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 "2010 AIME II Problems/Problem 9"

m (→‎Solution)
m
 
(15 intermediate revisions by 10 users not shown)
Line 1: Line 1:
−
== Problem 9 ==
+
__TOC__
−
Let <math>ABCDEF</math> be a regular hexagon. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the midpoints of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, and <math>AF</math>, respectively. The segments <math>\overbar{AH}</math>, <math>\overbar{BI}</math>, <math>\overbar{CJ}</math>, <math>\overbar{DK}</math>, <math>\overbar{EL}</math>, and <math>\overbar{FG}</math> bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of <math>ABCDEF</math> be expressed as a fraction <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+
 
 +
== Problem ==
 +
Let <math>ABCDEF</math> be a [[regular polygon|regular]] [[hexagon]]. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the [[midpoint]]s of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, and <math>AF</math>, respectively. The [[segment]]s <math>\overline{AH}</math>, <math>\overline{BI}</math>, <math>\overline{CJ}</math>, <math>\overline{DK}</math>, <math>\overline{EL}</math>, and <math>\overline{FG}</math> bound a smaller regular hexagon. Let the [[ratio]] of the area of the smaller hexagon to the area of <math>ABCDEF</math> be expressed as a fraction <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.
  
 
==Solution==
 
==Solution==
−
===diagram===
 
 
<center><asy>
 
<center><asy>
 
defaultpen(0.8pt+fontsize(12pt));
 
defaultpen(0.8pt+fontsize(12pt));
Line 39: Line 40:
  
  
−
label('A',A, E);
+
label('$A$',A,(1,0));
−
label('B',B,NE);
+
label('$B$',B,NE);
−
label('C',C,NW);
+
label('$C$',C,NW);
−
label('D',D, W);
+
label('$D$',D, W);
−
label('E',E,SW);
+
label('$E$',E,SW);
−
label('F',F,SE);
+
label('$F$',F,SE);
−
label('G',G,NE);
+
label('$G$',G,NE);
−
label('H',H, N);
+
label('$H$',H, (0,1));
−
label('I',I,NW);
+
label('$I$',I,NW);
−
label('J',J,SW);
+
label('$J$',J,SW);
−
label('K',K, S);
+
label('$K$',K, S);
−
label('L',L,SE);
+
label('$L$',L,SE);
−
label('M',M);
+
label('$M$',M);
−
label('N',N);
+
label('$N$',N);
−
label('O',(0,0));
+
label('$O$',(0,0),NE); dot((0,0));
 
</asy></center>
 
</asy></center>
  
−
Let <math>M</math> be the intersection of <math>\overline{AG}</math> and <math>\overline{BI}</math>
+
Let <math>M</math> be the intersection of <math>\overline{AH}</math> and <math>\overline{BI}</math>.
  
−
and <math>N</math> be the intersection of <math>\overline{BI}</math> and <math>\overline{CJ}</math>.
+
Let <math>N</math> be the intersection of <math>\overline{BI}</math> and <math>\overline{CJ}</math>.
  
 
Let <math>O</math> be the center.
 
Let <math>O</math> be the center.
−
 
−
 
  
 
===Solution 1===
 
===Solution 1===
 +
Without loss of generality, let <math>BC=2.</math>
  
−
Let <math>BC=2</math>
+
Note that <math>\angle BMH</math> is the vertical angle to an angle of the regular hexagon, so it has a measure of <math>120^\circ</math>.
−
 
 
−
Note that <math>\angle BMH</math> is the vertical angle to an angle of regular hexagon, thus, it is <math>120^\circ</math>.
 
  
 
Because <math>\triangle ABH</math> and <math>\triangle BCI</math> are rotational images of one another, we get that <math>\angle{MBH}=\angle{HAB}</math> and hence <math>\triangle ABH \sim \triangle BMH \sim \triangle BCI</math>.
 
