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

Answer:

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

Step-by-step 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 ...

  $x=\dfrac{3\sqrt{3}}{7\sqrt{3}+6}=\dfrac{3\sqrt{3}(7\sqrt{3}-6)}{(7\sqrt{3})^2-6^2}=\dfrac{3(21-6\sqrt{3})}{3(49-12)}\\\\\boxed{MN=\dfrac{21-6\sqrt{3}}{37}}\\\\RK=\dfrac{2\sqrt{3}}{3}MN\\\\\boxed{RK=\dfrac{14\sqrt{3}-12}{37}}$