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

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==Solution 4 (No intuition)==
 
==Solution 4 (No intuition)==
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Define points <math>EM \cap RC=Y,</math> and the foot of the altitude from <math>R</math> to <math>ED</math> as <math>X.</math> Suppose the median <math>EM</math> is equal to <math>m,</math> and the angle <math>\angle{MED}</math> as <math>\theta.</math> Using the length-of-median formulae, we get <math>m^2=1+\frac{\sqrt{3}}{4}.</math> By Law of Sines in <math>EMD,</math> <math>\frac{m}{\sin{60^{\circ}}} = \frac{\frac{1}{2}}{\sin\theta}.</math> This gives <math>\sin\theta = \frac{\sqrt{3}}{4m},</math> and solving for <math>\cos\theta</math> gives <math>\cos\theta = \sqrt{1-\frac{3}{16m^2}} = \frac{\sqrt{16m^2-3}}{4m}.</math> Using the established value of <math>m^2,</math> we have <math>\cos{\theta} = \frac{\sqrt{13+4\sqrt{3}}}{4m} = \frac{1+2\sqrt{3}}{4m}.</math> Thus, we have <math>\tan\theta = \frac{\sqrt{3}}{1+2\sqrt{3}}.</math> Noting that <math>\angle{XRC}=\theta</math> as well, this gives <math>CX = RX \tan \theta = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{1+2\sqrt{3}} = \frac{3}{2+4\sqrt{3}}.</math> Suppose <math>AE=n.</math> Now, using Pythagorean in triangle <math>AXR</math> gives  
+
Define the foot of the altitude from <math>R</math> to <math>ED</math> as <math>X.</math> Suppose the median <math>EM</math> is equal to <math>m,</math> and the angle <math>\angle{MED}</math> as <math>\theta.</math> Using the length-of-median formulae, we get <math>m^2=1+\frac{\sqrt{3}}{4}.</math> By Law of Sines in <math>EMD,</math> <math>\frac{m}{\sin{60^{\circ}}} = \frac{\frac{1}{2}}{\sin\theta}.</math> This gives <math>\sin\theta = \frac{\sqrt{3}}{4m},</math> and solving for <math>\cos\theta</math> gives <math>\cos\theta = \sqrt{1-\frac{3}{16m^2}} = \frac{\sqrt{16m^2-3}}{4m}.</math> Using the established value of <math>m^2,</math> we have <math>\cos{\theta} = \frac{\sqrt{13+4\sqrt{3}}}{4m} = \frac{1+2\sqrt{3}}{4m}.</math> Thus, we have <math>\tan\theta = \frac{\sqrt{3}}{1+2\sqrt{3}}.</math> Noting that <math>\angle{XRC}=\theta</math> as well, this gives <math>CX = RX \tan \theta = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{1+2\sqrt{3}} = \frac{3}{2+4\sqrt{3}}.</math> Suppose <math>AE=n.</math> Now, using Pythagorean in triangle <math>AXR</math> gives  
 
<cmath>\left(n + \frac{\sqrt{3}}{2}\right)^2 + \frac{3}{4} = \left(n + \frac{\sqrt{3}}{2} + \frac{3}{2+4\sqrt{3}}\right)^2</cmath>
 
<cmath>\left(n + \frac{\sqrt{3}}{2}\right)^2 + \frac{3}{4} = \left(n + \frac{\sqrt{3}}{2} + \frac{3}{2+4\sqrt{3}}\right)^2</cmath>
 
Solving, we have
 
Solving, we have

Revision as of 23:29, 17 February 2025

Problem 11

In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. 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 1

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 $D\left(\frac{1}{2}, 0\right)$, $E\left(-\frac{\sqrt{3}}{2}, 0\right)$, and $R\left(0, \frac{\sqrt{3}}{2}\right).$ $M =$ midpoint$(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}$, 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 $\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$, and $AE = \frac{7 - \sqrt{27}}{22}$, so the answer is $\boxed{056}$.

[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, W); label("$E$", e, SW); label("$C$", c, S); label("$O$", o, S); label("$D$", d, SE); label("$M$", m, NE); label("$R$", r, N); p = foot(a, r, c); label("$P$", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]

Solution 2

Let $MP = x.$ Meanwhile, since $\triangle R PM$ is similar to $\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\overline A\overline E$ and $\overline P\overline M$.

Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\overline R\overline O$, we get $ED = \frac{\sqrt{3}+1}{2}$. $CA=RA$, which equals $ED - x$

Using Law of Sine in $\triangle RED$, we find $RE$ = $\frac{\sqrt{6}}{2}$.

