Difference between revisions of "2013 AMC 12B Problems/Problem 24"
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<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> | ||
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+ | ==Solution 2 (Analytic)== | ||
+ | <center>[[File:2013 AMC 12B 24.jpg]]</center> | ||
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+ | 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. | ||
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+ | 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>. | ||
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+ | 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> | ||
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+ | 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> | ||
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+ | 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 07:25, 11 January 2015
Contents
[hide]Problem
Let be a triangle where is the midpoint of , and is the angle bisector of with on . Let be the intersection of the median and the bisector . In addition is equilateral with . What is ?
Solution
Let and . From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.
is the midpoint of side .
This implies that . Given that angle is degrees and angle is degrees, we can use the area formula to get
So, .....(1)
is angle bisector.
In the triangle , one has , therefore .....(2)
Furthermore, triangle is similar to triangle , so , therefore ....(3)
By (2) and (3) and the subtraction law of ratios, we get
Therefore , or . So .
Finally, using the law of cosine for triangle , we get
Solution 2 (Analytic)
Let us dilate triangle so that the sides of equilateral triangle are all equal to The purpose of this is to ease the calculations we make in the problem. Given this, we aim to find the length of segment so that we can un-dilate triangle by dividing each of its sides by . Doing so will make it so that , as desired, and doing so will allow us to get the length of , whose square is our final answer.
Let the foot of the altitude from to On the coordinate plane, position at , and make lie on the x-axis. Since points , , and , are collinear, must also lie on the x-axis. Additionally, since , , meaning that we can position point at . Now, notice that line has the equation and that line has the equation because angles and are both . We can then position at point and at point . Quickly note that, because is an angle bisector, must pass through the point .
We proceed to construct a system of equations. First observe that the midpoint of must lie on , with the equation . The coordinates of are , and we can plug in these coordinates into the equation of line , yielding that For our second equation, notice that line has equation . Midpoint must also lie on this line, and we can substitute coordinates again to get
Setting both equations equal to each other and multiplying both sides by , we have that , which in turn simplifies into when dividing the entire equation by Using the quadratic formula, we have that Then, the coordinates of are , and the coordinates of are Seeing that segment has half the length of side , we have that the length of is
Now, we divide each side length of by , and from this, will equal
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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