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Difference between revisions of "2013 AMC 12B Problems/Problem 24"
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Because <math>\triangle BXX'</math> is a <math>30, 60, 90</math> triangle, <math>XX' = \frac{b\sqrt{3}}{2}</math> and <math>BX' = \frac{b}{2}</math>. Thus, <math>D'A = b\sqrt{2} - \frac{b}{2}</math> and <math>DD' = XX' = \frac{b\sqrt{3}}{2}</math>. By the Pythagorean Theorem on <math>\triangle ADD'</math>,
 
Because <math>\triangle BXX'</math> is a <math>30, 60, 90</math> triangle, <math>XX' = \frac{b\sqrt{3}}{2}</math> and <math>BX' = \frac{b}{2}</math>. Thus, <math>D'A = b\sqrt{2} - \frac{b}{2}</math> and <math>DD' = XX' = \frac{b\sqrt{3}}{2}</math>. By the Pythagorean Theorem on <math>\triangle ADD'</math>,
  
−
<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{\frac{10-6\sqrt{2}}{7}}</cmath>.
+
<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>.
  
 
== See also ==
 
== See also ==

Revision as of 23:10, 22 April 2017

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 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 $\triangle ADD'$,

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

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 12 Problems and Solutions

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