1982 IMO Problems/Problem 5

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]

Solution 4

File:IMO1982P5.png $WLOG$ , consider $AB = BC = CD = DE = EF = FA = 1$ unit. Now,

$AB = \sqrt{AB^{2} + BC^{2} - 2 \dot AB \dot BC \dot cos \angle ABC} = \sqrt{3}$ (after all the simplifying, and substituting $cos 120$ = $\frac{-1}{2}$ ).

Now $\frac{AM}{AC} = \frac{CN}{CE}$ indicates $AM = CN$ . So let's go ahead and write $AM = CN = a$ and $CM = NE = \sqrt{3} - a$ . Applying the cosine rule, we get:

$BM = \sqrt{AM^{2} + AB^{2} - 2 \dot AM \dot AB \dot cos \angle BAM} = \sqrt{a^{2} - \sqrt{3} \dot a + 1}$ .

$MN = \sqrt{CM^2 + CN^2 - 2 \dot CM \dot CN \dot cos \angle MCN} = \sqrt{3 + a^{2} - 2\sqrt{3}\dot a + a^{2} - a\sqrt{3} + a^{2}} = \sqrt{3a^{2} - 3\sqrt{3}\dot a + 3}$ .

This means $MN = BM \dot \sqrt{3}$ (Error compiling LaTeX. Unknown error_msg).

B, M, and N are collinear if and only if $\angle CMN = \angle AMB$ .

$\implies sin\angle CMN = sin\angle AMB$ . Now by the law of Sines,

$\frac{a}{sin\angle CMN} = \frac{\sqrt{3}\dot BM}{\frac{\sqrt{3}}{2}} = 2BM$ .

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

So, $\frac{1}{sin\angle AMB} = \frac{a}{sin\angle CMN$ (Error compiling LaTeX. Unknown error_msg)

$\implies a = 1 = AM$ .

So since $r = \frac{AM}{AC} = \frac{1}{\sqrt{3}}$ .

So the answer is $r = \frac{1}{\sqrt{3}}$ .

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