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Difference between revisions of "2013 AMC 12A Problems/Problem 19"

(→‎Solution 2)
m (→‎Solution 1 (Diophantine PoP))
 
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Let CX=x, BX=y. Let the circle intersect AC at D and the diameter including AD intersect the circle again at E.
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== Problem==
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Use power of a point on point C to the circle centered at A.
 
  
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So CX*CB=CD*CE
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In <math> \bigtriangleup ABC </math>, <math> AB = 86 </math>, and <math> AC = 97 </math>. A circle with center <math> A </math> and radius <math> AB </math> intersects <math> \overline{BC} </math> at points <math> B </math> and <math> X </math>. Moreover <math> \overline{BX} </math> and <math> \overline{CX} </math> have integer lengths. What is <math> BC </math>?
−
x(x+y)=(97-86)(97+86)
 
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x(x+y)=3*11*61.
 
  
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Obviously x+y>x so we have three solution pairs for (x,x+y)=(1,2013),(3,671),(11,183),(33,61).
 
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By the Triangle Inequality, only x+y=61 yields a possible length of BX+CX=BC.
 
  
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Therefore, the answer is '''D) 61.'''
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<math> \textbf{(A)} \ 11 \qquad  \textbf{(B)} \ 28 \qquad  \textbf{(C)} \ 33 \qquad  \textbf{(D)} \ 61 \qquad  \textbf{(E)} \ 72 </math>
 +
 
 +
==Solution==
 +
===Solution 1 (Diophantine PoP)===
 +
 
 +
<asy>
 +
//Made by samrocksnature
 +
size(8cm);
 +
pair A,B,C,D,E,X;
 +
A=(0,0);
 +
B=(-53.4,-67.4);
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C=(0,-97);
 +
D=(0,-86);
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E=(0,86);
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X=(-29,-81);
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draw(circle(A,86));
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draw(E--C--B--A--X);
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label("$A$",A,NE);
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label("$B$",B,SW);
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label("$C$",C,S);
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label("$D$",D,NE);
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label("$E$",E,NE);
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label("$X$",X,dir(250));
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dot(A^^B^^C^^D^^E^^X);
 +
</asy>
 +
 
 +
Let circle <math>A</math> intersect <math>AC</math> at <math>D</math> and <math>E</math> as shown. We apply Power of a Point on point <math>C</math> with respect to circle <math>A.</math> This yields the diophantine equation
 +
 
 +
<cmath>CX \cdot CB = CD \cdot CE</cmath>
 +
<cmath>CX(CX+XB) = (97-86)(97+86)</cmath>
 +
<cmath>CX(CX+XB) = 3 \cdot 11 \cdot 61.</cmath>
 +
 
 +
Since lengths cannot be negative, we must have <math>CX+XB \ge CX.</math> This generates the four solution pairs for <math>(CX,CX+XB)</math>: <cmath>(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).</cmath>
 +
 
 +
However, by the Triangle Inequality on <math>\triangle ACX,</math> we see that <math>CX>13.</math> This implies that we must have <math>CX+XB= \boxed{\textbf{(D) }61}.</math>
 +
 
 +
(Solution by unknown, latex/asy modified majorly by samrocksnature)
  
 
===Solution 2===
 
===Solution 2===
−
Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86.
+
 
 +
Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meet the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>AC</math>.  Then <math>AZ = AY = 86</math>.  Using the Power of a Point, we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>.  We know that <math>p+q>p</math>, and that <math>p>13</math> by the triangle inequality on <math>\triangle ACX</math>.  Thus, we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>
 +
 
 +
===Solution 3===
 +
Let <math>x</math> represent <math>CX</math>, and let <math>y</math> represent <math>BX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB = AX = 86</math>.
 
Then by Stewart's Theorem,
 
Then by Stewart's Theorem,
  
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<math>x^2 + xy + 86^2 = 97^2</math>
 
<math>x^2 + xy + 86^2 = 97^2</math>
  
−
(Since <math>y</math> cannot be equal to 0, dividing both sides of the equation by <math>y</math> is allowed.)
+
(Since <math>y</math> cannot be equal to <math>0</math>, dividing both sides of the equation by <math>y</math> is allowed.)
  
 
<math>x(x+y) = (97+86)(97-86)</math>
 
<math>x(x+y) = (97+86)(97-86)</math>
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<math>x(x+y) = 2013</math>
 
<math>x(x+y) = 2013</math>
  
−
The prime factors of 2013 are 3, 11, and 61. Obviously <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal 33, and <math>x+y</math> must equal 61. <math>\boxed{D}</math>
+
The prime factors of <math>2013</math> are <math>3</math>, <math>11</math>, and <math>61</math>. Obviously, <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal <math>33</math>, and <math>x+y</math> must equal <math> \boxed{\textbf{(D) }61}</math>
 +
 
 +
==Video Solution by Richard Rusczyk==
 +
https://artofproblemsolving.com/videos/amc/2013amc12a/357
 +
 
 +
~dolphin7
 +
 
 +
==Video Solution==
 +
https://youtu.be/zxW3uvCQFls
 +
 
 +
~sugar_rush
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}
 +
 
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Number theory]]
 +
{{MAA Notice}}

Latest revision as of 14:24, 19 September 2021

Problem

In $\bigtriangleup ABC$, $AB = 86$, and $AC = 97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?


$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72$

Solution

Solution 1 (Diophantine PoP)

[asy] //Made by samrocksnature size(8cm); pair A,B,C,D,E,X; A=(0,0); B=(-53.4,-67.4); C=(0,-97); D=(0,-86); E=(0,86); X=(-29,-81); draw(circle(A,86)); draw(E--C--B--A--X); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,NE); label("$E$",E,NE); label("$X$",X,dir(250)); dot(A^^B^^C^^D^^E^^X); [/asy]

Let circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation

\[CX \cdot CB = CD \cdot CE\] \[CX(CX+XB) = (97-86)(97+86)\] \[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]

Since lengths cannot be negative, we must have $CX+XB \ge CX.$ This generates the four solution pairs for $(CX,CX+XB)$: \[(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).\]

However, by the Triangle Inequality on $\triangle ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB= \boxed{\textbf{(D) }61}.$

(Solution by unknown, latex/asy modified majorly by samrocksnature)

Solution 2

Let $BX = q$, $CX = p$, and $AC$ meet the circle at $Y$ and $Z$, with $Y$ on $AC$. Then $AZ = AY = 86$. Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$. We know that $p+q>p$, and that $p>13$ by the triangle inequality on $\triangle ACX$. Thus, we get that $BC = p+q = \boxed{\textbf{(D) }61}$

Solution 3

Let $x$ represent $CX$, and let $y$ represent $BX$. Since the circle goes through $B$ and $X$, $AB = AX = 86$. Then by Stewart's Theorem,

$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$

$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$

$x^2 + xy + 86^2 = 97^2$

(Since $y$ cannot be equal to $0$, dividing both sides of the equation by $y$ is allowed.)

$x(x+y) = (97+86)(97-86)$

$x(x+y) = 2013$

The prime factors of $2013$ are $3$, $11$, and $61$. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal $33$, and $x+y$ must equal $\boxed{\textbf{(D) }61}$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc12a/357

~dolphin7

Video Solution

https://youtu.be/zxW3uvCQFls

~sugar_rush

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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