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Difference between revisions of "1982 IMO Problems/Problem 5"

(→‎Solution 1)
(→‎Solution 3)
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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]
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==Solution 4==
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[[File: IMO1982P5.png]]
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<math>WLOG </math>, consider <math>AB = BC = CD = DE = EF = FA = 1 </math> unit. Now,
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<math>AB = \sqrt{AB^{2} + BC^{2} - 2 \dot AB \dot BC \dot cos \angle ABC} = \sqrt{3} </math>(after all the simplifying, and substituting <math>cos 120 </math> = <math>\frac{-1}{2} </math>).
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Now <math>\frac{AM}{AC} = \frac{CN}{CE} </math> indicates <math>AM = CN </math>. So let's go ahead and write <math>AM = CN = a </math> and <math>CM = NE = \sqrt{3} - a </math>. Applying the cosine rule, we get:
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<math>BM = \sqrt{AM^{2} + AB^{2} - 2 \dot AM \dot AB \dot cos \angle BAM} = \sqrt{a^{2} - \sqrt{3} \dot a + 1} </math>.
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<math>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}
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This means </math>MN = BM \dot \sqrt{3} <math>.
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B, M, and N are collinear if and only if </math>\angle CMN = \angle AMB <math>.
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</math>\implies sin\angle CMN = sin\angle AMB <math>. Now by the law of Sines,
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</math>\frac{a}{sin\angle CMN} = \frac{\sqrt{3}\dot BM}{\frac{\sqrt{3}}{2}} = 2BM <math>. 
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Now </math>\frac{1}{sin\angle AMB} = \frac{BM}{sin 30^{circ}} = 2BM <math>.
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So, </math>\frac{1}{sin\angle AMB} = \frac{a}{sin\angle CMN <math>
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</math>\implies a = 1 = AM<math>.
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So since </math>r = \frac{AM}{AC} = \frac{1}{\sqrt{3}} <math>.
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So the answer is </math>r = \frac{1}{\sqrt{3}} $.
  
 
== See Also == {{IMO box|year=1982|num-b=4|num-a=6}}
 
== See Also == {{IMO box|year=1982|num-b=4|num-a=6}}

Revision as of 11:37, 13 June 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 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$ (Error compiling LaTeX. Unknown error_msg)MN = BM \dot \sqrt{3} $.

B, M, and N are collinear if and only if$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{sin\angle AMB} = \frac{BM}{sin 30^{circ}} = 2BM $.

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

So since$ (Error compiling LaTeX. Unknown error_msg)r = \frac{AM}{AC} = \frac{1}{\sqrt{3}} $.

So the answer is$ (Error compiling LaTeX. Unknown error_msg)r = \frac{1}{\sqrt{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
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