there are no members left over. Find the maximum number of members this band can have.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
If <math>n > 14</math> then <math>n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21</math> and so <math>(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5</math>. If <math>n</math> is an [[integer]] there are no numbers which are 5 more than a [[perfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the number of students is <math>n(n + 7)</math> which must be 5 more than a perfect square, so <math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larger number can. Thus, the answer is <math>\boxed{294}</math>.<br />
<br />
=== Solution 2 ===<br />
Define the number of rows/columns of the square formation as <math>s</math>, and the number of rows of the rectangular formation <math>r</math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\sqrt{4s^2 + 69}</math> must be an [[integer]], say <math>x</math>. Then <math>4s^2 + 69 = x^2</math> and <math>(x + 2s)(x - 2s) = 69</math>. The factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The latter obviously can be discarded, so there are <math>21</math> rows and <math>21 - 7 = 14</math> columns, making the answer <math>294</math>.<br />
<br />
=== Solution 3 ===<br />
The number of members is <math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by <math>4</math> and [[completing the square|complete the square]] to get <math>4m^2+69=(2n+7)^2</math>. Thus, we have <math>69=((2n+7)+2m)((2n+7)-2m)</math>. Since we want to maximize <math>n</math>, set the first factor equal to <math>69</math> and the second equal to <math>1</math>. Solving gives <math>n=14</math>, so the answer is <math>14\cdot21=294</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=I|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_4&diff=1094702005 AIME I Problems/Problem 42019-08-29T20:42:31Z<p>Brudder: </p>
<hr />
<div>== Problem ==<br />
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are <math>5</math> members left over. The director realizes that if he arranges the group in a formation with <math>7</math> more rows than columns, there are no members left over. Find the maximum number of members this band can have.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
If <math>n > 14</math> then <math>n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21</math> and so <math>(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5</math>. If <math>n</math> is an [[integer]] there are no numbers which are 5 more than a [[perfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the number of students is <math>n(n + 7)</math> which must be 5 more than a perfect square, so <math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larger number can. Thus, the answer is <math>\boxed{294}</math>.<br />
<br />
=== Solution 2 ===<br />
Define the number of rows/columns of the square formation as <math>s</math>, and the number of rows of the rectangular formation <math>r</math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\sqrt{4s^2 + 69}</math> must be an [[integer]], say <math>x</math>. Then <math>4s^2 + 69 = x^2</math> and <math>(x + 2s)(x - 2s) = 69</math>. The factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The latter obviously can be discarded, so there are <math>21</math> rows and <math>21 - 7 = 14</math> columns, making the answer <math>294</math>.<br />
<br />
=== Solution 3 ===<br />
The number of members is <math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by <math>4</math> and [[completing the square|complete the square]] to get <math>4m^2+69=(2n+7)^2</math>. Thus, we have <math>69=((2n+7)+2m)((2n+7)-2m)</math>. Since we want to maximize <math>n</math>, set the first factor equal to <math>69</math> and the second equal to <math>1</math>. Solving gives <math>n=14</math>, so the answer is <math>14\cdot21=294</math>.<br />
<br />
=== Solution 4 (Guess And Check)===<br />
By reading the problem, we get the equation <math>m^2+5=n(n+7)</math>. Guess and check for integers yields <math>\boxed{294}</math><br />
== See also ==<br />
{{AIME box|year=2005|n=I|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_10&diff=1089792009 AIME II Problems/Problem 102019-08-19T19:20:13Z<p>Brudder: /* Diagram */</p>
<hr />
<div><br />
==Problem==<br />
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. <br />
<br />
==Diagram==<br />
<asy><br />
size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30));<br />
</asy><br />
-asjpz<br />
<br />
== Solution 1==<br />
<br />
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.<br />
<br />
==Solution 2==<br />
<br />
Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>.<br />
<br />
By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>:<br />
<br />
<cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\<br />
AD^2&=130-\dfrac{130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130(23^2-169)}{23^2}\\<br />
AD^2&=\dfrac{130(360)}{23^2}\\<br />
AD&=\dfrac{60\sqrt{13}}{23}\\<br />
\end{align*}</cmath><br />
<br />
and our final answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
==Solution 3==<br />
<br />
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.<br />
<br />
==Solution 4==<br />
<br />
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
(Solution by RootThreeOverTwo)<br />
<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_10&diff=1089782009 AIME II Problems/Problem 102019-08-19T19:19:55Z<p>Brudder: </p>
<hr />
<div><br />
==Problem==<br />
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. <br />
<br />
==Diagram==<br />
<asy><br />
size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30));<br />
</asy>v<br />
<br />
aspjz created this diagram<br />
<br />
<br />
== Solution 1==<br />
<br />
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.<br />
<br />
==Solution 2==<br />
<br />
Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>.<br />
<br />
By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>:<br />
<br />
<cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\<br />
AD^2&=130-\dfrac{130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130(23^2-169)}{23^2}\\<br />
AD^2&=\dfrac{130(360)}{23^2}\\<br />
AD&=\dfrac{60\sqrt{13}}{23}\\<br />
\end{align*}</cmath><br />
<br />
and our final answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
==Solution 3==<br />
<br />
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.<br />
<br />
==Solution 4==<br />
<br />
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
(Solution by RootThreeOverTwo)<br />
<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_10&diff=1089772009 AIME II Problems/Problem 102019-08-19T19:17:34Z<p>Brudder: /* Diagram */</p>
<hr />
<div><br />
==Problem==<br />
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. <br />
<br />
<br />
== Solution 1==<br />
<br />
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.<br />
<br />
==Solution 2==<br />
<br />
Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>.<br />
<br />
By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>:<br />
<br />
<cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\<br />
AD^2&=130-\dfrac{130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130(23^2-169)}{23^2}\\<br />
AD^2&=\dfrac{130(360)}{23^2}\\<br />
AD&=\dfrac{60\sqrt{13}}{23}\\<br />
\end{align*}</cmath><br />
<br />
and our final answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
==Solution 3==<br />
<br />
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.<br />
<br />
==Solution 4==<br />
<br />
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
(Solution by RootThreeOverTwo)<br />
<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_10&diff=1089762009 AIME II Problems/Problem 102019-08-19T19:17:06Z<p>Brudder: </p>
<hr />
<div><br />
==Problem==<br />
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. <br />
<br />
<br />
== Solution 1==<br />
<br />
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.<br />
<br />
==Solution 2==<br />
<br />
Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>.<br />
<br />
By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>:<br />
<br />
<cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\<br />
AD^2&=130-\dfrac{130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130(23^2-169)}{23^2}\\<br />
AD^2&=\dfrac{130(360)}{23^2}\\<br />
AD&=\dfrac{60\sqrt{13}}{23}\\<br />
\end{align*}</cmath><br />
<br />
and our final answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
==Solution 3==<br />
<br />
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.<br />
<br />
==Solution 4==<br />
<br />
After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>.<br />
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(Solution by RootThreeOverTwo)<br />
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==Diagram==<br />
[asy]size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30));[/asy]<br />
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This diagram is by aspjz<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_6&diff=1089691955 AHSME Problems/Problem 62019-08-19T03:07:19Z<p>Brudder: </p>
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<div>A merchant buys a number of oranges at <math>3</math> for <math>10</math> cents and an equal number at <math>5</math> for <math>20</math> cents. To "break even" he must sell all at:<br />
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<math>\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents}</math><br />
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Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges.<br />
That means that we are getting a total of <math>30</math> oranges for <math>(10\times5) + (20\times3)</math> cents.<br />
That comes to a total of <math>30</math> oranges for <math>110</math> cents. <math>110/30</math> = <math>11/3</math>. This leads us to <math>3</math> for <math>11</math> cents which is <math>\boxed{C}</math> and we are done.<br />
-Brudder</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_6&diff=1055251955 AHSME Problems/Problem 62019-04-28T15:20:03Z<p>Brudder: </p>
<hr />
<div>A merchant buys a number of oranges at <math>3</math> for <math>10</math> cents and an equal number at <math>5</math> for <math>20</math> cents. To "break even" he must sell all at:<br />
<br />
<math>\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents}</math><br />
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<br />
Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges.<br />
That means that we are getting a total of <math>30</math> oranges for <math>(10x5) + (20x3)</math> cents.<br />
That comes to a total of <math>30</math> oranges for <math>110</math> cents. <math>110/30</math> = <math>11/3</math>. This leads us to <math>3</math> for <math>11</math> cents which is <math>C</math> and we are done.<br />
-Brudder</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_6&diff=1055241955 AHSME Problems/Problem 62019-04-28T15:17:03Z<p>Brudder: Created page with "Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges. That means..."</p>
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<div>Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges.<br />
That means that we are getting a total of <math>30</math> oranges for <math>(10*5) + (20*3)</math> cents.<br />
That comes to a total of <math>30</math> oranges for <math>110</math> cents. <math>110/30</math> = <math>11/3</math> and we are done.<br />
-Brudder</div>Brudderhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_18&diff=1049602018 AMC 10A Problems/Problem 182019-03-26T01:46:22Z<p>Brudder: /* See Also */</p>
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<div>==Problem==<br />
How many nonnegative integers can be written in the form <cmath>a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,</cmath><br />
where <math>a_i\in \{-1,0,1\}</math> for <math>0\le i \le 7</math>?<br />
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<math>\textbf{(A) } 512 \qquad <br />
