On sides CB and CD of square ABCD are chosen points M and K so that the perimeter of triangle CMK equals double the side of the square. Find angle MAK.
It seems there should be a nice geometric construction, but I couldn't find it. Here's a trigonometric solution:
Let AB=1, CM=a, CK=b, MK=sqrt(a^2+b^2).
So we have a+b+sqrt(a^2+b^2)=2
sqrt(a^2+b^2)=2ab [note the rhs is positive]
a^2+b^2=4+a^2+b^24a4b+2ab
b=(2a2)/(a2)
tan(∠MAB)=1a, tan(∠KAD)=1b=a/(2a)
tan(∠MAB+∠KAD)= ((1a)+a/(2a)) / (1(1a)*a/(2a))
= (22a+a^2) / (22a+a^2)
=1
So ∠MAB+∠KAD = 45°
and therefore ∠MAK = 45°

Posted by Jer
on 20230902 17:26:48 