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3 equilateral triangles in a rectangle (Posted on 2020-02-28) Difficulty: 3 of 5
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.

  Submitted by Jer    
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Solution: (Hide)
Let M be the midpoint of AB and the perpendicular bisector of AB intersects EH at point K.

Let N be the midpoint of CH and the perpendicular bisector of CH intersects JM at point L.

Call CH=x and KL=a. CN=x/2. JK=2a.

IL=a*sqrt(3), IN=x*sqrt(3)/2, so LN=BM= x*sqrt(3)/2-a*sqrt(3).

JM=BM*sqrt(3)=3x/2-3a.

ML=BN=3x/2.

CB=2x.

Since CB=DA=12, EF=CH=6.

Note: There is a nice invariant here. The solution does not vary with the size of the congruent triangles.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: My attemptJer2020-03-14 17:39:47
SolutionMy attempttomarken2020-02-28 11:02:16
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