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 "1982 IMO Problems/Problem 5"
(→‎Solution 5)
 
Line 79: Line 79:
 
We can assign coordinates to solve this question. First, WLOG, let the side length of the hexagon be 1. We also know that each angle of the hexagon is <math>\frac{(6-2) \cdot 180}{6} = 120</math>. From Law of Cosines, we find <math>AC = \sqrt{1 + 1 - 2cos(120)} = \sqrt3</math>.
 
We can assign coordinates to solve this question. First, WLOG, let the side length of the hexagon be 1. We also know that each angle of the hexagon is <math>\frac{(6-2) \cdot 180}{6} = 120</math>. From Law of Cosines, we find <math>AC = \sqrt{1 + 1 - 2cos(120)} = \sqrt3</math>.
  
−
Now, let point <math>E</math> be located at <math>(0, 0)</math>. Since <math>AE = AC, A</math> is located at <math>(0, \sqrt3)</math> meaning <math>B</math> is located at <math>(1, \sqrt3)</math>. Using the ratios given to us, <math>CN = r\sqrt3</math> so <math>EN = (1-r)\sqrt3</math>. Let <math>O</math> be defined on line <math>ED</math> such that <math>EO</math> is perpendicular to <math>NO</math>. Since <math>\bigtriangleup ENO</math> is a <math>30-60-90</math> triangle, we can find that point N is located at <math>(\frac{3(1-r)}{2}, \frac{(1-r)\sqrt3}{2})</math>. Similarly, <math>M</math> would be located at <math>(\frac{3r}{2}, \frac{\sqrt3(1 - r)}{2})</math>.
+
Now, let point <math>E</math> be located at <math>(0, 0)</math>. Since <math>AE = AC,</math> <math>A</math> is located at <math>(0, \sqrt3)</math> meaning <math>B</math> is located at <math>(1, \sqrt3)</math>. Using the ratios given to us, <math>CN = r\sqrt3</math> so <math>EN = (1-r)\sqrt3</math>. Let <math>O</math> be defined on line <math>ED</math> such that <math>EO</math> is perpendicular to <math>NO</math>. Since <math>\bigtriangleup ENO</math> is a <math>30-60-90</math> triangle, we can find that point N is located at <math>(\frac{3(1-r)}{2}, \frac{(1-r)\sqrt3}{2})</math>. Similarly, <math>M</math> would be located at <math>(\frac{3r}{2}, \sqrt3(1 - \frac{r}{2}))</math>.
 +
 
 +
Since B, M, N, are collinear, they have the same slope. So,
 +
<cmath>\frac{\sqrt3 - \frac{(1-r)\sqrt3}{2}}{1 - \frac{3(1-r)}{2}} = \frac{\sqrt3 - \sqrt3(1 - \frac{r}{2})}{1 - \frac{3r}{2}}</cmath>
 +
<cmath>\frac{\sqrt3 + r\sqrt3}{3r - 1} = \frac{r\sqrt3}{2 - 3r}</cmath>
 +
<cmath>2\sqrt3 + 2r\sqrt3 - 3r\sqrt3 - 3r^2\sqrt3 = 3r^2\sqrt3 - r\sqrt3</cmath>
 +
<cmath>6r^2\sqrt3 = 2\sqrt3</cmath>
 +
<cmath>\boxed{r = \frac{\sqrt3}{3}}</cmath>
 +
~SID-DARTH-VATER

Latest revision as of 22:05, 1 December 2024

Problem

The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that\[{AM\over AC}={CN\over CE}=r.\]Determine $r$ if $B,M$ and $N$ are collinear.

Solution 1

O is the center of the regular hexagon. Then we clearly have $ABC\cong COA\cong EOC$. And therefore we have also obviously $ABM\cong AOM\cong CON$, as $\frac{AM}{AC} =\frac{CN}{CE}$. So we have $\angle{BMA} =\angle{AMO} =\angle{CNO}$ and $\angle{NOC} =\angle{ABM}$. Because of $\angle{AMO} =\angle{CNO}$ the quadrilateral $ONCM$ is cyclic. $\Rightarrow \angle{NOC} =\angle{NMC} =\angle{BMA}$. And as we also have $\angle{NOC} =\angle{ABM}$ we get $\angle{ABM} =\angle{BMA}$. $\Rightarrow AB=AM$. And as $AC=\sqrt{3} \cdot AB$ we get $r=\frac{AM}{AC} =\frac{AB}{\sqrt{3} \cdot AB} =\frac{1}{\sqrt{3}}$.

