Question

In a regular triangle ABC with side 1, two squares MNKL, RKPT are drawn such that points M, L, R are on the side AC (the order of the points on that side is as follows: A, M, L, R, C). Points P, T are on the side BC (the order of the points on that side is as follows: B, P, T, C). And the point N is on the side AB. Find the lengths of the sides of the two squares.

Answer 1

• MN = (21 -6√3)/37 ≈ 0.286694
• RK = (14√3 -12)/37 ≈ 0.331046

Explanation:

In the attached figure, we have defined LM to be length x. Then the other lengths on side AC are ...

AM = LR = x/√3

RC = (2/√3)RK = (2/√3)(2/√3)x = 4/3x

Then the sum of lengths along AC is ...

AC = AM +ML +LR +RC

1 = x(1/√3 +1 +1/√3 +4/3) = x(7/3 +2/√3) = x(7√3 +6)/(3√3)

Then the value of x is ...

`x=(3√(3))/(7√(3)+6)=(3√(3)(7√(3)-6))/((7√(3))^2-6^2)=(3(21-6√(3)))/(3(49-12))\\\\\boxed{MN=(21-6√(3))/(37)}\\\\RK=(2√(3))/(3)MN\\\\\boxed{RK=(14√(3)-12)/(37)}`