Difference between revisions of "1972 IMO Problems/Problem 2"
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| + | ==Problem== | ||
Prove that if n > 3; every quadrilateral that can be inscribed in a circle can | Prove that if n > 3; every quadrilateral that can be inscribed in a circle can | ||
be dissected into n quadrilaterals each of which is inscribable in a circle. | be dissected into n quadrilaterals each of which is inscribable in a circle. | ||
| − | + | ==Solution== | |
| + | Actually, <math>n\ge 4</math> should work. Our initial quadrilateral will be <math>ABCD</math>. | ||
| − | + | For <math>n=4</math>, we do this: | |
| + | |||
| + | Take <math>E\in AB,F\in CD,\ EF\|AD</math> with <math>E,F</math> sufficiently close to <math>A,D</math> respectively. Take <math>U\in AD,V\in EF</math> such that <math>AEVU</math> is an isosceles trapezoid, with <math>V</math> close enough to <math>F</math> (or <math>U</math> close enough to <math>D</math>) that we can find a circle passing through <math>U,D</math> (or <math>F,V</math>) which cuts the segments <math>UV,DF</math> in <math>X,Y</math>. Our four cyclic quadrilaterals are <math>BCFE,\ AEVU,\ VFYX,\ YXUD</math>. | ||
| + | |||
| + | For <math>n\ge 5</math> 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: [https://aops.com/community/p369049] | ||
| + | |||
| + | == See Also == {{IMO box|year=1972|num-b=1|num-a=3}} | ||
Revision as of 14:34, 29 January 2021
Problem
Prove that if n > 3; every quadrilateral that can be inscribed in a circle can be dissected into n quadrilaterals each of which is inscribable in a circle.
Solution
Actually, 
should work. 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 | ||