We got the three sides of $\triangle AER$. Now using the Law of Cosines on $\angle AER$. There we can equate $x$ and solve for it. We got $AE=x=\frac{\sqrt{3}-1}{4\sqrt{3}+2}$. Then rationalize the denominator, we get $AE = \frac{7 - \sqrt{27}}{22}$.

Solution 3

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 by midpoint theorem $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and therefore $AE = PM = \tfrac 12 CD$. [asy] unitsize(8cm); pair a, d, r, e, m, cm, c,p; d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m); draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed); pair[] PPAP = {a, d, r, e, m, c, p}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, E); label("$E$", e, S); label("$C$", c, S); label("$D$", d, SW); label("$M$", m, NW); label("$R$", r, N); label("$P$", p, NW); MA("60^\circ",black,c,d,m,0.07, black); [/asy] We can now use coordinates with $D(0,0)$ as origin and $DE$ along the $x$-axis.

Let $RD=4$ instead of $1$ (in the end we will scale down by $4$). Since $\angle D = 60^\circ$, we get $R(2,2\sqrt{3})$, and therefore $M(1, \sqrt{3})$.

We use sine-law in $\triangle RED$ to find the coordinates $E(2+2\sqrt{3}, 0)$:\[DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}.\] Since slope$(ME)= -\sqrt{3}/(1+2\sqrt{3})$, and $RC\perp ME$, it follows that slope$(RC)=(1+2\sqrt{3})/\sqrt{3}$. If $C(c,0)$ then we have\[\frac{2\sqrt{3}}{2-c}=\frac{1+2\sqrt{3}}{\sqrt{3}}\qquad \Longrightarrow\qquad c=\frac{4\sqrt{3}-4}{1+2\sqrt{3}}\] Now $\tfrac 12 CD = \tfrac 12c =(2\sqrt{3}-2)/(1+2\sqrt{3})= \tfrac 1{11}(14-6\sqrt{3})$.

Scaling down by $4$, we get $AE=\tfrac 1{22}(7-3\sqrt{3})$, so our answer is $7+27+22=056$.

Solution 4 (No intuition)

Define the foot of the altitude from $R$ to $ED$ as $X.$ Suppose the median $EM$ is equal to $m,$ and the angle $\angle{MED}$ as $\theta.$ Using the length-of-median formulae, we get $m^2=1+\frac{\sqrt{3}}{4}.$ By Law of Sines in $EMD,$ $\frac{m}{\sin{60^{\circ}}} = \frac{\frac{1}{2}}{\sin\theta}.$ This gives $\sin\theta = \frac{\sqrt{3}}{4m},$ and solving for $\cos\theta$ gives $\cos\theta = \sqrt{1-\frac{3}{16m^2}} = \frac{\sqrt{16m^2-3}}{4m}.$ Using the established value of $m^2,$ we have $\cos{\theta} = \frac{\sqrt{13+4\sqrt{3}}}{4m} = \frac{1+2\sqrt{3}}{4m}.$ Thus, we have $\tan\theta = \frac{\sqrt{3}}{1+2\sqrt{3}}.$ Noting that $\angle{XRC}=\theta$ as well, this gives $CX = RX \tan \theta = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{1+2\sqrt{3}} = \frac{3}{2+4\sqrt{3}}.$ Suppose $AE=n.$ Now, using Pythagorean in triangle $AXR$ gives \[\left(n + \frac{\sqrt{3}}{2}\right)^2 + \frac{3}{4} = \left(n + \frac{\sqrt{3}}{2} + \frac{3}{2+4\sqrt{3}}\right)^2\] Solving, we have \[\frac{3}{4} = \left(\frac{3}{2+4\sqrt{3}}\right) \left(2n + \sqrt{3} + \frac{3}{2+4\sqrt{3}}\right),\] \[2n + \sqrt{3} + \frac{3}{2+4\sqrt{3}} = \frac{1+2\sqrt{3}}{2},\] \[2n + \frac{3}{2+4\sqrt{3}} = \frac{1}{2}.\] Isolating $2n$ and rationalizing the denominator returns the final answer of $n=\frac{7-\sqrt{27}}{22},$ thus giving an answer of $\boxed{56}.$

Video Solution

https://youtu.be/muM8UcGKjHo?si=C6o7-C4DgB5i4yKv

~MathProblemSolvingSkills.com


See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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