This solution was posted and copyrighted by Kathi. The original thread for this problem can be found here: [1]

Solution 2

Let $X$ be the intersection of $AC$ and $BE$. $X$ is the mid-point of $AC$. Since $B$, $M$, and $N$ are collinear, then by Menelaus Theorem, $\frac{CN}{NE}\cdot\frac{EB}{BX}\cdot\frac{XM}{MC}=1$. Let the sidelength of the hexagon be $1$. Then $AC=CE=\sqrt{3}$. $\frac{CN}{NE}=\frac{CN}{CE-CN}=\frac{\frac{CN}{CE}}{1-\frac{CN}{CE}}=\frac{r}{1-r}$ $\frac{EB}{BX}=\frac{2}{\frac{1}{2}}=4$ $\frac{XM}{MC}=\frac{AM-AX}{AC-AM}=\frac{\frac{AM}{AC}-\frac{AX}{AC}}{1-\frac{AM}{AC}}=\frac{r-\frac{1}{2}}{1-r}$ Substituting them into the first equation yields $\frac{r}{1-r}\cdot\frac{4}{1}\cdot\frac{r-\frac{1}{2}}{1-r}=1$ $3r^2=1$ $\therefore r=\frac{\sqrt{3}}{3}$

This solution was posted and copyrighted by leepakhin. The original thread for this problem can be found here: [2]

Solution 3

Note $x=m(\widehat {EBM})$. From the relation $r=\frac{AM}{AC}=\frac{CN}{CE}$ results $\frac{MA}{MC}=\frac{NC}{NE}$, i.e.

$\frac{BA}{BC}\cdot \frac{\sin (60+x)}{\sin (60-x)}=\frac{BC}{BE}\cdot \frac{\sin (60-x)}{\sin x}$. Thus, $2\sin x\sin (60+x)=\sin^2(60-x)\Longrightarrow$ $2[\cos 60-\cos (60+2x)]=1-\cos (120-2x)\Longrightarrow \cos (60+2x)=0\Longrightarrow x=15^{\circ}.$

Therefore, $\frac{MA}{MC}=\frac{\sin 75}{\sin 45}=\frac{1+\sqrt 3}{2}$, i.e. $r=\frac{MA}{AC}=\frac{1+\sqrt 3}{3+\sqrt 3}=\frac{1}{\sqrt 3}\Longrightarrow r=\frac{\sqrt 3}{3}$

This solution was posted and copyrighted by Virgil. The original thread for this problem can be found hercommunity/p398343]

Solution 4

Let $AM = CN = a$. By the cosine rule,

$AC = \sqrt{AB^{2} + BC^{2} - 2 \cdot AB \cdot BC \cdot \cos \angle BAC}$

$= \sqrt{1 + 1 - 2 \cdot \cos 120^{\circ}}$

$= \sqrt{3}$.

$BM = \sqrt{a^{2} + 1 - 2a \cdot \cos 30^{\circ}}$

$= \sqrt{a^{2} - \sqrt{3} \cdot a + 1}$

$MN = \sqrt{(\sqrt{3} - a)^{2} + a^{2} - 2 \cdot (\sqrt{3} - a) \cdot a \cdot \cos \angle MCN}$

$= \sqrt{3 + a^{2} - 2\cdot \sqrt{3} \cdot a + a^{2} - \sqrt{3} \cdot a + a^{2}}$

$= \sqrt{3a^{2} - 3\sqrt{3}\cdot a + 3}$

$= BM \cdot \sqrt{3}$.

Now if B, M, and N are collinear, then $\angle AMB = \angle CMN$

$\implies \sin \angle AMB = \sin \angle CMN$.

By the law of Sines,

$\frac{1}{\sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM$

$\implies \sin \angle AMB = \frac{1}{2BM}$.

Also,

$\frac{a}{\sin \angle CMN} = \frac{\sqrt{3} \cdot BM}{\sin 60^{\circ}} = 2BM$

$\implies \sin \angle CMN = \frac{a}{2BM}$.

But $\sin \angle AMB = \sin \angle CMN$

$\implies \frac{1}{2BM} = \frac{a}{2BM}$, which means $a = 1$. So, r = $\frac{1}{\sqrt{3}}$.

Solution 5

We can assign coordinates to solve this question. First, WLOG, let the side length of the hexagon be 1. We also know that each angle of the hexagon is $\frac{(6-2) \cdot 180}{6} = 120$. From Law of Cosines, we find $AC = \sqrt{1 + 1 - 2cos(120)} = \sqrt3$.

Now, let point $E$ be located at $(0, 0)$. Since $AE = AC,$ $A$ is located at $(0, \sqrt3)$ meaning $B$ is located at $(1, \sqrt3)$. Using the ratios given to us, $CN = r\sqrt3$ so $EN = (1-r)\sqrt3$. Let $O$ be defined on line $ED$ such that $EO$ is perpendicular to $NO$. Since $\bigtriangleup ENO$ is a $30-60-90$ triangle, we can find that point N is located at $(\frac{3(1-r)}{2}, \frac{(1-r)\sqrt3}{2})$. Similarly, $M$ would be located at $(\frac{3r}{2}, \sqrt3(1 - \frac{r}{2}))$.

Since B, M, N, are collinear, they have the same slope. So, \[\frac{\sqrt3 - \frac{(1-r)\sqrt3}{2}}{1 - \frac{3(1-r)}{2}} = \frac{\sqrt3 - \sqrt3(1 - \frac{r}{2})}{1 - \frac{3r}{2}}\] \[\frac{\sqrt3 + r\sqrt3}{3r - 1} = \frac{r\sqrt3}{2 - 3r}\] \[2\sqrt3 + 2r\sqrt3 - 3r\sqrt3 - 3r^2\sqrt3 = 3r^2\sqrt3 - r\sqrt3\] \[6r^2\sqrt3 = 2\sqrt3\] \[\boxed{r = \frac{\sqrt3}{3}}\] ~SID-DARTH-VATER

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_IMO_Problems/Problem_5&oldid=236019"