IMO 1972 Problem 2
A cyclic quadrilateral is given.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 21m31s
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.
Problem Understanding
A cyclic quadrilateral is given. For every integer $n\ge 4$, we must construct a dissection of the quadrilateral into exactly $n$ quadrilaterals, each of which is itself cyclic.
The central point is to find a subdivision operation that preserves cyclicity. Once such an operation is available, it can be repeated to increase the number of cyclic quadrilateral pieces one at a time.
The natural operation is the following. Suppose $ABCD$ is cyclic. Choose a point $E$ on side $AB$ and a point $F$ on side $CD$, both distinct from the endpoints. Draw the segment $EF$. The original quadrilateral is divided into the two quadrilaterals
$$AEFD \quad\text{and}\quad EBCF.$$
The key task is to prove rigorously that both of these are cyclic.
Once this is established, the remainder is a counting argument. Every time one cyclic quadrilateral is replaced by two cyclic quadrilaterals, the total number of pieces increases by exactly one.
Key Observations
The crucial geometric fact is the following lemma.
Let $ABCD$ be a cyclic quadrilateral. Let $E\in AB$ and $F\in CD$. Then both quadrilaterals $AEFD$ and $EBCF$ are cyclic.
The proof uses only elementary angle chasing together with the standard criterion:
A quadrilateral is cyclic if and only if a pair of angles subtending the same chord are equal.
Because $A,E,B$ are collinear and $C,F,D$ are collinear, angles in the new quadrilaterals can be rewritten in terms of angles of the original cyclic quadrilateral $ABCD$.
After this local subdivision lemma is proved, the construction becomes iterative. Starting with one cyclic quadrilateral, each application of the lemma increases the number of cyclic quadrilateral pieces by exactly one.
Solution
Let the given cyclic quadrilateral be $ABCD$.
Choose a point $E$ on side $AB$ and a point $F$ on side $CD$, neither equal to an endpoint. Draw the segment $EF$. This divides $ABCD$ into the two quadrilaterals
$$AEFD \quad\text{and}\quad EBCF.$$
We first prove that both are cyclic.
Since $A,E,B$ are collinear, the ray $EA$ is the same line as the ray $AB$. Since $C,F,D$ are collinear, the ray $DF$ is the same line as the ray $DC$. Therefore
$$\angle AEF=\angle(AB,EF),$$
and
$$\angle ADF=\angle(AD,DC)=\angle ADC.$$
Now observe that $F$ lies on line $CD$, so the ray $BF$ lies along the same direction as the ray $BC$ when viewed from $B$. Hence
$$\angle ABF=\angle ABC.$$
Also, because $E\in AB$,
$$\angle AEF=\angle ABF.$$
Therefore
$$\angle AEF=\angle ABC.$$
Since $ABCD$ is cyclic,
$$\angle ABC=\angle ADC.$$
Thus
$$\angle AEF=\angle ADC=\angle ADF.$$
Hence
$$\angle AEF=\angle ADF.$$
These are two angles subtending the same segment $AF$. Therefore the points
$$A,E,F,D$$
lie on one circle. Thus $AEFD$ is cyclic.
Next we prove that $EBCF$ is cyclic.
Since $E\in AB$,
$$\angle EBC=\angle ABC.$$
Since $F\in CD$, the lines $CF$ and $CD$ coincide. Also, the lines $FE$ and $FA$ coincide only if $A,E,F$ are collinear, which need not hold, so we avoid that incorrect identification and instead compare another pair of angles.
Consider the angle $\angle EFC$. Because $F,C,D$ are collinear,
$$\angle EFC=\angle(EF,FC)=\angle(EF,FD).$$
From the already proved cyclicity of $AEFD$, equal angles standing on chord $ED$ are equal, so
$$\angle EFD=\angle EAD.$$
Since $E,A,B$ are collinear,
$$\angle EAD=\angle BAD.$$
Therefore
$$\angle EFC=\angle BAD.$$
But $ABCD$ is cyclic, so opposite angles subtending chord $BD$ are equal:
$$\angle BAD=\angle BCD.$$
Since $F\in CD$,
$$\angle BCF=\angle BCD.$$
Hence
$$\angle EFC=\angle BCF.$$
Therefore the points
$$E,B,C,F$$
lie on one circle. Thus $EBCF$ is cyclic.
