Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2 (Stewart's Theorem))
(→‎Solutions)
 
(31 intermediate revisions by 12 users not shown)
Line 4: Line 4:
  
 
In <math>\triangle ABC</math>, <math>AB = 86</math>, and <math>AC=97</math>.  A circle with center <math>A</math> and radius <math>AB</math> intersects <math>\overline{BC}</math> at points <math>B</math> and <math>X</math>.  Moreover <math>\overline{BX}</math> and <math>\overline{CX}</math> have integer lengths.  What is <math>BC</math>?
 
In <math>\triangle ABC</math>, <math>AB = 86</math>, and <math>AC=97</math>.  A circle with center <math>A</math> and radius <math>AB</math> intersects <math>\overline{BC}</math> at points <math>B</math> and <math>X</math>.  Moreover <math>\overline{BX}</math> and <math>\overline{CX}</math> have integer lengths.  What is <math>BC</math>?
−
 
−
 
  
 
<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 </math>
 
<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 </math>
−
[[Category: Introductory Geometry Problems]]
 
  
−
===Solution 1 (Power of a Point)===
+
==Solution 1 (Number Theoretic 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>.  <math>p</math> is 33.  Thus, 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 <math>3</math>, <math>11</math>, or <math>33</math>.  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>.
  
−
===Solution 2 (Stewart's Theorem)===
+
==Solution 2 ([[Stewart's Theorem]])==
−
[[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>.
Line 23: Line 19:
  
 
<math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math>
 
<math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math>
 +
 +
<math>x^2 y + xy^2 + 86^2 y = 97^2 y</math>
  
 
<math>x^2 + xy + 86^2 = 97^2</math>
 
<math>x^2 + xy + 86^2 = 97^2</math>
Line 34: Line 32:
 
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>.
  
−
===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>.
 
Use power of a point on point C to the circle centered at A.
 
Use power of a point on point C to the circle centered at A.
  
−
So <math>CX*CB=CD*CE=></math>
+
So <math>CX \cdot CB=CD \cdot CE \Rightarrow x(x+y)=(97-86)(97+86) \Rightarrow x(x+y)=3*11*61</math>.
−
<math>x(x+y)=(97-86)(97+86)=></math>
 
−
<math>x(x+y)=3*11*61</math>.
 
  
 
Obviously <math>x+y>x</math> so we have three solution pairs for <math>(x,x+y)=(1,2013),(3,671),(11,183),(33,61)</math>.
 
Obviously <math>x+y>x</math> so we have three solution pairs for <math>(x,x+y)=(1,2013),(3,671),(11,183),(33,61)</math>.
Line 47: Line 43:
  
 
Therefore, the answer is  <math> \boxed{\textbf{(D) }61}</math>.
 
Therefore, the answer is  <math> \boxed{\textbf{(D) }61}</math>.
 +
 +
==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 <math>ABD</math> and get two equations by the [[Pythagorean Theorem]].
 +
First, <math>h^2 + k^2 = 86^2</math>. Second, <math>h^2 + (k + m)^2 = 97^2</math>.
 +
Subtracting these two equations, we get <math>2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013</math>.
 +
We then add <math>k^2</math> to both sides to get <math>k^2 + 2km + m^2 = 2013 + k^2</math>.
 +
We then complete the square to get <math>(k + m)^2 = 2013 + k^2</math>. Because <math>k</math> and <math>m</math> are both integers, we get that <math>2013 + k^2</math> is a square number. Simple guess and check reveals that <math>k = 14</math>.
 +
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{(D)~61}</math>.
 +
 +
<math>\phantom{solution and diagram by bobjoe123}</math>
 +
 +
==Solution 5==
 +
 +
Let <math>E</math> be the foot of the altitude from <math>A</math> to <math>BX.</math> Since <math>\triangle ABX</math> is isosceles <math>AX=AB=86,EB=EX,</math> and the answer is <math>EC+EB=EC+EX.</math> <math>(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013</math> by the Pythagorean Theorem. Only <math>EC+EX=\boxed{(D)~61}</math> is a factor of <math>2013</math> such that <math>97>EC+EX>EC-EX=\frac{2013}{EC+EX}.</math>
 +
 +
~dolphin7
 +
 +
==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
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}}
 
{{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Number theory]]
 
{{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:02, 21 August 2023

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 (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,

$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$

$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$

$x^2 y + xy^2 + 86^2 y = 97^2 y$

$x^2 + xy + 86^2 = 97^2$

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

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

$\phantom{solution and diagram by bobjoe123}$

Solution 5

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_23&oldid=197297"