Dear colleagues and geometry enthusiasts,
I invite you to tackle an original aesthetic geometry problem:
A semicircle is circumscribed on the side of square ABCD, acting as its diameter.
Find: AP/PD
(BP is the tangent).
AP/PD=2R*ctg(2alpha)/(R*tg(alpha))=(ctg(alpha))^2-1, where alpha=
Sergey E. Gladun This is an excellent solution! Thank you.
3:1
AD=AB=KB=a
PB=PK=b
AP=a-b, PB=a+b
a^2+(a-b)^2=(a+b)^2
a=4b
AP=3b
AP:PB=3b:b
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