1972 IMO Problems/Problem 2

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 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, any 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.

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 $XY$ , so that $DUXY$ is inscribable.

Let us assume that $\angle D$ and $\angle DUV = \angle DAB$ are acute. One way of constructing the circle $DUXY$ is the following: Pick $U$ on $DA$ , pick $Y$ on $DF$ , and find the center $O$ of the circle $DUY$ as the intersection of the medians to $DY$ and $DU$ . Take $X$ to be the intersection of this circle with $UV$ . We want to show that we can choose $U$ and $Y$ so that $X$ will be between $U$ and $V$ .

Let us consider the median of $DF$ . This median intersects $DA$ someplace on the same side of $D$ as $A$ . Let us pick $U$ between $D$ and this point. This is the first condition for $U$ being close enough to $D$ . This choice ensures that the midpoint of $DU$ is in the region which is the strip between the median to $DF$ and the perpendicular to $DF$ at $D$ .

Now consider $UV$ , the median to $UV$ , and the perpendicular to $UV$ at $U$ . We want to make sure that $U$ is chosen so that the midpoint of $DU$ is in the region which is the strip between these two lines. For this, let us take $Z_1$ to be the midpoint of $UV$ and $Z_2$ be the intersection of the median to $UV$ with $DU$ . From the triangle $\triangle UZ_1Z_2$ , we have $\cos (\angle DUV) = \frac{UZ_1}{UZ_2}$ , so $UZ_2 = \frac{UV}{2} \cdot \frac{1}{\cos (\angle DUV)}$ . In order to satisfy the condition about the midpoint of $DU$ , it is enough to have $\frac{DU}{2} < UZ_2$ . This translates to $DU < \frac{UV}{\cos (\angle DUV)}$ . Note that $UV$ is of fixed slope and length because $UVEA$ has to be an isosceles trapezoid, so this is a second condition expressing how close $U$ has to be to $D$ .

With this choice of $U$ , the midpoint of $DU$ is in the region which is the intersection of the two strips. Now pick $Y \in DF$ close enough to $D$ , so that the intersection $O$ of the medians of $DY$ and $DU$ is in the same region. ( $Y$ has to be close enough to the foot of the perpendicular from the midpoint of $DU$ to $DF$ .) This ensures that the circle of center $O$ and radius $OD = OY = OU$ will intersect $UV$ at a point $X$ between $U$ and $V$ .

In the discussion above, we assumed $\angle D$ and $\angle DUV$ to be acute. The cases when either of these angles is not acute can be dealt with in a similar way. We will skip the discussion for these cases, and leave it to the interested reader to work out the details.

Solution 2

First, 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.

Let us assume that $\angle A$ is less or equal than all the other angles. We can assume this because otherwise, we would just need to re-label the vertices.

Choose a point $P$ inside $ABCD$ , and draw the parallel $PH$ to $AB$ , with $H \in DA$ . There will be restrictions on the position of $P$ , which we will discuss as we do the construction.

Choose $E \in AB$ so that $AEPH$ is an isosceles trapezoid. In order for this to be possible, $P$ has to be inside the angle $\angle ABB_1$ formed by drawing the line $BB_1$ such that $\angle ABB_1 = \angle A$ .

Next, choose $F \in BC$ so that $\angle PFB = \angle A$ . This will ensure that $EBFP$ is inscribable because $\angle PEB = \pi - \angle A$ . In order for $F$ to exist in $BC$ , $P$ has to be inside the angle $BCC_1$ formed by drawing the line $CC_1$ such that $\angle BCC_1 = \angle A$ .

Next choose $G \in CD$ such that $PG \parallel BC$ . This will ensure that $PFCG$ is an isosceles trapezoid because $\angle PFC = \pi - \angle A = \angle C$ . In order for $G$ to exist in $CD$ , $P$ has to be inside the angle $\angle CDD_1$ formed by drawing the line $DD_1$ such that $\angle CDD_1 = \angle A$ .

With these choices, it is easy to see that $GDHP$ is inscribable. Indeed, $\angle PHD = \angle A$ , and $\angle DGP = \pi - \angle CGP = \angle C = \pi - \angle A$ .

We now just have to show that the intersection of the interiors of angles $\angle ABB_1, \angle BCC_1, \angle CDD_1$ is not $\O$ . This follows immediately from noticing that $DD_1 \parallel BC$ .

We now proceed by induction on $n$ . For $n = 4$ we dissected the inscribable quadrilateral $ABCD$ in two inscribable rectangles, and $2 = 4 - 2$ isosceles trapezoids. Assume we dissected the inscribable quadrilateral $ABCD$ in two inscribable rectangles, and $k - 2$ isosceles trapezoids. Then we dissect one of the isosceles trapezoids in two isosceles trapezoids by a line parallel to the bases, thus obtaining the statement for $k + 1$ .

[Solution by pf02, March 2025]

Discuss the problem for n = 2, 3

I will only sketch proofs for these cases.

The case n = 2

In general, an inscribable quadrilateral can not be dissected into two inscribable quadrilaterals. In fact, the inscribable quadrilateral $ABCD$ can be dissected into two inscribable quadrilaterals if and only if $ABCD$ is an isosceles trapezoid.

To see this, assume we draw $EF$ as in the figure below, such that $ABEF$ is inscribable.

It follows that $\angle FEB = \pi - \angle A = \angle C$ , so $EF \parallel CD$ . Similarly, if $EFDC$ were inscribable, we would have $FE \parallel AB$ . So $ABCD$ would be a trapezoid. Being inscribable, it would have to be isosceles.

The case n = 3

This can be done, i.e. an inscribable quadrilateral can be dissected on three inscribable quadrilaterals. To see this, take $EF \parallel CD$ and $HG \parallel AB$ (equivalently, these say that $ABEF, GHDC$ are inscribable. It follows that $FEGH$ is inscribable as well. Indeed, $\angle FEG = \pi - \angle FEB = \angle A = \pi - C = \angle DHG = \pi - \angle FHG$ .

$\mathbf{Note}$ : This splitting does not yield an inductive method for $n > 3$ . For that, we still need to prove the case $n = 4$ .

1972 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions

Page

Toolbox

Search