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Difference between revisions of "1967 IMO Problems/Problem 1"
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==Solution==
 
==Solution==
 
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 <math>a</math> must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.
 
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 <math>a</math> must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.
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[http://i.imgur.com/jTwHZzq.png]*
 
  
 
To prove our conjecture we draw a parallelogram with <math>a=2</math> and draw a segment <math>DB</math> so that <math>\angle ADB=90^{\circ}</math>
 
To prove our conjecture we draw a parallelogram with <math>a=2</math> and draw a segment <math>DB</math> so that <math>\angle ADB=90^{\circ}</math>
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<math>1^2+x^2=2^2\\x=\sqrt{3}</math>
 
<math>1^2+x^2=2^2\\x=\sqrt{3}</math>
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[http://i.imgur.com/UjLs9oV.png]*
 
  
 
Using trigonometric functions we can compute:
 
Using trigonometric functions we can compute:
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Notice that as <math>\angle\alpha</math> decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as <math>\angle\alpha</math> increases, the value of (1) decreases below 2, confirming that (1) is only implied when <math>\Delta ABD</math> is acute.
 
Notice that as <math>\angle\alpha</math> decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as <math>\angle\alpha</math> increases, the value of (1) decreases below 2, confirming that (1) is only implied when <math>\Delta ABD</math> is acute.
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''*Images are to be used as guidance and are not drawn to scale.''
 
  
 
--[[User:Bjarnidk|Bjarnidk]] 02:16, 17 May 2013 (EDT)
 
--[[User:Bjarnidk|Bjarnidk]] 02:16, 17 May 2013 (EDT)

Revision as of 23:10, 9 December 2022

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)

Solution

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.

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:

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

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.

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

Solution can also be found here [1]

See Also

1967 IMO (Problems) • Resources
Preceded by
First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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