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Difference between revisions of "1967 IMO Problems/Problem 1"
m (Latexed it and provided the link for solution.)
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Let ''ABCD'' be a parallelogram with side lengths <math>AB=a, AD=1</math> and with <math>\angle BAD=\alpha</math>.
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Let <math>ABCD</math> be a parallelogram with side lengths <math>AB = a</math>, <math>AD = 1</math> and with <math>\angle BAD = \alpha</math>.
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If <math>\Delta ABD</math> is acute, prove that the four circles of radius 1 with centers A, B, C, D cover the parallelogram if and only if
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If <math>\Delta ABD</math> is acute, prove that the four circles of radius <math>1</math> with centers <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> cover the parallelogram if and only if
  
 
<math>a\leq \cos \alpha+\sqrt{3}\sin \alpha</math> (1)
 
<math>a\leq \cos \alpha+\sqrt{3}\sin \alpha</math> (1)
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--[[User:Bjarnidk|Bjarnidk]] 02:16, 17 May 2013 (EDT)
 
--[[User:Bjarnidk|Bjarnidk]] 02:16, 17 May 2013 (EDT)
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Solution can also be found here [https://artofproblemsolving.com/community/c6h21154p137323]

Revision as of 22:36, 1 August 2020

Let $ABCD$ be a parallelogram with side lengths $AB = a$, $AD = 1$ and with $\angle BAD = \alpha$. If $\Delta ABD$ is acute, prove that the four circles of radius $1$ with centers $A$, $B$, $C$, $D$ cover the parallelogram if and only if

$a\leq \cos \alpha+\sqrt{3}\sin \alpha$ (1)

To start our proof we draw a parallelogram with the requested sides. We notice that by drawing the circles with centers A, B, C, D that the length of $a$ must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.

[1]*

To prove our conjecture we draw a parallelogram with $a=2$ and draw a segment $DB$ so that $\angle ADB=90^{\circ}$

This is the parallelogram which we claim has the maximum length on $a$ and the highest value on any one angle.

We now have two triangles inside a parallelogram with lengths $1, 2$ and $x$, $x$ being segment $DB$. Using the Pythagorean theorem we conclude:

$1^2+x^2=2^2\\x=\sqrt{3}$

[2]*

Using trigonometric functions we can compute:

$cos\alpha=\frac{1}{2}\\sin\alpha=\frac{\sqrt{3}}{2}$

Notice that by applying the $arcsine$ and $arccos$ functions, we can conclude that our angle $\alpha=60^{\circ}$

To conclude our proof we make sure that our values match the required values for maximum length of $a$

$a\leq\cos\alpha+\sqrt{3}\sin\alpha\\\\a\leq\frac{1}{2}+\sqrt{3}\cdot \frac{\sqrt{3}}{2}\\\\a\leq 2$

Notice that as $\angle\alpha$ decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as $\angle\alpha$ increases, the value of (1) decreases below 2, confirming that (1) is only implied when $\Delta ABD$ is acute.

*Images are to be used as guidance and are not drawn to scale.

--Bjarnidk 02:16, 17 May 2013 (EDT)

Solution can also be found here [3]

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