Thursday, 20 June 2013
Two ways to Solve a Geometry Problem
Greg Ross at the Futility Closet is a constant source of entertaining information and frequently posts some really nice mathematical problems. Recently he posted the one above. I wanted to reprint it because it dramatically illustrates how simple problems can be made complex by the the failure to use all our mathematical perceptions. I have often conjectured that many students struggle with geometry because their vision is too literal. They see what is there, only what is there, and nothing more. And geometry, it seems, often asks us to see a line that isn't there, or a different orientation of the same problem, if we wish to see it at its most beautiful form.
I suppose the most difficult way to solve the problem would be to assign a unit length to the sides of one triangle in order to find its area. Proceed to find the radius of the circle, and use that to establish a side of the larger triangle from which the area can be found. Please don't make me do that!
Probably a quick way for a geometry student with a little appreciation of the relationship between lengths and areas in similar triangles would be to establish that from the common center of the objects the distance to a leg was half the distance to the opposite vertex. So the distance of a perpendicular segment from the common center to the parallel bases of the two triangles was in a relation of 1:2, allowing us to conclude the similar figures were in a ratio of 1:4.
As simple as that is, there is a calculating free method to solve the problem. Simply invert the inner triangle as shown below and ..."Behold" as Brahmagupta said,
That's how I WANT to learn to look at geometry problems. Still working on it.
I think I nice followup visualization challenge would be to ask the student to explain, without calculation, what would happen to the ratio of the two triangles if they were not equilateral and both remained similar and retained the circle as their respective incircle and excircle?