Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 2)
 
(15 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==Problem 11==
 
==Problem 11==
−
In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math>|RD|=1</math>. Let <math>M</math> be the midpoint of segment <math>\overline{RD}</math>. Point <math>C</math> lies on side <math>\overline{ED}</math> such that <math>\overline{RC}\perp\overline{EM}</math>. Extend segment <math>\overline{DE}</math> through <math>E</math> to point <math>A</math> such that <math>CA=AR</math>. Then <math>AE=\frac{a-\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer. Find <math>a+b+c</math>.
+
In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math>RD=1</math>. Let <math>M</math> be the midpoint of segment <math>\overline{RD}</math>. Point <math>C</math> lies on side <math>\overline{ED}</math> such that <math>\overline{RC}\perp\overline{EM}</math>. Extend segment <math>\overline{DE}</math> through <math>E</math> to point <math>A</math> such that <math>CA=AR</math>. Then <math>AE=\frac{a-\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer. Find <math>a+b+c</math>.
 +
 
 +
==Solution 1==
 +
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\left(\frac{1}{2}, 0\right)</math>, <math>E\left(-\frac{\sqrt{3}}{2}, 0\right)</math>, and <math>R\left(0, \frac{\sqrt{3}}{2}\right).</math> <math>M =</math> midpoint<math>(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)</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>
 +
 
 +
==Solution 2==
 +
 
 +
Let <math>MN = x.</math> Meanwhile, since <math>\triangle R PM</math> is similar to <math>\triangle RCD</math> (angle, side, and side- <math>RP</math> and <math>RC</math> ratio), <math>CD</math> must be 2<math>x</math>. Now, notice that <math>AE</math> is <math>x</math>, because of the parallel segments <math>\overline A\overline E</math> and <math>\overline P\overline M</math>.
 +
 
 +
Now we just have to calculate <math>ED</math>. Using the Law of Sines, or perhaps using altitude <math>\overline R\overline O</math>, we get <math>ED = \frac{\sqrt{3}+1}{2}</math>. <math>CA=RA</math>, which equals <math>ED - x</math>
 +
 
 +
Using Law of Sine in <math>\triangle RED</math>, we find <math>RE</math> = <math>\frac{\sqrt{6}}{2}</math>.
 +
 
 +
We got the three sides of <math>\triangle AER</math>. Now using the Law of Cosines on <math>\angle AER</math>. There we can equate <math>x</math> and solve for it. We got <math>AE=x=\frac{\sqrt{3}-1}{4\sqrt{3}+2}</math>. Then rationalize the denominator, we get <math>AE = \frac{7 - \sqrt{27}}{22}</math>.
 +
 
 +
==Solution 3==
 +
 
 +
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  <math>\triangle ARC</math> is isosceles, <math>P</math> is the midpoint of <math>\overline{CR}</math>, and by midpoint theorem <math>\overline{PM}\parallel\overline{CD}</math>. Thus, <math>APME</math> is a parallelogram and therefore <math>AE = PM = \tfrac 12 CD</math>.
 +
<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 <math>D(0,0)</math> as origin and <math>DE</math> along the <math>x</math>-axis.
 +
 
 +
Let <math>RD=4</math> instead of <math>1</math> (in the end we will scale down by <math>4</math>). Since <math>\angle D = 60^\circ</math>, we get <math>R(2,2\sqrt{3})</math>, and therefore <math>M(1, \sqrt{3})</math>.
 +
 
 +
We use sine-law in <math>\triangle RED</math> to find the coordinates <math>E(2+2\sqrt{3}, 0)</math>:<cmath>DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}.  </cmath>
 +
Since slope<math>(ME)= -\sqrt{3}/(1+2\sqrt{3})</math>, and <math>RC\perp ME</math>, it follows that slope<math>(RC)=(1+2\sqrt{3})/\sqrt{3}</math>. If <math>C(c,0)</math> then we have<cmath>\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}}</cmath>
 +
Now <math>\tfrac 12 CD = \tfrac 12c =(2\sqrt{3}-2)/(1+2\sqrt{3})= \tfrac 1{11}(14-6\sqrt{3})</math>.
 +
 
 +
Scaling down by <math>4</math>, we get <math>AE=\tfrac 1{22}(7-3\sqrt{3})</math>, so our answer is <math>7+27+22=056</math>.
 +
 
 +
 
 +
 
 +
== Video Solution ==
 +
https://youtu.be/muM8UcGKjHo?si=C6o7-C4DgB5i4yKv
 +
 
 +
~MathProblemSolvingSkills.com
  
−
==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>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 17:24, 20 January 2024

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 $MN = x.$ 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 $7+27+22=056$.


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_11&oldid=211584"