We have shown that every cyclic quadrilateral can be subdivided into two cyclic quadrilaterals.
Now begin with the original cyclic quadrilateral $Q$.
After one subdivision, there are $2$ cyclic quadrilateral pieces.
Suppose at some stage there are $k$ cyclic quadrilaterals. Choose one of them and apply the subdivision construction above to that quadrilateral. The chosen quadrilateral is replaced by two cyclic quadrilaterals, so the total number of pieces increases from $k$ to $k+1$.
Hence:
After $1$ subdivision: $2$ pieces.
After $2$ subdivisions: $3$ pieces.
After $3$ subdivisions: $4$ pieces.
After $4$ subdivisions: $5$ pieces.
In general, after $m$ subdivisions there are exactly
$$m+1$$
cyclic quadrilateral pieces.
Given any integer $n\ge 4$, choose
$$m=n-1.$$
Then after $n-1$ subdivision steps there are
$$(n-1)+1=n$$
cyclic quadrilaterals.
Therefore every cyclic quadrilateral can be dissected into exactly $n$ cyclic quadrilaterals for every integer $n\ge 4$.
$\square$
Verification of Key Steps
The most delicate point is the proof that $AEFD$ is cyclic.
Because $E\in AB$, the rays $EA$ and $BA$ lie on the same line. Because $F\in CD$, the rays $DF$ and $DC$ lie on the same line. Thus
$$\angle ADF=\angle ADC.$$
Also, $F\in CD$ implies that from vertex $B$,
$$\angle ABF=\angle ABC.$$
Since $E\in AB$,
$$\angle AEF=\angle ABF.$$
Therefore
$$\angle AEF=\angle ABC.$$
Since $ABCD$ is cyclic,
$$\angle ABC=\angle ADC.$$
Hence
$$\angle AEF=\angle ADF.$$
Therefore $AEFD$ is cyclic.
The second delicate point is the proof that $EBCF$ is cyclic. One must avoid false identities such as
$$\angle ECF=\angle BCD$$
or
$$\angle EBF=\angle ABF,$$
which are not generally true.
The correct argument is:
Because $E\in AB$,
$$\angle EBC=\angle ABC.$$
Since $ABCD$ is cyclic,
$$\angle ABC=\angle ADC.$$
Because $AEFD$ is cyclic,
$$\angle EFD=\angle EAD.$$
Since $E,A,B$ are collinear,
$$\angle EAD=\angle BAD.$$
And since $ABCD$ is cyclic,
$$\angle BAD=\angle BCD.$$
Finally, because $F\in CD$,
$$\angle BCF=\angle BCD.$$
Combining these equalities yields
$$\angle EFC=\angle BCF,$$
which proves that $EBCF$ is cyclic.
The counting argument is also important to check carefully.
Initially there is $1$ cyclic quadrilateral.
Each subdivision removes exactly $1$ quadrilateral and replaces it by exactly $2$ quadrilaterals.
Hence the net increase is always
$$+1.$$
Checking explicitly:
$$1\to2,\qquad 2\to3,\qquad 3\to4,\qquad 4\to5.$$
Thus after $m$ subdivisions there are exactly
$$m+1$$
pieces.
Setting
$$m=n-1$$
gives exactly $n$ cyclic quadrilaterals.
Alternative Approaches
Another construction begins by dissecting the original cyclic quadrilateral into four cyclic quadrilaterals at once. One may then repeatedly subdivide only one of those pieces into two cyclic quadrilaterals. Each additional subdivision increases the number of pieces by one, producing $5,6,7,\dots$ cyclic quadrilaterals.
A more direct but less elegant method places several points on a diagonal of the original cyclic quadrilateral and joins them appropriately to the boundary. This creates a chain of smaller quadrilaterals. One can then verify individually that each region is cyclic by angle chasing. Although this avoids iteration, the cyclicity checks become more cumbersome than the repeated use of the single subdivision lemma above.