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

m (→‎Solution 2 Stewart's Theorem)
m (Formatted solution headers)
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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
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===Solution 1 (Power of a Point)===
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==Solution 1 (Power of a Point)==
  
 
Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meets the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>AC</math>.  Then <math>AZ = AY = 86</math>.  Using the Power of a Point (Secant-Secant Power Theorem), we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>.  We know that <math>p+q>p</math>, so <math>p</math> is either 3,11, or 33.  We also know that <math>p>11</math> by the triangle inequality on <math>\triangle ACX</math>.  Thus, <math>p</math> is <math>33</math> so we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>.
 
Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meets the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>AC</math>.  Then <math>AZ = AY = 86</math>.  Using the Power of a Point (Secant-Secant Power Theorem), we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>.  We know that <math>p+q>p</math>, so <math>p</math> is either 3,11, or 33.  We also know that <math>p>11</math> by the triangle inequality on <math>\triangle ACX</math>.  Thus, <math>p</math> is <math>33</math> so we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>.
  
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===Solution 2 ([[Stewart's Theorem]])===
+
==Solution 2 ([[Stewart's Theorem]])==
  
 
Let <math>x</math> represent <math>CX</math>, and let <math>y</math> represent <math>BX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB = AX = 86</math>.
 
Let <math>x</math> represent <math>CX</math>, and let <math>y</math> represent <math>BX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB = AX = 86</math>.
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The prime factors of <math>2013</math> are <math>3</math>, <math>11</math>, and <math>61</math>. Obviously, <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal <math>33</math>, and <math>x+y</math> must equal <math> \boxed{\textbf{(D) }61}</math>.
 
The prime factors of <math>2013</math> are <math>3</math>, <math>11</math>, and <math>61</math>. Obviously, <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal <math>33</math>, and <math>x+y</math> must equal <math> \boxed{\textbf{(D) }61}</math>.
  
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===Solution 3===
+
==Solution 3==
  
 
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>.
 
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>.
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Therefore, the answer is  <math> \boxed{\textbf{(D) }61}</math>.
 
Therefore, the answer is  <math> \boxed{\textbf{(D) }61}</math>.
  
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===Solution 4===
+
==Solution 4==
 
<asy>
 
<asy>
 
unitsize(2);
 
unitsize(2);

Revision as of 20:58, 28 December 2019

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

Solution 2 (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 $\overline{BC} = \boxed{(D)~61}$.

$\phantom{solution and diagram by bobjoe123}$

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

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