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

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(→‎Solution)
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==Solution==
 
==Solution==
 
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 + 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{x(1 + 2\sqrt{3})}{\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>DC = \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>.
 
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 + 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{x(1 + 2\sqrt{3})}{\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>DC = \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>.
 +
 +
<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>
  
 
== See also ==
 
== See also ==

Revision as of 13:33, 31 December 2016

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

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 $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}$, 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]

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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