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

(→‎Solution 1 (Number theoretic power of a point))
m (→‎Solution 1 (Diophantine PoP))
 
(13 intermediate revisions by one other user not shown)
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==Solution==
 
==Solution==
−
===Solution 1 (Number theoretic power of a point)===
+
===Solution 1 (Diophantine PoP)===
  
 
<asy>
 
<asy>
−
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
+
//Made by samrocksnature
−
import graph; size(7cm);  
+
size(8cm);
−
real labelscalefactor = 0.5; /* changes label-to-point distance */
+
pair A,B,C,D,E,X;
−
pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */
+
A=(0,0);
−
pen dotstyle = black; /* point style */
+
B=(-53.4,-67.4);
−
real xmin = -156.86902060182828, xmax = 261.4826629156128, ymin = -137.6904841249097, ymax = 103.70826798306636; /* image dimensions */
+
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 <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
  
−
draw((0,0)--(-45.39592734786701,-73.0425203578517)--(10.963968250054833,-96.37837620655263)--cycle, linewidth(1));
+
<cmath>CX \cdot CB = CD \cdot CE</cmath>
−
/* draw figures */
+
<cmath>CX(CX+XB) = (97-86)(97+86)</cmath>
−
draw(circle((0,0), 86), linewidth(1));
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<cmath>CX(CX+XB) = 3 \cdot 11 \cdot 61.</cmath>
−
draw((-45.39592734786701,-73.0425203578517)--(10.963968250054833,-96.37837620655263), linewidth(1));
 
−
draw((0,0)--(-45.39592734786701,-73.0425203578517), linewidth(1));
 
−
draw((-45.39592734786701,-73.0425203578517)--(10.963968250054833,-96.37837620655263), linewidth(1));
 
−
draw((10.963968250054833,-96.37837620655263)--(0,0), linewidth(1));
 
−
draw((0,0)--(-19.525811335706223,-83.75406074741933), linewidth(1));
 
−
draw((0,0)--(-9.720631644378512,85.4488696264281), linewidth(1));
 
−
label("$x$",(-7.2236403357984384,-86.89843899811834),SE*labelscalefactor);
 
−
label("$y$",(-37.535022105012516,-75.42926751787513),SE*labelscalefactor);
 
−
/* dots and labels */
 
−
dot((0,0),linewidth(1pt) + dotstyle);
 
−
label("$A$", (0.9686250072323931,2.1241777294836823), NE * labelscalefactor);
 
−
dot((-45.39592734786701,-73.0425203578517),dotstyle);
 
−
label("$B$", (-47.36574051664951,-68.87545524345045), NE * labelscalefactor);
 
−
dot((10.963968250054833,-96.37837620655263),linewidth(1pt) + dotstyle);
 
−
label("$C$", (12.164720976041195,-94.27147780684612), NE * labelscalefactor);
 
−
dot((-19.525811335706223,-83.75406074741933),linewidth(1pt) + dotstyle);
 
−
label("$X$", (-15.68898119026363,-81.98307979229983), NE * labelscalefactor);
 
−
dot((9.720631644378512,-85.4488696264281),linewidth(1pt) + dotstyle);
 
−
label("$D$", (10.799343418869391,-83.34845734947163), NE * labelscalefactor);
 
−
dot((-9.720631644378512,85.4488696264281),linewidth(1pt) + dotstyle);
 
−
label("$E$", (-8.589017892970244,87.596812808439), NE * labelscalefactor);
 
−
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 
−
/* end of picture */
 
−
</asy>
 
−
 
 
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Let <math>CX=x, BX=y</math>. Let the circle intersect <math>AC</math> at <math>D</math> and the diameter including <math>AD</math> intersect the circle again at <math>E</math>.
 
−
Use power of a point on point C to the circle centered at A.
 
  
−
So <math>CX*CB=CD*CE=></math>
+
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>
−
<math>x(x+y)=(97-86)(97+86)=></math>
 
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<math>x(x+y)=3*11*61</math>.
 
  
−
Obviously <math>x+y>x</math> so we have three solution pairs for <math>(x,x+y)=(1,2013),(3,671),(11,183),(33,61)</math>.
+
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>  
−
By the Triangle Inequality, only<math> x+y=61</math> yields a possible length of <math>BX+CX=BC</math>.
 
  
−
Therefore, the answer is '''D) 61.'''
+
(Solution by unknown, latex/asy modified majorly by samrocksnature)
  
 
===Solution 2===
 
===Solution 2===

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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