Dynamic Geometry: P58
The diagram shows a blue 3-4-5 right triangle. Each circle is tangent to two sides of the triangles and moves at such a rate so that the centers meet at the triangle's incenter. Using the centers, we draw a black triangle. When the three circles meet at the orange point, the area of the black triangle can be expressed as
, where
and
are coprime positive integers. Find
.
The answer is 5.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
When the three circles meet at the orange point, let their radii be r , and label the diagram as follows:
Since △ F I H ∼ △ E A D , F I = E A ⋅ E D F H = 3 ⋅ 5 2 r = 5 6 r , and I H = A D ⋅ E D F H = 4 ⋅ 5 2 r = 5 8 r .
Since E D and A D are tangent to the yellow circle, H D is the angle bisector ∠ E D A , so tan ∠ H D C = tan 2 1 ∠ E D A = sin ∠ E D A 1 − cos ∠ E D A = 5 3 1 − 5 4 = 3 1 , which means C D = tan ∠ H D C H C = 3 1 r = 3 r .
Then A D = A B + B C + C D = r + 5 8 r + 3 r = 4 , which solves to r = 7 5 .
The area of the black triangle is then A = 2 1 ⋅ F I ⋅ I H = 2 1 ⋅ 5 6 r ⋅ 5 8 r = 2 1 ⋅ 5 6 ⋅ 7 5 ⋅ 5 8 ⋅ 7 5 = 4 9 2 4 , so p = 2 4 , q = 4 9 , and q − p = 5 .