Let ABCD be a rectangle with DA=12.
ABJ is an equilateral triangle with J inside ABCD.
CHI and EGF are congruent equilateral triangles with E and F on DA, H on CB, J on FG, and E, I, G, H collinear.
Find EF.
Geometry's not in my wheelhouse but figured I'd give this one a try.
Angles JAE and JBH are 30 degrees. Angle AEJ is 60 degrees, so angle AJE is 90 degrees. Angle AJB is 60 degrees, so angle BJH is 30 degrees.
So the triangle BJH is isosceles with angles 30-30-120. The length of the longest side JB, which is equal to the length of EF that we're trying to find, we'll call x. Note HC is also length x.
The length of the shorter side BH is x / sqrt(3). So the full length of BC = BH + HC = x + x/sqrt(3) = 12.
Solving for x I get 12 * sqrt(3) / (1 + sqrt(3)), or approximately 7.6077.
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Posted by tomarken
on 2020-02-28 11:02:16 |
My attempt
