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

Revision as of 19:33, 1 July 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 (Number theoretic power of a point)

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -156.86902060182828, xmax = 261.4826629156128, ymin = -137.6904841249097, ymax = 103.70826798306636; /* image dimensions */ draw((0,0)--(-45.39592734786701,-73.0425203578517)--(10.963968250054833,-96.37837620655263)--cycle, linewidth(1)); /* draw figures */ draw(circle((0,0), 86), linewidth(1)); draw((-45.39592734786701,-73.0425203578517)--(10.963968250054833,-96.37837620655263), linewidth(1)); draw((0,0)--(-45.39592734786701,-73.0425203578517), linewidth(1)); draw((-45.39592734786701,-73.0425203578517)--(10.963968250054833,-96.37837620655263), linewidth(1)); draw((10.963968250054833,-96.37837620655263)--(0,0), linewidth(1)); draw((0,0)--(-19.525811335706223,-83.75406074741933), linewidth(1)); draw((0,0)--(-9.720631644378512,85.4488696264281), linewidth(1)); label("$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$", (-47.36574051664951,-68.87545524345045), NW * labelscalefactor); dot((10.963968250054833,-96.37837620655263),linewidth(1pt) + dotstyle); label("$C$", (12.164720976041195,-94.27147780684612), SE * labelscalefactor); dot((-19.525811335706223,-83.75406074741933),linewidth(1pt) + dotstyle); label("$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$", (-8.589017892970244,87.596812808439), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $CX=x, BX=y$ . Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$ . Use power of a point on point C to the circle centered at A.

So $CX*CB=CD*CE=>$ $x(x+y)=(97-86)(97+86)=>$ $x(x+y)=3*11*61$ .

Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$ . By the Triangle Inequality, only $x+y=61$ yields a possible length of $BX+CX=BC$ .

Therefore, the answer is D) 61.

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

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

Page

Toolbox

Search