NEWTON AND MIDPOINTS OF DIAGONALS OF CIRCUMSCRIPTIBLE QUADRILATERALS TITU ANDREESCU, LUIS GONZALEZ AND COSMIN POHOATA Abstract. We generalize Newton's Theorem that the midpoints of the diagonals of a circumscriptible quadrilateral determine a line that passes through the center of the inscribed circle. The starting point of our paper is the following result due to Newton, which is perhaps one of the most classical theorems in quadrilateral geometry. Theorem 1 (Newton). In a circumscriptible quadrilateral ABCD, the midpoints of the diagonals and the center of the inscribed circle are collinear. Despite its simplicity, there aren't many proofs known for this magnificent little result. We refer to [1] for a more complete bibliography. By far the best way to look at it, nonetheless, is using the following Lemma, known as the Leon-Anne's Theorem, according to [1]. We record this proof below. Lemma 2. Let ABCD be a convex quadrilateral and let k2 be a (positive) real number. Then, the locus of the points P in plane for which the sum of areas SP AB + SPCD equals k2 is a line. 2 Proof of Lemma 2. Let P be a point satisfying SP AB + SPCD = k . If ABCD is a rectangle, then the result is easy. We leave it as an exercise for the reader. Otherwise, let us suppose WLOG that AB meets CD at X. Now, construct the points Y and Z on the rays XA and XD so that XY = AB and XZ = CD. We note that SP AB + SPCD = SPXY + SPXZ = SPZXY = SPYZ + SXYZ : Triangle XYZ is however fixed, as all points X, Y , Z are fixed (by construction). Thus, SXYZ is a constant, and so SPXY = (SP AB + SPCD) − SPXZ is a constant. But this just means that XY · δ(P; XY ) is constant, hence the locus of P is a line (at constant distance from the fixed line XY ). This proves the Lemma. This yields a really immediate proof Theorem 1, as advertised. Proof of Theorem 1. Note that 1 1 1 S + S = S + S = S ; MAB MCD 2 ABC 2 ADC 2 ABCD since M is the midpoint of AC. Similarly, we get that 1 1 1 S + S = S + S = S : NAB NCD 2 ABD 2 BCD 2 ABCD Furthermore, 1 S + S = r · (AB + CD) IAB ICD 2 1 = r · (AB + CD + DA + BC) 4 1 = (S + S + S + S ) 2 IAB IBC ICD IDA 1 = S ; 2 ABCD where the second equality holds by Pithot's theorem that a quadrilateral ABCD is cir- cumscriptile if and only if AB + CD = AD + BC. We refer to [3] for an easy proof. Mathematical Reflections 1 (2014) 2 Thus points M, N, I all belong to the locus that appears in the Lemma, in this case 1 associated with the constant 2 SABCD. Thus, they are collinear, and the proof is complete. The beauty is that this proof generalizes to prove something significatively more involved about convex quadrilaterals in general. To state the result, we record a simple theorem which we will prove synthetically. Theorem 2. Let ABCD be a cyclic quadrilateral and let HA, HB, HC , HD be the orthocenters of triangles BCD, CDA, DAB and ABC, respectively. Then, the segments AHA, BHB, CHC , DHD all have the same midpoint. Proof of Theorem 2. It clearly suffices to show that segments AHA and BHB have the same midpoint. To do this, we will argue that A, B, HA and HB are the ver- tices of a parallelogram. Indeed, note that AHB and BHA are parallel, since the lines are both perpendicular to CD. Also, as segments, they are both equal in length, since AHB = 2Rj cos CADj = 2Rj cos CBDj = BHA, where R is the circumradius of the cyclic quadrilateral ABCD. This proves that ABHAHB is a parallelogram, as claimed. The common midpoint of AHA, BHB, CHC , DHD described above is known in litera- ture as the anticenter of (the cyclic) quadrilateral ABCD and it has magnificient proper- ties. For example, it is also the common point of the nine-point circles of triangles BCD, CDA, DAB and ABC. There are many interesting alternative ways to describe this point, but for our purposes, the result from Theorem 2 will be enough. Mathematical Reflections 1 (2014) 3 We refer whatsover to [2] for a more complete list of properties of the anticenter and its