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Difference between revisions of "2014 AIME II Problems/Problem 11"

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==Solution==
 
==Solution==
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Let <math>P</math> be the foot of the perpendicular from <math>A</math> to <math>\overline{CR}</math>, so <math>\overline{AP}\parallel\overline{EM}</math>. Since triangle <math>ARC</math> is isosceles, <math>P</math> is the midpoint of <math>\overline{CR}</math>, and <math>\overline{PM}\parallel\overline{CD}</math>. Thus, <math>APME</math> is a parallelogram and <math>AE = PM = (CD)/2</math>. (someone incorporate LATEX please) We can then use coordinates. Let O be the foot of altitude RO and set O as the origin. Now we notice special right triangles! In particular, DO = 1/2 and EO = RO = √3/2, so D(1/2, 0), E(-√3/2, 0), and R(0, √3/2). M = midpoint(D, R) = (1/4, √3/4) and slope(ME) = √3/4 / (1/4 + √3/2) = √3 / (1 + 2√3), so slope(RC) = -(1 + 2√3)/√3. Instead of finding the equation of the line, we use the definition of slope: for every CO = x to the left, we go (1 + 2√3)/√3 * x = √3/2 up. Thus, x = 3/2 / (1 + 2√3) = 3 / (4√3 + 2) = 3(4√3 - 2) / 44 = (6√3 - 3) / 22. DO = 1/2 - x = 1/2 - (6√3 - 3)/22 = (14 - 6√3) / 22, and <math>AE = \frac{7 - \sqrt{27}}{22}</math>, so the answer is <math>\boxed{056}</math>.
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Let <math>P</math> be the foot of the perpendicular from <math>A</math> to <math>\overline{CR}</math>, so <math>\overline{AP}\parallel\overline{EM}</math>. Since triangle <math>ARC</math> is isosceles, <math>P</math> is the midpoint of <math>\overline{CR}</math>, and <math>\overline{PM}\parallel\overline{CD}</math>. Thus, <math>APME</math> is a parallelogram and <math>AE = PM = \frac{CD}{2}</math>. We can then use coordinates. Let <math>O</math> be the foot of altitude <math>RO</math> and set <math>O</math> as the origin. Now we notice special right triangles! In particular, <math>DO = \frac{1}{2}</math> and <math>EO = RO = \frac{\sqrt{3}}{2}</math>, so <math>D(\frac{1}{2}, 0)</math>, <math>E(-\frac{\sqrt{3}}{2}, 0)</math>, and <math>R(0, \frac{\sqrt{3}}{2}).</math> <math>M =</math> midpoint<math>(D, R) = (\frac{1}{4}, \frac{\sqrt{3}}{4})</math> and the slope of <math>ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + \frac{2}{\sqrt{3}}}</math>, so the slope of <math>RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.</math> Instead of finding the equation of the line, we use the definition of slope: for every <math>CO = x</math> to the left, we go <math>\frac{1 + 2\sqrt{3}}{x\sqrt{3}} = \frac{\sqrt{3}}{2}</math> up. Thus, <math>x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.</math> <math>DO = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}</math>, and <math>AE = \frac{7 - \sqrt{27}}{22}</math>, so the answer is <math>\boxed{056}</math>.

Revision as of 21:05, 20 May 2014

Problem 11

In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $\abs{RD}=1$ (Error compiling LaTeX. Unknown error_msg). Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$.

Solution

Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \frac{1}{2}$ and $EO = RO = \frac{\sqrt{3}}{2}$, so $D(\frac{1}{2}, 0)$, $E(-\frac{\sqrt{3}}{2}, 0)$, and $R(0, \frac{\sqrt{3}}{2}).$ $M =$ midpoint$(D, R) = (\frac{1}{4}, \frac{\sqrt{3}}{4})$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + \frac{2}{\sqrt{3}}}$, so the slope of $RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\frac{1 + 2\sqrt{3}}{x\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DO = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$, and $AE = \frac{7 - \sqrt{27}}{22}$, so the answer is $\boxed{056}$.

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