1967 IMO Problems/Problem 1

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. Also, we notice that if an angle exceeds 90 degrees (is no longer acute as requested), side $a$ can exceed 2. This is the interplay between the y- and the x-axis of our problem and also our conjecture.

[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)

Page

Toolbox

Search

1967 IMO Problems/Problem 1