$\triangle AOF$ is similar to $\triangle COD$, thus, $|\overline$$ Therefore, the orthocenter of the triangle must coincide with the center of the circles.Ĭonsider the following diagram, in which $O$ is both the orthocenter of $\triangle ABC$ and the center of the circles: Then by leaving the base of that altitude alone and moving that altitude to the center, we increase the altitude, and thereby the area of the triangle (as shown in black). Suppose that the center of the circles did not lie on one of the altitudes of $\triangle ABC$ (as shown in gray). And I'm also not seeing the "fancy geometry" that gives the angle $b$, nor, indeed, why angle $b$ is constant.Ĭould someone please explain what's happening here? I am happy with the expression for the triangle's area, and also with the differentiation and derivation of $C\cos c = B\cos b$.īut I don't see why the extended radii are perpendicular to the triangle's sides, which makes the centre of the concentric circles the orthocentre of the triangle. A little fancy geometry shows that $b$ is $225^o$ from This givesĪlso, from condition (2) extended radii are perpendicular to the triangle side. Position radius A on the positive x-axis at angle $a=0$ (no loss in generality). Let radii A,B, and C be at angles a,b, and c respectively. Here is a partial solution (not my own) which I have edited it a little for clarity. Given three concentric circles of radii 1, 2, and 3, respectively, find the maximum area of a triangle that has one vertex on each of the three circles. I have been attempting to solve this problem:
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