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| Pruned Triangle (Posted on 2017-07-28) |
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A paper triangle has its vertices cut off. Each cut is along a straight line
parallel to the side opposite the vertex and tangent to the triangle's incircle.
Prove that the triangle's inradius is equal to the sum of inradii of the
three triangles cut off.
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Submitted by Bractals
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Solution:
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Let ΔABC be the paper triangle with a, b, and c the lengths of the sides
opposite vertices A, B, and C respectively. Let r and s be the inradius and
semiperimeter. Let ha, hb, and hc be the altitudes through vertices A, B,
and C respectively.
Let B1C1 be the cut for vertex A with B1 on AB and C1 on AC,
A2C2 the
cut for vertex B with A2 on AB and C2 on BC, and
A3B3 the cut for vertex
C with A3 on AC and B3 on BC.
Let ra, rb, and rc be the inradii of ΔAB1C1,
ΔA2BC2, and ΔA3B3C respectively.
Clearly, ΔAB1C1 ~ ΔA2BC2 ~ ΔA3B3C ~ ΔABC.
Area(ΔABC) = r*s = 1/2*x*hx ⇒ 2*r/hx = x/s
rx / r = (hx - 2r)/hx = 1 - 2*r/hx = 1 - x/s = (s - x)/s
for x = a, b, and c.
Therefore,
ra + rb + rc =
r*(s - a)/s + r*(s - b)/s + r*(s - c)/s =
r*[ (s - a) + (s - b) + (s - c) ]/s =
r*[ 3*s - (a + b + c) ]/s =
r*[ 3*s - 2*s ]/s = r
QED
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