Because <math>\triangle ABH</math> and <math>\triangle BCI</math> are rotational images of one another, we get that <math>\angle{MBH}=\angle{HAB}</math> and hence <math>\triangle ABH \sim \triangle BMH \sim \triangle BCI</math>.
  
−
Using a simlar argument, <math>NI=MH</math>.
+
Using a similar argument, <math>NI=MH</math>, and
  
−
<math>MN=BI-NI-BM=BI-(BM+MH)</math>
+
<cmath>MN=BI-NI-BM=BI-(BM+MH).</cmath>
  
−
Applying law of cosine on <math>\triangle BCI</math>, <math>BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}</math>
+
Applying the [[Law of cosines]] on <math>\triangle BCI</math>, <math>BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}</math>
  
−
<math>\frac{BC+CI}{BI}=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH}</math>
+
<cmath>\begin{align*}\frac{BC+CI}{BI}&=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH} \\
 +
BM+MH&=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}} \\
 +
MN&=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}} \\
 +
\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}&=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}</cmath>
  
−
<math>BM+MH=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}}</math>
+
Thus, the answer is <math>4 + 7 = \boxed{011}.</math>
  
−
<math>MN=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}}</math>
+
===Solution 2 (Coordinate Bash)===
−
 
+
We can use coordinates. Let <math>O</math> be at <math>(0,0)</math> with <math>A</math> at <math>(1,0)</math>,  
−
<math>\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}</math>
 
−
 
 
−
Thus, answer is <math>\boxed{011}</math>
 
−
 
 
−
===Solution 2===
 
−
Let's coordinate bash this out.
 
−
 
 
−
Let <math>O</math> be at <math>(0,0)</math> with <math>A</math> be at <math>(1,0)</math>,  
 
  
 
then <math>B</math> is at <math>(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)</math>,  
 
then <math>B</math> is at <math>(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)</math>,  
Line 99: Line 92:
 
<math>D</math> is at <math>(\cos(180^\circ),\sin(180^\circ))=(-1,0)</math>,  
 
<math>D</math> is at <math>(\cos(180^\circ),\sin(180^\circ))=(-1,0)</math>,  
  
−
<math>H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right)</math>
+
<cmath>\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\
−
 
+
&I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}</cmath>
−
<math>I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)</math>
 
  
 
<br/>
 
<br/>
Line 117: Line 109:
 
Let's solve the system of equation to find <math>M</math>
 
Let's solve the system of equation to find <math>M</math>
  
−
<math>-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)</math>
+
<cmath>\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\
−
 
+
-5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\
−
<math>-5\sqrt{3}x=2\sqrt{3}x-\sqrt{3}</math>
+
x&=\frac{1}{7} \\
−
 
+
y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}</cmath>
−
<math>x=\frac{1}{7}</math>
 
  
−
<math>y=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}</math>
+
Finally,
  
−
<br/>
+
<cmath>\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\
 +
&\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}</cmath>
  
−
<math>\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}}</math>
+
Thus, the answer is <math>\boxed{011}</math>.
  
−
<math>\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}</math>
 
  
−
Thus, answer is <math>\boxed{011}</math>
+
Diagram (by dragoon)
 +
https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC82L2ZiNDljZmZiNjUzYWE4NGRmNmIwYTljMWQxZDU2ZDc1ZmNiMDQ3LmpwZWc=&rn=RDQ3ODA2RjUtMzlDNi00QzQ3LUE2OTYtMjlCQkE4NThDNkRBLmpwZWc=
  
−
P.S: Not too bad, isn't it?
+
==Solution 3==
 +
Use the diagram. Now notice that all of the "overlapping triangles" are congruent. You can use the AA similarity to see that the small triangles are similar to the large triangles. Now you can proceed as in Solution 1.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=8|num-a=10|n=II}}
 
{{AIME box|year=2010|num-b=8|num-a=10|n=II}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:22, 27 August 2023

