Thursday, May 28, 2020

A Triangle With a 45 Degrees Angle in Square : A very hard problem from Geometry


Let P and Q be the points where AN and AM intersect the diagonal BD, respectively. It is noted that QAN=QDN=45.Thus, AQND is a cyclic quadrilateral, therefore implying that AQN=90.

A Triangle With a 45 Degrees Angle in Square, proof 2

This in turn implies that ΔAQN is a right-isoceles triangle with AN as the hypotenuse. Thus, AQ=AN2.

The same argument can be employed to show that AP=AM2.

Now, since triangles APQ and AMN share a common vertex angle A, we have, Area(ΔAPQ)Area(ΔAMN)=AQAPAMAN. However, from the conclusions in the previous two paragraphs, we have AQAN=APAM=12.

This therefore implies that Area(ΔAPQ)Area(ΔAMN)=12, or equivalently, Area(ΔAPQ)Area(MNPQ)=1, as claimed by the problem.


The problem, due to V. Proizvolov, appeared in Kvant - a popular Russian magazine - (#1, 2004, M1895), with a solution in a later issue (#4, 2004).

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