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"

m (→‎Solution 1 (Diophantine PoP))
(→‎Solution 4)
 
(3 intermediate revisions by the same user not shown)
Line 63: Line 63:
  
 
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 4===
 +
Motivation and general line of reasoning: we use a law of cosines condition on triangle <math>ABX</math> and triangle <math>ABC</math> to derive some equivalent formations of the same quantity <math>\cos B</math>, which looks promising since it involves the desired length <math>BC</math>, as well as <math>BX</math> and <math>CX</math>.An intermediate step would be to use the integer condition and pay attention to the divisors of stuff.
 +
 +
Let <math>BX</math>=<math>x_1</math> and <math>CX</math>=<math>x_2</math>.
 +
 +
First we have
 +
<math>\cos(B)=\frac{86^2+x_{1}^2-86^2}{172x_{1}}</math>
 +
by applying the law of cosines to triangle <math>ABX</math>. Another equivalent formation of <math>\cos B</math> is
 +
<math>\frac{86^2+(x_{1}+x_2)^2-97^2}{172(x_{1}+x_2)}</math>.
 +
Now that we have the necessary ingredients, we can make a system of equations and deduce that
 +
<math>x_1=\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}</math>.
 +
 +
Don't lose focus by now-we try to find <math>x_1+x_2</math>, an integer value as given in the problem. To do this, we want the quantity
 +
<math>\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}</math>
 +
to 1) be an integer and 2) smaller than <math>x_1+x_2</math>. For the sake of conciseness in notation we let <math>M=x_1+x_2</math>, then <math>M+\frac{86^2-97^2}{M}</math> is an integer. Now recalling the fact that <math>(a+b)(a-b)=a^2-b^2</math>, we get that <math>\frac{183(-11)}{M}</math> must be an integer.
 +
 +
Now the prime factor decomposition of <math>183 \cdot -11</math> is  <math>(61)(3)(-11)</math>. Trying out all the possible integer values that divide this quantity, we get that the only viable option for <math>M</math> is 61 (verify that yourself!) Therefore the answer is <math> \boxed{\textbf{(D) }61}</math>. (Solution by CreamyCream123)
  
 
==Video Solution by Richard Rusczyk==
 
==Video Solution by Richard Rusczyk==

Latest revision as of 09:16, 3 December 2024

Problem

In $\bigtriangleup ABC$, $AB = 86$, 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

Solution 1 (Diophantine PoP)

[asy] //Made by samrocksnature size(8cm); pair A,B,C,D,E,X; A=(0,0); B=(-53.4,-67.4); C=(0,-97); D=(0,-86); E=(0,86); X=(-29,-81); draw(circle(A,86)); draw(E--C--B--A--X); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,NE); label("$E$",E,NE); label("$X$",X,dir(250)); dot(A^^B^^C^^D^^E^^X); [/asy]

Let circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation

\[CX \cdot CB = CD \cdot CE\] \[CX(CX+XB) = (97-86)(97+86)\] \[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]

Since lengths cannot be negative, we must have $CX+XB \ge CX.$ This generates the four solution pairs for $(CX,CX+XB)$: \[(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).\]

However, by the Triangle Inequality on $\triangle ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB= \boxed{\textbf{(D) }61}.$

(Solution by unknown, latex/asy modified majorly by samrocksnature)

Solution 2

Let $BX = q$, $CX = p$, and $AC$ meet the circle at $Y$ and $Z$, with $Y$ on $AC$. Then $AZ = AY = 86$. Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$. We know that $p+q>p$, and that $p>13$ by the triangle inequality on $\triangle ACX$. Thus, we get that $BC = p+q = \boxed{\textbf{(D) }61}$

Solution 3

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 4

Motivation and general line of reasoning: we use a law of cosines condition on triangle $ABX$ and triangle $ABC$ to derive some equivalent formations of the same quantity $\cos B$, which looks promising since it involves the desired length $BC$, as well as $BX$ and $CX$.An intermediate step would be to use the integer condition and pay attention to the divisors of stuff.

Let $BX$=$x_1$ and $CX$=$x_2$.

First we have $\cos(B)=\frac{86^2+x_{1}^2-86^2}{172x_{1}}$ by applying the law of cosines to triangle $ABX$. Another equivalent formation of $\cos B$ is $\frac{86^2+(x_{1}+x_2)^2-97^2}{172(x_{1}+x_2)}$. Now that we have the necessary ingredients, we can make a system of equations and deduce that $x_1=\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}$.

Don't lose focus by now-we try to find $x_1+x_2$, an integer value as given in the problem. To do this, we want the quantity $\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}$ to 1) be an integer and 2) smaller than $x_1+x_2$. For the sake of conciseness in notation we let $M=x_1+x_2$, then $M+\frac{86^2-97^2}{M}$ is an integer. Now recalling the fact that $(a+b)(a-b)=a^2-b^2$, we get that $\frac{183(-11)}{M}$ must be an integer.

Now the prime factor decomposition of $183 \cdot -11$ is $(61)(3)(-11)$. Trying out all the possible integer values that divide this quantity, we get that the only viable option for $M$ is 61 (verify that yourself!) Therefore the answer is $\boxed{\textbf{(D) }61}$. (Solution by CreamyCream123)

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc12a/357

~dolphin7

Video Solution

https://youtu.be/zxW3uvCQFls

~sugar_rush

See also

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_12A_Problems/Problem_19&oldid=236097"