Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1967 IMO Problems/Problem 1"
m
Line 2: Line 2:
 
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
 
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
  
−
<math>a\leq \cos \alpha+\sqrt{3}\sin \alpha</math>
+
<math>a\leq \cos \alpha+\sqrt{3}\sin \alpha</math> (1)
  
 
----
 
----
Line 14: Line 14:
 
Using the Pythagorean theorem we conclude:
 
Using the Pythagorean theorem we conclude:
  
−
<math>1^2+x^2=2^2\\x=\sqrt{3}</math> (1)
+
<math>1^2+x^2=2^2\\x=\sqrt{3}</math>
  
 
Using trigonometric functions we can compute:
 
Using trigonometric functions we can compute:

Revision as of 02:33, 17 May 2013

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 (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 $a$ can exceed 2. This is our conjecture.

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 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. 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)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1967_IMO_Problems/Problem_1&oldid=52764"