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

Revision as of 20:09, 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)

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

We have

$\begin{align*} CX \cdot CB &= CD \cdot C \\ CX(CX+XB) &= (97-86)(97+86) \\ CX(CX+XB) &= 3 \cdot 11 \cdot 61. \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

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

@@ Line 33: / Line 33: @@
 We have
-\begin{align*}
+<math>\begin{align*}
 CX \cdot CB &= CD \cdot C \\
 CX(CX+XB) &= (97-86)(97+86) \\
 CX(CX+XB) &= 3 \cdot 11 \cdot 61.
-\end{align*}
+\end{align*}</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>
+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>
-Since lengths cannot be negative, we must have <math>CX+XB \ge CX.</math> This generates the four solution pairs for <math>(CX,XB)</math>: <cmath>(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).</cmath>
+(Solution by unknown, latex/asy modified majorly by samrocksnature)
-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===

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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