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"

(Created page with "==Problem== The diagonals <math>AC</math> and <math>CE</math> of the regular hexagon <math>ABCDEF</math> are divided by inner points <math>M</math> and <math>N</math> respecti...")
 
Line 33: Line 33:
 
This solution was posted and copyrighted by Virgil Nicula. The original thread for this problem can be found here: [https://aops.com/community/p398343]
 
This solution was posted and copyrighted by Virgil Nicula. The original thread for this problem can be found here: [https://aops.com/community/p398343]
  
−
== See Also == {{IMO box|year=1979|num-b=4|num-a=6}}
+
== See Also == {{IMO box|year=1982|num-b=4|num-a=6}}

Revision as of 22:27, 29 January 2021

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 obviously 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 Nicula. The original thread for this problem can be found here: [3]

See Also

1982 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_IMO_Problems/Problem_5&oldid=144005"