1982 IMO Problems/Problem 5: Difference between revisions

Revision as of 11:02, 16 April 2022

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

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

1982 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions

Page

Toolbox

Search