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)=2-a-b [note the rhs is positive]
a^2+b^2=4+a^2+b^2-4a-4b+2ab
b=(2a-2)/(a-2)
tan(∠MAB)=1-a, tan(∠KAD)=1-b=a/(2-a)
tan(∠MAB+∠KAD)= ((1-a)+a/(2-a)) / (1-(1-a)*a/(2-a))
= (2-2a+a^2) / (2-2a+a^2)
=1
So ∠MAB+∠KAD = 45°
and therefore ∠MAK = 45°
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Posted by Jer
on 2023-09-02 17:26:48 |
Trig Solution
