Triangles.. ^_^
Geometry
Level
2
Triangle ABC has side lengths AB = 21, AC = 22 and BC = 20. Point D and E are on sides AB and AC, respectively such that DE is parallel to BC and DE passes through the incenter of ABC. Compute DE.
215/63
120/63
860/63
1
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1 solution
Solution 1:
Let r be the inradius, then
K(ABC) = r(22+21+20)/2 = 63r/2
Let h be the altitude of A relative to BC, then K(ABC) = h BC/2
So h BC/2 = 63r/2 => r/h = 20/63
The altitude of the incenter relative to BC = r. Therefore DE/BC = (h-r)/r
.=> DE = BC (h-r)/h = BC (1 - r/h) = 20(1 - 20/63) = 860/63 = 13 41/63
Solution 2:
Let O, r, s, h, K denote the incenter, inradius, semiperimeter, height on BC, and area of triangle ABC, respectively. Since s = (21+22+20)/3 = 63/2, r = K/s and h = 2K/BC, so r/h = BC/(2s) = 20/63. Since r is length of the perpendicular line segment from O to BC, so the height of triangle ADE on DE is h-r. Because triangles ADE and ABC are similar, their corresponding heights are in proportion, i.e. DE/BC = (h-r)/h = 1-r/h = 1-20/63 = 43/63. Therefore DE = (43/63)*BC = 860/63.