Difference between revisions of "2013 AMC 10A Problems/Problem 23"

Revision as of 01:47, 30 January 2018

The following problem is from both the 2013 AMC 12A #19 and 2013 AMC 10A #23, so both problems redirect to this page.

Problem

In $\triangle ABC$ , $AB = 86$ , 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 1 (Power of a Point)

Let $BX = q$ , $CX = p$ , and $AC$ meets the circle at $Y$ and $Z$ , with $Y$ on $AC$ . Then $AZ = AY = 86$ . Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q) = 11(183) = 11 * 3 * 61$ . We know that $p+q>p$ , so $p$ is either 3,11, or 33. We also know that $p>11$ by the triangle inequality on $\triangle ACX$ . $p$ is 33. Thus, we get that $BC = p+q = \boxed{\textbf{(D) }61}$ .

Solution 2 (Stewart's Theorem)

Stewart's Theorem

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 $\boxed{\textbf{(D) }61}$ .

Solution 3

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 $\boxed{\textbf{(D) }61}$ .

Solution 4

$[asy] unitsize(2); import olympiad; import graph; pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40); draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label("A",A,S); label("B",B,NE); label("C",C,S); label("D",D,NE); label("E",E,NE); label("86",(A+B)/2,NW); label("86",(A+D)/2,SE); label("97",(A+C)/2,S); label("h",(A+E)/2,N); label("k",(E+D)/2,NE); label("k",(B+E)/2,NE); label("m",(C+D)/2,NE); fill(anglemark(A,E,D,100),black); label("$90^\circ$",anglemark(A,E,D),3*S); [/asy]$

We first draw the height of isosceles triangle ABD, and get two equations by the Pythagorean Theorem. First, $h^2 + k^2 = 86^2$ . Second, $h^2 + (k + m)^2 = 97^2$ . Subtracting these two equations, we get $2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013$ . We then add $k^2$ to both sides to get $k^2 + 2km + m^2 = 2013 + k^2$ . We then complete the square to get $(k + m)^2 = 2013 + k^2$ . Because $k$ and $m$ are both integers, we get that $2013 + k^2$ is a square number. Simple guess and check reveals that $k = 14$ . Because $k$ equals $14$ , therefore $m = 33$ . We want $\overline{BC} = 2k + m$ , so we get that $\overline{BC} = \boxed{(B)~61}$ .

$[color=#eee]solution by bobjoe123[/color]$ (Error compiling LaTeX. Unknown error_msg)

@@ Line 95: / Line 95: @@
 Because <math>k</math> equals <math>14</math>, therefore <math>m = 33</math>. We want <math>\overline{BC} = 2k + m</math>, so we get that <math>\overline{BC} = \boxed{(B)~61}</math>.
-<math>\phantom{solution by bobjoe123}</math>
+<math>[color=#eee]solution by bobjoe123[/color]</math>
 ==See Also==

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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 10 Problems and Solutions

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