Difference between revisions of "1982 IMO Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
− | O is the center of the regular hexagon. Then we | + | 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> | ||
− | 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] |
− | == | + | ==Solution 4== |
+ | |||
+ | Let <math>AM = CN = a </math>. By the cosine rule, | ||
+ | |||
+ | <math>AC = \sqrt{AB^{2} + BC^{2} - 2 \cdot AB \cdot BC \cdot \cos \angle BAC} </math> | ||
+ | |||
+ | <math>= \sqrt{1 + 1 - 2 \cdot \cos 120^{\circ}} </math> | ||
+ | |||
+ | <math>= \sqrt{3} </math>. | ||
+ | |||
+ | <math>BM = \sqrt{a^{2} + 1 - 2a \cdot \cos 30^{\circ}} </math> | ||
+ | |||
+ | <math>= \sqrt{a^{2} - \sqrt{3} \cdot a + 1} </math> | ||
+ | |||
+ | <math>MN = \sqrt{(\sqrt{3} - a)^{2} + a^{2} - 2 \cdot (\sqrt{3} - a) \cdot a \cdot \cos \angle MCN} </math> | ||
+ | |||
+ | <math>= \sqrt{3 + a^{2} - 2\cdot \sqrt{3} \cdot a + a^{2} - \sqrt{3} \cdot a + a^{2}} </math> | ||
+ | |||
+ | <math>= \sqrt{3a^{2} - 3\sqrt{3}\cdot a + 3} </math> | ||
+ | |||
+ | <math>= BM \cdot \sqrt{3} </math>. | ||
+ | |||
+ | Now if B, M, and N are collinear, then <math>\angle AMB = \angle CMN </math> | ||
+ | |||
+ | <math>\implies \sin \angle AMB = \sin \angle CMN </math>. | ||
+ | |||
+ | By the law of Sines, | ||
+ | |||
+ | <math>\frac{1}{\sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM </math> | ||
+ | |||
+ | <math>\implies \sin \angle AMB = \frac{1}{2BM} </math>. | ||
+ | |||
+ | Also, | ||
+ | |||
+ | <math>\frac{a}{\sin \angle CMN} = \frac{\sqrt{3} \cdot BM}{\sin 60^{\circ}} = 2BM </math> | ||
+ | |||
+ | <math>\implies \sin \angle CMN = \frac{a}{2BM} </math>. | ||
+ | |||
+ | But <math>\sin \angle AMB = \sin \angle CMN </math> | ||
+ | |||
+ | <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 10:44, 16 June 2024
Problem
The diagonals and
of the regular hexagon
are divided by inner points
and
respectively, so that
Determine
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.
This solution was posted and copyrighted by Virgil. The original thread for this problem can be found hercommunity/p398343]
Solution 4
Let . By the cosine rule,
.
.
Now if B, M, and N are collinear, then
.
By the law of Sines,
.
Also,
.
But
, which means
. So, r = \frac{1}{\sqrt{3}} $.