Christmas Streak 81/88: Limits on Geometry #2

Geometry Level 5

There is a square A 1 B 1 C 1 D 1 \rm A_1B_1C_1D_1 with one of its side lengths 7. 7.

For every natural number n < N , n<N, conduct these:

1. Let E n , F n \rm E_n,~F_n be the midpoints of B n C n , C n D n \rm \overline{B_nC_n},~\overline{C_nD_n} respectively.

2. Let G n , B n + 1 \rm G_n,~B_{n+1} be the intersections of A n E n , A n C n \rm \overline{A_nE_n},~\overline{A_nC_n} with B n F n . \rm \overline{B_nF_n}.

3. Pick a point C n + 1 \rm C_{n+1} on C n D n \rm \overline{C_nD_n} such that A n C n \rm \overline{A_nC_n} is perpendicular to B n + 1 C n + 1 . \rm \overline{B_{n+1}C_{n+1}}.

4. Pick a point D n + 1 \rm D_{n+1} on A n D n \rm \overline{A_nD_n} and A n + 1 \rm A_{n+1} on A n C n \rm \overline{A_nC_n} such that B n + 1 C n + 1 \rm \overline{B_{n+1}C_{n+1}} is perpendicular to C n + 1 D n + 1 \rm \overline{C_{n+1}D_{n+1}} and parallel to D n + 1 A n + 1 . \rm \overline{D_{n+1}A_{n+1}}.

Find the area of the colored area as N N approaches infinity.


This problem is a part of <Christmas Streak 2017> series .


The answer is 16.8.

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1 solution

Miles Koumouris
Dec 23, 2017

The areas of the triangles in the largest square are 49 20 \tfrac{49}{20} , 98 15 \tfrac{98}{15} , and 49 12 \tfrac{49}{12} , which sum to 196 15 \tfrac{196}{15} . The geometric progression of the areas of the squares has common ratio 2 9 \tfrac{2}{9} , and so we calculate k = 0 ( ( 2 9 ) k × 196 15 ) = 84 5 = 16.8 . \sum_{k=0}^{\infty }\left(\left(\dfrac 29\right)^k\times \dfrac{196}{15}\right)=\dfrac{84}{5}=\boxed{16.8}.

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