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1972 IMO Problems/Problem 2

Revision as of 18:42, 26 March 2025 by Pf02 (talk | contribs)

Problem

Prove that if $n \geq 4$, every quadrilateral that can be inscribed in a circle can be dissected into $n$ quadrilaterals each of which is inscribable in a circle.

Solution

Our initial quadrilateral will be $ABCD$.

For $n=4$, we do this:

Take $E\in AB,F\in CD,\ EF\|AD$ with $E,F$ sufficiently close to $A,D$ respectively. Take $U\in AD,V\in EF$ such that $AEVU$ is an isosceles trapezoid, with $V$ close enough to $F$ (or $U$ close enough to $D$) that we can find a circle passing through $U,D$ (or $F,V$) which cuts the segments $UV,DF$ in $X,Y$. Our four cyclic quadrilaterals are $BCFE,\ AEVU,\ VFYX,\ YXUD$.

For $n\ge 5$ 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]


Remarks (added by pf02, March 2025)

The construction described in the solution above is correct (in the sense that it describes a legitimate way of dissecting an inscribable quadrilateral into four inscribable quadrilaterals). However, the solution is incomplete and sloppily written.

Below I will discuss and complete the solution given above. Then, I will give a second solution. And finally, I will discuss the cases when $n = 2, n = 3$.


Discussion and completion of the above solution

The first issue is the fact that a construction is described, but there is no proof, not even a hint, why the the quadrilaterals are inscribable. This is not obvious, and it needs a proof. I will give the proof below.

The second issue is the vagueness of "close enough" used twice in the proof. The first time it is used, "$E, F$ sufficiently close to $A, D$ respectively" is not needed (indeed, and segment $EF$ parallel to $AD$ would do), so there is no need to make this more precise. The second time it is used, namely "$V$ close enough to $F$ (or $U$ close enough to $D$) that we can find a circle passing through $U, D$ (or $F, V$)" is indeed needed, and it is not at all clear what "close enough" should be, or that this is at all possible. I will come back to this shortly.

The third issue is poor wording. We don't need to "add as many trapezoids as we want". We want to dissect the one isosceles trapezoid into as many isosceles trapezoids as we want by lines parallel to its bases.

Prob 1972 2.png

Before giving the missing details, let us remember that a quadrilateral $ABCD$ is inscribable if and only if a pair of opposing angles adds up to $\pi$, in other words $\angle A + \angle C = \pi$, or equivalently, $\angle B + \angle D = \pi$. In particular, any isosceles trapezoid is inscribable.

Now let us show that the four quadrilaterals are inscribable. It is easy to see that the first one, $EBCF$ is inscribable. Indeed, $\angle C + \angle A = \pi$. We know that $\angle A = \angle FEB$ because of parallelism, so $\angle C + \angle FEB = \pi$. The second one, $AEVU$ is an isosceles trapezoid by the choice of $VU$, so it is inscribable. The third one, $DUXY$ is inscribable by construction. It remains to be shown that $XYFV$ is inscribable.

We have $\angle YFV - \pi - \angle CFE = \angle B = \pi - \angle D = \angle YXU = \pi - \angle YXV$. This shows that $XYFV$ is inscribable.

Note that as suggested by the solution, we could have chosen $XY$ so that $XYFV$ is inscribable, in which case a similar argument would have shown that $DUXY$ is inscribable as well.

Let us now make precise what it means that $U$ should be close enough to $D$, or $V$ should be close enough to $F$, so that we can find $XY$.






TO BE CONTINUED. SAVING MID WAY, SO I DON'T LOOSE WORK DONE SO FAR.


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
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