Difference between revisions of "2013 AMC 12B Problems/Problem 24"

Latest revision as of 21:49, 20 October 2024

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 $BX^2$ ?

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

Solution 1

Let $BN=x$ and $NA=y$ . From the conditions, let's deduct some convenient conditions that seem 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

$2 = BC^2 = x^2 + (2x+y)^2 - x(2x+y) = 3x^2 + 3xy + y^2 = \left(5+3\sqrt{2}\right)x^2$

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

Solution 2 (Analytic)

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 $n = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 - \sqrt{2}.$ Here, we discard the positive root since $A$ must lie to the left of the y-axis. Then, the coordinates of $C$ 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 $\left(\frac{2}{\sqrt{10 + 6\sqrt{2}}}\right)^2 = \frac{2}{5+3\sqrt{2}} = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}.}$

Solution 3

By some angle-chasing, we find that $\triangle ANC \sim \triangle BXC$ . From here, construct a point $D$ on $AC$ such that $\triangle DXC \sim \triangle ANC$ . Now, let $BC = a$ , which means that $DM = a - 1$ and $AD = 2 - a$ , and let $BN = BX = XN = XD = DN = b$ . Note that we want to compute $b^2$ . Because $\triangle AND \sim \triangle DXM$ , we have:

$\frac{AN}{2-a} = \frac{b}{a-1} \implies AN = \frac{b(2-a)}{(a-1)}$

However, we have more similar triangles. In fact, going back to our original pair of similar triangles - $\triangle ANC$ and $\triangle BXC$ - gives us more similarity ratios:

$\frac{AN}{AC} = \frac{BX}{BC} \implies \frac{\frac{b(2-a)}{(a-1)}}{2} = \frac{b}{a} \implies a = \sqrt{2}$

Since we constructed point $D$ such that $DX$ is parallel to $AB$ , we now examine trapezoid $ABXD$ . From the variables that we already set up, we know that $AB = b + b\sqrt{2}, BX = XD = b$ , and $DA = 2 - \sqrt{2}$ . Let $X'$ denote the foot of the perpendicular from $X$ to base $AB$ and define $D'$ similarly.

Because $\triangle BXX'$ is a $30, 60, 90$ triangle, $XX' = \frac{b\sqrt{3}}{2}$ and $BX' = \frac{b}{2}$ . Thus, $D'A = b\sqrt{2} - \frac{b}{2}$ and $DD' = XX' = \frac{b\sqrt{3}}{2}$ . By the Pythagorean Theorem on $\triangle ADD'$ ,

$\left (b\sqrt{2} - \frac{b}{2} \right)^2 + \left(\frac{b\sqrt{3}}{2} \right)^2 = \left(2 - \sqrt{2} \right)^2 \implies b^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}$ .

Solution 4

Since $\triangle BXN$ is equilateral, let's assume the sides of them are all $a$ , and denote the length of $XM$ is $m$ . Since $CN$ bisects $\angle BCA$ , applying the angle bisector theorem and we can get $BC=\frac{a}{m}$ ; $AN=2m$ . Now applying LOC, we can get $(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1$ . We get $a^2+3m^2+3am=1$ . Now applying the Stewart theorem in $\triangle BAC$ , we can find that ${\frac{a^2}{m^2}+(a+2m)^2=2(1+(a+m)^2)}$ , after simplifying, we get ${\frac{a^2}{m^2}-a^2+2m^2=2}$ . After observation, the main key for this problem is $a^2$ , so we can solve $a$ in term of $m$ . Let's see the equation ${\frac{a^2}{m^2}-a^2+2m^2=2}$ , we can find that $a=\sqrt{2}m$ so $a^2=2m^2$ . Now back solving the first equation we can get that $a=\frac{-3m+\sqrt{4-3m^2}}{m}$ cuz the negative one can't work. After solving, we can get that $m^2=\frac{1}{5+3\sqrt{2}}$ so $a^2=2m^2$ and we get $a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}$ ~ bluesoul

Solution 5 (Similar Triangles)

Denote the length of $AN$ as $a$ and the length of $NB$ as $b.$

Let $M'$ be the midpoint of $\overline{BC} .$ Denote the intersection of $\overline{MM'}$ and $\overline{CN}$ as $X' .$ Note that $MX' = \frac12 AN = \frac{a}{2}$ and $CN = 2\cdot NX' .$ As $\overline{MM'} || \overline{AB},$ we have that $\triangle MXX' \sim \triangle BXN$ or $\triangle MXX'$ is equilateral and $XX' =MX' = \frac{a}{2}.$ Thus, $CN = 2b+a$ and $CX=a+b.$

Observe that

$\triangle BXC\sim \triangle ANC \implies \frac{BX}{XC} =\frac{AN}{NC} \implies \frac{b}{b+a} = \frac{a}{2b+a} \implies a = \sqrt 2 b.$

By the angle bisector theorem, we have that $BC=\frac{2b}{a} = \sqrt 2.$

We apply the Law of Cosines on $\triangle BXC$ as follows: $BC^2=BX^2+XC^2-2BX\cdot XC\cdot \cos120^\circ$ $\begin{align*} 2&=b^2+(\sqrt 2+1)^2b^2 +(\sqrt2 + 1)b^2 \\ &=b^2(5+3\sqrt 2) \end{align*}$ or $\boxed{b^2=\textbf{(A) } \frac{10-6\sqrt{2}}{7}}$

