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triangles within triangles

Expert replies
by gmatrix » Fri Oct 01, 2010 5:16 pm
Let E1 be an equilateral triangle of side "a". Another equilateral triangle E2 is formed by joining mid-points of the sides of the triangle E1.The same process is applied to E2 to form another equilateral triangle E3 and so on.If A1,A2,A3,....be the areas and P1,P2,P3,....be the perimeters of E1,E2,E3,.....respectively, then the ratio P1+P2+P3+...../A1+A2+A3+... equals:
1.square root 108/a
2.4 square root3/a
3.square root 72/a
4.square root 252/a

[spoiler]OA:later[/spoiler]
Life is all about ass; you're either covering it, laughing it off, kicking it, kissing it, busting it, trying to get a piece of it, or behaving like one.
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Source: — Problem Solving |

by mj78ind » Fri Oct 01, 2010 7:22 pm
dunno whether to categorize this as a pretty good 770 type question or something that will never appear since it uses GP.

Sum of the areas is a GP - {(sqrt3)/4}*a^2 , {(sqrt3)/4}*(1/4), {(sqrt3)/4}* (1/116)......so on

Sum of the perimeters: 3a, 3a/2, 3a/4, ..........

Sum of peri / sum of areas = [3a/(1-1/2)]/[(sqrt3)/4a^2/(1-1/4)]

This solves to = 18/a*sqrt3 = sqrt108/a .

[spoiler]Hence A[/spoiler]
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