Who is as Clever as a Canadian?

Question

The Canadian Open Math Challenge recently asked Canadian high school students to find the ratio of the inradii of two triangles ADP and CDP, given that ABCD is a parallelogram, that the circle inscribed in ABC is tangent to AC at P, that AD+DC=3(AC), and that AD=DP.

Can you match our Canadian students and find that ratio?

Update:

Here is a diagram:

http://www.flickr.com/photos/54423126@N02/82149235...

Not to scale of course!

Answer 1

Favorite Answer

I got 4/3. These Canadians are very clever indeed!

If we take |AD| = |DP| = 1, |DC| = a > 1 (i.e. up to similarity) and the property of each touching point of the incircle of ∆ABC being at distances to the vertices, equal to the semi-perimeter minus the opposing side, we can easily find

|AP| = (2a - 1)/3 and |PC| = (2 - a)/3 using |AC| = (1 + a)/3 (by the way 1< a < 5/4).

Now the calculation shows ∆ADP and ∆CDP have equal semi-perimeters, so the ratio in question is equal to the ratio of their areas, the latter in turn equal to the ratio

|AP| : |PC|. Dropping the perpendicular from D to AC and expressing the common altitude of the both aforementioned triangles from 2 right triangles with common leg we arrive to equation

10a² - a - 11 = 0, hence a = 1.1 and the rest is clear.