Problem

Let $ABCDEF$ be a regular hexagon. Let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$, respectively. The segments $\overline{AH}$, $\overline{BI}$, $\overline{CJ}$, $\overline{DK}$, $\overline{EL}$, and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

[asy] defaultpen(0.8pt+fontsize(12pt)); pair A,B,C,D,E,F; pair G,H,I,J,K,L; A=dir(0); B=dir(60); C=dir(120); D=dir(180); E=dir(240); F=dir(300); draw(A--B--C--D--E--F--cycle,blue); G=(A+B)/2; H=(B+C)/2; I=(C+D)/2; J=(D+E)/2; K=(E+F)/2; L=(F+A)/2; int i; for (i=0; i<6; i+=1) { draw(rotate(60*i)*(A--H),dotted); } pair M,N,O,P,Q,R; M=extension(A,H,B,I); N=extension(B,I,C,J); O=extension(C,J,D,K); P=extension(D,K,E,L); Q=extension(E,L,F,G); R=extension(F,G,A,H); draw(M--N--O--P--Q--R--cycle,red); label('$A$',A,(1,0)); label('$B$',B,NE); label('$C$',C,NW); label('$D$',D, W); label('$E$',E,SW); label('$F$',F,SE); label('$G$',G,NE); label('$H$',H, (0,1)); label('$I$',I,NW); label('$J$',J,SW); label('$K$',K, S); label('$L$',L,SE); label('$M$',M); label('$N$',N); label('$O$',(0,0),NE); dot((0,0)); [/asy]

Let $M$ be the intersection of $\overline{AH}$ and $\overline{BI}$.

Let $N$ be the intersection of $\overline{BI}$ and $\overline{CJ}$.

Let $O$ be the center.

Solution 1

Without loss of generality, let $BC=2.$

Note that $\angle BMH$ is the vertical angle to an angle of the regular hexagon, so it has a measure of $120^\circ$.

Because $\triangle ABH$ and $\triangle BCI$ are rotational images of one another, we get that $\angle{MBH}=\angle{HAB}$ and hence $\triangle ABH \sim \triangle BMH \sim \triangle BCI$.

Using a similar argument, $NI=MH$, and

\[MN=BI-NI-BM=BI-(BM+MH).\]

Applying the Law of cosines on $\triangle BCI$, $BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}$

\begin{align*}\frac{BC+CI}{BI}&=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH} \\ BM+MH&=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}} \\ MN&=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}} \\ \frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}&=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}

Thus, the answer is $4 + 7 = \boxed{011}.$

Solution 2 (Coordinate Bash)

We can use coordinates. Let $O$ be at $(0,0)$ with $A$ at $(1,0)$,

then $B$ is at $(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$,

$C$ is at $(\cos(120^\circ),\sin(120^\circ))=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$,

$D$ is at $(\cos(180^\circ),\sin(180^\circ))=(-1,0)$,

\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\ &I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}


Line $AH$ has the slope of $-\frac{\sqrt{3}}{2}$ and the equation of $y=-\frac{\sqrt{3}}{2}(x-1)$


Line $BI$ has the slope of $\frac{\sqrt{3}}{5}$ and the equation $y-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)$


Let's solve the system of equation to find $M$

\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\ -5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\ x&=\frac{1}{7} \\ y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}

Finally,

\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\ &\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}

Thus, the answer is $\boxed{011}$.


Diagram (by dragoon) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC82L2ZiNDljZmZiNjUzYWE4NGRmNmIwYTljMWQxZDU2ZDc1ZmNiMDQ3LmpwZWc=&rn=RDQ3ODA2RjUtMzlDNi00QzQ3LUE2OTYtMjlCQkE4NThDNkRBLmpwZWc=

Solution 3

Use the diagram. Now notice that all of the "overlapping triangles" are congruent. You can use the AA similarity to see that the small triangles are similar to the large triangles. Now you can proceed as in Solution 1.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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=2010_AIME_II_Problems/Problem_9&oldid=197541"