Difference between revisions of "1972 IMO Problems/Problem 2"
Mathboy100 (talk | contribs) (→Problem) |
Mathboy100 (talk | contribs) m (→Solution) |
||
| Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
| − | + | Our initial quadrilateral will be <math>ABCD</math>. | |
For <math>n=4</math>, we do this: | For <math>n=4</math>, we do this: | ||
Latest revision as of 17:15, 7 December 2022
Problem
Prove that if 
, every quadrilateral that can be inscribed in a circle can be dissected into 
quadrilaterals each of which is inscribable in a circle.
Solution
Our initial quadrilateral will be 
.
For 
, we do this:
Take 
with 
sufficiently close to 
respectively. Take 
such that 
is an isosceles trapezoid, with 
close enough to 
(or 
close enough to 
) that we can find a circle passing through 
(or 
) which cuts the segments 
in 
. Our four cyclic quadrilaterals are 
.
For 
we do the exact same thing as above, but now, since we have an isosceles trapezoid, we can add as many trapezoids as we want by dissecting the one trapezoid with lines parallel to its bases.
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]
See Also
| 1972 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
