2013 AMC 10A Problems/Problem 23

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 (Number Theoretic 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 $\triangle ACX$ . Thus, $p$ is $33$ so we get that $BC = p+q = \boxed{\textbf{(D) }61}$ .

Solution 2

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 $CX \cdot CB=CD \cdot CE \Rightarrow x(x+y)=(97-86)(97+86) \Rightarrow x(x+y)=3*11*61$ .

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 3

$[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 $\overline{BC} = \boxed{(D)~61}$ .

$\phantom{solution and diagram by bobjoe123}$

Solution 4

Let $E$ be the foot of the altitude from $A$ to $BX.$ Since $\triangle ABX$ is isosceles $AX=AB=86,EB=EX,$ and the answer is $EC+EB=EC+EX.$ $(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013$ by the Pythagorean Theorem. Only $EC+EX=\boxed{(D)~61}$ is a factor of $2013$ such that $97>EC+EX>EC-EX=\frac{2013}{EC+EX}.$

~dolphin7

Solution 5

We can apply Stewarts to $\triangle{ABC}$ with cevian $AX.$ If we let $BX = b$ and $CX = c,$ We can write the equation, $bc(b+c) + 86^2(b+c) = 97^2(b) + 86^2(c).$ Now we do a clever set of changes to factor, $bc(b+c) = 97^2(b) + 86^2(c) - 86^2(b+c) = 97^2(b)+86^2(c-(b+c)).$ $bc(b+c)=97^2(b)+86^2(-b) = b(97^2-86^2).$ $c(b+c) = 97^2 - 86^2 = 11\cdot 183 = 3 \cdot 11 \cdot 61.$ Now we can just use the factors of $3 \cdot 11 \cdot 61$ to find what $c$ is and an allowed value of $b+c.$ Using triangle inequality we have $11 < BC = b+c < 183.$ Therefore applying these restrictions the only value of $b+c$ that works is $\boxed{61}$ with corresponding values of $c=33$ and $b = 28.$

~mathkiddus

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=f1nxu8MWWKc

Video Solution by OmegaLearn

https://youtu.be/NsQbhYfGh1Q?t=2692

~ pi_is_3.14

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

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