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 12A Problems/Problem 19"

(→‎Solution 2)
Line 11: Line 11:
 
Therefore, the answer is '''D) 61.'''
 
Therefore, the answer is '''D) 61.'''
  
−
===Solution 2===
+
 
 +
==Solution 2==
 +
Let <math>h</math> be the perpendicular from <math>B</math> to <math>AC</math>, <math>AX=x</math>, <math>XC=y</math>, then by Pythagorean Theorem,
 +
 
 +
<math>h^2 + (x/2)^2 = 86^2</math>
 +
 
 +
<math>h^2 + (x/2 + y)^2 = 97^2</math>
 +
 
 +
Subtracting the two equations, we get <math>(x+y)y = (97-86)(97+86)</math>,
 +
 
 +
then the rest is similar to the above solution by power of points.
 +
 
 +
==Solution 3==
 
Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86.
 
Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86.
 
Then by Stewart's Theorem,
 
Then by Stewart's Theorem,

Revision as of 12:22, 12 February 2013

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 D) 61.


Solution 2

Let $h$ be the perpendicular from $B$ to $AC$, $AX=x$, $XC=y$, then by Pythagorean Theorem,

$h^2 + (x/2)^2 = 86^2$

$h^2 + (x/2 + y)^2 = 97^2$

Subtracting the two equations, we get $(x+y)y = (97-86)(97+86)$,

then the rest is similar to the above solution by power of points.

Solution 3

Let $x$ represent $BX$, and let $y$ represent $CX$. 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 61. $\boxed{D}$


-Solution 2 by fowlmaster

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_19&oldid=51044"