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 12B Problems/Problem 24"

(→‎Solution)
(→‎Solution)
Line 36: Line 36:
  
 
<cmath>x^2 = \frac{2}{5+3\sqrt{2}} = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}}.</cmath>
 
<cmath>x^2 = \frac{2}{5+3\sqrt{2}} = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}}.</cmath>
 +
 +
==Solution 2 (Analytic)==
 +
<center>[[File:2013 AMC 12B 24.jpg]]</center>
 +
 +
 +
Let us dilate triangle <math>ABC</math> so that the sides of equilateral triangle <math>BXN</math> are all equal to <math>2.</math> The purpose of this is to ease the calculations we make in the problem. Given this, we aim to find the length of segment <math>AM</math> so that we can un-dilate triangle <math>ABC</math> by dividing each of its sides by <math>AM</math>. Doing so will make it so that <math>AM = 1</math>, as desired, and doing so will allow us to get the length of <math>BN</math>, whose square is our final answer.
 +
 +
Let <math>O</math> the foot of the altitude from <math>B</math> to <math>NX.</math> On the coordinate plane, position <math>O</math> at <math>(0, 0)</math>, and make <math>NX</math> lie on the x-axis. Since points <math>N</math>, <math>X</math>, and <math>C</math>, are collinear, <math>C</math> must also lie on the x-axis. Additionally, since <math>NX = 2</math>, <math>OB = \sqrt{3}</math>, meaning that we can position point <math>B</math> at <math>(0, \sqrt{3})</math>. Now, notice that line <math>\overline{AB}</math> has the equation <math>y = \sqrt{3}x + \sqrt{3}</math> and that line <math>\overline{BM}</math> has the equation <math>y = -\sqrt{3}x + \sqrt{3}</math> because angles <math>BNX</math> and <math>BXN</math> are both <math>60^{\circ}</math>. We can then position <math>A</math> at point <math>(n, \sqrt{3}(n + 1))</math> and <math>C</math> at point <math>(p, 0)</math>. Quickly note that, because <math>CN</math> is an angle bisector, <math>AC</math> must pass through the point <math>(0, -\sqrt{3})</math>.
 +
 +
We proceed to construct a system of equations. First observe that the midpoint <math>M</math> of <math>AC</math> must lie on <math>BM</math>, with the equation <math>y = -\sqrt{3}x + \sqrt{3}</math>. The coordinates of <math>M</math> are <math>\left(\frac{p + n}{2}, \frac{\sqrt{3}}{2}(n + 1)\right)</math>, and we can plug in these coordinates into the equation of line <math>BM</math>, yielding that <cmath>\frac{\sqrt{3}}{2}(n + 1) = -\sqrt{3}(\frac{p + n}{2}) + \sqrt{3} \implies n + 1 = -p - n + 2 \implies p = -2n + 1.</cmath> For our second equation, notice that line <math>AC</math> has equation <math>y = \frac{\sqrt{3}}{p}x - \sqrt{3}</math>. Midpoint <math>M</math> must also lie on this line, and we can substitute coordinates again to get <cmath>\frac{\sqrt{3}}{2}(n + 1) = \frac{\sqrt{3}}{p}(\frac{p + n}{2}) - \sqrt{3} \implies n + 1 = \frac{p + n}{p} - 2 \implies n + 1 = \frac{n}{p} - 1</cmath> <cmath>\implies p = \frac{n}{n + 2}.</cmath>
 +
 +
Setting both equations equal to each other and multiplying both sides by <math>(n + 2)</math>, we have that <math>-2n^2 - 4n + n + 2 = n \implies -2n^2 - 4n + 2 = 0</math>, which in turn simplifies into <math>0 = n^2 + 2n - 1</math> when dividing the entire equation by <math>-2.</math> Using the quadratic formula, we have that <cmath>n = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 - \sqrt{2}.</cmath> Then, the coordinates of <math>p</math> are <math>(3 + 2\sqrt{2}, 0)</math>, and the coordinates of <math>A</math> are <math>(-1 - \sqrt{2}, -\sqrt{6}).</math> Seeing that segment <math>AM</math> has half the length of side <math>AC</math>, we have that the length of <math>AM</math> is <cmath>\frac{\sqrt{(3 + 2\sqrt{2} - (-1 - \sqrt{2}))^2 + (\sqrt{6})^2}}{2} = \frac{\sqrt{16 + 24\sqrt{2} + 18 + 6}}{2} = \sqrt{10 + 6\sqrt{2}}.</cmath>
 +
 +
Now, we divide each side length of <math>\triangle ABC</math> by <math>AM</math>, and from this, <math>BN^2</math> will equal <math>\left(\frac{2}{\sqrt{10 + 6\sqrt{2}}}\right)^2 = \frac{2}{5+3\sqrt{2}} = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}.}</math>
  
 
== See also ==
 
== See also ==

Revision as of 08:25, 11 January 2015

Problem

Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$, and $\overline{CN}$ is the angle bisector of $\angle{ACB}$ with $N$ on $\overline{AB}$. Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$. In addition $\triangle BXN$ is equilateral with $AC=2$. What is $BN^2$?

$\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}$

Solution

Let $BN=x$ and $NA=y$. From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.


