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

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}} $.

@@ Line 61: / Line 61: @@
 By the law of Sines,
-<math>\frac{1}{s\in \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM </math>
+<math>\frac{1}{\sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM </math>
 <math>\implies \sin \angle AMB = \frac{1}{2BM} </math>.

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