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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...")
 
(→‎Solution 4)
 
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==Solution 1==
 
==Solution 1==
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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>.
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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>.
 
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>.
  
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<math>2[\cos 60-\cos (60+2x)]=1-\cos (120-2x)\Longrightarrow \cos (60+2x)=0\Longrightarrow x=15^{\circ}.</math>
 
<math>2[\cos 60-\cos (60+2x)]=1-\cos (120-2x)\Longrightarrow \cos (60+2x)=0\Longrightarrow x=15^{\circ}.</math>
  
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Therefore, <math>\frac{MA}{MC}=\frac{\sin 75}{\sin 45}=\frac{1+\sqrt 3}{2}</math>, i.e. <math>r=\frac{MA}{AC}=\frac{1+\sqrt 3}{3+\sqrt 3}=\frac{1}{\sqrt 3}\Longrightarrow r=\frac{\sqrt 3}{3}.</math>
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Therefore, <math>\frac{MA}{MC}=\frac{\sin 75}{\sin 45}=\frac{1+\sqrt 3}{2}</math>, i.e. <math>r=\frac{MA}{AC}=\frac{1+\sqrt 3}{3+\sqrt 3}=\frac{1}{\sqrt 3}\Longrightarrow r=\frac{\sqrt 3}{3}</math>
  
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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]
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This solution was posted and copyrighted by Virgil. The original thread for this problem can be found hercommunity/p398343]
  
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== See Also == {{IMO box|year=1979|num-b=4|num-a=6}}
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==Solution 4==
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 +
Let <math>AM = CN = a </math>. By the cosine rule,
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<math>AC = \sqrt{AB^{2} + BC^{2} - 2 \cdot AB \cdot BC \cdot \cos \angle BAC} </math>
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<math>= \sqrt{1 + 1 - 2 \cdot \cos 120^{\circ}} </math>
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<math>= \sqrt{3} </math>.
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<math>BM = \sqrt{a^{2} + 1 - 2a \cdot \cos 30^{\circ}} </math>
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<math>= \sqrt{a^{2} - \sqrt{3} \cdot a + 1} </math>
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<math>MN = \sqrt{(\sqrt{3} - a)^{2} + a^{2} - 2 \cdot (\sqrt{3} - a) \cdot a \cdot \cos \angle MCN} </math>
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<math>= \sqrt{3 + a^{2} - 2\cdot \sqrt{3} \cdot a + a^{2} - \sqrt{3} \cdot a + a^{2}} </math>
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<math>= \sqrt{3a^{2} - 3\sqrt{3}\cdot a + 3} </math>
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<math>= BM \cdot \sqrt{3} </math>.
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Now if B, M, and N are collinear, then <math>\angle AMB = \angle CMN </math>
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<math>\implies \sin \angle AMB = \sin \angle CMN </math>.
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By the law of Sines,
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<math>\frac{1}{\sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM </math>
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<math>\implies \sin \angle AMB = \frac{1}{2BM} </math>.
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Also,
 +
 
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<math>\frac{a}{\sin \angle CMN} = \frac{\sqrt{3} \cdot BM}{\sin 60^{\circ}} = 2BM </math>
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<math>\implies \sin \angle CMN = \frac{a}{2BM} </math>.
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But <math>\sin \angle AMB = \sin \angle CMN </math>
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<math>\implies \frac{1}{2BM} = \frac{a}{2BM} </math>, which means <math>a = 1 </math>. So, r = \frac{1}{\sqrt{3}} $.

Latest revision as of 09:44, 16 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. 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}} $.

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