 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  3 equilateral triangles in a rectangle (Posted on 2020-02-28) 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 No Rating 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. Subject Author Date re: My attempt Jer 2020-03-14 17:39:47 My attempt tomarken 2020-02-28 11:02:16 Please log in:

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