$M$ is the midpoint of side $AC$.

This implies that $[ABX]=[CBX]$. Given that angle $ABX$ is $60$ degrees and angle $BXC$ is $120$ degrees, we can use the area formula to get

\[\frac{1}{2}(x+y)x \frac{\sqrt{3}}{2} = \frac{1}{2} x \cdot CX \frac{\sqrt{3}}{2}\]

So, $x+y=CX$ .....(1)


$CN$ is angle bisector.

In the triangle $ABC$, one has $BC/AC=x/y$, therefore $BC=2x/y$.....(2)

Furthermore, triangle $BCN$ is similar to triangle $MCX$, so $BC/CM=CN/CX$, therefore $BC = (CX+x)/CX = (2x+y)/(x+y)$....(3)

By (2) and (3) and the subtraction law of ratios, we get

\[BC=2x/y = (2x+y)/(y+x) = y/x\]

Therefore $2x^2=y^2$, or $y=\sqrt{2}x$. So $BC = 2x/(\sqrt{2}x) = \sqrt{2}$.

Finally, using the law of cosine for triangle $BCN$, we get

\[2 = BC^2 = x^2 + (2x+y)^2 - x(2x+y) = 3x^2 + 3xy + y^2 = (5+3\sqrt{2})x^2\]

\[x^2 = \frac{2}{5+3\sqrt{2}} = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}}.\]

Solution 2 (Analytic)

2013 AMC 12B 24.jpg


Let us dilate triangle $ABC$ so that the sides of equilateral triangle $BXN$ are all equal to $2.$ The purpose of this is to ease the calculations we make in the problem. Given this, we aim to find the length of segment $AM$ so that we can un-dilate triangle $ABC$ by dividing each of its sides by $AM$. Doing so will make it so that $AM = 1$, as desired, and doing so will allow us to get the length of $BN$, whose square is our final answer.

Let $O$ the foot of the altitude from $B$ to $NX.$ On the coordinate plane, position $O$ at $(0, 0)$, and make $NX$ lie on the x-axis. Since points $N$, $X$, and $C$, are collinear, $C$ must also lie on the x-axis. Additionally, since $NX = 2$, $OB = \sqrt{3}$, meaning that we can position point $B$ at $(0, \sqrt{3})$. Now, notice that line $\overline{AB}$ has the equation $y = \sqrt{3}x + \sqrt{3}$ and that line $\overline{BM}$ has the equation $y = -\sqrt{3}x + \sqrt{3}$ because angles $BNX$ and $BXN$ are both $60^{\circ}$. We can then position $A$ at point $(n, \sqrt{3}(n + 1))$ and $C$ at point $(p, 0)$. Quickly note that, because $CN$ is an angle bisector, $AC$ must pass through the point $(0, -\sqrt{3})$.

We proceed to construct a system of equations. First observe that the midpoint $M$ of $AC$ must lie on $BM$, with the equation $y = -\sqrt{3}x + \sqrt{3}$. The coordinates of $M$ are $\left(\frac{p + n}{2}, \frac{\sqrt{3}}{2}(n + 1)\right)$, and we can plug in these coordinates into the equation of line $BM$, yielding that \[\frac{\sqrt{3}}{2}(n + 1) = -\sqrt{3}(\frac{p + n}{2}) + \sqrt{3} \implies n + 1 = -p - n + 2 \implies p = -2n + 1.\] For our second equation, notice that line $AC$ has equation $y = \frac{\sqrt{3}}{p}x - \sqrt{3}$. Midpoint $M$ must also lie on this line, and we can substitute coordinates again to get \[\frac{\sqrt{3}}{2}(n + 1) = \frac{\sqrt{3}}{p}(\frac{p + n}{2}) - \sqrt{3} \implies n + 1 = \frac{p + n}{p} - 2 \implies n + 1 = \frac{n}{p} - 1\] \[\implies p = \frac{n}{n + 2}.\]

Setting both equations equal to each other and multiplying both sides by $(n + 2)$, we have that $-2n^2 - 4n + n + 2 = n \implies -2n^2 - 4n + 2 = 0$, which in turn simplifies into $0 = n^2 + 2n - 1$ when dividing the entire equation by $-2.$ Using the quadratic formula, we have that \[n = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 - \sqrt{2}.\] Then, the coordinates of $p$ are $(3 + 2\sqrt{2}, 0)$, and the coordinates of $A$ are $(-1 - \sqrt{2}, -\sqrt{6}).$ Seeing that segment $AM$ has half the length of side $AC$, we have that the length of $AM$ is \[\frac{\sqrt{(3 + 2\sqrt{2} - (-1 - \sqrt{2}))^2 + (\sqrt{6})^2}}{2} = \frac{\sqrt{16 + 24\sqrt{2} + 18 + 6}}{2} = \sqrt{10 + 6\sqrt{2}}.\]

Now, we divide each side length of $\triangle ABC$ by $AM$, and from this, $BN^2$ will equal $\left(\frac{2}{\sqrt{10 + 6\sqrt{2}}}\right)^2 = \frac{2}{5+3\sqrt{2}} = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}.}$

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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_12B_Problems/Problem_24&oldid=66827"