# Dissection of Cyclic Quadrilateral What Is This About? A Mathematical Droodle

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Explanation The idea for the applet and the question it attempts to illustrate came up while trying to solve a problem from an outstanding collection by T. Andreescu and R. Gelca (Cut a cyclic quadrilateral into n > 4 cyclic quadrilaterals.) We discuss that problem elsewhere.

Assuming all the cuts are straight lines joining point P to the points M, N, Q, R on the sides of quadrilateral ABCD, one per side, what is the locus of points P for which such a dissection is possible?

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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The applet serves to illustrate the fact that, for any P, points M, N, Q, R always exist on the side lines of ABCD such the quadrilaterals AQPR, DNPQ, CMPN, and BRPM are cyclic. Indeed, the feet of the perpendiculars from P to the side lines of the quadrilateral fit the bill. Alternatively, let's fix N on CD. By moving M along BC we can achieve the identity

∠CNP + ∠CMP = 180°.

Next, by moving R along AB, we can get

∠BMP + ∠BRP = 180°.

Note that since ∠BMP is supplementary to ∠CMP, ∠BRP = ∠CMP and ∠ARP = ∠BMP = ∠CNP. For this reason, when Q is found on AD that makes a cyclic quadrilateral AQPR, the quadrilateral DNPQ will be automatically cyclic.

For a given P, we may start with N anywhere on CD. But, for a fixed N, the other points M, N, and R are found uniquely. For what N are they located in the interiors of the respective sides of the quadrilateral? And for which P the set of N's that answers that question is not empty?

Following is Nathan Bowler's response:

The rays PM, PN, PQ and PR are at fixed angles to one another, and must lie inside the angles CPB, DPC, APD and BPA respectively. Say AD and BC intersect at X, and AB and CD intersect at Y. Without loss of generality, A lies between X and D and B lies between A and Y. We must have

∠BPA <= ∠MPQ = π - ∠AXB,

so P lies outside the circumcircle of ABX. Similarly P lies outside the circumcircle of BCY. Suppose we are given any P lying outside these two circles.

For any choices of M on BC and R on AB, let Q be the point on DA with ∠MPQ = π - ∠AXB, and N the unique point on CD with ∠NPR = π - ∠BYC. We say a choice of M is suitable if both of M and Q lie in the interiors of the respective sides of the quadrilateral ABCD; the definition of suitability for R is similar. Now choose M' and R' as close to B as possible so that both choices are suitable. At most one of M', R' is not B; without loss of generality M'. Then

 ∠M'PR' ≤ ∠M'PQ' - ∠BPD ≤ π - ∠CXD - ∠XCD = ∠CDA.

On the other hand, choose M'' and R'' as far from B as possible so that both choices are suitable. We have either M'' = C or Q'' = A, and also R'' = A or N'' = C. This gives 4 possible values for ∠M''PR'':

∠CPA, ∠M''PQ'', ∠N''PR'' or ∠M''PQ'' + ∠N''PR'' - ∠CPA,

and in all of these cases ∠M''PR'' >= ∠CDA. The last case is least obvious. We have

 ∠M''PQ'' + ∠N''PR'' - ∠CPA = (π - ∠BXD) + (π - ∠XBY) - (π - ∠XBY) = ∠XDY = ∠CDA

So we may choose M between M' and M'' and R between R' and R'' so that ∠MPR = ∠CDA, and then M, N, Q and R meet all of the conditions in the problem. To summarise, a necessary and sufficient condition is that P should lie outside the circumcircles of ABX and BXY.

### References

1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.4.4 (p. 15) 