~ASAB

Solution 6

$\angle BXC = \angle ANC$ , $\angle BCX = \angle ACN$ , $\triangle BCX \sim \triangle ACN$ , $\frac{CX}{CN} = \frac{BC}{AC}$

$\angle MBC = \angle BAC$ , $\angle BCM = \angle ACB$ , $\triangle BCM \sim \triangle ACB$ , $\frac{BC}{AC} = \frac{CM}{BC}$ , $BC = \sqrt{2}$

Let $BX = x$ , $CN = CX + x$ , $\frac{CX}{CX + x} = \frac{\sqrt{2}}{2}$ , $2CX = CX \sqrt{2} + x \sqrt{2}$ , $CX = x(\sqrt{2} + 1)$

$BC^2 = BX^2 + CX^2 - 2 \cdot BX \cdot CX \cdot \cos 120^\circ$

$2 = x^2(\sqrt{2} + 1)^2 + x^2 + x^2(\sqrt{2} + 1) = 5x^2 + 3x^2 \sqrt{2}$

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

~isabelchen

Video Solution by MOP 2024

https://youtu.be/y0s6OTQ7KfI

~r00tsOfUnity

@@ Line 4: / Line 4: @@
 <math>\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}</math>
-==Solution==
+==Solution 1==
 Let <math>BN=x</math> and <math>NA=y</math>. From the conditions, let's deduct some convenient conditions that seem sufficient to solve the problem.
@@ Line 66: / Line 66: @@
 <cmath>\left (b\sqrt{2} - \frac{b}{2} \right)^2 + \left(\frac{b\sqrt{3}}{2} \right)^2 = \left(2 - \sqrt{2} \right)^2 \implies b^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</cmath>.
+==Solution 4 ==
+Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\frac{a^2}{m^2}+(a+2m)^2=2(1+(a+m)^2)}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2</math> and we get <math>a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</math>~ bluesoul
+==Solution 5 (Similar Triangles)==
+Denote the length of <math>AN</math> as <math>a</math> and the length of <math>NB</math> as <math>b.</math>
+Let <math>M'</math> be the midpoint of <math>\overline{BC} .</math> Denote the intersection of <math>\overline{MM'}</math> and <math>\overline{CN}</math> as <math>X' .</math> Note that <math>MX' = \frac12 AN = \frac{a}{2}</math> and <math>CN = 2\cdot NX' .</math> As <math>\overline{MM'} || \overline{AB},</math> we have that <math>\triangle MXX' \sim \triangle BXN</math> or <math>\triangle MXX'</math> is equilateral and <math>XX' =MX' = \frac{a}{2}.</math> Thus, <math>CN = 2b+a</math> and <math>CX=a+b.</math>
+Observe that
+<cmath>\triangle BXC\sim \triangle ANC \implies \frac{BX}{XC} =\frac{AN}{NC} \implies \frac{b}{b+a} = \frac{a}{2b+a} \implies a = \sqrt 2 b.</cmath>
+By the angle bisector theorem, we have that <math>BC=\frac{2b}{a} = \sqrt 2.</math>
+We apply the Law of Cosines on <math>\triangle BXC</math> as follows:
+<cmath>
+BC^2=BX^2+XC^2-2BX\cdot XC\cdot \cos120^\circ
+</cmath>
+<cmath>\begin{align*}
+&=b^2+(\sqrt 2+1)^2b^2 +(\sqrt2 + 1)b^2 \\
+&=b^2(5+3\sqrt 2)
+\end{align*}</cmath>
+or
+<cmath>\boxed{b^2=\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</cmath>
+~ASAB
+==Solution 6==
+[[File:2013AMC12BProblem24Solution6.png|center|500px]]
+<math>\angle BXC = \angle ANC</math>, <math>\angle BCX = \angle ACN</math>, <math>\triangle BCX \sim \triangle ACN</math>, <math>\frac{CX}{CN} = \frac{BC}{AC}</math>
+<math>\angle MBC = \angle BAC</math>, <math>\angle BCM = \angle ACB</math>, <math>\triangle BCM \sim \triangle ACB</math>, <math>\frac{BC}{AC} = \frac{CM}{BC}</math>, <math>BC = \sqrt{2}</math>
+Let <math>BX = x</math>, <math>CN = CX + x</math>, <math>\frac{CX}{CX + x} = \frac{\sqrt{2}}{2}</math>, <math>2CX = CX \sqrt{2} + x \sqrt{2}</math>, <math>CX = x(\sqrt{2} + 1)</math>
+<math>BC^2 = BX^2 + CX^2 - 2 \cdot BX \cdot CX \cdot \cos 120^\circ</math>
+<math>2 = x^2(\sqrt{2} + 1)^2 + x^2 + x^2(\sqrt{2} + 1) = 5x^2 + 3x^2 \sqrt{2}</math>
+<math>x^2 = \frac{2}{5 + 3 \sqrt{2} } = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+==Video Solution by MOP 2024==
+https://youtu.be/y0s6OTQ7KfI
+~r00tsOfUnity
 == 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

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