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Difference between revisions of "1967 IMO Problems/Problem 1"

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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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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 (a radius for each circle) or the circles will not meet and thus not cover the parallelogram. Also, we notice that if an angle exceeds 90 degrees (is no longer acute as requested), side <math>a</math> can exceed 2. This is our conjecture.
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==Solution==
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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 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>cos\alpha=\frac{1}{2}\\sin\alpha=\frac{\sqrt{3}}{2}</math>
 
<math>cos\alpha=\frac{1}{2}\\sin\alpha=\frac{\sqrt{3}}{2}</math>
  
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Notice that our angle <math>\alpha=60^{\circ}</math>
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Notice that by applying the <math>arcsine</math> and <math>arccos</math> functions, we can conclude that our angle <math>\alpha=60^{\circ}</math>
  
 
To conclude our proof we make sure that our values match the required values for maximum length of <math>a</math>
 
To conclude our proof we make sure that our values match the required values for maximum length of <math>a</math>
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<math>a\leq\cos\alpha+\sqrt{3}\sin\alpha\\\\a\leq\frac{1}{2}+\sqrt{3}\cdot \frac{\sqrt{3}}{2}\\\\a\leq 2</math>
 
<math>a\leq\cos\alpha+\sqrt{3}\sin\alpha\\\\a\leq\frac{1}{2}+\sqrt{3}\cdot \frac{\sqrt{3}}{2}\\\\a\leq 2</math>
  
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Notice that as <math>\angle\alpha</math> decreases, the value of (1) increases beyond 2. 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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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.
  
 
--[[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]
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== See Also == {{IMO box|year=1967|before=First question|num-a=2}}

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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