Problem of the Month – July, August 2020

Let A₁A₂A₃A₄ be a circumscribed quadrilateral with the perimeter p₁ and with the sum of
its diagonals k₁ and let B₁B₂B₃B₄ be a circumscribed quadrilateral with the perimeter p₂
and with the sum of its diagonals k₂. Given

p₁² + p₂² = (k₁ + k₂)² 

prove that A₁A₂A₃A₄ and B₁B₂B₃B₄ are congruent squares.

 

Correct Solutions by,

• Toshihiro Shimizu, Kawasaki, Japan
• Haruhisa   Oda, University of Tokyo, Japan
• Alper Balcı, METU
• Mehmet Özgün Cihangir, METU
• Roger Bengtsson, Lund, Sweden
• Emirhan Yağcıoğlu, Özel Ankara Matematik ve Fen Bilimleri Anadolu Lisesi
• Vedat Deveci, İstanbul
• İbrahim Berkay Ceylan, Giresun Fen Lisesi
• Chui Tsz Fung, Hong Kong
• Kadyr Annabayev, Turkmen State Institute of Arhcitecture and Construction,  Aşkabat
• Ahmed Said Gençkol, Ankara Atatürk Anadolu Lisesi
• Mehmet Koca, Denizli Koleji

Solution: http://www.fen.bilkent.edu.tr/~cvmath/Problem/2008a.pdf