Incenter and a half
Take an isosceles triangle and its inradius .
Split the triangle by its altitude. The inradius of one of these smaller triangles is .
If = what is the ratio of base to height in the original triangle?
The answer is 1.5.
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
let the sides be a , a , b . the height of the triangle would then be ( a 2 − b 2 / 4 ) . using r s = A → r 1 2 2 a + b = 2 b h , r 2 2 a + b / 2 + c = 4 b h → a + b / 2 + c 2 a + b r 2 r 1 = 2 → r 2 r 1 = 2 a + b 2 a + b + 2 h = 1 + 2 a + b 2 h for the sake of generalization, lets say the ratio of inradius is \dfrac{\gamma), then ( γ − 1 ) ( 2 a + b ) = 2 h → 2 ( γ − 1 ) a = 2 h − ( γ − 1 ) b → ( γ − 1 ) b 2 + 4 h 2 = 2 h − ( γ − 1 ) b let k be the ratio of base to hieght, then ( γ − 1 ) 2 ( k 2 + 4 ) = 4 − 4 ( γ − 1 ) k + ( γ − 1 ) 2 k 2 → 4 γ ( γ − 2 ) = − 4 ( γ − 1 ) k → k = γ − 1 γ ( 2 − γ ) for γ = 3 / 2 the answer follows.