



 
Hi Ann, We started with an equilateral triangle of side length a, so the length of PR is a.
The triangle PFR is a 306090 triangle so the side lengths ore in the proportions 1:2:√3. Since the length of PR is a the length of FR is a/2 and the length of PF is √3 a/2. Let the radius of the first circle, the one with centre C be r_{1}. Since the triangle PCA is a right triangle I can use Pythagoras' theorem to write
Expand and simplify to get r_{1} = a/(2 √3). Now turn your attention to triangle EGR and the next circle. The length of BR is equal to the length of PF which is √3 a/2. Thus the length of DR is √3 a/2  2r_{1} = a/(2 √3). This is the height of the triangle EGR. Notice that this is one third of the height of the triangle PQR. Thus the radius r_{2 }of the second circle is one third of r_{1}. Likewise the radius r_{3} of the third circle (not on the diagram) is one third of r_{2}. Thus the radius of the n+1^{st} circle is
Hence the area of the n+1^{st} circle is
The area you want is thus
This sum is a geometric series which your daughter will be able to evaluate. When we did it we got the final answer to be 11 π/24. Penny and Sue
Ann, The answer to this question touches on ideas about infinity that are well beyond grade 9. I assume that the procedure intends to inscribe the largest possible circle in all the gaps between circles of the previous step (so you use one circle in step 1, 3 circles in step 2, 9 circles in step 3, 27 circles in step 4, and so on). In this way the new inscribed circles are tangent to the sides of each empty triangular region of the previous step, where some of these regions have curved sides. The area covered by the circles equals the area of the whole triangle, namely half base times height
Intuitively, every positive piece of area is eventually covered by a circle. (The fact that infinitely many points remain uncovered is what makes the problem advanced. It turns out that these missing infinitely many points add nothing to the area covered by the triangle that is outside the circles.) Chris
 


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 