\textbf{(B) } 729 \qquad <br />
\textbf{(C) } 1094 \qquad <br />
\textbf{(D) } 3281 \qquad <br />
\textbf{(E) } 59,048 </math><br />
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==Solution 1==<br />
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This looks like balanced ternary, in which all the integers with absolute values less than <math>\frac{3^n}{2}</math> are represented in <math>n</math> digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of <math>|x|=3280.5</math>, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are <math>3280+1=\boxed{3281}</math> integers or <math>\boxed{D}</math>.<br />
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==Solution 2==<br />
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Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all <math>a_i=0</math>. The total number of ways to pick <math>a_i</math> from <math>i=0, 1, 2, 3, ... 7</math> is <math>3^8=6561</math>. <math>\frac{6561-1}{2}=3280</math> gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives <math>6561-3280=\boxed{3281}</math>. (RegularHexagon)<br />
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==Solution 3==<br />
Note that the number of total possibilities (ignoring the conditions set by the problem) is <math>3^8=6561</math>. So, E is clearly unrealistic. <br />
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Note that if <math>a_7</math> is 1, then it's impossible for <cmath>a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,</cmath> to be negative. Therefore, if <math>a_7</math> is 1, there are <math>3^7=2187</math> possibilities. (We also must convince ourselves that these <math>2187</math> different sets of coefficients must necessarily yield <math>2187</math> different integer results.)<br />
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As A, B, and C are all less than 2187, the answer must be <math>\boxed{(D) 3281}</math><br />
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==Solution 4==<br />
Note that we can do some simple casework:<br />
If <math>a_7=1</math>, then we can choose anything for the other 7 variables, so this give us <math>3^7</math>.<br />
If <math>a_7=0</math> and <math>a_6=1</math>, then we can choose anything for the other 6 variables, giving us <math>3^6</math>.<br />
If <math>a_7=0</math>, <math>a_6=0</math>, and <math>a_5=1</math>, then we have <math>3^5</math>.<br />
Continuing in this vein, we have <math>3^7+3^6+\cdots+3^1+3^0</math> ways to choose the variables' values, except we have to add 1<br />
because we haven't counted the case where all variables are 0. So our total sum is <math>\boxed{(D) 3281}</math>.<br />
Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.<br />
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==Solution 5==<br />
The key is to realize that this question is basically taking place in <math>a\in\{0,1,2\}</math> if each value of <math>a</math> was increased by <math>1</math>, essentially making it into base <math>3</math>. Then the range would be from <math>0\cdot3^7+</math> <math>0\cdot3^6+</math> <math>0\cdot3^5+</math> <math>0\cdot3^4+</math> <math>0\cdot3^3+</math> <math>0\cdot3^2+</math> <math>0\cdot3^1+</math> <math>0\cdot3^0=</math> <math>0</math> to <math>2\cdot3^7+</math> <math>2\cdot3^6+</math> <math>2\cdot3^5+</math> <math>2\cdot3^4+</math> <math>2\cdot3^3+</math> <math>2\cdot3^2+</math> <math>2\cdot3^1+</math> <math>2\cdot3^0=</math> <math>3^8-1=</math> <math>6561-1=</math> <math>6560</math>, yielding <math>6561</math> different values. Since the distribution for all <math>a_i\in \{-1,0,1\}</math> the question originally gave is symmetrical, we retain the <math>3280</math> positive integers and one <math>0</math> but discard the <math>3280</math> negative integers. Thus, we are left with the answer, <math>\boxed{\textbf{(D)} 3281}\qquad</math>. ∎ --anna0kear<br />
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==Solution 6==<br />
First, set <math>a_i=0</math> for all <math>i\geq1</math>. The range would be the integers for which <math>[-1,1]</math>. If <math>a_i=0</math> for all <math>i\geq2</math>, our set expands to include all integers such that <math>-4\leq\mathbb{Z}\leq4</math>. Similarly, when <math>i\geq3</math> we get <math>-13\leq\mathbb{Z}\leq13</math>, and when <math>i\geq4</math> the range is <math>-40\leq\mathbb{Z}\leq40</math>. The pattern continues until we reach <math>i=7</math>, where <math>-3280\leq\mathbb{Z}\leq3280</math>. Because we are only looking for positive integers, we filter out all <math>\mathbb{Z}<0</math>, leaving us with all integers between <math>0\leq\mathbb{Z}\leq3280</math>, inclusive. The answer becomes <math>\boxed{(D)3281}</math>. ∎ --anna0kear<br />
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==Solution 7==<br />
To get the number of integers, we can get the highest positive integer that can be represented using <cmath>a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,</cmath><br />
where <math>a_i\in \{-1,0,1\}</math> for <math>0\le i \le 7</math>.<br />
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Note that the least nonnegative integer that can be represented is <math>0</math>, when all <math>a_i=0</math>. The highest number will be the number when all <math>a_i=1</math>. That will be <cmath>3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}</cmath> <cmath>=3280</cmath><br />
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Therefore, there are <math>3280</math> positive integers and <math>(3280+1)</math> nonnegative integers (while including <math>0</math>) that can be represented. Our answer is <math>\boxed{\textbf{(D) } 3281}</math><br />
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~OlutosinNGA<br />
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==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}}<br />
{{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}<br />
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[[Category: Intermediate Number Theory Problems]]</div>Brudder