Difference between revisions of "1982 IMO Problems/Problem 5"
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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> | ||
− | 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} | + | 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> |
− | This solution was posted and copyrighted by Virgil | + | This solution was posted and copyrighted by Virgil. The original thread for this problem can be found hercommunity/p398343] |
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Let <math>AM = CN = a </math>. By the cosine rule, | 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>AC = \sqrt{AB^{2} + BC^{2} - 2 \cdot AB \cdot BC \cdot cos \angle BAC} |
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= \sqrt{1 + 1 - 2 cos 120^{\circ}} | = \sqrt{1 + 1 - 2 cos 120^{\circ}} | ||
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= \sqrt{3} </math>. | = \sqrt{3} </math>. | ||
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<math>BM = \sqrt{a^{2} + 1 - 2a \cdot cos 30^{\circ}} | <math>BM = \sqrt{a^{2} + 1 - 2a \cdot cos 30^{\circ}} | ||
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= \sqrt{a^{2} - \sqrt{3} \cdot a + 1} </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>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{3 + a^{2} - 2\cdot \sqrt{3} \cdot a + a^{2} - \sqrt{3} \cdot a + a^{2}} | ||
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= \sqrt{3a^{2} - 3\sqrt{3}\cdot a + 3} | = \sqrt{3a^{2} - 3\sqrt{3}\cdot a + 3} | ||
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= BM \cdot \sqrt{3} </math>. | = BM \cdot \sqrt{3} </math>. | ||
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Now if B, M, and N are collinear, then <math>\angle AMB = \angle CMN | Now if B, M, and N are collinear, then <math>\angle AMB = \angle CMN | ||
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\implies sin \angle AMB = sin \angle CMN </math>. | \implies sin \angle AMB = sin \angle CMN </math>. | ||
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By the law of Sines, | By the law of Sines, | ||
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<math>\frac{1}{sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM | <math>\frac{1}{sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM | ||
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\implies sin \angle AMB = \frac{1}{2BM} </math>. | \implies sin \angle AMB = \frac{1}{2BM} </math>. | ||
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Also, | Also, | ||
<math>\frac{a}{sin \angle CMN} = \frac{\sqrt{3} \cdot BM}{sin 60^{\circ}} = 2BM | <math>\frac{a}{sin \angle CMN} = \frac{\sqrt{3} \cdot BM}{sin 60^{\circ}} = 2BM | ||
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\implies sin \angle CMN = \frac{a}{2BM} </math>. | \implies sin \angle CMN = \frac{a}{2BM} </math>. | ||
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But <math>sin \angle AMB = sin \angle CMN | But <math>sin \angle AMB = sin \angle CMN | ||
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\implies \frac{1}{2BM} = \frac{a}{2BM} </math>, which means <math>a = 1 </math>. So, r = \frac{1}{\sqrt{3}} $. | \implies \frac{1}{2BM} = \frac{a}{2BM} </math>, which means <math>a = 1 </math>. So, r = \frac{1}{\sqrt{3}} $. | ||
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Revision as of 05:20, 15 June 2024
Contents
Problem
The diagonals and of the regular hexagon are divided by inner points and respectively, so thatDetermine if and are collinear.
Solution 1
O is the center of the regular hexagon. Then we clearly have . And therefore we have also obviously , as . So we have and . Because of the quadrilateral is cyclic. . And as we also have we get . . And as we get .
This solution was posted and copyrighted by Kathi. The original thread for this problem can be found here: [1]
Solution 2
Let be the intersection of and . is the mid-point of . Since , , and are collinear, then by Menelaus Theorem, . Let the sidelength of the hexagon be . Then . Substituting them into the first equation yields
This solution was posted and copyrighted by leepakhin. The original thread for this problem can be found here: [2]
Solution 3
Note . From the relation results , i.e.
. Thus,
Therefore, , i.e. $r=\frac{MA}{AC}=\frac{1+\sqrt 3}{3+\sqrt 3}=\frac{1}{\sqrt 3}\Longrightarrow r=\frac{\sqrt 3}{3}<\math>
This solution was posted and copyrighted by Virgil. The original thread for this problem can be found hercommunity/p398343]
Let <math>AM = CN = a$ (Error compiling LaTeX. Unknown error_msg). By the cosine rule,
$AC = \sqrt{AB^{2} + BC^{2} - 2 \cdot AB \cdot BC \cdot cos \angle BAC}
= \sqrt{1 + 1 - 2 cos 120^{\circ}}
= \sqrt{3}$ (Error compiling LaTeX. Unknown error_msg).
$BM = \sqrt{a^{2} + 1 - 2a \cdot cos 30^{\circ}}
= \sqrt{a^{2} - \sqrt{3} \cdot a + 1}$ (Error compiling LaTeX. Unknown error_msg)
$MN = \sqrt{(\sqrt{3} - a)^{2} + a^{2} - 2 \cdot (\sqrt{3} - a) \cdot a \cdot cos \angle MCN}
= \sqrt{3 + a^{2} - 2\cdot \sqrt{3} \cdot a + a^{2} - \sqrt{3} \cdot a + a^{2}}
= \sqrt{3a^{2} - 3\sqrt{3}\cdot a + 3}
= BM \cdot \sqrt{3}$ (Error compiling LaTeX. Unknown error_msg).
Now if B, M, and N are collinear, then $\angle AMB = \angle CMN
\implies sin \angle AMB = sin \angle CMN$ (Error compiling LaTeX. Unknown error_msg).
By the law of Sines,
$\frac{1}{sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM
\implies sin \angle AMB = \frac{1}{2BM}$ (Error compiling LaTeX. Unknown error_msg).
Also,
$\frac{a}{sin \angle CMN} = \frac{\sqrt{3} \cdot BM}{sin 60^{\circ}} = 2BM
\implies sin \angle CMN = \frac{a}{2BM}$ (Error compiling LaTeX. Unknown error_msg).
But $sin \angle AMB = sin \angle CMN
\implies \frac{1}{2BM} = \frac{a}{2BM}$ (Error compiling LaTeX. Unknown error_msg), which means . So, r = \frac{1}{\sqrt{3}} $.