2013 AMC 12B Problems/Problem 24

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

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)

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 \[n = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 - \sqrt{2}.\] 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

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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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