Question P.420: The radii of the circles circumscribing (Exercise 66) the qu......

Suppose the given quadrilateral is MNPQ, and suppose the quadri-laterals formed by its internal and external angle bisectors are ABCD and A^{\prime}B^{\prime}C^{\prime}D^{\prime}  respectively (fig. t420).

Lemma: Quadrilaterals ABCD, A^{\prime}B^{\prime}C^{\prime}D^{\prime}  have the same interior angles.

Proof: Since AM, A^{\prime}M bisect {\widehat{Q M T}},~{\widehat{Q M N}} respectively, and since the last two angles are supplementary, we have A^{\prime}M\perp\ A M. Likewise, A^{\prime}Q\perp\ A Q, so quadrilateral A Q A^{\prime}M is cyclic. This means that angles {\widehat{Q A M}},\ {\widehat{Q A^{\prime}M}} are sup-plementary (80), and therefore {\widehat{Q A M}},\ {\widehat{B^{\prime}A^{\prime}D^{\prime}}} are supplementary. But we know, from exercise 66, that quadrilateral A^{\prime}B^{\prime}C^{\prime}D^{\prime} is also cyclic, so {\widehat{B^{\prime}A^{\prime}D}}\ supplements {\widehat{B^{\prime}C^{\prime}D^{\prime}}}, and \widehat{Q A M}=\widehat{B^{\prime}C^{\prime}D^{\prime}}. We can show that the other three angles are equal in pairs in just the same way.

Note. Students can be reminded that this does not make the quadrilaterals similar. Any two rectangles have pairs of equal angles, but may not be similar.

Note also that the angles are not in ‘corresponding’ positions in figure t420: angles {\widehat{A}},{\widehat{A}^{\prime}} are supplementary, not equal.

We turn now to the assertion itself.

We know from exercise 66 that ABCD is cyclic. And it is clear from the construction of ABCD by intersecting exterior angle bisectors, that MNP Q is the quadrilateral investigated in exercise 362b: it has the minimal perimeter of any quadrilateral inscribed in ABCD, and in particular, its sides make equal angles with those of ABCD. So we can use various relations derived in the solution to exercise 362b. (We note that these relationships are far from obvious: the solution to that exercise involved some difficult computation.)

We have, from that discussion:

(1)              R · (MN + NP + P Q + QM) = AB · CD + AD · BC,

in which R is the radius of the circle circumscribing quadrilateral ABCD. Also drawing on the arguments in the solution to exercise 362b, we have (from equation (2) and others following it, in that solution):

2R^{\prime}\cdot M N=B^{\prime}M\cdot C^{\prime}D^{\prime}+B^{\prime}N\cdot D^{\prime}A^{\prime}

2R^{\prime}\cdot N P=C^{\prime}N\cdot D^{\prime}A^{\prime}+C^{\prime}P\cdot A^{\prime}B^{\prime}

2R^{\prime}\cdot P Q=D^{\prime}P\cdot A^{\prime}B^{\prime}+D^{\prime}Q\cdot B^{\prime}C^{\prime}

2R^{\prime}\cdot Q M=A^{\prime}Q\cdot B^{\prime}C^{\prime}+A^{\prime}M\cdot C^{\prime}D^{\prime},

where {{R}}^{\prime} is the circumradius of quadrilateral A^{\prime}B^{\prime}C^{\prime}D^{\prime}.

From these relationships, and Ptolemy’s theorem 237 we have
2R^{\prime}(M N-N P+P Q-Q M)\;=\;A^{\prime}B^{\prime}(D^{\prime}P-C^{\prime}P)+B^{\prime}C^{\prime}(D^{\prime}Q-A^{\prime}Q)\,+ C^{\prime}D^{\prime}(B^{\prime}M-A^{\prime}M)+D^{\prime}A^{\prime}(B^{\prime}N-C^{\prime}N)=2(A^{\prime}B^{\prime}.C^{\prime}D^{\prime}+B^{\prime}C^{\prime}.D^{\prime}A^{\prime})=2A^{\prime}C^{\prime}.B^{\prime}D^{\prime}

Combining this with (1) gives us

(1)          {\frac{R(M N+N P+P Q+Q M)}{R^{\prime}(M N-N P+P Q-Q M)}}={\frac{A C\cdot B D}{A^{\prime}C^{\prime}\cdot B^{\prime}D^{\prime}}}

We now relate the product on the right hand side of this equation to R,\ R^{\prime}. From the note in 251 we have (using absolute value for area),

4R · |ABD| = AB · BD · DA;

4R^{\prime}\cdot|A^{\prime}B^{\prime}D^{\prime}|=A^{\prime}B^{\prime}\cdot B^{\prime}D^{\prime}\cdot D^{\prime}A^{\prime},

so that

(2)        \frac{R}{R^{\prime}}\cdot\frac{|A B D|}{|A^{\prime}B^{\prime}D^{\prime}|}=\frac{A B\cdot B D\cdot A D}{A^{\prime}B^{\prime}\cdot B^{\prime}D^{\prime}\cdot D^{\prime}A^{\prime}}.

Then 256, together with our lemma, gives us {\frac{|A B D|}{|A^{\prime} B^{\prime} D^{\prime}|}}={\frac{A B{\cdot}A D}{A^{\prime}B^{\prime}\cdot A^{\prime}D^{\prime}}}. Comparing  this with (2), we see that

{\frac{R}{R^{\prime}}}\cdot{\frac{A B\cdot A D}{A^{\prime}B^{\prime}\cdot A^{\prime}D^{\prime}}}={\frac{A B\cdot B D\cdot A D}{A^{\prime}B^{\prime}\cdot B^{\prime}D^{\prime}\cdot D^{\prime}A^{\prime}}},

or

{\frac{R}{R^{\prime}}}={\frac{B D}{B^{\prime}D^{\prime}}}.

Reasoning analogously from triangles ACD, A^{\prime}C^{\prime}D^{\prime} , we find that {\begin{array}{l}{{\frac{R}{R^{\prime}}}={\frac{A C}{A^{\prime}C^{\prime}}}.}\end{array}}\, Combining these results with the result in (1), we have:

\frac{R(M N+N P+P Q+Q M)}{R^{\prime}(M N-N P+P Q-Q M)}=\frac{R^{2}}{R^{\prime2}}.

and a bit of algebra shows that this last equation is equivalent to the announced result.