The trick with this question was not to make life difficult by working values out. As a consequence some of you managed much more elegant solutions than others. Here is an example based on that offered by Shu Cao (very well done). Jeongmin Lee, Rebecca Bartram (The Mount School, York) and Andrei Lazanu (School 205, Bucharest) should also be congratulated.


Using Pythagoras < quote/ > theorem in ABC = = = a
amp;=
amp; Ö(b2 + c2)
Area of semicircle on BC
amp;=
amp; p a2
4
amp;=
amp; S1
Area of semicircle on AC
amp;=
amp; p b2
4
amp;=
amp; S2
Area of semicircle on AB
amp;=
amp; p c2
4
amp;=
amp; S3
Area of crescents
amp;=
amp; S2 + S3 + Area ABC - S1
amp;=
amp; p( b2
4
 )2 + p c2
4
 + Area ABC - p a2
4
amp;=
amp; p b2
4
 + p c2
4
 - p a2
4
 + Area ABC
amp;=
amp; p
4
 ×( b2 + c2 - a2 ) + Area ABC
amp;=
amp; 0 + Area ABC
amp;=
amp; Area ABC

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