generalization, the Poncelet point of an arbitrary convex quadrilateral. We are now ready to state the main result of our paper. Theorem 3. Let ABCD be a convex quadrilateral and let M, N be the midpoints of diagonals AC and BD, respectively. Let XYZT be the quadrilateral determined by the internal angle bisectors of its four angles. Then, XYZT is cyclic and M, N and the anticenter of XYZT are collinear. Proof of Theorem 3. Obviously, to see that XYZT is cyclic is immediate. Just note that ◦ \YXT = 180 − (\XDC + \XCD) 1 1 = 180◦ − ( ADC + BCD) 2\ 2\ 1 1 = 180◦ − (180◦ − BAD − ABC) 2\ 2\ 1 1 = BAD + ABC 2\ 2\ = \ZAB + \ZBA ◦ = 180 − \YZT: Mathematical Reflections 1 (2014) 4 It is for the second part that we have to do all the work. Let P be the intersection of BC and AD and let Q be the intersection of AB and CD. WLOG, we consider the configuration where X and T are the incenters of PCD and QDA and Y , Z are the excenters of QBC, P AB againts vertices Q and P , respectively (as in the diagram). For convenience, label \P = \AP B. In triangles P AB and PCD, recall the well-known identities P PZ · AB P PX · CD cot = ; cot = : 2 ZA · ZB 2 XD · XC It follows that P cot · (XD · XC + ZA · ZB) 2 = PZ · AB + PX · DC P P = PZ · 2 cos · PX + AB − PX · 2 cos · PZ − DC : (1) 2 2 Also, recall that in triangles P AB and PCD we have that P PA + PB + AB P PD + PC − DC cos = (2); cos = : (3) 2 2 · PZ 2 2 · PX Therefore, combining (2) and (3) with (1) gives us P cot · (XD · XC + ZA · ZB) 2 = PZ · (PD + PC − DC + AB) − PX · (PA + PB + AB − DC); hence P PZ ·(PD+PC)−PX ·(PA+PB) = XZ ·(DC −AB)+cot ·(XD·XC +ZA·ZB): (4) 2 On the other hand, note that 1 1 P S = XD · XC · sin CXD = XD · XC · sin XDC 2 2 2 1 1 P S = ZA · ZB · sin AZB = ZA · ZB · sin ZAB 2 2 2 1 P S = S + S = PZ · (PD + PC) · sin P DZC P DZ PCZ 2 2 1 P S = S + S = PX · (PA + PB) · sin : P AXB P AX P BX 2 2 Combining these four latter expressions with (4) yields 1 P XZ · (DC − AB) · sin + S + S 2 2 XDC ZAB = SP DZC − SP AXB = SABCD − SXAB − SZCD: (5) Let E, F be the orthocenters of triangles XYT and ZYT . Then, according to Theorem 2, EXZF is a parallelogram with diagonal intersection K = XF \ ZE; the anticenter of (cyclic) quadrilateral XYZT . The angle between EX and CD clearly equals \XYZ, and similarly the angle between ZF and AB equals \XYZ. Mathematical Reflections 1 (2014) 5 Hence 1 S = S − S = EX · DC · sin XYZ DECX EDC XDC 2 1 S = S − S = EX · AB · sin XYZ; AF BZ ZAB F AB 2 thus 1 S + S = EX(DC − AB) · sin XYZ + S + S : EDC F AB 2 XDC ZAB But cos YZT sin P EX = XZ · = XZ · 2 ; sin XYZ sin XYZ hence 1 P S + S = XZ(DC − AB) · sin + S + S (6): EDC F AB 2 2 XDC ZAB From (5) and (6) we get SEDC + SF AB + SXAB + SZCD = SABCD: (7) Now, since K is the midpoint of FX and EZ, we get 2SKAB = SXAB + SF AB ; 2SKDC = SEDC + SZDC ; which yields 2SKAB + 2SKDC = SXAB + SF AB + SEDC + SZDC : (8) Finally, from (7) and (8), we conclude that 1 S + S = S : KAB KCD 2 ABCD Therefore, by the Lemma, we conclude that K is on the line MN of the convex ABCD. This completes the proof of Theorem 3. Clearly, when ABCD is circumscriptible, the internal angle bisectors of the angles of the quadrilateral are concurrent at the center of the inscribed circle; thus X = Y = Z = T = K; thus, we recover Newton's Theorem! References [1] A. Bogomolny, Newton's Theorem: What is it? A Mathematical Droodle, http://www.cut- the-knot.org/Curriculum/Geometry/NewtonTheorem.shtml#explanation. [2] D. Grinberg, Poncelet points and antigonal conjugates, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109112?ml=1. [3] R. A. Johnson, Advanced Euclidean Geometry, Dover, 2007. Titu Andreescu, University of Texas at Dallas, Richardson, TX, USA. E-mail address: [email protected] Luis Gonzalez: 5 Julio Avenue, Maracaibo, Venezuela. E-mail address: [email protected] Cosmin Pohoata: 3533 Frist Center Mailbox, Princeton University, Princeton, NJ, USA. E-mail address: [email protected] Mathematical Reflections 1 (2014) 6.